Enumeration of complementary-dual cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^t}$-codes

TL;DR

This paper systematically enumerates all complementary-dual cyclic codes over finite fields, considering various trace bilinear forms, providing a comprehensive classification relevant for coding theory applications.

Contribution

It introduces a complete enumeration method for complementary-dual cyclic codes over finite fields using different trace bilinear forms.

Findings

Enumeration formulas for all such codes

Classification based on trace bilinear forms

Explicit counts for specific parameters

Abstract

Let $\mathbb{F}_q$ denote the finite field of order $q,$ $n$ be a positive integer coprime to $q$ and $t \geq 2$ be an integer. In this paper, we enumerate all the complementary-dual cyclic $\mathbb{F}_q$-linear $\mathbb{F}_{q^t}$-codes of length $n$ by placing $\ast$, ordinary and Hermitian trace bilinear forms on $\mathbb{F}_{q^t}^n.$

Equations

$$N=\mathfrak{R}^{\gcd(n,2)}\displaystyle\prod\limits_{i\in\mathfrak{F}}\left\{2+\sum\limits_{r=1}^{t-1}q^{\frac{r(t-r)d_{i}}{2}}\prod\limits_{j=0}^{r-1}\biggl{(}\frac{q^{(t-j)\frac{d_{i}}{2}}-(-1)^{t-j}}{q^{(r-j)\frac{d_{i}}{2}}-(-1)^{r-j}}\biggr{)}\right\}\prod\limits_{h\in\mathfrak{M}}\left\{2+\sum\limits_{\ell=1}^{t-1}q^{\ell(t-\ell)d_{h}}{t\brack\ell}_{q^{d_{h}}}\right\},\vspace{-2mm}$$

$$\mathfrak{U}_{k,t}=H_{\mathfrak{r}_{1},t-2\mathfrak{r}_{1}}H_{\mathfrak{r}_{2},t-2-2\mathfrak{r}_{2}}\cdots H_{\mathfrak{r}_{\frac{k-1}{2}},t-k+3-2\mathfrak{r}_{\frac{k-1}{2}}}\bigl{(}q^{(t-k+1)d_{i}}-1-I_{\mathfrak{r}_{\frac{k+1}{2}},t-k+1-2\mathfrak{r}_{\frac{k+1}{2}}}\bigr{)}.\vspace{-3mm}$$

$$t-2(\ell-1)-2\mathfrak{r}_{\ell}=\left\{\begin{array}[]{ll}0&\text{if }t\text{ is even;}\\ 1&\text{if }t\text{ is odd.}\end{array}\right.\vspace{-5mm}$$

$$N_{i,k}=\frac{{\mathfrak{U}}_{k,t}}{{\mathfrak{U}}_{k,k}}=q^{\frac{k(t-k)d_i}{2}}\prod _{j=0}^{k-1}\left(\frac{q^{\frac{(t-j)d_i}{2}}-(-1{)}^{t-j}}{q^{\frac{(k-j)d_i}{2}}-(-1{)}^{k-j}}\right).\text{vspace}-2mm$$

$$2\lambda (X)\kappa (X)+{\left[w(X),w(X)\right]}_{\delta }=\theta (X){\left[r(X),r(X)\right]}_{\delta }.\text{vspace}-2mm$$

$$N_i=\sum _{k=0}^t{N}_{i,k}.\text{vspace}-3mm$$

$$\mathfrak{w}=\left\{\begin{array}[]{cl}\frac{t-1}{2}&\text{if }t\text{ is odd};\\ \frac{t-2}{2}&\text{if either }t\text{ is even and }q\equiv 1~{}(\text{mod }4)\text{ or }t\equiv 0~{}(\text{mod }4)\text{ and }q\equiv 3~{}(\text{mod }4);\\ \frac{t}{2}&\text{if }t\equiv 2~{}(\text{mod }4)\text{ and }q\equiv 3~{}(\text{mod }4).\end{array}\right.\vspace{-2mm}$$

$${\mathfrak{Q}}_{\frac{k-1}{2},\mathfrak{w}}=H_{\mathfrak{w},t-2\mathfrak{w}}{H}_{\mathfrak{w}-1,t-2\mathfrak{w}}\cdots H_{\mathfrak{w}-\frac{(k-1)}{2}+1,t-2\mathfrak{w}}\left(q^{t-k+1}-1-I_{\mathfrak{w}-\frac{(k-1)}{2},t-2\mathfrak{w}}\right).\text{vspace}-1mm$$

$$\mathfrak{Q}_{\frac{k-1}{2},\mathfrak{w}}=\left\{\begin{array}[]{ll}q^{\frac{2tk-(k+1)^{2}}{4}}(q^{\frac{t}{2}}-1)(q-1)\prod\limits_{j=1}^{(k-1)/2}(q^{t-2j}-1)&\hbox{if $\mathfrak{w}=\frac{t}{2}$ and $k$ is odd; }\\ q^{\frac{2t(k+1)-k(k+4)+1}{4}}(q-1)\prod\limits_{j=0}^{(k-3)/2}(q^{t-2j-1}-1)&\hbox{if $\mathfrak{w}=\frac{t-1}{2}$ and $k$ is odd;}\\ q^{\frac{2tk-(k+1)^{2}}{4}}(q^{\frac{t}{2}}+1)(q-1)\prod\limits_{j=1}^{(k-1)/2}(q^{t-2j}-1)&\hbox{ if $\mathfrak{w}=\frac{t-2}{2}$ and $k$ is odd}.\end{array}\right.\vspace{-2mm}$$

$${\mathfrak{Q}}_{\frac{k-1}{2}}=H_{\frac{k-1}{2},1}{H}_{\frac{k-3}{2},1}\cdots H_{1,1}(q-1)=q^{\frac{(k-1{)}^2}{4}}(q-1)\prod _{j=0}^{(k-3)/2}(q^{k-2j-1}-1).\text{vspace}-1mm$$

$$\displaystyle N_{i,k}=\frac{\mathfrak{Q}_{\frac{k-1}{2},\mathfrak{w}}}{\mathfrak{Q}_{\frac{k-1}{2}}}=\left\{\begin{array}[]{ll}\displaystyle q^{\frac{tk-k^{2}-1}{2}}(q^{\frac{t}{2}}-1){(t-2)/2\brack(k-1)/2}_{q^{2}}&\hbox{if $\mathfrak{w}=\frac{t}{2}$ and $k$ is odd;}\vspace{1mm}\\ \displaystyle q^{\frac{(t-k)(k+1)}{2}}{(t-1)/2\brack(k-1)/2}_{q^{2}}&\hbox{if $\mathfrak{w}=\frac{t-1}{2}$ and $k$ is odd;}\vspace{1mm}\\ \displaystyle q^{\frac{tk-k^{2}-1}{2}}(q^{\frac{t}{2}}+1){(t-2)/2\brack(k-1)/2}_{q^{2}}&\hbox{if $\mathfrak{w}=\frac{t-2}{2}$ and $k$ is odd.}\par\end{array}\right.\vspace{-2mm}$$

$$N_{i,k}=R_{i,k}+S_{i,k}.\text{vspace}-2mm$$

$$\mathbb{A}_{k,\mathfrak{w}}=\left\{\begin{array}[]{ll}\frac{q^{t-k}(q-1)^{2}(q^{\frac{t-k}{2}}-1)(q^{\frac{t-k+2}{2}}-1)}{2}&\hbox{if $\mathfrak{w}=\frac{t}{2};$ }\vspace{1mm}\\ \frac{q^{t-k}(q-1)^{2}(q^{t-k+1}-1)}{2}&\hbox{if $\mathfrak{w}=\frac{t-1}{2};$}\vspace{1mm}\\ \frac{q^{t-k}(q-1)^{2}(q^{\frac{t-k}{2}}+1)(q^{\frac{t-k+2}{2}}+1)}{2}&\hbox{ if $\mathfrak{w}=\frac{t-2}{2}$}.\end{array}\right.\vspace{-2mm}$$

$${\mathfrak{Q}}_{\frac{k-2}{2},\mathfrak{w}}=H_{\mathfrak{w},t-2\mathfrak{w}}{H}_{\mathfrak{w}-1,t-2\mathfrak{w}}\cdots H_{\mathfrak{w}-\frac{(k-2)}{2}+1,t-2\mathfrak{w}}{\mathbb{A}}_{k,\mathfrak{w}}.\text{vspace}-8mm$$

$$\mathfrak{Q}_{\frac{k-2}{2},\mathfrak{w}}=\left\{\begin{array}[]{ll}\frac{q^{\frac{k(2t-k-2)}{4}}(q^{\frac{t}{2}}-1)(q^{\frac{t-k}{2}}-1)(q-1)^{2}}{2}\prod\limits_{j=1}^{(k-2)/2}(q^{t-2j}-1)&\hbox{if $\mathfrak{w}=\frac{t}{2}$ and $k$ is even; }\\ \frac{q^{\frac{k(2t-k-2)}{4}}(q-1)^{2}}{2}\prod\limits_{j=0}^{(k-2)/2}(q^{t-2j-1}-1)&\hbox{if $\mathfrak{w}=\frac{t-1}{2}$ and $k$ is even;}\\ \frac{q^{\frac{k(2t-k-2)}{4}}(q^{\frac{t}{2}}+1)(q^{\frac{t-k}{2}}+1)(q-1)^{2}}{2}\prod\limits_{j=1}^{(k-2)/2}(q^{t-2j}-1)&\hbox{if $\mathfrak{w}=\frac{t-2}{2}$ and $k$ is even.}\\ \end{array}\right.\vspace{-2mm}$$

$${\mathfrak{Q}}_{\frac{k-2}{2}}=H_{\frac{k-2}{2},2}{H}_{\frac{k-4}{2},2}\cdots H_{1,2}(q^2-1)(q-1)=q^{\frac{k(k-2)}{4}}(q^{\frac{k}{2}}+1)(q-1{)}^2\prod _{j=1}^{(k-2)/2}(q^{k-2j}-1).\text{vspace}-2mm$$

$$\displaystyle R_{i,k}=\left\{\begin{array}[]{ll}\displaystyle\frac{q^{\frac{k(t-k)}{2}}(q^{\frac{k}{2}}-1)(q^{\frac{t-k}{2}}-1)}{2(q^{\frac{t}{2}}+1)}{t/2\brack k/2}_{q^{2}}&\hbox{if $\mathfrak{w}=\frac{t}{2}$ and $k$ is even;}\vspace{1mm}\\ \displaystyle\frac{q^{\frac{k(t-k)}{2}}(q^{\frac{k}{2}}-1)}{2}{(t-1)/2\brack k/2}_{q^{2}}&\hbox{if $\mathfrak{w}=\frac{t-1}{2}$ and $k$ is even;}\vspace{1mm}\\ \displaystyle\frac{q^{\frac{k(t-k)}{2}}(q^{\frac{k}{2}}-1)(q^{\frac{t-k}{2}}+1)}{2(q^{\frac{t}{2}}-1)}{t/2\brack k/2}_{q^{2}}&\hbox{if $\mathfrak{w}=\frac{t-2}{2}$ and $k$ is even}.\end{array}\right.\vspace{-2mm}$$

$${\mathfrak{Q}}_{\frac{k}{2},\mathfrak{w}}=H_{\mathfrak{w},t-2\mathfrak{w}}{H}_{\mathfrak{w}-1,t-2\mathfrak{w}}\cdots H_{\mathfrak{w}-\frac{(k-2)}{2},t-2\mathfrak{w}}.\text{vspace}-1mm$$

$$\mathfrak{Q}_{\frac{k}{2},\mathfrak{w}}=\left\{\begin{array}[]{ll}q^{\frac{k(2t-k-2)}{4}}(q^{\frac{t}{2}}-1)(q^{\frac{t-k}{2}}+1)\prod\limits_{j=1}^{(k-2)/2}(q^{t-2j}-1)&\hbox{if $\mathfrak{w}=\frac{t}{2}$ and $k$ is even; }\\ q^{\frac{k(2t-k-2)}{4}}\prod\limits_{j=0}^{(k-2)/2}(q^{t-2j-1}-1)&\hbox{if $\mathfrak{w}=\frac{t-1}{2}$ and $k$ is even;}\\ q^{\frac{k(2t-k-2)}{4}}(q^{\frac{t}{2}}+1)(q^{\frac{t-k}{2}}-1)\prod\limits_{j=1}^{(k-2)/2}(q^{t-2j}-1)&\hbox{if $\mathfrak{w}=\frac{t-2}{2}$ and $k$ is even}.\end{array}\right.\vspace{-2mm}$$

$${\mathfrak{Q}}_{\frac{k}{2}}=H_{\frac{k}{2},0}{H}_{\frac{k-2}{2},0}\cdots H_{1,0}=2q^{\frac{k(k-2)}{4}}(q^{\frac{k}{2}}-1)\prod _{j=1}^{(k-2)/2}(q^{k-2j}-1).\text{vspace}-5mm$$

$$S_{i,k}=\left\{\begin{array}[]{ll}\displaystyle\frac{q^{\frac{k(t-k)}{2}}(q^{\frac{k}{2}}+1)(q^{\frac{t-k}{2}}+1)}{2(q^{\frac{t}{2}}+1)}{t/2\brack k/2}_{q^{2}}&\hbox{if $\mathfrak{w}=\frac{t}{2}$ and $k$ is even; }\\ \displaystyle\frac{q^{\frac{k(t-k)}{2}}(q^{\frac{k}{2}}+1)}{2}{(t-1)/2\brack k/2}_{q^{2}}&\hbox{if $\mathfrak{w}=\frac{t-1}{2}$ and $k$ is even;}\\ \displaystyle\frac{q^{\frac{k(t-k)}{2}}(q^{\frac{k}{2}}+1)(q^{\frac{t-k}{2}}-1)}{2(q^{\frac{t}{2}}-1)}{t/2\brack k/2}_{q^{2}}&\hbox{if $\mathfrak{w}=\frac{t-2}{2}$ and $k$ is even}.\end{array}\right.\vspace{-1mm}$$

$$N_{i,k}=\left\{\begin{array}[]{ll}\displaystyle q^{\frac{k(t-k)}{2}}{t/2\brack k/2}_{q^{2}}&\hbox{if $\mathfrak{w}=\frac{t}{2}$ and $k$ is even; }\vspace{1mm}\\ \displaystyle q^{\frac{k(t-k+1)}{2}}{(t-1)/2\brack k/2}_{q^{2}}&\hbox{if $\mathfrak{w}=\frac{t-1}{2}$ and $k$ is even;}\vspace{1mm}\\ \displaystyle q^{\frac{k(t-k)}{2}}{t/2\brack k/2}_{q^{2}}&\hbox{if $\mathfrak{w}=\frac{t-2}{2}$ and $k$ is even}.\end{array}\right.\vspace{-2mm}$$

$$N_{0,k}=\left\{\begin{array}[]{ll}q^{\frac{k(t-k+1)}{2}}{(t-1)/2\brack k/2}_{q^{2}}&\hbox{if $k$ is even;}\\ q^{\frac{(t-k)(k+1)}{2}}{(t-1)/2\brack(k-1)/2}_{q^{2}}&\hbox{if $k$ is odd.}\par\end{array}\right.\vspace{-1mm}$$

$$\text{det }\mathfrak{G}(x_1,x_2,\cdots ,x_{k-2},1+x_{k-1},\alpha +x_k)=\text{det }\mathfrak{G}(x_1,x_2,\cdots ,x_{k-2})+\text{det }\mathfrak{G}(x_1,x_2,\cdots ,x_k).\text{vspace}-3mm$$

$$\text{det }\mathfrak{G}(x_1,x_2,\cdots ,x_k)=\text{det }\mathfrak{G}(x_1,x_2,\cdots ,x_{k-2})\text{det }\mathfrak{G}(x_{k-1},x_k).\text{vspace}-2.5mm$$

$$\text{vspace}-1mm{q}^{\frac{(k-2)(t-k)}{2}}\left\{(q^{t-k}-1)(q^{t-k}-q^{t-k-1})-q^{t-k-1}(q^{t-k}-1)\right\}{\left[\genfrac{}{}{0ex}{}{(t-2)/2}{(k-2)/2}\right]}_{q^2}\text{vspace}-10mm$$

$$=q^{\frac{(k-2)(t-k)}{2}}{q}^{t-k-1}(q^{t-k}-1)(q-2){\left[\genfrac{}{}{0ex}{}{(t-2)/2}{(k-2)/2}\right]}_{q^2}.\text{vspace}-1mm$$

$$N_{0,k}=\left\{\begin{array}[]{ll}q^{\frac{tk-k^{2}-2}{2}}\Big{\{}(q^{k}+q-1){(t-2)/2\brack k/2}_{q^{2}}+(q^{t-k+1}-q^{t-k}+1){(t-2)/2\brack(k-2)/2}_{q^{2}}\Big{\}}&\text{if }k\text{ is even; }\vspace{3mm}\\ q^{\frac{tk-k^{2}+t-1}{2}}{(t-2)/2\brack(k-1)/2}_{q^{2}}&\text{if }k\text{ is odd.}\par\end{array}\right.\vspace{-1.2mm}$$

$$\displaystyle\bigl{(}a(X)+b(X)\bigr{)}+\bigl{(}c(X)+d(X)\bigr{)}$$

$$\displaystyle\bigl{(}u(X)+v(X)\bigr{)}\bigl{(}a(X)+b(X)\bigr{)}$$

$$\big{[}a(X)+b(X),c(X)+d(X)\big{]}_{\delta}=\big{[}a(X),d(X)\big{]}_{\delta}+\big{[}b(X),c(X)\big{]}_{\delta}\in\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}\vspace{-2mm}$$

Full text

Keywords: Sesquilinear forms; Witt decomposition; Formed spaces.

2000 Mathematics Subject Classification: 94B15

1 Introduction

Cyclic codes form an important class of linear codes having a rich algebraic structure. Their algebraic properties enable one to effectively detect or correct errors using linear shift registers. Self-dual, self-orthogonal and complementary-dual codes constitute three important classes of cyclic codes, which have been studied and enumerated by Conway et al. [2], Huffman [10]-[11], Jia et al. [12] and Pless and Sloane [15]. Linear codes are further generalized to additive codes, which have recently attracted a lot of attention due to their connection with quantum error-correcting codes.

Calderbank et al. [1] introduced and studied additive codes of length $n$ over the finite field $\mathbb{F}_{4}.$ They further studied their dual codes with respect to the trace inner product on $\mathbb{F}_{4}^{n}.$ In the same work, they related the problem of finding quantum error-correcting codes with that of finding self-orthogonal additive codes over $\mathbb{F}_{4}.$ Huffman [7] and [8] provided a canonical form decomposition of cyclic additive codes over $\mathbb{F}_{4}.$ Using this decomposition, he further studied and enumerated these codes. Besides this, he obtained the number of self-orthogonal and self-dual cyclic additive codes of length $n$ over $\mathbb{F}_{4}$ with respect to the trace inner product on $\mathbb{F}_{4}^{n}.$ In order to further explore the properties of cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-codes, Huffman [9] generalized the theory developed in [7] and viewed cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-codes of length $n$ as $\mathbb{F}_{q}[X]/\left<X^{n}-1\right>$-submodules of the quotient ring $\mathbb{F}_{q^{t}}[X]/\left<X^{n}-1\right>,$ where $\gcd(n,q)=1$ and $t\geq 2$ is an integer. He also determined the number of cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-codes of length $n$ and studied their dual codes with respect to the ordinary and Hermitian trace bilinear forms on $\mathbb{F}_{q^{t}}^{n}.$ In addition to this, he determined bases of all the self-orthogonal and self-dual cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{2}}$-codes with respect to these two bilinear forms on $\mathbb{F}_{q^{2}}^{n}.$ Furthermore, for any integer $t\geq 2,$ he enumerated all the self-dual and self-orthogonal cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-codes of length $n$ with respect to these two bilinear forms. In a recent work [16], we introduced and studied a new trace bilinear form, denoted by $\ast,$ on $\mathbb{F}_{q^{t}}^{n}$ for any integer $t\geq 2$ satisfying $t\not\equiv 1(\text{mod }p),$ where $p$ is the characteristic of the finite field $\mathbb{F}_{q}.$ We also observed that the $\ast$ bilinear form on $\mathbb{F}_{q^{t}}^{n}$ coincides with the trace inner product considered by Calderbank et al. [1] when $q=t=2$ and Hermitian trace inner product considered by Ezerman et al. [5] when $q$ is even and $t=2,$ which are well-studied inner products in coding theory. Further, by placing the bilinear form $\ast$ on $\mathbb{F}_{q^{t}}^{n},$ we enumerated all the self-orthogonal and self-dual cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-codes of length $n,$ provided $\gcd(n,q)=1.$ Besides this, we explicitly determined bases of all the complementary-dual cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{2}}$-codes of length $n$ with respect to $\ast,$ ordinary and Hermitian trace bilinear forms on $\mathbb{F}_{q^{2}}^{n}$ and enumerated these three classes of codes, where $\gcd(n,q)=1.$

The main goal of this paper is to enumerate all the complementary-dual cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-codes of length $n$ with respect to $\ast$, ordinary and Hermitian trace bilinear forms on $\mathbb{F}_{q^{t}}^{n},$ where $q$ is a prime power, $n$ is a positive integer coprime to $q$ and $t\geq 2$ is an integer. Although our enumeration technique is based upon the theory of cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-codes developed by Huffman [7], it is quite different from the technique employed by Huffman [7] in the enumeration of self-orthogonal and self-dual cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-codes.

This paper is organized as follows: In Section 2, we state some preliminaries that are needed to derive our main result. In Section 3, we determine the number of complementary-dual cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-codes of length $n$ with respect to $\ast,$ ordinary and Hermitian trace bilinear forms on $\mathbb{F}_{q^{t}}^{n},$ where $q$ is a prime power, $n$ is a positive integer with $\gcd(n,q)=1$ and $t\geq 2$ is an integer (Theorem 3.1).

2 Some preliminaries

In this section, we will state some basic definitions and results that are needed to derive our main result.

Throughout this paper, let $q$ be a power of the prime $p,$ $\mathbb{F}_{q}$ denote the finite field with $q$ elements, $n$ be a positive integer coprime to $p$ and $t\geq 2$ be an integer. Let $\mathcal{R}_{n}^{(q)}$ and $\mathcal{R}_{n}^{(q^{t})}$ denote the quotient rings $\mathbb{F}_{q}[X]/\langle X^{n}-1\rangle$ and $\mathbb{F}_{q^{t}}[X]/\langle X^{n}-1\rangle$ respectively, where $X$ is an indeterminate over $\mathbb{F}_{p}$ and over any extension field of $\mathbb{F}_{p}.$ As $\gcd(n,q)=1,$ by Maschke’s Theorem, both the rings $\mathcal{R}_{n}^{(q)}$ and $\mathcal{R}_{n}^{(q^{t})}$ are semi-simple, and hence can be written as direct sums of minimal ideals. More explicitly, if $\{\ell_{0}=0,\ell_{1},\cdots,\ell_{s-1}\}$ is a complete set of representatives of $q$-cyclotomic cosets modulo $n,$ then $X^{n}-1=f_{0}(X)f_{1}(X)\cdots f_{s-1}(X)$ is the factorization of $X^{n}-1$ into monic irreducible polynomials over $\mathbb{F}_{q}$ with $f_{i}(X)=\displaystyle\prod_{k\in C_{\ell_{i}}^{(q)}}(X-\eta^{k})$ for $0\leq i\leq s-1,$ where $C_{\ell_{i}}^{(q)}~{}(0\leq i\leq s-1)$ is the $q$-cyclotomic coset modulo $n$ containing the integer $\ell_{i}$ and $\eta$ is a primitive $n$th root of unity over $\mathbb{F}_{q}.$ Therefore for $0\leq i\leq s-1,$ if $\mathcal{K}_{i}$ is the (minimal) ideal of $\mathcal{R}_{n}^{(q)}$ generated by the polynomial $(X^{n}-1)/f_{i}(X),$ then it is easy to see that $\mathcal{R}_{n}^{(q)}=\mathcal{K}_{0}\oplus\mathcal{K}_{1}\oplus\cdots\oplus\mathcal{K}_{s-1},$ where $\mathcal{K}_{i}\mathcal{K}_{j}=\{0\}$ for all $i\neq j$ and $\mathcal{K}_{i}\simeq\mathbb{F}_{q^{d_{i}}}$ with $d_{i}$ as the cardinality of $C_{\ell_{i}}^{(q)}.$ Next to write $\mathcal{R}_{n}^{(q^{t})}$ as a direct sum of minimal ideals, we further factorize the polynomials $f_{0}(X),f_{1}(X),\cdots,f_{s-1}(X)$ into monic irreducible polynomials over $\mathbb{F}_{q^{t}}.$ To do this, by Lemma 1 of Huffman [9], we see that for $0\leq i\leq s-1,$ $C_{\ell_{i}}^{(q)}=C_{\ell_{i}}^{(q^{t})}\cup C_{\ell_{i}q}^{(q^{t})}\cup\cdots\cup C_{\ell_{i}q^{g_{i}-1}}^{(q^{t})}$ with $g_{i}=\gcd(d_{i},t),$ which led to the following factorization of $f_{i}(X)$ over $\mathbb{F}_{q^{t}}$: $f_{i}(X)=M_{i,0}(X)M_{i,1}(X)\cdots M_{i,g_{i}-1}(X),$ where $M_{i,j}(X)=\displaystyle\prod_{k\in C_{\ell_{i}q^{j}}^{(q^{t})}}(X-\eta^{k})$ for $0\leq j\leq g_{i}-1.$ Therefore, if for each $i$ and $j,$ $\mathcal{I}_{i,j}$ is the (minimal) ideal of $\mathcal{R}_{n}^{(q^{t})}$ generated by $(X^{n}-1)/M_{i,j}(X),$ then one can show that $\mathcal{R}_{n}^{(q^{t})}=\displaystyle\bigoplus_{i=0}^{s-1}\bigoplus_{j=0}^{g_{i}-1}\mathcal{I}_{i,j},$ where $\mathcal{I}_{i,j}\mathcal{I}_{i^{\prime},j^{\prime}}=\{0\}$ for all $(i,j)\neq(i^{\prime},j^{\prime})$ and $\mathcal{I}_{i,j}\simeq\mathbb{F}_{q^{tD_{i}}}$ with $D_{i}=\frac{d_{i}}{g_{i}}.$ Further, it is easy to see that $\mathcal{J}_{i}=\mathcal{I}_{i,0}\oplus\mathcal{I}_{i,1}\oplus\cdots\oplus\mathcal{I}_{i,g_{i}-1}$ is a vector space of dimension $t$ over $\mathcal{K}_{i}$ for $0\leq i\leq s-1.$

In order to study the containment $\mathcal{R}_{n}^{(q)}\subset\mathcal{R}_{n}^{(q^{t})},$ Huffman [9] defined ring automorphisms $\tau_{q^{u},v}:\mathcal{R}_{n}^{(q^{r})}\rightarrow\mathcal{R}_{n}^{(q^{r})}$ as \tau_{q^{u},v}\biggl{(}\sum\limits_{i=0}^{n-1}a_{i}X^{i}\biggr{)}=\sum\limits_{i=0}^{n-1}a_{i}^{q^{u}}X^{vi} for any integer $r\geq 1,$ where $u,v$ are integers satisfying $0\leq u\leq r,$ $1\leq v\leq n$ and $\gcd(v,n)=1.$ When $r=t,$ by Lemma 2 of Huffman [9], we see that $\tau_{q^{u},v}$ permutes minimal ideals $\mathcal{I}_{i,j}$’s of the ring $\mathcal{R}_{n}^{(q^{t})}.$ It is easy to see that for every $i~{}(1\leq i\leq s-1),$ there exists a unique integer $i^{\prime}~{}(1\leq i^{\prime}\leq s-1)$ satisfying $C_{-\ell_{i}}^{(q)}=C_{\ell_{i^{\prime}}}^{(q)}.$ This gives rise to a permutation $\mu$ of the set $\{0,1,2,\cdots,s-1\},$ defined by $C_{-\ell_{i}}^{(q)}=C_{\ell_{\mu(i)}}^{(q)}$ for $0\leq i\leq s-1.$ Note that $\mu(0)=0$ and $\mu(\mu(i))=i$ for $0\leq i\leq s-1.$ That is, $\mu$ is either the identity permutation or is a product of transpositions. Note that when $n$ is even, there exists an integer $i^{\#}$ satisfying $0\leq i^{\#}\leq s-1$ and $C_{\ell_{i^{\#}}}^{(q)}=C_{\frac{n}{2}}^{(q)}=\{\frac{n}{2}\},$ as $q$ is odd. From this, we see that $C_{-\ell_{i^{\#}}}^{(q)}=C_{\ell_{i^{\#}}}^{(q)},$ which implies that $\mu(i^{\#})=i^{\#}.$

Now an $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-code $\mathcal{C}$ of length $n$ is defined as an $\mathbb{F}_{q}$-linear subspace of $\mathbb{F}_{q^{t}}^{n}.$ Further, the code $\mathcal{C}$ is said to be cyclic if $(c_{0},c_{1},\cdots,c_{n-1})\in\mathcal{C}$ implies that $(c_{n-1},c_{0},c_{1},\cdots,c_{n-2})\in\mathcal{C}.$ Throughout this paper, we shall identify each vector $a=(a_{0},a_{1},\cdots,a_{n-1})\in\mathbb{F}_{q^{t}}^{n}$ with the representative $a(X)=\sum\limits_{i=0}^{n-1}a_{i}X^{i}$ of the coset $a(X)+\left<X^{n}-1\right>\in\mathcal{R}_{n}^{(q^{t})},$ and perform addition and multiplication of its representatives modulo $X^{n}-1.$ Under this identification, one can easily observe that the cyclic shift $\sigma(a)=(a_{n-1},a_{0},a_{1},\cdots,a_{n-2})$ of $a\in\mathbb{F}_{q^{t}}^{n}$ is identified with $Xa(X)\in\mathcal{R}_{n}^{(q^{t})}.$ In view of this, every cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-code of length $n$ can be viewed as an $\mathcal{R}_{n}^{(q)}$-submodule of $\mathcal{R}_{n}^{(q^{t})}.$ Huffman [9] studied dual codes of cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-codes of length $n$ with respect to ordinary and Hermitian trace bilinear forms on $\mathbb{F}_{q^{t}}^{n},$ which are defined as follows:

Let $\text{Tr}_{q,t}:\mathbb{F}_{q^{t}}\rightarrow\mathbb{F}_{q}$ be the trace map defined as $\text{Tr}_{q,t}(\alpha)=\sum\limits_{j=0}^{t-1}\alpha^{q^{j}}$ for each $\alpha\in\mathbb{F}_{q^{t}}.$ It is well-known that $\text{Tr}_{q,t}$ is an $\mathbb{F}_{q}$-linear, surjective map with kernel of size $q^{t-1}$ (see [14, Th. 2.23]). Then for any integer $t\geq 2,$ the ordinary trace inner product on $\mathbb{F}_{q^{t}}^{n}$ is a map $(\cdot,\cdot)_{0}:\mathbb{F}_{q^{t}}^{n}\times\mathbb{F}_{q^{t}}^{n}\rightarrow\mathbb{F}_{q},$ defined as $(a,b)_{0}=\sum\limits_{j=0}^{n-1}\text{Tr}_{q,t}(a_{j}b_{j})\text{ ~{}for ~{}all~{} }a=(a_{0},a_{1},\cdots,a_{n-1}),~{}b=(b_{0},b_{1},\ldots,b_{n-1})\in\mathbb{F}_{q^{t}}^{n},$ and the map $\left[\cdot,\cdot\right]_{0}:\mathcal{R}_{n}^{(q^{t})}\times\mathcal{R}_{n}^{(q^{t})}\rightarrow\mathcal{R}_{n}^{(q)}$ is defined as $\left[a(X),b(X)\right]_{0}=\sum\limits_{w=0}^{t-1}\tau_{q^{w},1}\left(a(X)\tau_{1,-1}(b(X))\right)\text{ ~{}for~{} all~{}}a(X),~{}b(X)\in\mathcal{R}_{n}^{(q^{t})}.$ The Hermitian trace inner product on $\mathbb{F}_{q^{t}}^{n}$ is defined only for even integers $t\geq 2,$ which can be written as $t=2^{a}m,$ where $a\geq 1$ and $m$ is odd. It is easy to see that there exists an element $\gamma\in\mathbb{F}_{q^{2^{a}}}$ satisfying $\gamma+\gamma^{q^{2^{a-1}}}=0.$ Then the Hermitian trace inner product on $\mathbb{F}_{q^{t}}^{n}$ is a map $(\cdot,\cdot)_{\gamma}:\mathbb{F}_{q^{t}}^{n}\times\mathbb{F}_{q^{t}}^{n}\rightarrow\mathbb{F}_{q},$ defined as $(a,b)_{\gamma}=\sum\limits_{j=0}^{n-1}\text{Tr}_{q,t}(\gamma a_{j}b_{j}^{q^{t/2}})\text{ ~{}for~{}all~{} }a=(a_{0},a_{1},\cdots,a_{n-1}),~{}b=(b_{0},b_{1},\ldots,b_{n-1})\in\mathbb{F}_{q^{t}}^{n},$ and the map $\left[\cdot,\cdot\right]_{\gamma}:\mathcal{R}_{n}^{(q^{t})}\times\mathcal{R}_{n}^{(q^{t})}\rightarrow\mathcal{R}_{n}^{(q)}$ is defined as $\left[a(X),b(X)\right]_{\gamma}=\sum\limits_{w=0}^{t-1}\tau_{q^{w},1}\left(\gamma a(X)\tau_{q^{t/2},-1}(b(X))\right)\text{ ~{}for~{}all~{} }a(X),~{}b(X)\in\mathcal{R}_{n}^{(q^{t})}.$ By Lemma 6 of Huffman [9], we see that $\left[\cdot,\cdot\right]_{0}$ is a non-degenerate, reflexive and Hermitian $\tau_{1,-1}$-sesquilinear form on $\mathcal{R}_{n}^{(q^{t})},$ while $\left[\cdot,\cdot\right]_{\gamma}$ is a non-degenerate, reflexive and skew-Hermitian $\tau_{1,-1}$-sesquilinear form on $\mathcal{R}_{n}^{(q^{t})}.$ Sharma and Kaur [16] studied dual codes of cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-codes with respect to the bilinear form $\ast$ on $\mathbb{F}_{q^{t}}^{n},$ which is defined only for integers $t\geq 2$ satisfying $t\not\equiv 1(\text{mod }p).$ In order to define the form $\ast$, we see that for any integer $t\geq 2$ satisfying $t\not\equiv 1(\text{mod }p),$ the map $\phi:\mathbb{F}_{q^{t}}\rightarrow\mathbb{F}_{q^{t}},$ defined as $\phi(\alpha)=\alpha^{q}+\alpha^{q^{2}}+\cdots+\alpha^{q^{t-1}}\text{ for each }\alpha\in\mathbb{F}_{q^{t}},$ is an $\mathbb{F}_{q}$-linear vector space automorphism. Now the $\ast$-bilinear form on $\mathbb{F}_{q^{t}}^{n}$ is a map $(\cdot,\cdot)_{\ast}:\mathbb{F}_{q^{t}}^{n}\times\mathbb{F}_{q^{t}}^{n}\rightarrow\mathbb{F}_{q},$ defined as $(a,b)_{\ast}=\sum\limits_{j=0}^{n-1}\text{Tr}_{q,t}(a_{j}\phi(b_{j}))\text{ ~{}for~{}all~{} }a=(a_{0},a_{1},\cdots,a_{n-1}),~{}b=(b_{0},b_{1},\ldots,b_{n-1})\text{ ~{}in~{}}\mathbb{F}_{q^{t}}^{n},$ and the map $\left[\cdot,\cdot\right]_{\ast}:\mathcal{R}_{n}^{(q^{t})}\times\mathcal{R}_{n}^{(q^{t})}\rightarrow\mathcal{R}_{n}^{(q)}$ is defined as \left[a(X),b(X)\right]_{\ast}=\displaystyle\sum_{u=0}^{t-1}\tau_{q^{u},1}\biggl{(}a(X)\displaystyle\sum_{w=1}^{t-1}\tau_{q^{w},-1}\bigl{(}b(X)\bigr{)}\biggr{)}\text{ ~{}for~{}all~{} }a(X),~{}b(X)\in\mathcal{R}_{n}^{(q^{t})}. By Lemma 3.3 of Sharma and Kaur [16], we see that $\left[\cdot,\cdot\right]_{\ast}$ is a non-degenerate, reflexive and Hermitian $\tau_{1,-1}$-sesquilinear form on $\mathcal{R}_{n}^{(q^{t})}.$

Next for $\delta\in\{\ast,0,\gamma\},$ throughout this paper, let $\mathbb{T}_{\delta}$ be defined as (i) the set of all integers $t\geq 2$ satisfying $t\not\equiv 1(\text{mod }p)$ when $\delta=\ast,$ (ii) the set of all integers $t\geq 2$ when $\delta=0,$ and (iii) the set of all even integers $t\geq 2$ when $\delta=\gamma.$ Now for each $\delta\in\{\ast,0,\gamma\}$ with $t\in\mathbb{T}_{\delta},$ the $\delta$-dual code of $\mathcal{C}$ is defined as $\mathcal{C}^{\perp_{\delta}}=\{v\in\mathbb{F}_{q^{t}}^{n}:\left(v,c\right)_{\delta}=0\text{ for all }c\in\mathcal{C}\}.$ It is easy to see that the dual code $\mathcal{C}^{\perp_{\delta}}$ is also an $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-code of length $n.$ We also note that if the code $\mathcal{C}$ is cyclic, then its dual code $\mathcal{C}^{\perp_{\delta}}$ is also cyclic. In fact, if $\mathcal{C}\subseteq\mathcal{R}_{n}^{(q^{t})}$ is any cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-code, then one can easily view its dual code $\mathcal{C}^{\perp_{\delta}}\subseteq\mathcal{R}_{n}^{(q^{t})}$ as the dual code of $\mathcal{C}$ with respect to the form $\left[\cdot,\cdot\right]_{\delta}$ on $\mathcal{R}_{n}^{(q^{t})}.$ Further, by Theorem 4 of Huffman [9], we see that $\mathcal{C}=\mathcal{C}_{0}\oplus\mathcal{C}_{1}\oplus\cdots\oplus\mathcal{C}_{s-1}$ and $\mathcal{C}^{\perp_{\delta}}=\mathcal{C}_{0}^{(\delta)}\oplus\mathcal{C}_{1}^{(\delta)}\oplus\cdots\oplus\mathcal{C}_{s-1}^{(\delta)},$ where $\mathcal{C}_{i}=\mathcal{C}\cap\mathcal{J}_{i}$ and $\mathcal{C}_{i}^{(\delta)}=\mathcal{C}^{\perp_{\delta}}\cap\mathcal{J}_{i}$ are $\mathcal{K}_{i}$-subspaces of $\mathcal{J}_{i}$ for $0\leq i\leq s-1.$ In addition, by Theorem 7 of Huffman [9] and Theorem 4.1 of Sharma and Kaur [16], we have $\mathcal{C}_{\mu(i)}^{(\delta)}=\{a(X)\in\mathcal{J}_{\mu(i)}:\left[a(X),c(X)\right]_{\delta}=0\text{ for all }c(X)\in\mathcal{C}_{i}\}$ and $\text{dim}_{\mathcal{K}_{\mu(i)}}\mathcal{C}_{\mu(i)}^{(\delta)}=t-\text{dim}_{\mathcal{K}_{i}}\mathcal{C}_{i}$ for $0\leq i\leq s-1,$ (throughout this paper, $\text{dim}_{F}V$ denotes the dimension of a finite-dimensional vector space $V$ over the field $F$). Now for $\delta\in\{\ast,0,\gamma\},$ the code $\mathcal{C}$ is said to be $\delta$-complementary-dual if it satisfies $\mathcal{C}\cap\mathcal{C}^{\perp_{\delta}}=\{0\}.$ For more details, one may refer to Huffman [9] and Sharma and Kaur [16].

Henceforth, we shall follow the same notations and terminology as in Section 2.

3 Determination of the number of $\delta$-complementary-dual cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-codes

Throughout this paper, let $n$ be a positive integer, $q$ be a power of the prime $p$ with $\gcd(n,q)=1$ and $t\in\mathbb{T}_{\delta}$ for $\delta\in\{\ast,0,\gamma\}.$ In this section, we shall count all the $\delta$-complementary-dual cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-codes of length $n$ for each $\delta\in\{\ast,0,\gamma\}.$ For this, we first recall that the permutation $\mu$ of the set $\{0,1,2,\cdots,s-1\}$ is either the identity or is a product of transpositions, and it satisfies $\mu(0)=0$ and $\mu(i^{\#})=i^{\#}$ (provided $n$ is even). Let $\mathfrak{I}=\{0\}$ if $n$ is odd and $\mathfrak{I}=\{0,i^{\#}\}$ if $n$ is even. Let $\mathfrak{F}$ denote the set consisting of all the fixed points of $\mu$ excluding the points of $\mathfrak{I}$ and $\mathfrak{M}$ denote the set consisting of exactly one element from each of the transpositions in $\mu.$ Throughout this paper, we shall denote the $q$-binomial coefficient or Gaussian binomial coefficient by ${\cdot\brack\cdot}_{q},$ which is defined as ${a\brack b}_{q}=\prod\limits_{i=0}^{b-1}\left(\frac{q^{a}-q^{i}}{q^{b}-q^{i}}\right),$ where $a,b$ are integers satisfying $1\leq b\leq a.$ From now onwards, we shall denote the number of isotropic vectors and hyperbolic pairs in a reflexive and non-degenerate formed space of dimension $n$ and Witt index $m$ by $I_{m,n-2m}$ and $H_{m,n-2m},$ respectively.

In the following theorem, we enumerate all the $\delta$-complementary-dual cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-codes of length $n$ for each $\delta\in\{\ast,0,\gamma\}.$

Theorem 3.1**.**

Let $q$ be a power of the prime $p$ and $n$ be a positive integer with $\gcd(n,q)=1.$ Then for $\delta\in\{\ast,0,\gamma\}$ with $t\in\mathbb{T}_{\delta},$ the number $N$ of distinct $\delta$-complementary-dual cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-codes of length $n$ is given by

[TABLE]

where $\mathfrak{R}$ equals

• •

$2+\displaystyle\sum\limits_{\stackrel{{\scriptstyle k=2}}{{k\equiv 0(\text{mod }2)}}}^{t-1}q^{\frac{k(t-k)}{2}}{t/2\brack k/2}_{q^{2}}$* when either $\delta=\ast$ and both $t,q$ are even or $\delta=\gamma$ and $t$ is even;*

• •

$\displaystyle 2+\sum\limits_{\stackrel{{\scriptstyle k=1}}{{k\equiv 0(\text{mod }2)}}}^{t-1}q^{\frac{k(t-k+1)}{2}}{(t-1)/2\brack k/2}_{q^{2}}+\sum\limits_{\stackrel{{\scriptstyle k=1}}{{k\equiv 1(\text{mod }2)}}}^{t-1}q^{\frac{(t-k)(k+1)}{2}}{(t-1)/2\brack(k-1)/2}_{q^{2}}\vspace{2mm}$* when $\delta\in\{\ast,0\}$ and both $t,q$ are odd;*

• •

$\displaystyle 2+\sum\limits_{\stackrel{{\scriptstyle k=1}}{{k\equiv 0(\text{mod }2)}}}^{t-1}q^{\frac{k(t-k)}{2}}{t/2\brack k/2}_{q^{2}}+\sum\limits_{\stackrel{{\scriptstyle k=1}}{{k\equiv 1(\text{mod }2)}}}^{t-1}q^{\frac{tk-k^{2}-1}{2}}(q^{\frac{t}{2}}+1){(t-2)/2\brack(k-1)/2}_{q^{2}}\vspace{2mm}$* when $\delta\in\{\ast,0\}$ with either $t$ even and $q\equiv 1(\text{mod }4)$ or $t\equiv 0(\text{mod }4)$ and $q\equiv 3(\text{mod }4).$*

• •

$\displaystyle 2+\sum\limits_{\stackrel{{\scriptstyle k=1}}{{k\equiv 0(\text{mod }2)}}}^{t-1}q^{\frac{k(t-k)}{2}}{t/2\brack k/2}_{q^{2}}+\sum\limits_{\stackrel{{\scriptstyle k=1}}{{k\equiv 1(\text{mod }2)}}}^{t-1}q^{\frac{tk-k^{2}-1}{2}}(q^{\frac{t}{2}}-1){(t-2)/2\brack(k-1)/2}_{q^{2}}\vspace{2mm}$* when $\delta\in\{\ast,0\},$ $t\equiv 2(\text{mod }4)$ and $q\equiv 3(\text{mod }4).$*

• •

$2+\displaystyle\sum\limits_{\stackrel{{\scriptstyle k=1}}{{k\equiv 0(\text{mod }2)}}}^{t-1}q^{\frac{k(t-k+1)}{2}}{(t-1)/2\brack k/2}_{q^{2}}+\sum\limits_{\stackrel{{\scriptstyle k=1}}{{k\equiv 1(\text{mod }2)}}}^{t-1}q^{\frac{(t-k)(k+1)}{2}}{(t-1)/2\brack(k-1)/2}_{q^{2}}\vspace{2mm}$* when $\delta=0,$ $q$ is even and $t$ is odd;*

• •

\displaystyle 2+\sum\limits_{\stackrel{{\scriptstyle k=1}}{{k\equiv 0(\text{mod }2)}}}^{t-1}q^{\frac{tk-k^{2}-2}{2}}\Big{\{}(q^{k}+q-1){(t-2)/2\brack k/2}_{q^{2}}+(q^{t-k+1}-q^{t-k}+1){(t-2)/2\brack(k-2)/2}_{q^{2}}\Big{\}}+\\ \sum\limits_{\stackrel{{\scriptstyle k=1}}{{k\equiv 1(\text{mod }2)}}}^{t-1}q^{\frac{tk-k^{2}+t-1}{2}}{(t-2)/2\brack(k-1)/2}_{q^{2}}\vspace{2mm}* when $\delta=0$ and both $q,t$ are even.*

To prove the above theorem, we first observe the following:

Proposition 3.1**.**

Let $q$ be a power of the prime $p$ and $n$ be a positive integer coprime to $q.$ Then for $\delta\in\{\ast,0,\gamma\}$ with $t\in\mathbb{T}_{\delta},$ the following hold:

1. (a)

Let $\mathcal{C}\subseteq\mathcal{R}_{n}^{(q^{t})}$ be a cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-code of length $n.$ Let us write $\mathcal{C}=\mathcal{C}_{0}\oplus\mathcal{C}_{1}\oplus\cdots\oplus\mathcal{C}_{s-1}$ and $\mathcal{C}^{\perp_{\delta}}=\mathcal{C}_{0}^{(\delta)}\oplus\mathcal{C}_{1}^{(\delta)}\oplus\cdots\oplus\mathcal{C}_{s-1}^{(\delta)},$ where $\mathcal{C}_{i}=\mathcal{C}\cap\mathcal{J}_{i}$ and $\mathcal{C}_{i}^{(\delta)}=\mathcal{C}^{\perp_{\delta}}\cap\mathcal{J}_{i}$ for $0\leq i\leq s-1.$ Then the code $\mathcal{C}$ is ${\delta}$-complementary-dual if and only if $\mathcal{C}_{i}\cap\mathcal{C}_{i}^{(\delta)}=\{0\}$ for $0\leq i\leq s-1.$ 2. (b)

For each $i\in\mathfrak{F}\cup\mathfrak{I},$ let $N_{i}$ denote the number of $\mathcal{K}_{i}$-subspaces $\mathcal{C}_{i}$ of $\mathcal{J}_{i}$ satisfying $\mathcal{C}_{i}\cap\mathcal{C}_{i}^{(\delta)}=\{0\}.$ For each $h\in\mathfrak{M},$ let $N_{h}$ denote the number of pairs $(\mathcal{C}_{h},\mathcal{C}_{\mu(h)})$ with $\mathcal{C}_{h}$ as a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{C}_{\mu(h)}$ as a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}$ satisfying $\mathcal{C}_{h}\cap\mathcal{C}_{h}^{(\delta)}=\{0\}$ and $\mathcal{C}_{\mu(h)}\cap\mathcal{C}_{\mu(h)}^{(\delta)}=\{0\}.$ Then the total number of distinct $\delta$-complementary-dual cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-codes of length $n$ is given by $\text{N}=\prod\limits_{i\in\mathfrak{F}\cup\mathfrak{I}}N_{i}\prod\limits_{h\in\mathfrak{M}}N_{h}.$

Proof.

Proof is trivial. ∎

In view of the above proposition, we see that to enumerate all $\delta$-complementary-dual cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-codes of length $n$ for each $\delta\in\{\ast,0,\gamma\},$ we need to determine the following:

• •

For each $i\in\mathfrak{F}\cup\mathfrak{I},$ the number $N_{i}$ of distinct $\mathcal{K}_{i}$-subspaces $\mathcal{C}_{i}$ of $\mathcal{J}_{i}$ satisfying $\mathcal{C}_{i}\cap\mathcal{C}_{i}^{(\delta)}=\{0\}.$

• •

For each $h\in\mathfrak{M},$ the number $N_{h}$ of pairs $\left(\mathcal{C}_{h},\mathcal{C}_{\mu(h)}\right)$ with $\mathcal{C}_{h}$ as a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{C}_{\mu(h)}$ as a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}$ satisfying $\mathcal{C}_{h}\cap\mathcal{C}_{h}^{(\delta)}=\{0\}$ and $\mathcal{C}_{\mu(h)}\cap\mathcal{C}_{\mu(h)}^{(\delta)}=\{0\}.$

First of all, we shall determine the number $N_{i}$ for each $i\in\mathfrak{F}\cup\mathfrak{I}.$ For this, we need the following lemma:

Lemma 3.1**.**

Let $i\in\mathfrak{F}\cup\mathfrak{I}$ be fixed. For each $\delta\in\{\ast,0,\gamma\},$ let $\left[\cdot,\cdot\right]_{\delta}\restriction_{\mathcal{J}_{i}\times\mathcal{J}_{i}}$ denote the restriction of the $\tau_{1,-1}$-sesquilinear form $\left[\cdot,\cdot\right]_{\delta}$ to $\mathcal{J}_{i}\times\mathcal{J}_{i}.$ Then the following hold.

1. (a)

For each $\delta\in\{\ast,0,\gamma\},$ the form $\left[\cdot,\cdot\right]_{\delta}\restriction_{\mathcal{J}_{i}\times\mathcal{J}_{i}}$ is reflexive and non-degenerate. 2. (b)

For $\delta\in\{\ast,0\},$ the form $\left[\cdot,\cdot\right]_{\delta}\restriction_{\mathcal{J}_{i}\times\mathcal{J}_{i}}$ is Hermitian when $i\in\mathfrak{F}$ and is symmetric when $i\in\mathfrak{I}.$ For $\delta=\gamma,$ the form $\left[\cdot,\cdot\right]_{\gamma}\restriction_{\mathcal{J}_{i}\times\mathcal{J}_{i}}$ is skew-Hermitian when $i\in\mathfrak{F}$ and is alternating when $i\in\mathfrak{I}.$

Proof.

It follows from Lemmas 6 and 13 of Huffman [9] and Lemma 4.3 of Sharma and Kaur [16]. ∎

Next we make the following observation.

Remark 3.1**.**

Let $\mathcal{C}$ be a cyclic $\mathbb{F}_{q}$-linear $\mathbb{F}_{q^{t}}$-code of length $n.$ For each $i\in\mathfrak{F}\cup\mathfrak{I}$ and $\delta\in\{\ast,0,\gamma\},$ if $\mathcal{C}_{i}=\mathcal{C}\cap\mathcal{J}_{i},$ then by Lemma 3.1, the $\mathcal{K}_{i}$-subspace $\mathcal{C}_{i}^{(\delta)}=\mathcal{C}^{\perp_{\delta}}\cap\mathcal{J}_{i}=\{a(X)\in\mathcal{J}_{i}:\left[a(X),c(X)\right]_{\delta}=0\text{ for all }c(X)\in\mathcal{C}_{i}\}$ can be viewed as the orthogonal complement of $\mathcal{C}_{i}$ with respect to the restricted form $\left[\cdot,\cdot\right]_{\delta}\restriction_{\mathcal{J}_{i}\times\mathcal{J}_{i}}.$ From this, it follows that a $\mathcal{K}_{i}$-subspace $\mathcal{C}_{i}$ of $\mathcal{J}_{i}$ satisfies $\mathcal{C}_{i}\cap\mathcal{C}_{i}^{(\delta)}=\{0\}$ if and only if $\mathcal{C}_{i}$ is a non-degenerate $\mathcal{K}_{i}$-subspace of $\left(\mathcal{J}_{i},\left[\cdot,\cdot\right]_{\delta}\restriction_{\mathcal{J}_{i}\times\mathcal{J}_{i}}\right).$ Therefore the number $N_{i}$ equals the number of non-degenerate $\mathcal{K}_{i}$-subspaces of $\left(\mathcal{J}_{i},\left[\cdot,\cdot\right]_{\delta}\restriction_{\mathcal{J}_{i}\times\mathcal{J}_{i}}\right)$ for each $i\in\mathfrak{F}\cup\mathfrak{I}$ and $\delta\in\{\ast,0,\gamma\}.$

3.1 Determination of the number $N_{i}$ when $i\in\mathfrak{F}$

Throughout this section, we assume that $i\in\mathfrak{F}.$ Here we recall that $\mathcal{J}_{i}$ is a $t$-dimensional vector space over $\mathcal{K}_{i}\simeq\mathbb{F}_{q^{d_{i}}}.$ Further, by Lemma 10(i) of Huffman [9], we see that the integer $d_{i}$ is even. Then in the following proposition, we determine the number $N_{i}$ for each $\delta\in\{\ast,0,\gamma\}.$

Proposition 3.2**.**

Let $i\in\mathfrak{F}$ be fixed. Then for each $\delta\in\{\ast,0,\gamma\},$ the number $N_{i}$ of distinct $\mathcal{K}_{i}$-subspaces $\mathcal{C}_{i}$ of $\mathcal{J}_{i}$ satisfying $\mathcal{C}_{i}\cap\mathcal{C}_{i}^{(\delta)}=\{0\}$ is given by $N_{i}=2+\displaystyle\sum\limits_{k=1}^{t-1}q^{\frac{k(t-k)d_{i}}{2}}\prod\limits_{j=0}^{k-1}\left(\frac{q^{\frac{(t-j)d_{i}}{2}}-(-1)^{t-j}}{q^{\frac{(k-j)d_{i}}{2}}-(-1)^{k-j}}\right).$

Proof.

To prove this, for each integer $k~{}(0\leq k\leq t),$ let $N_{i,k}$ denote the number of $k$-dimensional $\mathcal{K}_{i}$-subspaces $\mathcal{C}_{i}$ of $\mathcal{J}_{i}$ satisfying $\mathcal{C}_{i}\cap\mathcal{C}_{i}^{(\delta)}=\{0\}.$ Then we have $N_{i}=\sum\limits_{k=0}^{t}N_{i,k}.$

When $k=0,$ we have $\mathcal{C}_{i}=\{0\},$ which gives $\mathcal{C}_{i}^{(\delta)}=\mathcal{J}_{i}$ and thus $\mathcal{C}_{i}\cap\mathcal{C}_{i}^{(\delta)}=\{0\}$ holds. This implies that $N_{i,0}=1.$ Moreover, by Lemma 3.1(a), we see that $\left(\mathcal{J}_{i},\left[\cdot,\cdot\right]_{\delta}\restriction_{\mathcal{J}_{i}\times\mathcal{J}_{i}}\right)$ is a $t$-dimensional reflexive and non-degenerate space over $\mathcal{K}_{i}$. This implies that when $\mathcal{C}_{i}=\mathcal{J}_{i},$ we have $\mathcal{C}_{i}^{(\delta)}=\{0\},$ and thus $\mathcal{C}_{i}\cap\mathcal{C}_{i}^{(\delta)}=\{0\}$ holds. This gives $N_{i,t}=1.$ So from now onwards, we assume that $1\leq k\leq t-1$ and we assert that $N_{i,k}=\displaystyle q^{\frac{k(t-k)d_{i}}{2}}\prod\limits_{j=0}^{k-1}\left(\frac{q^{\frac{(t-j)d_{i}}{2}}-(-1)^{t-j}}{q^{\frac{(k-j)d_{i}}{2}}-(-1)^{k-j}}\right)$ for $1\leq k\leq t-1.$ To prove this assertion, by Remark 3.1, we note that the number $N_{i,k}$ is equal to the number of $k$-dimensional non-degenerate $\mathcal{K}_{i}$-subspaces of $\mathcal{J}_{i}$ for $1\leq k\leq t-1.$ Now we shall consider the following two cases separately: **I. ** $\delta\in\{\ast,0\}$ and **II. ** $\delta=\gamma.$

Case I. Let $\delta\in\{\ast,0\}.$ Here by Lemma 3.1, we see that $\left[\cdot,\cdot\right]_{\delta}\restriction_{\mathcal{J}_{i}\times\mathcal{J}_{i}}$ is a reflexive, non-degenerate and Hermitian form, i.e., $\left(\mathcal{J}_{i},\left[\cdot,\cdot\right]_{\delta}\restriction_{\mathcal{J}_{i}\times\mathcal{J}_{i}}\right)$ is a unitary space over $\mathcal{K}_{i}.$

First let $k$ be odd. By [18, p. 116], we see that any $k$-dimensional non-degenerate (and hence unitary) $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}$ has a Witt decomposition of the form $\langle a_{1}(X),b_{1}(X)\rangle\perp\cdots\perp\langle a_{\frac{k-1}{2}}(X),b_{\frac{k-1}{2}}(X)\rangle\perp\langle w(X)\rangle,$ where \big{(}a_{\ell}(X),b_{\ell}(X)\big{)} is a hyperbolic pair in $\mathcal{J}_{i}$ for $1\leq\ell\leq\frac{k-1}{2}$ and $w(X)$ is an anisotropic vector in $\mathcal{J}_{i}.$ Now in order to determine the number of $k$-dimensional non-degenerate (and hence unitary) $\mathcal{K}_{i}$-subspaces of $\mathcal{J}_{i},$ we shall count all Witt bases of the type $\{a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-1}{2}}(X),b_{\frac{k-1}{2}}(X),w(X)\}$ in $\mathcal{J}_{i}.$ For this, we note that $(a_{1}(X),b_{1}(X))$ is a hyperbolic pair in $\mathcal{J}_{i},$ so by Corollary 10.6 of [18], it has $H_{\mathfrak{r}_{1},t-2\mathfrak{r}_{1}}$ choices, where $\mathfrak{r}_{1}$ is the Witt index of $\mathcal{J}_{i}.$ Now since $\langle a_{1}(X),b_{1}(X)\rangle$ is a non-degenerate $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i},$ working as in Proposition 2.9 of [6], we can write $\mathcal{J}_{i}=\langle a_{1}(X),b_{1}(X)\rangle\perp\langle a_{1}(X),b_{1}(X)\rangle^{\perp_{\delta}},$ where $\langle a_{1}(X),b_{1}(X)\rangle^{\perp_{\delta}}$ is a $(t-2)$-dimensional non-degenerate $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}.$ Next we need to choose the second hyperbolic pair $(a_{2}(X),b_{2}(X))$ from the unitary $\mathcal{K}_{i}$-subspace $\langle a_{1}(X),b_{1}(X)\rangle^{\perp_{\delta}}$ of $\mathcal{J}_{i}.$ By Corollary 10.6 of [18], we see that $(a_{2}(X),b_{2}(X))$ has $H_{\mathfrak{r}_{2},t-2-2\mathfrak{r}_{2}}$ choices, where $\mathfrak{r}_{2}$ is the Witt index of $\langle a_{1}(X),b_{1}(X)\rangle^{\perp_{\delta}}.$ Continuing like this, we see that the $(\frac{k-1}{2})$th hyperbolic pair $(a_{\frac{k-1}{2}}(X),b_{\frac{k-1}{2}}(X))$ has to be chosen from the $(t-k+3)$-dimensional unitary $\mathcal{K}_{i}$-subspace $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-3}{2}}(X),b_{\frac{k-3}{2}}(X)\rangle^{\perp}$ of $\mathcal{J}_{i},$ so by Corollary 10.6 of [18], the hyperbolic pair $(a_{\frac{k-1}{2}}(X),b_{\frac{k-1}{2}}(X))$ has $H_{\mathfrak{r}_{\frac{k-1}{2}},t-k+3-2\mathfrak{r}_{\frac{k-1}{2}}}$ choices, where $\mathfrak{r}_{\frac{k-1}{2}}$ is the Witt index of $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-3}{2}},b_{\frac{k-3}{2}}\rangle^{\perp_{\delta}}.$ Next we need to choose an anisotropic vector $w(X)$ from the $(t-k+1)$-dimensional unitary $\mathcal{K}_{i}$-subspace $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-1}{2}},b_{\frac{k-1}{2}}\rangle^{\perp_{\delta}}$ of $\mathcal{J}_{i}.$ By Lemma 10.4 of [18], we see that $w(X)$ has $q^{(t-k+1)d_{i}}-1-I_{\mathfrak{r}_{\frac{k+1}{2}},t-k+1-2\mathfrak{r}_{\frac{k+1}{2}}}$ choices, where $\mathfrak{r}_{\frac{k+1}{2}}$ is the Witt index of $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-1}{2}},b_{\frac{k-1}{2}}\rangle^{\perp_{\delta}}.$ Therefore the number of choices for a Witt basis of cardinality $k$ in $\mathcal{J}_{i}$ is given by

[TABLE]

By [18, p. 116], for $1\leq\ell\leq\frac{k+1}{2},$ we note that

[TABLE]

Now using Lemma 10.4 and Corollary 10.6 of [18], we see that $\mathfrak{U}_{k,t}=q^{\frac{(2t-k-1)kd_{i}}{4}}(q^{\frac{d_{i}}{2}}-1)\prod\limits_{j=0}^{k-1}(q^{\frac{(t-j)d_{i}}{2}}-(-1)^{t-j}).$ Next working in a similar manner as above and using Lemma 10.4 and Corollary 10.6 of [18], we see that the number of Witt bases of a $k$-dimensional unitary $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}$ is given by $\mathfrak{U}_{k,k}=H_{\frac{k-1}{2},1}H_{\frac{k-3}{2},1}\cdots H_{1,1}(q^{d_{i}}-1)=q^{\frac{k(k-1)d_{i}}{4}}(q^{\frac{d_{i}}{2}}-1)\prod\limits_{j=0}^{k-1}(q^{\frac{(k-j)d_{i}}{2}}-(-1)^{k-j}).$ Therefore the number of distinct $k$-dimensional non-degenerate (and hence unitary) $\mathcal{K}_{i}$-subspaces of $\mathcal{J}_{i}$ is given by

[TABLE]

Next let $k$ be even. Here by [18, p. 116], we see that any $k$-dimensional non-degenerate (and hence unitary) $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}$ has a Witt decomposition of the form $\langle a_{1}(X),b_{1}(X)\rangle\perp\cdots\perp\langle a_{\frac{k}{2}}(X),b_{\frac{k}{2}}(X)\rangle,$ where \big{(}a_{\ell}(X),b_{\ell}(X)\big{)} is a hyperbolic pair in $\mathcal{J}_{i}$ for $1\leq\ell\leq\frac{k}{2}.$ Now working in a similar manner as in the previous case, the assertion follows.

Case II. Let $\delta=\gamma.$ By Lemma 3.1, we see that $\left[\cdot,\cdot\right]_{\gamma}\restriction_{\mathcal{J}_{i}\times\mathcal{J}_{i}}$ is a reflexive, non-degenerate and skew-Hermitian form. In this case, we will first transform the skew-Hermitian form $\left[\cdot,\cdot\right]_{\gamma}\restriction_{\mathcal{J}_{i}\times\mathcal{J}_{i}}$ to a Hermitian form $\left[\cdot,\cdot\right]_{\gamma}^{\prime}$ on $\mathcal{J}_{i}$ as follows:

By Lemma 11(iii) of Huffman [9], we note that $\tau_{1,-1}$ is not the identity map on $\mathcal{K}_{i},$ so there exists $\mathfrak{u}(X)\in\mathcal{K}_{i}$ such that \tau_{1,-1}\bigl{(}\mathfrak{u}(X)\bigr{)}\neq\mathfrak{u}(X). Then we define a map $\left[\cdot,\cdot\right]_{\gamma}^{\prime}:\mathcal{J}_{i}\times\mathcal{J}_{i}\rightarrow\mathcal{K}_{i}$ as $\left[c(X),d(X)\right]_{\gamma}^{\prime}=\mathfrak{v}(X)\left[c(X),d(X)\right]_{\gamma}$ for all $c(X),d(X)\in\mathcal{J}_{i},$ where \mathfrak{v}(X)=\mathfrak{u}(X)-\tau_{1,-1}\bigl{(}\mathfrak{u}(X)\bigr{)}\neq 0. It is easy to see that $\left[\cdot,\cdot\right]_{\gamma}^{\prime}$ is a reflexive, non-degenerate and Hermitian form on $\mathcal{J}_{i}.$ Next we observe that if two vectors in $\mathcal{J}_{i}$ are orthogonal with respect to $\left[\cdot,\cdot\right]_{\gamma}\restriction_{\mathcal{J}_{i}\times\mathcal{J}_{i}},$ then those two vectors are also orthogonal with respect to $\left[\cdot,\cdot\right]_{\gamma}^{\prime}$ and vice versa. This implies that any non-degenerate $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}$ with respect to $\left[\cdot,\cdot\right]_{\gamma}\restriction_{\mathcal{J}_{i}\times\mathcal{J}_{i}}$ is also non-degenerate with respect to $\left[\cdot,\cdot\right]_{\gamma}^{\prime}$ and vice versa. So the number $N_{i}$ equals the number of non-degenerate $\mathcal{K}_{i}$-subspaces of $\mathcal{J}_{i}$ with respect to $\left[\cdot,\cdot\right]_{\gamma}^{\prime}.$ Now as $(\mathcal{J}_{i},\left[\cdot,\cdot\right]_{\gamma}^{\prime})$ is a unitary space over $\mathcal{K}_{i}$ having dimension $t,$ working in a similar manner as in case I, the assertion follows. This proves the proposition. ∎

3.2 Determination of the number $N_{i}$ when $i\in\mathfrak{I}$

In this section, we will determine the number $N_{i}$ for each $i\in\mathfrak{I}$ and $\delta\in\{\ast,0,\gamma\}.$ Here we recall that $\mathcal{K}_{i}\simeq\mathbb{F}_{q},$ as $d_{i}=1$ for each $i\in\mathfrak{I}.$

In the following proposition, we suppose that either $\delta=\gamma$ or $\delta=\ast$ and $q$ is even, and determine the number $N_{i}$ of $\mathcal{K}_{i}$-subspaces $\mathcal{C}_{i}$ of $\mathcal{J}_{i}$ satisfying $\mathcal{C}_{i}\cap\mathcal{C}_{i}^{(\delta)}=\{0\}$ for each $i\in\mathfrak{I}.$

Proposition 3.3**.**

Suppose that either $\delta=\gamma$ or $\delta=\ast$ and $q$ is even. For $i\in\mathfrak{I},$ the number $N_{i}$ of distinct $\mathcal{K}_{i}$-subspaces $\mathcal{C}_{i}$ of $\mathcal{J}_{i}$ satisfying $\mathcal{C}_{i}\cap\mathcal{C}_{i}^{(\delta)}=\{0\}$ is given by $N_{i}=2+\displaystyle\sum\limits_{\stackrel{{\scriptstyle k=1}}{{k\equiv 0(\text{mod }2)}}}^{t-1}q^{\frac{k(t-k)}{2}}{t/2\brack k/2}_{q^{2}}.$

Proof.

In order to prove this, for each integer $k~{}(0\leq k\leq t),$ let $N_{i,k}$ denote the number of $k$-dimensional $\mathcal{K}_{i}$-subspaces $\mathcal{C}_{i}$ of $\mathcal{J}_{i}$ satisfying $\mathcal{C}_{i}\cap\mathcal{C}_{i}^{(\delta)}=\{0\}.$ Then we have $N_{i}=\sum\limits_{k=0}^{t}N_{i,k}.$

First we observe that $N_{i,0}=N_{i,t}=1.$ So from now onwards, throughout this proof, we assume that $1\leq k\leq t-1.$ Here we assert that for $1\leq k\leq t-1,$ the number $N_{i,k}=0$ when $k$ is odd and $N_{i,k}=q^{\frac{k(t-k)}{2}}{t/2\brack k/2}_{q^{2}}$ when $k$ is even. To prove this assertion, by Remark 3.1, we note that the number $N_{i,k}$ is equal to the number of $k$-dimensional non-degenerate $\mathcal{K}_{i}$-subspaces of $\mathcal{J}_{i}$ for $1\leq k\leq t-1.$ Now we shall distinguish the cases, I. $\delta=\gamma$ and II. $\delta=\ast$ and $q$ is even.

Case I. Let $\delta=\gamma.$ Here by Lemma 3.1, we see that $\left[\cdot,\cdot\right]_{\gamma}\restriction_{\mathcal{J}_{i}\times\mathcal{J}_{i}}$ is a reflexive, non-degenerate and alternating form, i.e., $(\mathcal{J}_{i},\left[\cdot,\cdot\right]_{\gamma}\restriction_{\mathcal{J}_{i}\times\mathcal{J}_{i}})$ is a symplectic space over $\mathcal{K}_{i}.$ In view of this, we see that the number $N_{i,k}$ equals the number of distinct $k$-dimensional non-degenerate (and hence symplectic) $\mathcal{K}_{i}$-subspaces of $\mathcal{J}_{i}.$ By [18, p. 69], we see that $N_{i,k}=0$ if $k$ is odd. So we need to determine the number $N_{i,k}$ when $k\geq 2$ is an even integer. For this, by [18, p. 69], we see that when $k$ is even, any $k$-dimensional non-degenerate (and hence symplectic) $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}$ has a Witt decomposition of the form $\langle a_{1}(X),b_{1}(X)\rangle\perp\cdots\perp\langle a_{\frac{k}{2}}(X),b_{\frac{k}{2}}(X)\rangle,$ where each $(a_{\ell}(X),b_{\ell}(X))$ is a hyperbolic pair in $\mathcal{J}_{i}.$ Further, working in a similar manner as in Proposition 3.2 and using [18, pp. 69-70], we get $N_{i,k}=q^{\frac{k(t-k)}{2}}{t/2\brack k/2}_{q^{2}}$ for each even integer $k.$

Case II. Let $\delta=\ast$ and $q$ be even. In this case, $t$ is even and $n$ is odd, and hence $i=0$ only. By Huffman [9, p. 264], we see that $\mathcal{J}_{0}=\{aM(X):a\in\mathbb{F}_{q^{t}}\}\simeq\mathbb{F}_{q^{t}}$ and $\mathcal{K}_{0}=\{rM(X):r\in\mathbb{F}_{q}\}\simeq\mathbb{F}_{q},$ where $M(X)=1+X+X^{2}+\cdots+X^{n-1}.$ We next observe that \left[aM(X),bM(X)\right]_{\ast}=n\text{Tr}_{q,t}\bigl{(}a\phi(b)\bigr{)}M(X)=n(a,b)_{\ast}M(X) for all $aM(X),bM(X)\in\mathcal{J}_{0}.$ Since $M(X)\neq 0,$ we observe that if two vectors $aM(X),bM(X)\in\mathcal{J}_{0}$ are orthogonal with respect to $\left[\cdot,\cdot\right]_{\ast}\restriction_{\mathcal{J}_{0}\times\mathcal{J}_{0}},$ then the corresponding vectors $a,b\in\mathbb{F}_{q^{t}}$ are orthogonal with respect to $(\cdot,\cdot)_{\ast}$ on $\mathbb{F}_{q^{t}}$ and vice versa. So the number of distinct $k$-dimensional non-degenerate $\mathcal{K}_{0}$-subspaces of $\mathcal{J}_{0}$ with respect to $\left[\cdot,\cdot\right]_{\ast}\restriction_{\mathcal{J}_{0}\times\mathcal{J}_{0}}$ is equal to the number of distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ with respect to $(\cdot,\cdot)_{\ast}$ on $\mathbb{F}_{q^{t}}.$ Further, by Lemma 3.2 of Sharma and Kaur [16], we see that when $q$ is even, $(\cdot,\cdot)_{\ast}$ is a reflexive, non-degenerate and alternating form on $\mathbb{F}_{q^{t}},$ i.e., $(\mathbb{F}_{q^{t}},(\cdot,\cdot)_{\ast})$ is a symplectic space over $\mathbb{F}_{q}$ having dimension $t.$ Now working in a similar manner as in case I, we obtain $N_{0,k}=q^{\frac{k(t-k)}{2}}{t/2\brack k/2}_{q^{2}}$ when $k$ is even and $N_{0,k}=0$ when $k$ is odd. This proves the proposition. ∎

In the following proposition, we suppose that $\delta\in\{\ast,0\}$ and $q$ is odd, and determine the number of $\mathcal{K}_{i}$-subspaces $\mathcal{C}_{i}$ of $\mathcal{J}_{i}$ satisfying $\mathcal{C}_{i}\cap\mathcal{C}_{i}^{(\delta)}=\{0\}$ for each $i\in\mathfrak{I}.$

Proposition 3.4**.**

Let $\delta\in\{\ast,0\}$ and $q$ be an odd prime power. For $i\in\mathfrak{I},$ the number $N_{i}$ of distinct $\mathcal{K}_{i}$-subspaces $\mathcal{C}_{i}$ of $\mathcal{J}_{i}$ satisfying $\mathcal{C}_{i}\cap\mathcal{C}_{i}^{(\delta)}=\{0\}$ is given by the following:

1. (a)

$N_{i}=\displaystyle 2+\sum\limits_{\stackrel{{\scriptstyle k=1}}{{k\equiv 0(\text{mod }2)}}}^{t-1}q^{\frac{k(t-k+1)}{2}}{(t-1)/2\brack k/2}_{q^{2}}+\sum\limits_{\stackrel{{\scriptstyle k=1}}{{k\equiv 1(\text{mod }2)}}}^{t-1}q^{\frac{(t-k)(k+1)}{2}}{(t-1)/2\brack(k-1)/2}_{q^{2}}$* if $t$ is odd.* 2. (b)

$N_{i}=\displaystyle 2+\sum\limits_{\stackrel{{\scriptstyle k=1}}{{k\equiv 0(\text{mod }2)}}}^{t-1}q^{\frac{k(t-k)}{2}}{t/2\brack k/2}_{q^{2}}+\sum\limits_{\stackrel{{\scriptstyle k=1}}{{k\equiv 1(\text{mod }2)}}}^{t-1}q^{\frac{tk-k^{2}-1}{2}}(q^{\frac{t}{2}}+1){(t-2)/2\brack(k-1)/2}_{q^{2}}\vspace{2mm}$* if either $t$ is even and $q\equiv 1(\text{mod }4)$ or $t\equiv 0(\text{mod }4)$ and $q\equiv 3(\text{mod }4).$* 3. (c)

$N_{i}=\displaystyle 2+\sum\limits_{\stackrel{{\scriptstyle k=1}}{{k\equiv 0(\text{mod }2)}}}^{t-1}q^{\frac{k(t-k)}{2}}{t/2\brack k/2}_{q^{2}}+\sum\limits_{\stackrel{{\scriptstyle k=1}}{{k\equiv 1(\text{mod }2)}}}^{t-1}q^{\frac{tk-k^{2}-1}{2}}(q^{\frac{t}{2}}-1){(t-2)/2\brack(k-1)/2}_{q^{2}}\vspace{2mm}$* if $t\equiv 2(\text{mod }4)$ and $q\equiv 3(\text{mod }4).$*

To prove this proposition, we see, by Lemma 3.1, that when $\delta\in\{\ast,0\}$ and $q$ is an odd prime power, $\left[\cdot,\cdot\right]_{\delta}\restriction_{\mathcal{J}_{i}\times\mathcal{J}_{i}}$ is a reflexive, non-degenerate and symmetric form, i.e., $\left(\mathcal{J}_{i},\left[\cdot,\cdot\right]_{\delta}\restriction_{\mathcal{J}_{i}\times\mathcal{J}_{i}}\right)$ is an orthogonal space over $\mathcal{K}_{i}.$ Furthermore, since $q$ is odd, the map $Q_{i}:\mathcal{J}_{i}\rightarrow\mathcal{K}_{i},$ defined by $Q_{i}(u(X))=\frac{1}{2}\left[u(X),u(X)\right]_{\delta}$ for each $u(X)\in\mathcal{J}_{i},$ is a quadratic map, and hence $\left(\mathcal{J}_{i},Q_{i}\right)$ is a non-degenerate quadratic space over $\mathcal{K}_{i}.$ Now to prove Proposition 3.4, we shall first prove the following lemma:

Lemma 3.2**.**

Let $\delta\in\{\ast,0\},$ $q$ be an odd prime power and $k\geq 1$ be an odd integer. Let $U$ be a $k$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}$ having a Witt decomposition of the form $\langle m_{1}(X),n_{1}(X)\rangle\perp\langle m_{2}(X),n_{2}(X)\rangle\perp\cdots\perp\langle m_{\frac{k-1}{2}}(X),n_{\frac{k-1}{2}}(X)\rangle\perp\langle r(X)\rangle,$ where $(m_{\ell}(X),n_{\ell}(X))$ is a hyperbolic pair in $U$ for $1\leq\ell\leq\frac{k-1}{2}$ and $r(X)$ is a non-singular vector in $U.$ Then the following hold.

• (a)

The number $\mathsf{L}_{k}$ of non-singular vectors $u(X)\in U$ with $\left<u(X)\right>$ isometric to $\langle r(X)\rangle$ is given by $\mathsf{L}_{k}=\frac{q^{\frac{k-1}{2}}(q-1)(q^{\frac{k-1}{2}}+1)}{2}.$

• (b)

The number $\mathsf{M}_{k}$ of non-singular vectors $u(X)\in U$ with $\left<u(X)\right>$ not isometric to $\langle r(X)\rangle$ is given by $\mathsf{M}_{k}=\frac{q^{\frac{k-1}{2}}(q-1)(q^{\frac{k-1}{2}}-1)}{2}.$

Proof.

• (a)

First of all, we observe that for each odd integer $k\geq 1,$ the number $\mathsf{L}_{k}$ is equal to the number of pairs $\left(u(X),\theta(X)\right),$ where $u(X)$ is a non-singular vector in $U$ and $\theta(X)$ is a non-zero square in $\mathcal{K}_{i}$ satisfying $\left[u(X),u(X)\right]_{\delta}=\theta(X)\left[r(X),r(X)\right]_{\delta}.$ Now to determine the number $\mathsf{L}_{k},$ we will apply induction on an odd integer $k\geq 1.$

When $k=1,$ we have $U=\left<r(X)\right>.$ From this, it follows trivially that $\mathsf{L}_{1}=q-1,$ i.e., the result holds for $k=1.$ Next we assume that $k\geq 3$ is an odd integer and the result holds for all $(k-2)$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspaces of $\mathcal{J}_{i}.$ Now let $U$ be a $k$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}.$ As $k\geq 3,$ by Lemma 7.3 and Theorem 11.2 of [18], there exists a hyperbolic pair in $U,$ say $(e(X),f(X)).$ Further by Proposition 2.9 of [6], we write $U=\langle e(X),f(X)\rangle\perp\langle e(X),f(X)\rangle^{\perp_{\delta}},$ where $\langle e(X),f(X)\rangle^{\perp_{\delta}}$ is a $(k-2)$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspace of $U.$ In view of this, any element $u(X)\in U$ can be written as $u(X)=\lambda(X)e(X)+\kappa(X)f(X)+w(X),$ where $\lambda(X),\kappa(X)\in\mathcal{K}_{i}$ and $w(X)\in\langle e(X),f(X)\rangle^{\perp_{\delta}}.$ Now we shall count all pairs $(u(X),\theta(X))$ with $u(X)$ as a non-singular vector in $U$ and $\theta(X)$ as a non-zero square in $\mathcal{K}_{i}$ satisfying $\left[u(X),u(X)\right]_{\delta}=\theta(X)\left[r(X),r(X)\right]_{\delta}.$ For this, we see that $\left[u(X),u(X)\right]_{\delta}=\theta(X)\left[r(X),r(X)\right]_{\delta}$ if and only if

[TABLE]

Here also, we shall distinguish the following two cases: I. $\lambda(X)\kappa(X)=0$ and II. $\lambda(X)\kappa(X)\neq 0.$

I. Let $\lambda(X)\kappa(X)=0.$ Then by (1), we have $\left[w(X),w(X)\right]_{\delta}=\theta(X)\left[r(X),r(X)\right]_{\delta},$ that is, the $\mathcal{K}_{i}$-subspaces $\langle w(X)\rangle$ and $\langle r(X)\rangle$ are isometric. As $w(X)\in\left<e(X),f(X)\right>^{\perp},$ by applying the induction hypothesis, we see that the pair $(w(X),\theta(X))$ has precisely $\mathsf{L}_{k-2}$ relevant choices. Moreover, the pair $(\lambda(X),\kappa(X))$ has $2q-1$ choices. Thus in this case, there are precisely $(2q-1)\mathsf{L}_{k-2}$ number of pairs $(u(X),\theta(X)),$ where $u(X)$ is a non-singular vector of $U$ and $\theta(X)$ is a non-zero square in $\mathcal{K}_{i}$ satisfying $\left[u(X),u(X)\right]_{\delta}=\theta(X)\left[r(X),r(X)\right]_{\delta}.$

II. Next let $\lambda(X)\kappa(X)\neq 0.$ Then by (1), we get $\left[w(X),w(X)\right]_{\delta}\neq\theta(X)\left[r(X),r(X)\right]_{\delta},$ that is, the $\mathcal{K}_{i}$-subspace $\langle w(X)\rangle$ is not isometric to $\langle r(X)\rangle.$ By applying the induction hypothesis, we see that the number of pairs $(w(X),\theta(X))$ with $\theta(X)$ as a non-zero square in $\mathcal{K}_{i}$ and $w(X)\in\left<e(X),f(X)\right>^{\perp}$ satisfying $\left[w(X),w(X)\right]_{\delta}\neq\theta(X)\left[r(X),r(X)\right]_{\delta}$ is given by $\frac{q^{k-2}(q-1)}{2}-\mathsf{L}_{k-2}.$ In view of (1), we see that $\lambda(X)$ is fixed for a given choice of $\theta(X),w(X)$ and $\kappa(X).$ Also as $\kappa(X)\neq 0,$ it has $q-1$ choices. Thus in this case, there are precisely (q-1)\bigg{(}\frac{q^{k-2}(q-1)}{2}-\mathsf{L}_{k-2}\bigg{)} number of pairs $(u(X),\theta(X))$ with $u(X)$ as a non-singular vector of $U$ and $\theta(X)$ as a non-zero square in $\mathcal{K}_{i}$ satisfying $\left[u(X),u(X)\right]_{\delta}=\theta(X)\left[r(X),r(X)\right]_{\delta}.$

Therefore from the cases I and II above, we see that the number $\mathsf{L}_{k}$ of pairs $(u(X),\theta(X))$ with $u(X)$ as a non-singular vector in $U$ and $\theta(X)$ as a non-zero square in $\mathcal{K}_{i}$ satisfying $\left[u(X),u(X)\right]_{\delta}=\theta(X)\left[r(X),r(X)\right]_{\delta},$ is given by \mathsf{L}_{k}=(2q-1)\mathsf{L}_{k-2}+(q-1)\bigg{(}\frac{q^{k-2}(q-1)}{2}-\mathsf{L}_{k-2}\bigg{)}. Further, after a simple computation and on substituting the value of $\mathsf{L}_{1}=q-1,$ we obtain $\mathsf{L}_{k}=\frac{q^{\frac{k-1}{2}}(q-1)(q^{\frac{k-1}{2}}+1)}{2},$ which proves (a).

• (b)

To prove this, we observe that for each odd integer $k\geq 1,$ the number $\mathsf{L}_{k}+\mathsf{M}_{k}$ is equal to the number of non-singular vectors in $U.$ Furthermore, by Theorem 11.5 of [18], we see that the number of non-singular vectors in $U$ is given by $q^{k}-1-I_{\frac{k-1}{2},1}=q^{k}-q^{k-1},$ where $I_{\frac{k-1}{2},1}$ denotes the number of isotropic vectors in $U$ having Witt index $\frac{k-1}{2}.$ From this and using part (a), we obtain $\mathsf{M}_{k}=q^{k}-q^{k-1}-\mathsf{L}_{k}=\frac{q^{\frac{k-1}{2}}(q-1)(q^{\frac{k-1}{2}}-1)}{2},$ which proves (b).

∎

Proof of Proposition 3.4. To prove this, for each integer $k~{}(0\leq k\leq t),$ let $N_{i,k}$ denote the number of $k$-dimensional $\mathcal{K}_{i}$-subspaces $\mathcal{C}_{i}$ of $\mathcal{J}_{i}$ satisfying $\mathcal{C}_{i}\cap\mathcal{C}_{i}^{(\delta)}=\{0\}.$ Then we have

[TABLE]

Now we proceed to determine the number $N_{i,k}$ for each integer $k,$ $0\leq k\leq t.$ Towards this, it is easy to observe that $N_{i,0}=N_{i,t}=1.$ So we assume $1\leq k\leq t-1$ from now onwards. Here by Remark 3.1, we note that the number $N_{i,k}$ is equal to the number of $k$-dimensional non-degenerate $\mathcal{K}_{i}$-subspaces of $\mathcal{J}_{i}$ for $1\leq k\leq t-1.$ Besides this, we recall that $\left(\mathcal{J}_{i},Q_{i}\right)$ is a non-degenerate quadratic space over $\mathcal{K}_{i}\simeq\mathbb{F}_{q},$ where the quadratic map $Q_{i}:\mathcal{J}_{i}\rightarrow\mathcal{K}_{i}$ is defined as $Q_{i}(u(X))=\frac{1}{2}\left[u(X),u(X)\right]_{\delta}$ for each $u(X)\in\mathcal{J}_{i}.$ Then we observe the following:

1. ($\square$)

The discriminant of every 2-dimensional anisotropic $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}$ is a non-square in $\mathcal{K}_{i}.$ As a consequence, every 2-dimensional anisotropic $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}$ is non-degenerate. 2. ($\blacksquare$)

Every 2-dimensional anisotropic $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}$ has an orthogonal basis, i.e., it has a basis of the type $\{u(X),v(X)\}$ with $\left[u(X),v(X)\right]_{\delta}=0.$

Furthermore, by Proposition 1 of Drees et al. [3], we see that the determinant of $\left(\mathcal{J}_{i},Q_{i}\right)$ is a non-square in $\mathcal{K}_{i}.$ From this, by [18, p. 138] and assertion ($\square$), it is easy to observe (see Huffman [9, p. 279]) that the Witt index $\mathfrak{w}$ of $\left(\mathcal{J}_{i},Q_{i}\right)$ is given by

[TABLE]

Now in order to determine the number $N_{i,k},$ by [18, p. 138], we see that any $k$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}$ has a Witt decomposition of the form $\langle a_{1}(X),b_{1}(X)\rangle\perp\cdots\perp\langle a_{\mathfrak{m}_{k}}(X),b_{\mathfrak{m}_{k}}(X)\rangle\perp W_{k},$ where each $(a_{\ell}(X),b_{\ell}(X))$ is a hyperbolic pair in $\mathcal{J}_{i},$ $\mathfrak{m}_{k}$ is its Witt index and $W_{k}$ is an anisotropic $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}$ with $\text{dim}_{\mathcal{K}_{i}}W_{k}=k-2\mathfrak{m}_{k}\leq 2.$ By [18, p. 138], we also see that $\mathfrak{m}_{k}=\frac{k-1}{2}$ when $k$ is odd and $\mathfrak{m}_{k}$ is either $\frac{k-2}{2}$ or $\frac{k}{2}$ (depending upon the isometry class of $\mathcal{J}_{i})$ when $k$ is even. We shall first determine the number $\mathfrak{Q}_{\mathfrak{m}_{k},\mathfrak{w}}$ of Witt bases of the type $\{a_{1}(X),b_{1}(X),\cdots,a_{\mathfrak{m}_{k}}(X),b_{\mathfrak{m}_{k}}(X)\}\cup\beta_{W_{k}}$ with $\beta_{W_{k}}$ as a basis of the anisotropic $\mathcal{K}_{i}$-subspace $W_{k}$ of $\mathcal{J}_{i},$ where the basis $\beta_{W_{k}}$ is orthogonal when $\text{dim}_{\mathcal{K}_{i}}W_{k}=2$ (or equivalently, $\mathfrak{m}_{k}=\frac{k-2}{2}$). For this, we shall distinguish the following two cases: **I. ** $k$ is odd and **II. ** $k$ is even.

I. Let $k$ be odd. Here, by [18, p. 138], we have $\mathfrak{m}_{k}=\frac{k-1}{2},$ which implies that $\text{dim}_{\mathcal{K}_{i}}W_{k}=1.$ So in this case, any $k$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}$ has a Witt decomposition of the form $\langle a_{1}(X),b_{1}(X)\rangle\perp\cdots\perp\langle a_{\frac{k-1}{2}}(X),b_{\frac{k-1}{2}}(X)\rangle\perp\langle w(X)\rangle,$ where $(a_{\ell}(X),b_{\ell}(X))$ is a hyperbolic pair in $\mathcal{J}_{i}$ for $1\leq\ell\leq\frac{k-1}{2}$ and $w(X)$ is a non-singular vector of $\mathcal{J}_{i}.$ We first note that as $(a_{1}(X),b_{1}(X))$ is a hyperbolic pair in $\mathcal{J}_{i},$ by [18, p. 141], it has $H_{\mathfrak{w},t-2\mathfrak{w}}$ choices, where $\mathfrak{w}$ is the Witt index of $\mathcal{J}_{i}.$ Further by Proposition 2.9 of [6], we write $\mathcal{J}_{i}=\langle a_{1}(X),b_{1}(X)\rangle\perp\langle a_{1}(X),b_{1}(X)\rangle^{\perp_{\delta}},$ where $\langle a_{1}(X),b_{1}(X)\rangle^{\perp_{\delta}}$ is a $(t-2)$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}.$ Now we choose the second hyperbolic pair $(a_{2}(X),b_{2}(X))$ from the quadratic space $\langle a_{1}(X),b_{1}(X)\rangle^{\perp_{\delta}}.$ By applying Witt’s cancellation theorem, we see that the space $\langle a_{1}(X),b_{1}(X)\rangle^{\perp_{\delta}}$ has Witt index $\mathfrak{w}-1.$ By [18, p. 141], again, we see that the second hyperbolic pair $(a_{2}(X),b_{2}(X))$ has $H_{\mathfrak{w}-1,t-2\mathfrak{w}}$ choices. Continuing like this, we see that the $(\frac{k-1}{2})$th hyperbolic pair $(a_{\frac{k-1}{2}}(X),b_{\frac{k-1}{2}}(X))$ can be chosen in $H_{\mathfrak{w}-\frac{(k-3)}{2},t-2\mathfrak{w}}$ number of ways. Next to count all possible choices for the basis $\beta_{W_{k}}$ of a 1-dimensional anisotropic $\mathcal{K}_{i}$-subspace $W_{k}$ of $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-1}{2}}(X),b_{\frac{k-1}{2}}(X)\rangle^{\perp_{\delta}},$ we need to choose a non-singular vector $w(X)$ from the $(t-k+1)$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspace $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-1}{2}}(X),b_{\frac{k-1}{2}}(X)\rangle^{\perp_{\delta}}$ of $\mathcal{J}_{i}.$ By applying Witt’s cancellation theorem, we see that the Witt index of $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-1}{2}}(X),b_{\frac{k-1}{2}}(X)\rangle^{\perp_{\delta}}$ is $\mathfrak{w}-\frac{(k-1)}{2}.$ Now by Theorem 11.5 of [18], we see that the number of non-singular vectors in $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-1}{2}}(X),b_{\frac{k-1}{2}}(X)\rangle^{\perp_{\delta}}$ is given by $q^{t-k+1}-1-I_{\mathfrak{w}-\frac{(k-1)}{2},t-k+1-2(\mathfrak{w}-\frac{(k-1)}{2})}.$ From this, it follows that when $k$ is odd, the number of Witt bases of the form $\{a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-1}{2}}(X),b_{\frac{k-1}{2}}(X)\}\cup\beta_{W_{k}}$ and having cardinality $k$ in $\mathcal{J}_{i}$ is given by

[TABLE]

This, by [18, pp. 140-141], gives

[TABLE]

Next when $k$ is odd, let $\mathfrak{Q}_{\frac{k-1}{2}}$ denote the number of Witt bases of a $k$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}.$ Here working in a similar manner as above and by [18, pp. 138-141], we obtain

[TABLE]

Therefore when $k$ is odd, using (5) and (6), we obtain

[TABLE]

II. Next let $k$ be even. Here by [18, p. 138], we see that $\mathfrak{m}_{k}$ is either $\frac{k-2}{2}$ or $\frac{k}{2}.$ For each even integer $k~{}(1\leq k\leq t-1),$ let $R_{i,k}$ denote the number of $k$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspaces of $\mathcal{J}_{i}$ having Witt index as $\mathfrak{m}_{k}=\frac{k-2}{2}$ and $S_{i,k}$ denote the number of $k$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspaces of $\mathcal{J}_{i}$ having Witt index as $\mathfrak{m}_{k}=\frac{k}{2}.$ In view of this, we have

[TABLE]

Now we shall distinguish the following two cases: **A. ** $\mathfrak{m}_{k}=\frac{k-2}{2}$ and **B. ** $\mathfrak{m}_{k}=\frac{k}{2}.$

A. First let $\mathfrak{m}_{k}=\frac{k-2}{2}.$ In this case, by [18, p. 138], we see that any $k$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}$ has a Witt decomposition of the form $\langle a_{1}(X),b_{1}(X)\rangle\perp\cdots\perp\langle a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle\perp W_{k},$ where each $(a_{\ell}(X),b_{\ell}(X))$ is a hyperbolic pair in $\mathcal{J}_{i}$ and $W_{k}$ is a 2-dimensional anisotropic $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}.$ We note that as $(a_{1}(X),b_{1}(X))$ is a hyperbolic pair in $\mathcal{J}_{i},$ by [18, p. 141], it has $H_{\mathfrak{w},t-2\mathfrak{w}}$ choices, where $\mathfrak{w}$ is the Witt index of $\mathcal{J}_{i}.$ Further by Proposition 2.9 of [6], we write $\mathcal{J}_{i}=\langle a_{1}(X),b_{1}(X)\rangle\perp\langle a_{1}(X),b_{1}(X)\rangle^{\perp_{\delta}},$ where $\langle a_{1}(X),b_{1}(X)\rangle^{\perp_{\delta}}$ is a $(t-2)$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}.$ Continuing like this, we see that the $(\frac{k-2}{2})$th hyperbolic pair $(a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X))$ can be chosen in $H_{\mathfrak{w}-\frac{(k-4)}{2},t-2\mathfrak{w}}$ number of ways. In order to determine the number $\mathfrak{Q}_{\frac{k-2}{2},\mathfrak{w}},$ we need to determine the number $\mathbb{A}_{k,\mathfrak{w}}$ of orthogonal bases of all 2-dimensional anisotropic $\mathcal{K}_{i}$-subspaces of the $(t-k+2)$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspace $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ of $\mathcal{J}_{i}.$ For this, by applying Witt’s cancellation theorem, we see that the space $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ has Witt index as $\mathfrak{w}-\frac{(k-2)}{2}.$ Now we assert that

[TABLE]

To prove assertion (9), we will consider the following three cases separately: (i) $\mathfrak{w}=\frac{t}{2},$ (ii) $\mathfrak{w}=\frac{t-1}{2}$ and (iii) $\mathfrak{w}=\frac{t-2}{2}.$

(i) First let $\mathfrak{w}=\frac{t}{2}.$ Here we assert the following:

1. (a)

The number of orthogonal bases of the type $\{u(X),v(X)\}$ with $u(X),v(X)$ as non-singular vectors of $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}},$ is given by $q^{\frac{3(t-k)}{2}}(q-1)^{2}(q^{\frac{t-k+2}{2}}-1).$ 2. (b)

The number of orthogonal bases of the type $\{u(X),v(X)\}$ with $u(X),v(X)$ as non-singular vectors of $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ and $\left<u(X),v(X)\right>$ containing a singular vector, is given by $\frac{q^{t-k}(q-1)^{2}(q^{\frac{t-k+2}{2}}-1)(q^{\frac{t-k}{2}}+1)}{2}.$

To prove (a), we need to count orthogonal bases of the type $\{u(X),v(X)\}$ with $u(X),v(X)$ as non-singular vectors of $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}.$ For this, we see that as $\mathfrak{w}=\frac{t}{2},$ by Witt’s cancellation theorem, the Witt index of $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ is $\frac{t}{2}-\frac{(k-2)}{2}=\frac{t-k+2}{2}.$ Now as $u(X)$ is a non-singular vector in $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}},$ by Theorem 11.5 of [18], we see that $u(X)$ has $q^{t-k+2}-1-I_{\frac{(t-k+2)}{2},0}$ choices. Further, by applying Proposition 2.9 of [6], we write $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}=\langle u(X)\rangle\perp\langle u(X)\rangle^{\perp_{\delta}},$ where $\langle u(X)\rangle^{\perp_{\delta}}$ is a $(t-k+1)$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspace of $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}.$ Now we will choose another non-singular vector $v(X)$ from the space $\langle u(X)\rangle^{\perp_{\delta}}.$ As $u(X)$ is non-singular, we see that $u(X)\not\in\langle u(X)\rangle^{\perp_{\delta}}.$ Moreover, since $t-k+1$ is odd, by [18, p. 138], we see that the Witt index of $\langle u(X)\rangle^{\perp_{\delta}}$ is $\frac{t-k}{2}.$ Then we see that $v(X)$ has $q^{t-k+1}-1-I_{\frac{t-k}{2},1}$ choices. From this and using Theorem 11.5 of [18], we see that the number of choices for the basis $\{u(X),v(X)\}$ with $u(X),v(X)$ as non-singular vectors of $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ and satisfying $\left[u(X),v(X)\right]_{\delta}=0,$ is given by $(q^{t-k+2}-1-I_{\frac{(t-k+2)}{2},0})(q^{t-k+1}-1-I_{\frac{(t-k)}{2},1})=q^{\frac{3(t-k)}{2}}(q-1)^{2}(q^{\frac{t-k+2}{2}}-1).$

To prove (b), we see that if there exists a singular vector, say $r_{1}(X),$ in a 2-dimensional non-degenerate $\mathcal{K}_{i}$-subspace $\langle u(X),v(X)\rangle$ of $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ satisfying Q_{i}\big{(}u(X)\big{)}\neq 0,~{}Q_{i}\big{(}v(X)\big{)}\\ \neq 0 and $\left[u(X),v(X)\right]_{\delta}=0,$ then by Lemma 7.3 of [18], we have $\langle u(X),v(X)\rangle=\langle r_{1}(X),r_{2}(X)\rangle,$ where $(r_{1}(X),r_{2}(X))$ is a hyperbolic pair in the $\mathcal{K}_{i}$-subspace $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ of $\mathcal{J}_{i}.$ By [18, p. 138], we see that the space $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ contains $H_{\frac{t-k+2}{2},0}$ hyperbolic pairs. On the other hand, by [18, p. 138] again, we see that there are precisely $H_{1,0}$ distinct hyperbolic pairs in a 2-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}$ containing a singular vector. Thus the number of distinct 2-dimensional non-degenerate $\mathcal{K}_{i}$-subspaces of $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ containing a singular vector is given by $\frac{H_{\frac{t-k+2}{2},0}}{H_{1,0}},$ which equals $\frac{q^{t-k}(q^{\frac{t-k+2}{2}}-1)(q^{\frac{t-k}{2}}+1)}{2(q-1)}$ by [18, p. 141]. Further, we observe that within each such non-degenerate quadratic $\mathcal{K}_{i}$-subspace $\langle u(X),v(X)\rangle=\left<r_{1}(X),r_{2}(X)\right>$ of $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ containing a singular vector, there are precisely $(q-1)^{3}$ distinct choices for a basis of the type $\{u(X),v(X)\}$ satisfying Q_{i}\big{(}u(X)\big{)}\neq 0,~{}Q_{i}\big{(}v(X)\big{)}\neq 0 and $\left[u(X),v(X)\right]_{\delta}=0.$ Thus there are precisely $\Delta_{k}=\frac{q^{t-k}(q-1)^{3}(q^{\frac{t-k+2}{2}}-1)(q^{\frac{t-k}{2}}+1)}{2(q-1)}$ choices for a basis of the type $\{u(X),v(X)\}$ with $u(X),v(X)\in\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ satisfying Q_{i}\big{(}u(X)\big{)}\neq 0,~{}Q_{i}\big{(}v(X)\big{)}\neq 0, $\left[u(X),v(X)\right]_{\delta}=0$ and $\langle u(X),v(X)\rangle$ containing a singular vector.

Now from assertions (a) and (b), it follows that the number of orthogonal bases of 2-dimensional anisotropic $\mathcal{K}_{i}$-subspaces of $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ is given by $\mathbb{A}_{k,\frac{t}{2}}=q^{\frac{3(t-k)}{2}}(q-1)^{2}(q^{\frac{t-k+2}{2}}\\ -1)-\Delta_{k}=\frac{q^{t-k}(q-1)^{2}(q^{\frac{t-k+2}{2}}-1)(q^{\frac{t-k}{2}}-1)}{2},$ which proves assertion (9) when $\mathfrak{w}=\frac{t}{2}.$

(ii) Next let $\mathfrak{w}=\frac{t-1}{2}.$ Here we assert the following:

1. (c)

The number of orthogonal bases of the type $\{u(X),v(X)\}$ with $u(X),v(X)$ as non-singular vectors of $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ is given by $q^{t-k}(q-1)^{2}(q^{t-k+1}-1).$ 2. (d)

The number of orthogonal bases of the type $\{u(X),v(X)\}$ with $u(X),v(X)$ as non-singular vectors of $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ and $\left<u(X),v(X)\right>$ containing a singular vector, is given by $\frac{q^{t-k}(q-1)^{2}(q^{t-k+1}-1)}{2}.$

To prove this assertion, as $\mathfrak{w}=\frac{t-1}{2},$ by [18, p. 138], we write $\mathcal{J}_{i}=\langle r_{1}(X),s_{1}(X)\rangle\perp\cdots\perp\langle r_{\frac{t-1}{2}}(X),s_{\frac{t-1}{2}}(X)\rangle\perp\langle\eta(X)\rangle,$ where $(r_{\ell}(X),s_{\ell}(X))$ is a hyperbolic pair in $\mathcal{J}_{i}$ for $1\leq\ell\leq\frac{t-1}{2}$ and $\eta(X)$ is a non-singular vector of $\mathcal{J}_{i}.$

In order to prove (c), by applying Witt’s cancellation theorem, we see that the Witt index of the $\mathcal{K}_{i}$-subspace $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ of $\mathcal{J}_{i}$ is $\frac{(t-1)}{2}-\frac{(k-2)}{2}=\frac{t-k+1}{2}.$ Now we need to choose a non-singular vector $u(X)$ from the space $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}.$ Here the following two cases arise:

• •

the non-singular vector $u(X)\in\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ is such that the $\mathcal{K}_{i}$-subspace $\langle u(X)\rangle$ is isometric to $\langle\eta(X)\rangle,$ and

• •

the non-singular vector $u(X)\in\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ is such that the $\mathcal{K}_{i}$-subspace $\langle u(X)\rangle$ is not isometric to $\langle\eta(X)\rangle.$

First of all, we choose a non-singular vector $u(X)$ from $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ such that $\langle u(X)\rangle$ is isometric to $\langle\eta(X)\rangle.$ In this case, by Lemma 3.2(a), we see that $u(X)$ has $\mathsf{L}_{t-k+2}$ choices. Next for each such choice of $u(X),$ by applying Proposition 2.9 of [6], we write $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),\\ b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}=\langle u(X)\rangle\perp\langle u(X)\rangle^{\perp_{\delta}},$ where $\langle u(X)\rangle^{\perp_{\delta}}$ is a $(t-k+1)$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspace of $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}.$ Here as $\langle u(X)\rangle$ and $\langle\eta(X)\rangle$ are isometric, by Witt’s cancellation theorem, we see that the Witt index of $\langle u(X)\rangle^{\perp_{\delta}}$ is $\frac{t-k+1}{2}.$ Now we will choose another non-singular vector $v(X)$ from the space $\langle u(X)\rangle^{\perp_{\delta}}.$ As $u(X)$ is non-singular, we see that $u(X)\not\in\langle u(X)\rangle^{\perp_{\delta}}.$ By [18, p. 138], we observe that $v(X)$ has $q^{t-k+1}-1-I_{\frac{(t-k+1)}{2},0}$ relevant choices. Therefore by Theorem 11.5 of [18] and Lemma 3.2(a), the number of choices for the basis $\{u(X),v(X)\}$ with $\left<u(X)\right>$ isometric to $\left<\eta(X)\right>$ and satisfying Q_{i}\big{(}u(X)\big{)}\neq 0,~{}Q_{i}\big{(}v(X)\big{)}\neq 0 and $\left[u(X),v(X)\right]_{\delta}=0,$ is given by $\mathsf{L}_{t-k+2}(q^{t-k+1}-1-I_{\frac{(t-k+1)}{2},0})=\frac{q^{t-k}(q-1)^{2}(q^{t-k+1}-1)}{2}.$

Next we choose a non-singular vector $u(X)$ from $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ such that $\langle u(X)\rangle$ is not isometric to $\langle\eta(X)\rangle.$ Here by Lemma 3.2(b), such a non-singular vector $u(X)$ has $\mathsf{M}_{t-k+2}$ choices. Now by applying Proposition 2.9 of [6], we write $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}=\langle u(X)\rangle\perp\langle u(X)\rangle^{\perp_{\delta}},$ where $\langle u(X)\rangle^{\perp_{\delta}}$ is a $(t-k+1)$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspace of $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}.$ Here as the $\mathcal{K}_{i}$-subspaces $\langle u(X)\rangle$ and $\langle\eta(X)\rangle$ of $\mathcal{J}_{i}$ are not isometric, we see, by Witt’s cancellation theorem, that the Witt index of $\langle u(X)\rangle^{\perp_{\delta}}$ is $\frac{t-k-1}{2}.$ Now we will choose another non-singular vector $v(X)$ from the space $\langle u(X)\rangle^{\perp_{\delta}}.$ As $u(X)$ is non-singular, we see that $u(X)\not\in\langle u(X)\rangle^{\perp_{\delta}}.$ By [18, p. 138], we see that $v(X)$ has $q^{t-k+1}-1-I_{\frac{(t-k-1)}{2},2}$ choices. Therefore, by Theorem 11.5 of [18] and Lemma 3.2(b), the number of choices for the basis $\{u(X),v(X)\}$ with $\left<u(X)\right>$ not isometric to $\left<\eta(X)\right>$ and satisfying Q_{i}\big{(}u(X)\big{)}\neq 0,~{}Q_{i}\big{(}v(X)\big{)}\neq 0 and $\left[u(X),v(X)\right]_{\delta}=0,$ is given by $\mathsf{M}_{t-k+2}(q^{t-k+1}-1-I_{\frac{(t-k-1)}{2},2})=\frac{q^{t-k}(q-1)^{2}(q^{t-k+1}-1)}{2}.$

From the above, we see that the number of choices for the basis $\{u(X),v(X)\}$ with $u(X),v(X)\in\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ satisfying Q_{i}\big{(}u(X)\big{)}\neq 0,~{}Q_{i}\big{(}v(X)\big{)}\neq 0 and $\left[u(X),v(X)\right]_{\delta}=0,$ is given by $\frac{q^{t-k}(q-1)^{2}(q^{t-k+1}-1)}{2}+\frac{q^{t-k}(q-1)^{2}(q^{t-k+1}-1)}{2}=q^{t-k}(q-1)^{2}(q^{t-k+1}-1),$ which proves (c).

Next working in a similar way as in assertion (b) of case (i), assertion (d) follows.

Now from assertions (c) and (d), we see that the number of orthogonal bases of 2-dimensional anisotropic $\mathcal{K}_{i}$-subspaces of $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ is given by $\mathbb{A}_{k,\frac{t-1}{2}}=q^{t-k}(q-1)^{2}(q^{t-k+1}-1)-\frac{q^{t-k}(q-1)^{2}(q^{t-k+1}-1)}{2}=\frac{q^{t-k}(q-1)^{2}(q^{t-k+1}-1)}{2},$ which proves assertion (9) when $\mathfrak{w}=\frac{t-1}{2}.$

(iii) Next suppose that $\mathfrak{w}=\frac{t-2}{2}.$ Here working in a similar manner as in case (i), we observe the following:

1. (e)

The number of orthogonal bases of the type $\{u(X),v(X)\}$ with $u(X),v(X)$ as non-singular vectors of $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ is given by $q^{\frac{3(t-k)}{2}}(q-1)^{2}(q^{\frac{t-k+2}{2}}+1).$ 2. (f)

The number of orthogonal bases of the type $\{u(X),v(X)\}$ with $u(X),v(X)$ as non-singular vectors of $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ and $\left<u(X),v(X)\right>$ containing a singular vector, is given by $\frac{q^{t-k}(q-1)^{2}(q^{\frac{t-k+2}{2}}+1)(q^{\frac{t-k}{2}}-1)}{2}.$

Now from (e) and (f), it follows that the number of orthogonal bases of 2-dimensional anisotropic $\mathcal{K}_{i}$-subspaces of $\langle a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}}$ is given by $\mathbb{A}_{k,\frac{t-2}{2}}=q^{\frac{3(t-k)}{2}}(q-1)^{2}(q^{\frac{t-k+2}{2}}+1)-\frac{q^{t-k}(q-1)^{2}(q^{\frac{t-k+2}{2}}+1)(q^{\frac{t-k}{2}}-1)}{2}=\frac{q^{t-k}(q-1)^{2}(q^{\frac{t-k+2}{2}}+1)(q^{\frac{t-k}{2}}+1)}{2},$ which proves assertion (9) when $\mathfrak{w}=\frac{t-2}{2}.$

Next we see that the number of Witt bases of the form $\{a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\}\cup\beta_{W_{k}}$ with $\beta_{W_{k}}$ as an orthogonal basis of a 2-dimensional anisotropic $\mathcal{K}_{i}$-subspace $W_{k}$ of $\langle a_{1}(X),b_{1}(X),\cdots,\\ a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\rangle^{\perp_{\delta}},$ is given by

[TABLE]

From this, using (9), (10) and [18, p. 141], we get

[TABLE]

Next let $\mathfrak{Q}_{\frac{k-2}{2}}$ denote the number of Witt bases of a $k$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}$ of the form $\{a_{1}(X),b_{1}(X),\cdots,a_{\frac{k-2}{2}}(X),b_{\frac{k-2}{2}}(X)\}\cup\beta_{W_{k}},$ where $\beta_{W_{k}}$ is an orthogonal basis of the 2-dimensional anisotropic $\mathcal{K}_{i}$-subspace $W_{k}$ of $\mathcal{J}_{i}.$ Here also, working in a similar manner as above and by [18, pp. 138-141], we obtain

[TABLE]

Therefore the number $R_{i,k}$ of $k$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspaces of $\mathcal{J}_{i}$ having Witt index as $\mathfrak{m}_{k}=\frac{k-2}{2},$ is given by $R_{i,k}=\frac{\mathfrak{Q}_{\frac{k-2}{2},\mathfrak{w}}}{\mathfrak{Q}_{\frac{k-2}{2}}}.$ From this and using (11) and (12), we obtain

[TABLE]

B. Next let $\mathfrak{m}_{k}=\frac{k}{2}.$ Here by [18, p. 138], any $k$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}$ has a Witt decomposition of the form $\langle a_{1}(X),b_{1}(X)\rangle\perp\cdots\perp\langle a_{\frac{k}{2}}(X),b_{\frac{k}{2}}(X)\rangle,$ where $(a_{\ell}(X),b_{\ell}(X))$ is a hyperbolic pair in $\mathcal{J}_{i}$ for $1\leq\ell\leq\frac{k}{2}.$ Now working as in case A, we see that the number of Witt bases of the form $\{a_{1}(X),b_{1}(X),\cdots,a_{\frac{k}{2}}(X),b_{\frac{k}{2}}(X)\}$ in $\mathcal{J}_{i}$ is given by

[TABLE]

This, by [18, p. 141], gives

[TABLE]

Next when $\mathfrak{m}_{k}=\frac{k}{2},$ let $\mathfrak{Q}_{\frac{k}{2}}$ denote the number of Witt bases of a $k$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspace of $\mathcal{J}_{i}$ having Witt index as $\frac{k}{2}.$ Here working in a similar manner as above and by [18, pp. 138-141], we obtain

[TABLE]

Therefore the number $S_{i,k}$ of $k$-dimensional non-degenerate $\mathcal{K}_{i}$-subspaces of $\mathcal{J}_{i}$ having Witt index as $\frac{k}{2},$ is given by $S_{i,k}=\frac{\mathfrak{Q}_{\frac{k}{2},\mathfrak{w}}}{\mathfrak{Q}_{\frac{k}{2}}}.$ From this and using (15) and (16), we obtain

[TABLE]

Now on substituting the values of $R_{i,k}$ and $S_{i,k}$ from (13) and (17) in (8), we obtain

[TABLE]

Further, on substituting the values of $N_{i,k}$ from (7) and (18) (accordingly as $k$ is even or odd) in (2) and using equation (3), the desired result follows. $\hfill\Box$

By closely looking at the proof of Proposition 3.4, we deduce the following divisibility results:

Corollary 3.1**.**

Let $q$ be an odd prime power and $\lambda,\mu$ be positive integers satisfying $\mu\leq\lambda.$ Then we have the following:

• (a)

If $q\equiv 3(\text{mod }4)$ and $\lambda$ is odd, then both $\displaystyle\frac{(q^{\mu}+1)(q^{\lambda-\mu}+1)}{2(q^{\lambda}+1)}{\lambda\brack\mu}_{q^{2}}$ and $\displaystyle\frac{(q^{\mu}-1)(q^{\lambda-\mu}-1)}{2(q^{\lambda}+1)}{\lambda\brack\mu}_{q^{2}}$ are integers.

• (b)

If either $q\equiv 3(\text{mod }4)$ and $\lambda$ is even or $q\equiv 1(\text{mod }4),$ then both $\displaystyle\frac{(q^{\mu}+1)(q^{\lambda-\mu}-1)}{2(q^{\lambda}-1)}{\lambda\brack\mu}_{q^{2}}$ and

$\displaystyle\frac{(q^{\mu}-1)(q^{\lambda-\mu}+1)}{2(q^{\lambda}-1)}{\lambda\brack\mu}_{q^{2}}$* are integers.*

Proof.

• (a)

Let $i\in\mathfrak{I},$ $\delta\in\{\ast,0\},$ $q\equiv 3(\text{mod }4)$ and $\dim_{\mathcal{K}_{i}}\mathcal{J}_{i}=2\lambda,$ where $\lambda$ is an odd integer. Then by (3), we see that the Witt index of $\mathcal{J}_{i}$ is $\lambda.$ Further, by (13), we see that the number of $(2\mu)$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspaces of $\mathcal{J}_{i}$ having Witt index as $(\mu-1)$ is given by $R_{i,2\mu}=q^{2\mu(\lambda-\mu)}\frac{(q^{\mu}-1)(q^{\lambda-\mu}-1)}{2(q^{\lambda}+1)}{\lambda\brack\mu}_{q^{2}},$ from which it follows that $\frac{(q^{\mu}-1)(q^{\lambda-\mu}-1)}{2(q^{\lambda}+1)}{\lambda\brack\mu}_{q^{2}}$ is an integer. Moroever, using (17), we see that the number of $(2\mu)$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspaces of $\mathcal{J}_{i}$ having Witt index as $\mu$ is given by $S_{i,2\mu}=q^{2\mu(\lambda-\mu)}\frac{(q^{\mu}+1)(q^{\lambda-\mu}+1)}{2(q^{\lambda}+1)}{\lambda\brack\mu}_{q^{2}},$ which implies that $\frac{(q^{\mu}+1)(q^{\lambda-\mu}+1)}{2(q^{\lambda}+1)}{\lambda\brack\mu}_{q^{2}}$ is an integer.

• (b)

Let $i\in\mathfrak{I},$ $\delta\in\{\ast,0\}$ and $\dim_{\mathcal{K}_{i}}\mathcal{J}_{i}=2\lambda,$ where $\lambda$ is an even integer when $q\equiv 3(\text{mod }4)$ and $\lambda\geq 1$ is any integer when $q\equiv 1(\text{mod }4).$ Here by (3), we see that the Witt index of $\mathcal{J}_{i}$ is $(\lambda-1).$ Further, by (13), we see that the number of $(2\mu)$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspaces of $\mathcal{J}_{i}$ having Witt index as $(\mu-1)$ is given by $R_{i,2\mu}=q^{2\mu(\lambda-\mu)}\frac{(q^{\mu}-1)(q^{\lambda-\mu}+1)}{2(q^{\lambda}-1)}{\lambda\brack\mu}_{q^{2}},$ from which it follows that $\frac{(q^{\mu}-1)(q^{\lambda-\mu}+1)}{2(q^{\lambda}-1)}{\lambda\brack\mu}_{q^{2}}$ is an integer. Moroever, using (17), we see that the number of $(2\mu)$-dimensional non-degenerate quadratic $\mathcal{K}_{i}$-subspaces of $\mathcal{J}_{i}$ having Witt index as $\mu$ is given by $S_{i,2\mu}=q^{2\mu(\lambda-\mu)}\frac{(q^{\mu}+1)(q^{\lambda-\mu}-1)}{2(q^{\lambda}-1)}{\lambda\brack\mu}_{q^{2}},$ which implies that $\frac{(q^{\mu}+1)(q^{\lambda-\mu}-1)}{2(q^{\lambda}-1)}{\lambda\brack\mu}_{q^{2}}$ is an integer.

∎

Next we shall consider the case $\delta=0$ and $q$ is even. As $\gcd(n,q)=1,$ $n$ must be odd, which implies that $\mathfrak{I}=\{0\},$ i.e., $i=0$ is the only choice. In the following proposition, we determine the number $N_{0}$ of $\mathcal{K}_{0}$-subspaces $\mathcal{C}_{0}$ of $\mathcal{J}_{0}$ satisfying $\mathcal{C}_{0}\cap\mathcal{C}_{0}^{(0)}=\{0\}.$

Proposition 3.5**.**

Let $\delta=0$ and $q$ be an even prime power. Then the number $N_{0}$ of $\mathcal{K}_{0}$-subspaces $\mathcal{C}_{0}$ of $\mathcal{J}_{0}$ satisfying $\mathcal{C}_{0}\cap\mathcal{C}_{0}^{(0)}=\{0\}$ is given by

1. (a)

$\displaystyle N_{0}=2+\sum\limits_{\stackrel{{\scriptstyle k=1}}{{k\equiv 0(\text{mod }2)}}}^{t-1}q^{\frac{k(t-k+1)}{2}}{(t-1)/2\brack k/2}_{q^{2}}+\sum\limits_{\stackrel{{\scriptstyle k=1}}{{k\equiv 1(\text{mod }2)}}}^{t-1}q^{\frac{(t-k)(k+1)}{2}}{(t-1)/2\brack(k-1)/2}_{q^{2}}$* if $t$ is odd.* 2. (b)

\displaystyle N_{0}=2+\sum\limits_{\stackrel{{\scriptstyle k=1}}{{k\equiv 0(\text{mod }2)}}}^{t-1}q^{\frac{tk-k^{2}-2}{2}}\Big{\{}(q^{k}+q-1){(t-2)/2\brack k/2}_{q^{2}}+(q^{t-k+1}-q^{t-k}+1){(t-2)/2\brack(k-2)/2}_{q^{2}}\Big{\}}+\sum\limits_{\stackrel{{\scriptstyle k=1}}{{k\equiv 1(\text{mod }2)}}}^{t-1}q^{\frac{tk-k^{2}+t-1}{2}}{(t-2)/2\brack(k-1)/2}_{q^{2}}* if $t$ is even.*

Proof.

To prove this, for each integer $k~{}(0\leq k\leq t),$ let $N_{0,k}$ denote the number of $k$-dimensional $\mathcal{K}_{0}$-subspaces $\mathcal{C}_{0}$ of $\mathcal{J}_{0}$ satisfying $\mathcal{C}_{0}\cap\mathcal{C}_{0}^{(0)}=\{0\}.$ Then we have $N_{0}=\sum\limits_{k=0}^{t}N_{0,k}.$ For this, we first observe that $N_{0,0}=N_{0,t}=1.$ So from now onwards, we assume that $1\leq k\leq t-1.$ By Remark 3.1, we note that the number $N_{0,k}$ is equal to the number of $k$-dimensional non-degenerate $\mathcal{K}_{0}$-subspaces of $\mathcal{J}_{0}$ for $1\leq k\leq t-1.$ To determine the numbers $N_{0,k}$ for $1\leq k\leq t-1,$ by Huffman [9, p. 264], we see that $\mathcal{J}_{0}=\{aM(X):a\in\mathbb{F}_{q^{t}}\}\simeq\mathbb{F}_{q^{t}}$ and $\mathcal{K}_{0}=\{rM(X):r\in\mathbb{F}_{q}\}\simeq\mathbb{F}_{q},$ where $M(X)=1+X+X^{2}+\cdots+X^{n-1}\neq 0.$ In view of this, it is easy to observe that $\left[aM(X),bM(X)\right]_{0}=n\text{Tr}_{q,t}(ab)M(X)=n(a,b)_{0}M(X)$ for all $a,b\in\mathbb{F}_{q^{t}}.$ From this, we observe that if two vectors $aM(X),bM(X)\in\mathcal{J}_{0}$ are orthogonal with respect to $\left[\cdot,\cdot\right]_{0}\restriction_{\mathcal{J}_{0}\times\mathcal{J}_{0}},$ then the corresponding vectors $a,b\in\mathbb{F}_{q^{t}}$ are also orthogonal with respect to ordinary trace bilinear form $(\cdot,\cdot)_{0}$ on $\mathbb{F}_{q^{t}}$ and vice versa. So for each integer $k~{}(1\leq k\leq t-1),$ the number $N_{0,k}$ of $k$-dimensional non-degenerate $\mathcal{K}_{0}$-subspaces of $\mathcal{J}_{0}$ with respect to $\left[\cdot,\cdot\right]_{0}\restriction_{\mathcal{J}_{0}\times\mathcal{J}_{0}}$ is equal to the number of $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ with respect to $(\cdot,\cdot)_{0}$ on $\mathbb{F}_{q^{t}}.$

In order to determine the number of $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ with respect to $(\cdot,\cdot)_{0},$ let $\mathcal{V}_{0}=\{x\in\mathbb{F}_{q^{t}}:\text{Tr}_{q,t}(x)=0\},$ i.e., $\mathcal{V}_{0}$ equals kernel of the trace map $\text{Tr}_{q,t}.$ Then it is well-known that $\mathcal{V}_{0}$ is an $\mathbb{F}_{q}$-subspace of $\mathbb{F}_{q^{t}}$ having dimension $t-1.$ Now we will distinguish the following two cases: $t$ is odd and $t$ is even.

(a) Let $t$ be odd. Here we see that $1\not\in\mathcal{V}_{0}$ and $(v,1)_{0}=0$ for all $v\in\mathcal{V}_{0},$ which implies that $\mathbb{F}_{q^{t}}=\mathcal{V}_{0}\perp\langle 1\rangle.$ Furthermore, by Huffman [9, p. 280], we see that $(\cdot,\cdot)_{0}$ is a reflexive, non-degenerate and alternating form on $\mathcal{V}_{0},$ i.e., $(\mathcal{V}_{0},(\cdot,\cdot)_{0}\restriction_{\mathcal{V}_{0}\times\mathcal{V}_{0}})$ is a symplectic space over $\mathbb{F}_{q}.$

We now proceed to count all $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ with respect to $(\cdot,\cdot)_{0}.$ To do so, we first observe that any $k$-dimensional $\mathbb{F}_{q}$-subspace of $\mathbb{F}_{q^{t}}$ is I. either contained in $\mathcal{V}_{0},$ or II. contained in $\mathbb{F}_{q^{t}}$ but not in $\mathcal{V}_{0},$ i.e., it is of the type $\langle x_{1},x_{2},\cdots,x_{k-1},1+x_{k}\rangle$ with $x_{j}\in\mathcal{V}_{0}\setminus\{0\}$ for $1\leq j\leq k-1$ and $x_{k}\in\mathcal{V}_{0}.$

**I.  ** As $(\mathcal{V}_{0},(\cdot,\cdot)_{0}\restriction_{\mathcal{V}_{0}\times\mathcal{V}_{0}})$ is a symplectic space, by [18, p. 69], we see that a $k$-dimensional non-degenerate (and hence symplectic) $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{0}$ exists if and only if $k$ is even. Further, working in a similar manner as in Proposition 3.3 and using the fact that $\text{dim}_{\mathbb{F}_{q}}\mathcal{V}_{0}=t-1,$ we see that the number of $k$-dimensional non-degenerate (and hence symplectic) $\mathbb{F}_{q}$-subspaces of $\mathcal{V}_{0}$ is given by $q^{\frac{k(t-k-1)}{2}}{(t-1)/2\brack k/2}_{q^{2}}$ when $k$ is even.

**II.  ** We will next count all $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the type $\langle x_{1},x_{2},\cdots,x_{k-1},1+x_{k}\rangle,$ where $x_{j}\in\mathcal{V}_{0}\setminus\{0\}$ for $1\leq j\leq k-1$ and $x_{k}\in\mathcal{V}_{0}.$

First let $k$ be even. Here when $x_{k}=0,$ by applying Theorem 5.1.1 of [17], we see that the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-1},1\rangle$ of $\mathbb{F}_{q^{t}}$ is degenerate. Next let $x_{k}\neq 0.$ In this case, by Theorem 5.1.1 of [17] again, we note that $\langle x_{1},x_{2},\cdots,x_{k-1},1+x_{k}\rangle$ is a $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of $\mathbb{F}_{q^{t}}$ if and only if $\langle x_{1},x_{2},\cdots,x_{k-1},x_{k}\rangle$ is a $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{0}.$ Now working similarly as in Proposition 3.3 again, we see that the number of distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathcal{V}_{0}$ is given by $q^{\frac{k(t-k-1)}{2}}{(t-1)/2\brack k/2}_{q^{2}}.$ Further, we observe that given any $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{0},$ there are precisely $q^{k}-1$ distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the type $\left<x_{1},x_{2},\cdots,x_{k-1},1+x_{k}\right>$ with $x_{j}$’s in $\mathcal{V}_{0}\setminus\{0\}.$ Therefore when $k$ is even, the number of distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the type $\left<x_{1},x_{2},\cdots,x_{k-1},1+x_{k}\right>$ with $x_{j}$’s in $\mathcal{V}_{0}\setminus\{0\},$ is given by $q^{\frac{k(t-k-1)}{2}}(q^{k}-1){(t-1)/2\brack k/2}_{q^{2}}.$

Next let $k$ be odd. Here using Theorem 5.1.1 of [17], we see that when $x_{k}=0,$ $\langle x_{1},x_{2},\cdots,x_{k-1},1\rangle$ is a $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of $\mathbb{F}_{q^{t}}$ if and only if $\langle x_{1},x_{2},\cdots,x_{k-1}\rangle$ is a $(k-1)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{0}.$ Now as $(\mathcal{V}_{0},(\cdot,\cdot)_{0}\restriction_{\mathcal{V}_{0}\times\mathcal{V}_{0}})$ is a symplectic space over $\mathbb{F}_{q},$ working similarly as in Proposition 3.3, we see that the number of $(k-1)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathcal{V}_{0}$ is $q^{\frac{(k-1)(t-k)}{2}}{(t-1)/2\brack(k-1)/2}_{q^{2}},$ which is equal to the number of distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the type $\langle x_{1},x_{2},\cdots,x_{k-1},1\rangle$ with $x_{j}\in\mathcal{V}_{0}\setminus\{0\}$ for $1\leq j\leq k-1.$ Next let $x_{k}$ be non-zero. In this case, applying Theorem 5.1.1 of [17] again, we see that $\langle x_{1},x_{2},\cdots,x_{k-1},1+x_{k}\rangle$ is a $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of $\mathbb{F}_{q^{t}}$ if and only if $\langle x_{1},x_{2},\cdots,x_{k-1}\rangle$ is a $(k-1)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{0}.$ Further, working in a similar way as in Proposition 3.3 again, we see that the number of $(k-1)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathcal{V}_{0}$ is $q^{\frac{(k-1)(t-k)}{2}}{(t-1)/2\brack(k-1)/2}_{q^{2}}.$ Further, for a given $(k-1)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace $\left<x_{1},x_{2},\cdots,x_{k-1}\right>$ of $\mathcal{V}_{0},$ we need to count all choices for $x_{k}\in\mathcal{V}_{0}\setminus\{0\}$ that give rise to distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces $\left<x_{1},x_{2},\cdots,x_{k-1},1+x_{k}\right>$ of $\mathbb{F}_{q^{t}}.$ To do this, using Proposition 2.9 of [6], we write $\mathcal{V}_{0}=\langle x_{1},x_{2},\cdots,x_{k-1}\rangle\perp\langle x_{1},x_{2},\cdots,x_{k-1}\rangle^{\perp_{0}},$ where the $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-1}\rangle^{\perp_{0}}$ of $\mathcal{V}_{0}$ has dimension $t-k.$ Therefore each $x_{k}\in\mathcal{V}_{0}$ can be uniquely written as $x_{k}=w+\widetilde{w},$ where $w\in\langle x_{1},x_{2},\cdots,x_{k-1}\rangle$ and $\widetilde{w}\in\langle x_{1},x_{2},\cdots,x_{k-1}\rangle^{\perp_{0}}.$ It is easy to see that the subspace $\langle x_{1},x_{2},\cdots,x_{k-1},1+w+\widetilde{w}\rangle=\langle x_{1},x_{2},\cdots,x_{k-1},1+\widetilde{w}\rangle,$ where $\widetilde{w}\in\langle x_{1},x_{2},\cdots,x_{k-1}\rangle^{\perp_{0}}.$ We further observe that each $x_{k}\in\langle x_{1},x_{2},\cdots,x_{k-1}\rangle^{\perp_{\delta}}\setminus\{0\}$ gives rise to a distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace $\left<x_{1},x_{2},\cdots,x_{k-1},1+x_{k}\right>$ of $\mathbb{F}_{q^{t}},$ and thus there are precisely $q^{t-k}-1$ relevant choices for $x_{k}.$ Therefore when $k$ is odd, the number of distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the type $\langle x_{1},x_{2},\cdots,x_{k-1},1+x_{k}\rangle$ with $x_{j}\in\mathcal{V}_{0}\setminus\{0\}$ for $1\leq j\leq k,$ is given by $q^{\frac{(k-1)(t-k)}{2}}(q^{t-k}-1){(t-1)/2\brack(k-1)/2}_{q^{2}}.$

On combining all the above cases, when $t$ is odd, for $1\leq k\leq t-1,$ we get

[TABLE]

(b) ** Let $t$ be even. Here we have $\text{Tr}_{q,t}(1)=0,$ which implies that $1\in\mathcal{V}_{0}.$ Since the trace map $\text{Tr}_{q,t}$ is non-zero, there exists $\alpha\in\mathbb{F}_{q^{t}}$ such that $\text{Tr}_{q,t}(\alpha)\neq 0.$ Without any loss of generality, we assume that $\text{Tr}_{q,t}(\alpha)=1.$ Now we define $\mathcal{V}_{1}=\mathcal{V}_{0}\cap\alpha^{-1}\mathcal{V}_{0}.$ It is clear that $1\not\in\mathcal{V}_{1}.$ By Huffman [9, p. 281], we see that $\mathcal{V}_{1}$ is an $\mathbb{F}_{q}$-subspace of $\mathbb{F}_{q^{t}}$ having dimension $t-2$ and $\mathcal{V}_{0}=\mathcal{V}_{1}\oplus\langle 1\rangle,$ which implies that $\mathbb{F}_{q^{t}}=\mathcal{V}_{1}\oplus\langle 1\rangle\oplus\langle\alpha\rangle.$ Further, we observe that $(\cdot,\cdot)_{0}\restriction_{\mathcal{V}_{1}\times\mathcal{V}_{1}}$ is a reflexive, non-degenerate and alternating form, that is, $(\mathcal{V}_{1},(\cdot,\cdot)_{0}\restriction_{\mathcal{V}_{1}\times\mathcal{V}_{1}})$ is a symplectic space over $\mathbb{F}_{q}.$ In this case, for $1\leq k\leq t-1,$ we see that any $k$-dimensional $\mathbb{F}_{q}$-subspace of $\mathbb{F}_{q^{t}}$ is III. either contained in $\mathcal{V}_{1},$ or ** IV. contained in $\mathcal{V}_{1}\oplus\langle 1\rangle$ but not in $\mathcal{V}_{1},$ or ** V.** contained in $\mathcal{V}_{1}\oplus\langle\alpha\rangle$ but not in $\mathcal{V}_{1},$ or VI. contained in $\mathbb{F}_{q^{t}}=\mathcal{V}_{1}\oplus\langle 1\rangle\oplus\langle\alpha\rangle$ but not in any of the spaces $\mathcal{V}_{1},$ $\mathcal{V}_{1}\oplus\langle 1\rangle$ and $\mathcal{V}_{1}\oplus\langle\alpha\rangle.$

**III.  ** To begin with, we shall first enumerate all $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ that are contained in the symplectic space $\mathcal{V}_{1}.$ Here we must have $1\leq k\leq t-2.$ When $k$ is odd, by [18, p. 69], we see that there does not exist any $k$-dimensional non-degenerate (and hence symplectic) $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{1}.$ When $k$ is even, working as in Proposition 3.3, it is easy to see that the number of $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathcal{V}_{1}$ is given by $q^{\frac{k(t-k-2)}{2}}{(t-2)/2\brack k/2}_{q^{2}}.$

**IV.  ** Next we proceed to enumerate all $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ that are contained in $\mathcal{V}_{1}\oplus\langle 1\rangle$ but not in $\mathcal{V}_{1}.$ For this, we observe that any such $k$-dimensional $\mathbb{F}_{q}$-subspace of $\mathbb{F}_{q^{t}}$ is of the type $\langle x_{1},x_{2},\cdots,x_{k-1},1+x_{k}\rangle,$ where $x_{j}\in\mathcal{V}_{1}\setminus\{0\}$ for $1\leq j\leq k-1$ and $x_{k}\in\mathcal{V}_{1}.$ Here by applying Theorem 5.1.1 of [17], it is easy to observe that the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-1},1+x_{k}\rangle$ of $\mathbb{F}_{q^{t}}$ is degenerate when either $x_{k}=0$ or $x_{k}\neq 0$ and $k$ is odd. Further, when $k$ is even and $x_{k}\neq 0,$ the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-1},1+x_{k}\rangle$ of $\mathbb{F}_{q^{t}}$ is non-degenerate if and only if the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k}\rangle$ of $\mathcal{V}_{1}$ is non-degenerate. Now when $k$ is even, working in a similar manner as in case II of part (a), we see that the number of distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the type $\langle x_{1},x_{2},\cdots,x_{k-1},1+x_{k}\rangle$ with $x_{j}\in\mathcal{V}_{1}\setminus\{0\}$ for $1\leq j\leq k,$ is given by $q^{\frac{k(t-k-2)}{2}}(q^{k}-1){(t-2)/2\brack k/2}_{q^{2}}.$

**V.  ** Next we shall enumerate all $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ that are contained in $\mathcal{V}_{1}\oplus\langle\alpha\rangle$ but not in $\mathcal{V}_{1}.$ To do so, we observe that any such $k$-dimensional $\mathbb{F}_{q}$-subspace of $\mathbb{F}_{q^{t}}$ is of the type $\langle x_{1},x_{2},\cdots,x_{k-1},\alpha+x_{k}\rangle,$ where $x_{j}\in\mathcal{V}_{1}\setminus\{0\}$ for $1\leq j\leq k-1$ and $x_{k}\in\mathcal{V}_{1}.$

First let $k$ be even. Here when $x_{k}=0,$ by applying Theorem 5.1.1 of [17], it is easy to observe that the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-1},\alpha\rangle$ of $\mathbb{F}_{q^{t}}$ is degenerate. When $x_{k}\neq 0,$ the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-1},\alpha+x_{k}\rangle$ of $\mathbb{F}_{q^{t}}$ is non-degenerate if and only if the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k}\rangle$ of $\mathcal{V}_{1}$ is non-degenerate. Now working as in case II of part (a), we see that when $k$ is even, the number of distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the type $\langle x_{1},x_{2},\cdots,x_{k-1},\alpha+x_{k}\rangle$ with $x_{j}$’s in $\mathcal{V}_{1}\setminus\{0\},$ is given by $q^{\frac{k(t-k-2)}{2}}(q^{k}-1){(t-2)/2\brack k/2}_{q^{2}}.$

Next let $k$ be odd. Here using Theorem 5.1.1 of [17], we see that when $x_{k}=0,$ the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-1},\alpha\rangle$ of $\mathbb{F}_{q^{t}}$ is non-degenerate if and only if the $(k-1)$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-1}\rangle$ of $\mathcal{V}_{1}$ is non-degenerate. Now as $\left(\mathcal{V}_{1},\left(\cdot,\cdot\right)_{0}\restriction_{\mathcal{V}_{1}\times\mathcal{V}_{1}}\right)$ is a symplectic space, working similarly as in Proposition 3.3, we see that the number of such spaces is given by $q^{\frac{(k-1)(t-k-1)}{2}}{(t-2)/2\brack(k-1)/2}_{q^{2}}.$ When $x_{k}\neq 0,$ by applying Theorem 5.1.1 of [17], we see that $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-1},\alpha+x_{k}\rangle$ of $\mathbb{F}_{q^{t}}$ is non-degenerate if and only if the $(k-1)$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-1}\rangle$ of $\mathcal{V}_{1}$ is non-degenerate. Now working as in case II of part (a), we see that the number of distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of the type $\langle x_{1},x_{2},\cdots,x_{k-1},\alpha+x_{k}\rangle$ with $x_{j}$’s in $\mathcal{V}_{1}\setminus\{0\},$ is given by $q^{\frac{(k-1)(t-k-1)}{2}}(q^{t-k-1}-1){(t-2)/2\brack(k-1)/2}_{q^{2}}.$

**VI.  ** We next proceed to enumerate all $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ that are not contained in any of the subspaces $\mathcal{V}_{1},$ $\mathcal{V}_{1}\oplus\langle 1\rangle$ and $\mathcal{V}_{1}\oplus\langle\alpha\rangle$ of $\mathbb{F}_{q^{t}}.$ Towards this, we first observe that any such $k$-dimensional $\mathbb{F}_{q}$-subspace of $\mathbb{F}_{q^{t}}$ is of the following two types:

• •

$\langle x_{1},x_{2},\cdots,x_{k-1},1+\lambda\alpha+x_{k}\rangle,$ where $\lambda\in\mathbb{F}_{q}\setminus\{0\},$ $x_{j}\in\mathcal{V}_{1}\setminus\{0\}$ for $1\leq j\leq k-1$ and $x_{k}\in\mathcal{V}_{1}.$

• •

$\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+x_{k}\rangle,$ where $k\geq 2,$ $x_{j}\in\mathcal{V}_{1}\setminus\{0\}$ for $1\leq j\leq k-2$ and $x_{k-1},x_{k}\in\mathcal{V}_{1}.$

To begin with, we shall first count all $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of the type $\langle x_{1},x_{2},\cdots,\\ x_{k-1},1+\lambda\alpha+x_{k}\rangle,$ where $\lambda\in\mathbb{F}_{q}\setminus\{0\},$ $x_{j}\in\mathcal{V}_{1}\setminus\{0\}$ for $1\leq j\leq k-1$ and $x_{k}\in\mathcal{V}_{1}.$

First let $k$ be even. When $x_{k}=0,$ by applying Theorem 5.1.1 of [17], we see that the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-1},1+\lambda\alpha\rangle$ of $\mathbb{F}_{q^{t}}$ is degenerate for each $\lambda\in\mathbb{F}_{q}\setminus\{0\}.$ When $x_{k}\neq 0,$ by applying Theorem 5.1.1 of [17] again, we see that for each $\lambda\in\mathbb{F}_{q}\setminus\{0\},$ the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-1},1+\lambda\alpha+x_{k}\rangle$ of $\mathbb{F}_{q^{t}}$ is non-degenerate if and only if the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k}\rangle$ of $\mathcal{V}_{1}$ is non-degenerate. Now as $\left(\mathcal{V}_{1},\left(\cdot,\cdot\right)_{0}\restriction_{\mathcal{V}_{1}\times\mathcal{V}_{1}}\right)$ is a symplectic space, working similarly as in Proposition 3.3, we see that the number of $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathcal{V}_{1}$ is given by $q^{\frac{k(t-k-2)}{2}}{(t-2)/2\brack k/2}_{q^{2}}.$ Further, it is easy to observe that each $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{1}$ gives rise to precisely $(q^{k}-1)(q-1)$ distinct $k$-dimensional $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the type $\langle x_{1},x_{2},\cdots,x_{k-1},1+\lambda\alpha+x_{k}\rangle$ with $\lambda\in\mathbb{F}_{q}\setminus\{0\}$ and $x_{j}\in\mathcal{V}_{1}\setminus\{0\}$ for $1\leq j\leq k.$ Therefore when $k$ is even, the number of distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the type $\langle x_{1},x_{2},\cdots,x_{k-1},1+\lambda\alpha+x_{k}\rangle$ with $x_{j}\in\mathcal{V}_{1}\setminus\{0\}$ for $1\leq j\leq k$ and $\lambda\in\mathbb{F}_{q}\setminus\{0\},$ is given by $q^{\frac{k(t-k-2)}{2}}(q^{k}-1)(q-1){(t-2)/2\brack k/2}_{q^{2}}.$

Next let $k$ be odd. Here we observe, by Theorem 5.1.1 of [17], that when $x_{k}=0,$ the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-1},1+\lambda\alpha\rangle$ of $\mathbb{F}_{q^{t}}$ is non-degenerate if and only if the $(k-1)$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-1}\rangle$ of $\mathcal{V}_{1}$ is non-degenerate for each $\lambda\in\mathbb{F}_{q}\setminus\{0\}.$ Now working as in Proposition 3.3, we see that the number of $(k-1)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of the symplectic space $\mathcal{V}_{1}$ is given by $q^{\frac{(k-1)(t-k-1)}{2}}{(t-2)/2\brack(k-1)/2}_{q^{2}}.$ Further, we observe that each $(k-1)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{1}$ gives rise to precisely $(q-1)$ distinct $\mathbb{F}_{q}$-subspaces of the type $\langle x_{1},x_{2},\cdots,x_{k-1},1+\lambda\alpha\rangle$ with $x_{j}$’s in $\mathcal{V}_{1}\setminus\{0\}$ and $\lambda\in\mathbb{F}_{q}\setminus\{0\}.$ Therefore when $x_{k}=0,$ the number of distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the type $\langle x_{1},x_{2},\cdots,x_{k-1},1+\lambda\alpha\rangle$ with $x_{j}$ in $\mathcal{V}_{1}\setminus\{0\}$ for $1\leq j\leq k-1$ and $\lambda\in\mathbb{F}_{q}\setminus\{0\},$ is given by $q^{\frac{(k-1)(t-k-1)}{2}}(q-1){(t-2)/2\brack(k-1)/2}_{q^{2}}$ when $k$ is odd. Next let $x_{k}$ be non-zero. Here using Theorem 5.1.1 of [17], we see that for each $\lambda\in\mathbb{F}_{q}\setminus\{0\},$ the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-1},1+\lambda\alpha+x_{k}\rangle$ of $\mathbb{F}_{q^{t}}$ is non-degenerate if and only if the $(k-1)$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-1}\rangle$ of $\mathcal{V}_{1}$ is non-degenerate. Now as $\left(\mathcal{V}_{1},\left(\cdot,\cdot\right)_{0}\restriction_{\mathcal{V}_{1}\times\mathcal{V}_{1}}\right)$ is a symplectic space, working similarly as in Proposition 3.3, we see that the number of $(k-1)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathcal{V}_{1}$ is given by $q^{\frac{(k-1)(t-k-1)}{2}}{(t-2)/2\brack(k-1)/2}_{q^{2}}.$ Further, for a given $(k-1)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-1}\rangle$ of $\mathcal{V}_{1},$ we need to count all choices for $x_{k}\in\mathcal{V}_{1}\setminus\{0\}$ that give rise to distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces $\left<x_{1},x_{2},\cdots,x_{k-1},1+\lambda\alpha+x_{k}\right>$ of $\mathbb{F}_{q^{t}}$ for each $\lambda\in\mathbb{F}_{q}\setminus\{0\}.$ To determine the number of choices for such an element $x_{k}\in\mathcal{V}_{1},$ we note that given any $(k-1)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-1}\rangle$ of $\mathcal{V}_{1},$ by Proposition 2.9 of [6], we can write $\mathcal{V}_{1}=\langle x_{1},x_{2},\cdots,x_{k-1}\rangle\perp\langle x_{1},x_{2},\cdots,x_{k-1}\rangle^{\perp_{0}},$ where $\langle x_{1},x_{2},\cdots,x_{k-1}\rangle^{\perp_{0}}$ is a non-degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{1}$ having dimension $t-k-1.$ In view of this, each $x_{k}\in\mathcal{V}_{1}$ can be uniquely written as $x_{k}=v+\widetilde{v},$ where $v\in\langle x_{1},x_{2},\cdots,x_{k-1}\rangle$ and $\widetilde{v}\in\langle x_{1},x_{2},\cdots,x_{k-1}\rangle^{\perp_{0}}.$ It is easy to see that the subspace $\langle x_{1},x_{2},\cdots,x_{k-1},1+\lambda\alpha+v+\widetilde{v}\rangle=\langle x_{1},x_{2},\cdots,x_{k-1},1+\lambda\alpha+\widetilde{v}\rangle,$ where $\widetilde{v}\in\langle x_{1},x_{2},\cdots,x_{k-1}\rangle^{\perp_{0}}.$ We further observe that for every $\lambda\in\mathbb{F}_{q}\setminus\{0\}$, each non-zero $x_{k}\in\langle x_{1},x_{2},\cdots,x_{k-1}\rangle^{\perp_{0}}$ gives rise to a distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-1},1+\lambda\alpha+x_{k}\rangle$ of $\mathbb{F}_{q^{t}},$ and thus there are precisely $q^{t-k-1}-1$ relevant choices for $x_{k}.$ Moreover, each choice of $\lambda\in\mathbb{F}_{q}\setminus\{0\}$ gives rise to a distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of the type $\langle x_{1},x_{2},\cdots,x_{k-1},1+\lambda\alpha+x_{k}\rangle.$ Therefore the number of distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the type $\langle x_{1},x_{2},\cdots,x_{k-1},1+\lambda\alpha+x_{k}\rangle$ with $x_{j}\in\mathcal{V}_{1}\setminus\{0\}$ for $1\leq j\leq k$ and $\lambda\in\mathbb{F}_{q}\setminus\{0\},$ is given by $q^{\frac{(k-1)(t-k-1)}{2}}(q-1)(q^{t-k-1}-1){(t-2)/2\brack(k-1)/2}_{q^{2}}$ when $k$ is odd.

Henceforth we assume that $k\geq 2.$ Now we shall enumerate all $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the form $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+x_{k}\rangle$ with $k\geq 2,$ where $x_{j}\in\mathcal{V}_{1}\setminus\{0\}$ for $1\leq j\leq k-2$ and $x_{k-1},x_{k}\in\mathcal{V}_{1}.$ Here also, we shall distinguish the cases: A. $k$ is even and B. $k$ is odd.

Case A. First let $k$ be even. When $x_{k-1}=x_{k}=0,$ by Theorem 5.1.1 of [17], we see that the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2},1,\alpha\rangle$ of $\mathbb{F}_{q^{t}}$ is non-degenerate if and only if the $(k-2)$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2}\rangle$ of $\mathcal{V}_{1}$ is non-degenerate. Now as $\left(\mathcal{V}_{1},\left(\cdot,\cdot\right)_{0}\restriction_{\mathcal{V}_{1}\times\mathcal{V}_{1}}\right)$ is a symplectic space, working similarly as in Proposition 3.3, we see that the number of such spaces is given by $q^{\frac{(k-2)(t-k)}{2}}{(t-2)/2\brack(k-2)/2}_{q^{2}}.$ When $x_{k-1}\neq 0$ and $x_{k}=0,$ by Theorem 5.1.1 of [17], we see that the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha\rangle$ of $\mathbb{F}_{q^{t}}$ is non-degenerate if and only if the $(k-2)$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2}\rangle$ of $\mathcal{V}_{1}$ is non-degenerate. Now working as in case II of part (a), we see that the number of distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the type $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha\rangle$ with $x_{j}$’s in $\mathcal{V}_{1}\setminus\{0\},$ is given by $q^{\frac{(k-2)(t-k)}{2}}(q^{t-k}-1){(t-2)/2\brack(k-2)/2}_{q^{2}}.$ When $x_{k-1}=0$ and $x_{k}\neq 0,$ by Theorem 5.1.1 of [17], we see that the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2},1,\alpha+x_{k}\rangle$ of $\mathbb{F}_{q^{t}}$ is non-degenerate if and only if the $(k-2)$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2}\rangle$ of $\mathcal{V}_{1}$ is non-degenerate. Now working as in case II of part (a), we see that the number of distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the type $\langle x_{1},x_{2},\cdots,x_{k-2},1,\alpha+x_{k}\rangle$ with $x_{j}$’s in $\mathcal{V}_{1}\setminus\{0\},$ is given by $q^{\frac{(k-2)(t-k)}{2}}(q^{t-k}-1){(t-2)/2\brack(k-2)/2}_{q^{2}}.$ Next suppose that both $x_{k-1},x_{k}\in\mathcal{V}_{1}$ are linearly independent over $\mathbb{F}_{q}.$

For any even integer $k\geq 2,$ if $\mathfrak{G}(x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+x_{k}),$ $\mathfrak{G}(x_{1},x_{2},\cdots,x_{k-2})$ and $\mathfrak{G}(x_{1},x_{2},\cdots,x_{k})$ are Gram matrices of $\left<x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+x_{k}\right>,$ $\left<x_{1},x_{2},\cdots,x_{k-2}\right>$ and $\left<x_{1},x_{2},\cdots,x_{k}\right>$ respectively, then we observe that

[TABLE]

This, by Theorem 5.1.1 of [17], implies that the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+x_{k}\rangle$ of $\mathbb{F}_{q^{t}}$ is non-degenerate if and only if either

1. (i)

the $(k-2)$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2}\rangle$ of $\mathcal{V}_{1}$ is non-degenerate and the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k}\rangle$ of $\mathcal{V}_{1}$ is degenerate, or 2. (ii)

the $(k-2)$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2}\rangle$ of $\mathcal{V}_{1}$ is degenerate and the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k}\rangle$ of $\mathcal{V}_{1}$ is non-degenerate, or 3. (iii)

both the $\mathbb{F}_{q}$-subspaces $\langle x_{1},x_{2},\cdots,x_{k-2}\rangle$ and $\langle x_{1},x_{2},\cdots,x_{k}\rangle$ of $\mathcal{V}_{1}$ are non-degenerate with $\text{det }\mathfrak{G}(x_{1},\\ x_{2},\cdots,x_{k})\neq\text{det }\mathfrak{G}(x_{1},x_{2},\cdots,x_{k-2}).$

We also observe that the condition $\text{det }\mathfrak{G}(x_{1},x_{2},\cdots,x_{k})\neq\text{det }\mathfrak{G}(x_{1},x_{2},\cdots,x_{k-2})$ trivially holds in the cases (i) and (ii).

(i) We shall first enumerate all $k$-dimensional $\mathbb{F}_{q}$-subspaces $\left<x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+x_{k}\right>$ of $\mathbb{F}_{q^{t}}$ such that $\langle x_{1},x_{2},\cdots,x_{k-2}\rangle$ is a $(k-2)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{1}$ and $\langle x_{1},x_{2},\cdots,x_{k}\rangle$ is a $k$-dimensional degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{1}.$ For this, as $\left(\mathcal{V}_{1},\left(\cdot,\cdot\right)_{0}\restriction_{\mathcal{V}_{1}\times\mathcal{V}_{1}}\right)$ is a symplectic space, working similarly as in Proposition 3.3, we see that the number of $(k-2)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathcal{V}_{1}$ is $q^{\frac{(k-2)(t-k)}{2}}{(t-2)/2\brack(k-2)/2}_{q^{2}}.$ Now for each such $(k-2)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace $\left<x_{1},x_{2},\cdots,x_{k-2}\right>$ of $\mathcal{V}_{1},$ we need to count all choices for linearly independent vectors $x_{k-1},x_{k}\in\mathcal{V}_{1}\setminus\{0\}$ such that the $k$-dimensional $\mathbb{F}_{q}$-subspace $\left<x_{1},x_{2},\cdots,x_{k}\right>$ of $\mathcal{V}_{1}$ is degenerate and that give rise to distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+x_{k}\rangle$ of $\mathbb{F}_{q^{t}}.$ For this, by Proposition 2.9 of [6], we write $\mathcal{V}_{1}=\langle x_{1},x_{2},\cdots,x_{k-2}\rangle\perp\langle x_{1},x_{2},\cdots,x_{k-2}\rangle^{\perp_{0}},$ where the dimension of $\langle x_{1},x_{2},\cdots,x_{k-2}\rangle^{\perp_{0}}$ is $t-k.$ In view of this, both $x_{k-1},x_{k}\in\mathcal{V}_{1}\setminus\{0\}$ can be uniquely written as $x_{k-1}=w_{1}+\widetilde{w_{1}}$ and $x_{k}=w_{2}+\widetilde{w_{2}},$ where $w_{1},w_{2}\in\langle x_{1},x_{2},\cdots,x_{k-2}\rangle$ and $\widetilde{w_{1}},\widetilde{w_{2}}\in\langle x_{1},x_{2},\cdots,x_{k-2}\rangle^{\perp_{0}}.$ It is easy to observe that $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+x_{k}\rangle=\langle x_{1},x_{2},\cdots,x_{k-2},1+w_{1}+\widetilde{w_{1}},\alpha+w_{2}+\widetilde{w_{2}}\rangle=\langle x_{1},x_{2},\cdots,x_{k-2},1+\widetilde{w_{1}},\alpha+\widetilde{w_{2}}\rangle.$ We further observe that each pair $(x_{k-1},x_{k})$ of linearly independent vectors in $\langle x_{1},x_{2},\cdots,x_{k-2}\rangle^{\perp_{0}}$ gives rise to a distinct $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+x_{k}\rangle$ of $\mathbb{F}_{q^{t}}$ with $x_{j}$’s in $\mathcal{V}_{1}\setminus\{0\}.$ It is easy to see that $\langle x_{1},x_{2},\cdots,x_{k}\rangle=\langle x_{1},x_{2},\cdots,x_{k-2}\rangle\perp\langle x_{k-1},x_{k}\rangle.$ This implies that $\text{det }\mathfrak{G}(x_{1},x_{2},\cdots,x_{k})=\text{det }\mathfrak{G}(x_{1},x_{2},\cdots,x_{k-2})~{}\text{det }\mathfrak{G}(x_{k-1},x_{k}).$ Therefore by Theorem 5.1.1 of [17], we see that the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k}\rangle$ of $\mathcal{V}_{1}$ is degenerate if and only if $\text{Tr}_{q,t}(x_{k-1}x_{k})=0$ if and only if $x_{k}\in\left<x_{1},x_{2},\cdots,x_{k-1}\right>^{\perp_{0}}.$ As $x_{k-1}\in\langle x_{1},x_{2},\cdots,x_{k-2}\rangle^{\perp_{0}}\setminus\{0\},$ it has $q^{t-k}-1$ choices. Further, we note that $x_{k}\in\langle x_{1},x_{2},\cdots,x_{k-2},x_{k-1}\rangle^{\perp_{0}},$ whose dimension is $t-k-1$ by Proposition 2.4 of [6]. Also as $x_{k-1}\in\langle x_{1},x_{2},\cdots,x_{k-2},x_{k-1}\rangle^{\perp_{0}},$ we see that $x_{k}$ has $q^{t-k-1}-q$ choices, which implies that the pair $(x_{k-1},x_{k})$ has $(q^{t-k}-1)(q^{t-k-1}-q)$ choices. Therefore for an even integer $k\geq 4,$ the number of distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the form $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+x_{k}\rangle$ such that $\langle x_{1},x_{2},\cdots,x_{k-2}\rangle$ is a $(k-2)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{1}$ and $\langle x_{1},x_{2},\cdots,x_{k}\rangle$ is a $k$-dimensional degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{1},$ is given by $q^{\frac{(k-2)(t-k)}{2}}(q^{t-k}-1)(q^{t-k-1}-q){(t-2)/2\brack(k-2)/2}_{q^{2}}.$

(ii) Next we will enumerate all $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+x_{k}\rangle$ of $\mathbb{F}_{q^{t}}$ such that $\langle x_{1},x_{2},\cdots,x_{k}\rangle$ is a $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{1}$ and $\langle x_{1},x_{2},\cdots,x_{k-2}\rangle$ is a $(k-2)$-dimensional degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{1}.$ For this, as $\left(\mathcal{V}_{1},\left(\cdot,\cdot\right)_{0}\restriction_{\mathcal{V}_{1}\times\mathcal{V}_{1}}\right)$ is a symplectic space, working similarly as in Proposition 3.3, the number of $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathcal{V}_{1}$ is $q^{\frac{k(t-k-2)}{2}}{(t-2)/2\brack k/2}_{q^{2}}.$ Working as in Proposition 3.3 again, we see that every $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace $\left<x_{1},x_{2},\cdots,x_{k}\right>$ of $\mathcal{V}_{1}$ has precisely $q^{k-2}{k/2\brack(k-2)/2}_{q^{2}}$ distinct $(k-2)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces, which implies that the number of $(k-2)$-dimensional degenerate $\mathbb{F}_{q}$-subspaces of the $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace $\left<x_{1},x_{2},\cdots,x_{k}\right>$ of $\mathcal{V}_{1}$ is given by $\left\{{k\brack k-2}_{q}-q^{k-2}{k/2\brack(k-2)/2}_{q^{2}}\right\}.$ Furthermore, we observe that each such $(k-2)$-dimensional degenerate $\mathbb{F}_{q}$-subspace $\left<y_{1},y_{2},\cdots,y_{k-2}\right>$ of $\left<x_{1},x_{2},\cdots,x_{k}\right>$ gives rise to precisely $(q^{2}-1)(q^{2}-q)$ distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the form $\langle y_{1},y_{2},\cdots,y_{k-2},1+y_{k-1},\alpha+y_{k}\rangle$ satisfying $\langle y_{1},y_{2},\cdots,y_{k}\rangle=\langle x_{1},x_{2},\cdots,x_{k}\rangle.$ Therefore for an even integer $k\geq 4,$ the number of distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the form $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+x_{k}\rangle$ such that $\langle x_{1},x_{2},\cdots,x_{k-2}\rangle$ is a $(k-2)$-dimensional degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{1}$ and $\langle x_{1},x_{2},\cdots,x_{k}\rangle$ is a $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{1},$ is given by $(q^{2}-1)(q^{2}-q)q^{\frac{k(t-k-2)}{2}}{(t-2)/2\brack k/2}_{q^{2}}\left\{{k\brack k-2}_{q}-q^{k-2}{k/2\brack(k-2)/2}_{q^{2}}\right\}.$

(iii) Finally, we will enumerate all $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+\nolinebreak x_{k}\rangle$ of $\mathbb{F}_{q^{t}}$ such that $\langle x_{1},x_{2},\cdots,x_{k}\rangle$ is a $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{1}$ and $\langle x_{1},x_{2},\cdots,x_{k-2}\rangle$ is a $(k-2)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{1}$ with $\text{det }\mathfrak{G}(x_{1},x_{2},\cdots,x_{k})\neq\text{det }\mathfrak{G}(x_{1},x_{2},\cdots,x_{k-2}).$ To do so, we see that as $\left(\mathcal{V}_{1},\left(\cdot,\cdot\right)_{0}\restriction_{\mathcal{V}_{1}\times\mathcal{V}_{1}}\right)$ is a $(t-2)$-dimensional symplectic space over $\mathbb{F}_{q},$ working similarly as in Proposition 3.3, the number of distinct $(k-2)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathcal{V}_{1}$ is $q^{\frac{(k-2)(t-k)}{2}}{(t-2)/2\brack(k-2)/2}_{q^{2}}.$ Next for each $(k-2)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace $\left<x_{1},x_{2},\cdots,x_{k-2}\right>$ of $\mathcal{V}_{1},$ we need to determine choices for linearly independent vectors $x_{k-1},x_{k}$ in $\mathcal{V}_{1}$ such that the $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k}\rangle$ of $\mathcal{V}_{1}$ is non-degenerate and $\text{det }\mathfrak{G}(x_{1},x_{2},\cdots,x_{k})\neq\text{det }\mathfrak{G}(x_{1},x_{2},\cdots,x_{k-2}).$ For this, by Proposition 2.9 of [6], we write $\mathcal{V}_{1}=\langle x_{1},x_{2},\cdots,x_{k-2}\rangle\perp\langle x_{1},x_{2},\cdots,x_{k-2}\rangle^{\perp_{0}},$ where the dimension of $\langle x_{1},x_{2},\cdots,x_{k-2}\rangle^{\perp_{0}}$ is $t-k.$ So we need to choose vectors $x_{k-1}$ and $x_{k}$ from $\langle x_{1},x_{2},\cdots,x_{k-2}\rangle^{\perp_{0}},$ and each pair $(x_{k-1},x_{k})$ of linearly independent vectors in $\langle x_{1},x_{2},\cdots,x_{k-2}\rangle^{\perp_{0}}$ gives rise to a distinct $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+x_{k}\rangle$ of $\mathbb{F}_{q^{t}}$ with $x_{j}$’s in $\mathcal{V}_{1}\setminus\{0\}.$ It is easy to see that $\langle x_{1},x_{2},\cdots,x_{k}\rangle=\langle x_{1},x_{2},\cdots,x_{k-2}\rangle\perp\langle x_{k-1},x_{k}\rangle,$ which gives

[TABLE]

From this, it follows that the $k$-dimensional $\mathbb{F}_{q}$-subspace $\left<x_{1},x_{2},\cdots,x_{k}\right>$ of $\mathcal{V}_{1}$ is non-degenerate if and only if $\text{det }\mathfrak{G}(x_{k-1},x_{k})=\text{Tr}_{q,t}(x_{k-1}x_{k})^{2}\neq 0$ if and only if $x_{k}\not\in\left<x_{1},x_{2},\cdots,x_{k-1}\right>^{\perp_{0}}.$ Besides this, we note that $\text{det }\mathfrak{G}(x_{1},x_{2},\cdots,x_{k})\neq\text{det }\mathfrak{G}(x_{1},x_{2},\cdots,x_{k-2})$ if and only if $\text{det }\mathfrak{G}(x_{k-1},x_{k})=\text{Tr}_{q,t}(x_{k-1}x_{k})^{2}\neq 1$ if and only if $(x_{k-1},x_{k})$ is not a hyperbolic pair in $\langle x_{1},x_{2},\cdots,x_{k-2}\rangle^{\perp_{0}}.$ First of all, we shall count all pairs $(x_{k-1},x_{k})$ of linearly independent vectors in $\langle x_{1},x_{2},\cdots,x_{k-2}\rangle^{\perp_{0}}$ satisfying $x_{k}\not\in\left<x_{1},x_{2},\cdots,x_{k-1}\right>^{\perp_{0}}.$ For this, as $x_{k-1}\in\langle x_{1},x_{2},\cdots,x_{k-2}\rangle^{\perp_{0}}\setminus\{0\},$ it has $q^{t-k}-1$ choices. Further, we note, by Proposition 2.4 of [6], that the dimension of $\langle x_{1},x_{2},\cdots,x_{k-2},x_{k-1}\rangle^{\perp_{0}}$ is $t-k-1.$ From this, we see that $x_{k}$ has $q^{t-k}-q^{t-k-1}$ choices. This implies that the number of pairs $(x_{k-1},x_{k})$ such that both the $\mathbb{F}_{q}$-subspaces $\left<x_{1},x_{2},\cdots,x_{k-2}\right>$ and $\left<x_{1},x_{2},\cdots,x_{k}\right>$ of $\mathcal{V}_{1}$ are non-degenerate, is given by $(q^{t-k}-1)(q^{t-k}-q^{t-k-1}).$ On the other hand, by [18, p. 70], we see that the number of hyperbolic pairs in $\langle x_{1},x_{2},\cdots,x_{k-2}\rangle^{\perp_{0}}$ is $H_{\frac{t-k}{2}},0=q^{t-k-1}(q^{t-k}-1),$ which implies that the pair $(x_{k-1},x_{k})$ has $(q^{t-k}-1)(q^{t-k}-q^{t-k-1})-H_{\frac{t-k}{2}},0=(q^{t-k}-1)(q^{t-k}-q^{t-k-1})-q^{t-k-1}(q^{t-k}-1)$ relevant choices. Therefore the number of distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the type $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+x_{k}\rangle$ such that $\left<x_{1},x_{2},\cdots,x_{k-2}\right>$ is a $(k-2)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{1}$ and $\left<x_{1},x_{2},\cdots,x_{k}\right>$ is a $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{1}$ with $\text{det }\mathfrak{G}(x_{1},x_{2},\cdots,x_{k})\neq\text{det }\mathfrak{G}(x_{1},x_{2},\cdots,x_{k-2}),\vspace{-2mm}$ is given by

[TABLE]

[TABLE]

On combining cases (i)-(iii), we see that the total number of $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the type $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+x_{k}\rangle$ with $x_{1},x_{2},\cdots,x_{k}\in\mathcal{V}_{1}$ as linearly independent vectors over $\mathbb{F}_{q},$ is given by $q^{\frac{k(t-k-2)}{2}}(q^{k+1}-q)(q^{k-2}-1){(t-2)/2\brack k/2}_{q^{2}}+q^{\frac{(k-2)(t-k)}{2}}(q^{t-k}-1)(q^{t-k}-q^{t-k-1}-q){(t-2)/2\brack(k-2)/2}_{q^{2}}.$

Next let $x_{k-1},x_{k}\in\mathcal{V}_{1}\setminus\{0\}$ be linearly dependent over $\mathbb{F}_{q}.$ In this case, the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+x_{k}\rangle$ of $\mathbb{F}_{q^{t}}$ is equal to the $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},1+\lambda\alpha\rangle$ for some $\lambda\in\mathbb{F}_{q}\setminus\{0\}.$ Further, for each $\lambda\in\mathbb{F}_{q}\setminus\{0\},$ the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},1+\lambda\alpha\rangle$ of $\mathbb{F}_{q^{t}}$ is non-degenerate if and only if the $(k-2)$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2}\rangle$ of $\mathcal{V}_{1}$ is non-degenerate. Now working as in case II of part (a) and noting that each $\lambda\in\mathbb{F}_{q}\setminus\{0\}$ gives rise to a distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},1+\lambda\alpha\rangle$ of $\mathbb{F}_{q^{t}},$ the number of distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the type $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},1+\lambda\alpha\rangle$ with $x_{j}$’s in $\mathcal{V}_{1}\setminus\{0\}$ and $\lambda\in\mathbb{F}_{q}\setminus\{0\},$ is given by $q^{\frac{(k-2)(t-k)}{2}}(q^{t-k}-1)(q-1){(t-2)/2\brack(k-2)/2}_{q^{2}}.$

Case B. Let $k$ be odd. When $x_{k-1}=x_{k}=0,$ by Theorem 5.1.1 of [17], we see that the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2},1,\alpha\rangle$ of $\mathbb{F}_{q^{t}}$ is degenerate. When $x_{k-1}\neq 0$ and $x_{k}=0,$ by Theorem 5.1.1 of [17], we see that the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha\rangle$ of $\mathbb{F}_{q^{t}}$ is non-degenerate if and only if the $(k-1)$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-1}\rangle$ of $\mathcal{V}_{1}$ is non-degenerate. Now working as in case II of part (a), we see that the number of distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the type $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha\rangle$ with $x_{j}$’s in $\mathcal{V}_{1}\setminus\{0\},$ is given by $q^{\frac{(k-1)(t-k-1)}{2}}(q^{k-1}-1){(t-2)/2\brack(k-1)/2}_{q^{2}}.$ When $x_{k-1}=0$ and $x_{k}\neq 0,$ by Theorem 5.1.1 of [17], we see that the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2},1,\alpha+x_{k}\rangle$ of $\mathbb{F}_{q^{t}}$ is degenerate. Let $x_{k-1},x_{k}\in\mathcal{V}_{1}$ be linearly independent over $\mathbb{F}_{q}.$ Here by Theorem 5.1.1 of [17], we see that the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+x_{k}\rangle$ of $\mathbb{F}_{q^{t}}$ is non-degenerate if and only if the $(k-1)$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-1}\rangle$ of $\mathcal{V}_{1}$ is non-degenerate. Now as $\left(\mathcal{V}_{1},\left(\cdot,\cdot\right)_{0}\restriction_{\mathcal{V}_{1}\times\mathcal{V}_{1}}\right)$ is a symplectic space, working similarly as in Proposition 3.3, we see that the number of $(k-1)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathcal{V}_{1}$ is given by $q^{\frac{(k-1)(t-k-1)}{2}}{(t-2)/2\brack(k-1)/2}_{q^{2}}.$ Further, we see that each $(k-1)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{1}$ gives rise to precisely $(q^{k-1}-1)$ distinct $(k-1)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of the type $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1}\rangle$ with $x_{j}\in\mathcal{V}_{1}\setminus\{0\}$ for $1\leq j\leq k-1.$ Now for a given $(k-1)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of the type $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1}\rangle$ with $x_{j}\in\mathcal{V}_{1}\setminus\{0\}$ for $1\leq j\leq k-1,$ we need to count all choices for $x_{k}\in\mathcal{V}_{1}\setminus\{0\}$ that give rise to distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces $\left<x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+x_{k}\right>$ of $\mathbb{F}_{q^{t}}.$ For this, as $\langle x_{1},x_{2},\cdots,x_{k-1}\rangle$ is a non-degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{1},$ by Proposition 2.9 of [6], we write $\mathcal{V}_{1}=\langle x_{1},x_{2},\cdots,x_{k-1}\rangle\perp\langle x_{1},x_{2},\cdots,x_{k-1}\rangle^{\perp_{0}},$ where $\langle x_{1},x_{2},\cdots,x_{k-1}\rangle^{\perp_{0}}$ is a non-degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{1}$ having dimension $t-k-1.$ In view of this, each $x_{k}\in\mathcal{V}_{1}\setminus\{0\}$ can be uniquely written as $x_{k}=\sum\limits_{\ell=1}^{k-1}a_{\ell}x_{\ell}+\omega,$ where $\omega\in\langle x_{1},x_{2},\cdots,x_{k-1}\rangle^{\perp_{0}}$ and $a_{\ell}\in\mathbb{F}_{q}$ for $1\leq\ell\leq k-1.$ Then it is easy to see that $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+x_{k}\rangle=\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+\sum\limits_{\ell=1}^{k-1}a_{\ell}x_{\ell}+\omega\rangle=\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+a_{k-1}x_{k-1}+\omega\rangle,$ where $\omega\in\langle x_{1},x_{2},\cdots,x_{k-1}\rangle^{\perp_{0}}.$ This implies that each $x_{k}=a_{k-1}x_{k-1}+\omega$ with $\omega\in\langle x_{1},x_{2},\cdots,x_{k-1}\rangle^{\perp_{0}}\setminus\{0\}$ and $a_{k-1}\in\mathbb{F}_{q}$ gives rise to a distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+x_{k}\rangle$ of $\mathbb{F}_{q^{t}}.$ Further, we observe that $x_{k}=a_{k-1}x_{k-1}+\omega\neq 0.$ This is because, if $x_{k}=a_{k-1}x_{k-1}+\omega=0,$ then $\omega=a_{k-1}x_{k-1}\in\langle x_{1},x_{2},\cdots,x_{k-1}\rangle\perp\langle x_{1},x_{2},\cdots,x_{k-1}\rangle^{\perp_{0}}=\{0\}.$ This is a contradiction, as $x_{k-1},x_{k}$ are linearly independent vectors in $\mathcal{V}_{1}.$ From this, we observe that $x_{k}$ has precisely $q(q^{t-k-1}-1)=(q^{t-k}-q)$ relevant choices. Therefore when $k$ is odd, the number of distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the type $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+x_{k}\rangle$ with $x_{j}$’s in $\mathcal{V}_{1}\setminus\{0\},$ is given by $q^{\frac{(k-1)(t-k-1)}{2}}(q^{k-1}-1)(q^{t-k}-q){(t-2)/2\brack(k-1)/2}_{q^{2}}.$ Next let $x_{k-1},x_{k}\in\mathcal{V}_{1}\setminus\{0\}$ be linearly dependent over $\mathbb{F}_{q}.$ In this case, the $k$-dimensional $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},\alpha+x_{k}\rangle$ of $\mathbb{F}_{q^{t}}$ is equal to the $\mathbb{F}_{q}$-subspace $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},1+\lambda\alpha\rangle$ for some $\lambda\in\mathbb{F}_{q}\setminus\{0\}.$ Further, by Theorem 5.1.1 of [17], we see that $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},1+\lambda\alpha\rangle$ is a $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of $\mathbb{F}_{q^{t}}$ if and only if $\langle x_{1},x_{2},\cdots,x_{k-1}\rangle$ is a $(k-1)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{1}$ for each $\lambda\in\mathbb{F}_{q}\setminus\{0\}.$ Now as $\left(\mathcal{V}_{1},\left(\cdot,\cdot\right)_{0}\restriction_{\mathcal{V}_{1}\times\mathcal{V}_{1}}\right)$ is a symplectic space, working similarly as in Proposition 3.3, we see that the number of $(k-1)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathcal{V}_{1}$ is $q^{\frac{(k-1)(t-k-1)}{2}}{(t-2)/2\brack(k-1)/2}_{q^{2}}.$ Further, we observe that each $(k-1)$-dimensional non-degenerate $\mathbb{F}_{q}$-subspace of $\mathcal{V}_{1}$ gives rise to precisely $(q^{k-1}-1)(q-1)$ distinct k-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of the type $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},1+\lambda\alpha\rangle$ with $x_{j}\in\mathcal{V}_{1}\setminus\{0\}$ for $1\leq j\leq k-1$ and $\lambda\in\mathbb{F}_{q}\setminus\{0\}.$ Therefore when $k$ is odd, the number of distinct $k$-dimensional non-degenerate $\mathbb{F}_{q}$-subspaces of $\mathbb{F}_{q^{t}}$ of the type $\langle x_{1},x_{2},\cdots,x_{k-2},1+x_{k-1},1+\lambda\alpha\rangle$ with $x_{j}$’s in $\mathcal{V}_{1}\setminus\{0\}$ and $\lambda\in\mathbb{F}_{q}\setminus\{0\},$ is given by $(q^{k-1}-1)q^{\frac{(k-1)(t-k-1)}{2}}(q-1){(t-2)/2\brack(k-1)/2}_{q^{2}}.$

Now on combining all the above cases, when $t$ is even, for $1\leq k\leq t-1,$ we get

[TABLE]

This completes the proof of the proposition. ∎

3.3 Determination of the number $N_{h}$ when $h\in\mathfrak{M}$

In the following proposition, we consider the case $h\in\mathfrak{M}$ and enumerate all pairs $\left(\mathcal{C}_{h},\mathcal{C}_{\mu(h)}\right)$ with $\mathcal{C}_{h}$ as a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{C}_{\mu(h)}$ as a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}$ satisfying $\mathcal{C}_{h}\cap\mathcal{C}_{h}^{(\delta)}=\{0\}$ and $\mathcal{C}_{\mu(h)}\cap\mathcal{C}_{\mu(h)}^{(\delta)}=\{0\}$ for each $\delta\in\{\ast,0,\gamma\}.$

Proposition 3.6**.**

Let $h\in\mathfrak{M}$ be fixed. For $\delta\in\{\ast,0,\gamma\},$ the number of pairs $\left(\mathcal{C}_{h},\mathcal{C}_{\mu(h)}\right)$ with $\mathcal{C}_{h}$ as a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{C}_{\mu(h)}$ as a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}$ satisfying $\mathcal{C}_{h}\cap\mathcal{C}_{h}^{(\delta)}=\{0\}$ and $\mathcal{C}_{\mu(h)}\cap\mathcal{C}_{\mu(h)}^{(\delta)}=\{0\},$ is given by $\displaystyle N_{h}=\displaystyle 2+\sum\limits_{k=1}^{t-1}q^{kd_{h}(t-k)}{t\brack k}_{q^{d_{h}}}.$

To prove the above proposition, let $h\in\mathfrak{M}$ be fixed from now onwards. Throughout this section, we shall represent elements of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}$ as $a(X)+b(X),$ where $a(X)\in\mathcal{J}_{h}$ and $b(X)\in\mathcal{J}_{\mu(h)}.$ On similar lines, we shall represent elements of $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$ as $u(X)+v(X),$ where $u(X)\in\mathcal{K}_{h}$ and $v(X)\in\mathcal{K}_{\mu(h)}.$ We shall view $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}$ as a $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-module under the following operations:

[TABLE]

for each $a(X)+b(X),~{}c(X)+d(X)\in\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}$ and $u(X)+v(X)\in\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}.$

Next for each $\delta\in\{\ast,0,\gamma\},$ we observe that

[TABLE]

for all $a(X)+b(X),~{}c(X)+d(X)\in\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}.$ For each $h\in\mathfrak{M}$ and $\delta\in\{\ast,0,\gamma\},$ let \big{[}\cdot,\cdot\big{]}_{\delta}\restriction_{\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}\times\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}} denote the restriction of the form \big{[}\cdot,\cdot\big{]}_{\delta} to $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}\times\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}.$ Then we make the following observation.

Lemma 3.3**.**

For each $h\in\mathfrak{M},$ the following hold.

1. (a)

$\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}$* is a free $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-module of rank *$t.$ 2. (b)

For each $\delta\in\{\ast,0,\gamma\},$ the form \big{[}\cdot,\cdot\big{]}_{\delta}\restriction_{\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}\times\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}} is reflexive and non-degenerate. 3. (c)

For each $\delta\in\{\ast,0,\gamma\},$ the form \big{[}\cdot,\cdot\big{]}_{\delta}\restriction_{\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}\times\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}} is Hermitian when $\delta\in\{\ast,0\}$ and is skew-Hermitian when $\delta=\gamma.$

Proof.

Proof is trivial. ∎

It is easy to see that both $\mathcal{J}_{h}$ and $\mathcal{J}_{\mu(h)}$ can be viewed as $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodules of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}$ with respect to operations defined by (19) and (20). From this, we observe that if $\mathcal{C}_{h}$ is a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{C}_{\mu(h)}$ is a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)},$ then $\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}$ is a $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}.$ Further, if $\text{dim}_{\mathcal{K}_{h}}\mathcal{C}_{h}=\ell$ and $\text{dim}_{\mathcal{K}_{\mu(h)}}\mathcal{C}_{\mu(h)}=r,$ then working as in Theorem 6 of Huffman [9], we see that the size of the smallest generating set of $\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}$ is $\max\{\ell,r\}.$ Further, for each $\delta\in\{\ast,0,\gamma\},$ the orthogonal complement of $\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)},$ denoted by \bigl{(}\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}\bigr{)}^{\perp_{\delta}}, is defined as \bigl{(}\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}\bigr{)}^{\perp_{\delta}}=\{a(X)+b(X)\in\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}:\left[a(X)+b(X),u(X)+v(X)\right]_{\delta}=0\text{ for all }u(X)+v(X)\in\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}\}. It is easy to see that \bigl{(}\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}\bigr{)}^{\perp_{\delta}} is also a $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}.$ Further, the $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule $\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}$ of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}$ is said to be non-degenerate if the restricted form $\left[\cdot,\cdot\right]_{\delta}\restriction_{\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}\times\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}}$ is non-degenerate, which is equivalent to saying that \bigl{(}\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}\bigr{)}\cap\bigl{(}\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}\bigr{)}^{\perp_{\delta}}=\{0\}. The $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule $\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}$ of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}$ is said to be Hermitian (skew-Hermitian) if the restricted form $\left[\cdot,\cdot\right]_{\delta}\restriction_{\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}\times\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}}$ is Hermitian (skew-Hermitian). We will say that \bigl{(}\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)},\left[\cdot,\cdot\right]_{\delta}\restriction_{\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}\times\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}}\bigr{)} is a Hermitian (skew-Hermitian) $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-space if $\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}$ is a non-degenerate and Hermitian (skew-Hermitian) $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}.$ In view of Lemma 3.3, we see that $\left(\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)},\left[\cdot,\cdot\right]_{\delta}\restriction_{\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}\times\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}}\right)$ is a Hermitian $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-space when $\delta\in\{\ast,0\}$ and is a skew-Hermitian $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-space when $\delta=\gamma.$

In the following lemma, we observe that for each $\delta\in\{\ast,0,\gamma\},$ if $\mathcal{C}_{h}$ is a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{C}_{\mu(h)}$ is a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}$ satisfying $\mathcal{C}_{h}\cap\mathcal{C}_{h}^{(\delta)}=\{0\}$ and $\mathcal{C}_{\mu(h)}\cap\mathcal{C}_{\mu(h)}^{(\delta)}=\{0\},$ then $\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}$ is a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}$ and vice versa.

Lemma 3.4**.**

Let $h\in\mathfrak{M}$ be fixed. Let $\mathcal{C}_{h}$ be a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{C}_{\mu(h)}$ be a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}.$ Then for each $\delta\in\{\ast,0,\gamma\},$ the following hold.

1. (a)

\bigr{(}\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}\bigl{)}^{\perp_{\delta}}=\mathcal{C}_{h}^{(\delta)}\oplus\mathcal{C}_{\mu(h)}^{(\delta)}.** 2. (b)

$\mathcal{C}_{h}\cap\mathcal{C}_{h}^{(\delta)}=\{0\}$* and $\mathcal{C}_{\mu(h)}\cap\mathcal{C}_{\mu(h)}^{(\delta)}=\{0\}$ if and only if (\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)})\cap\bigl{(}\mathcal{C}_{h}^{(\delta)}\oplus\mathcal{C}_{\mu(h)}^{(\delta)}\bigr{)}=\{0\}.* 3. (c)

$\mathcal{C}_{h}\cap\mathcal{C}_{h}^{(\delta)}=\{0\}$* and $\mathcal{C}_{\mu(h)}\cap\mathcal{C}_{\mu(h)}^{(\delta)}=\{0\}$ if and only if (\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)})\cap\bigl{(}\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}\bigr{)}^{\perp_{\delta}}=\{0\}.*

Proof.

Proof is trivial. ∎

Next we observe that any $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule $\mathfrak{N}$ of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}$ can be written as $\mathfrak{N}=\mathfrak{N}_{h}\oplus\mathfrak{N}_{\mu(h)},$ where $\mathfrak{N}_{h}=\mathfrak{N}\cap\mathcal{J}_{h}$ is a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathfrak{N}_{\mu(h)}=\mathfrak{N}\cap\mathcal{J}_{\mu(h)}$ is a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}.$ In view of this and Lemma 3.4, we see that for each $h\in\mathfrak{M}$ and $\delta\in\{\ast,0,\gamma\},$ the number of pairs \big{(}\mathcal{C}_{h},\mathcal{C}_{\mu(h)}\big{)} with $\mathcal{C}_{h}$ as a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{C}_{\mu(h)}$ as a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}$ satisfying $\mathcal{C}_{h}\cap\mathcal{C}_{h}^{(\delta)}=\{0\}$ and $\mathcal{C}_{\mu(h)}\cap\mathcal{C}_{\mu(h)}^{(\delta)}=\{0\}$ is equal to the number of non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodules of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}$ with respect to $\left[\cdot,\cdot\right]_{\delta}\restriction_{\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}\times\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}}.$ Further, we make the following observation.

Remark 3.2**.**

When $\delta=\gamma,$ by Lemma 3.3, we see that $\left[\cdot,\cdot\right]_{\gamma}$ is a non-degenerate and skew-Hermitian form on $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}.$ In this case, we will first transform the skew-Hermitian form $\left[\cdot,\cdot\right]_{\gamma}$ to a Hermitian form $\left[\cdot,\cdot\right]_{\gamma^{(H)}}$ as follows: Let $\vartheta(X)\in\mathcal{K}_{h}\setminus\{0\}.$ Then the corresponding element $\chi(X)=\vartheta(X)-\tau_{1,-1}(\vartheta(X))\in\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}\setminus\{0\}$ satisfies \tau_{1,-1}\big{(}\chi(X)\big{)}=-\chi(X). Now we define the form $\left[\cdot,\cdot\right]_{\gamma^{(H)}}$ on $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}\times\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}$ as $\left[\alpha_{1}(X)+\alpha_{2}(X),\beta_{1}(X)+\beta_{2}(X)\right]_{\gamma^{(H)}}=\chi(X)\left[\alpha_{1}(X)+\alpha_{2}(X),\beta_{1}(X)+\beta_{2}(X)\right]_{\gamma}$ for all $\alpha_{1}(X)+\alpha_{2}(X),\beta_{1}(X)+\beta_{2}(X)\in\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}.$ It is easy to see that $\left[\cdot,\cdot\right]_{\gamma^{(H)}}$ is a non-degenerate and Hermitian form on $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}.$ Further, as $\vartheta(X)$ and $\tau_{1,-1}(\vartheta(X))$ are non-zero elements of $\mathcal{K}_{h}$ and $\mathcal{K}_{\mu(h)}$ respectively, we observe that if two elements in $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}$ are orthogonal with respect to $\left[\cdot,\cdot\right]_{\gamma},$ then those two elements are also orthogonal with respect to $\left[\cdot,\cdot\right]_{\gamma^{(H)}}$ and vice versa. This implies that if $\mathcal{C}_{h}$ is a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{C}_{\mu(h)}$ is a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)},$ then the dual code \big{(}\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}\big{)}^{\perp_{\gamma}} of the $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule $\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}$ of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}$ with respect to $\left[\cdot,\cdot\right]_{\gamma}\restriction_{\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}\times\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}}$ coincides with the dual code \bigl{(}\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}\bigr{)}^{\perp_{\gamma^{(H)}}}=\{a(X)+b(X)\in\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}:\left[a(X)+b(X),c(X)+d(X)\right]_{\gamma^{(H)}}=0\text{ for all }c(X)+d(X)\in\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}\} of $\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}$ with respect to $\left[\cdot,\cdot\right]_{\gamma^{(H)}}.$ Therefore the number of non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodules of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}$ with respect to $\left[\cdot,\cdot\right]_{\gamma}\restriction_{\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}\times\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}}$ is equal to the number of non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodules of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}$ with respect to $\left[\cdot,\cdot\right]_{\gamma^{(H)}}.$ In view of this, we will consider the Hermitian and non-degenerate form $\left[\cdot,\cdot\right]_{\gamma^{(H)}}$ on $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}\times\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}$ in lieu of the skew-Hermitian and non-degenerate form $\left[\cdot,\cdot\right]_{\gamma}\restriction_{\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}\times\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}}$ from now onwards.

In order to enumerate the number $N_{h}$ for each $h\in\mathfrak{M},$ we will first prove that if $\mathcal{C}_{h}$ is a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{C}_{\mu(h)}$ is a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}$ such that $\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}$ is a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}$ with respect to $\left[\cdot,\cdot\right]_{\delta}$ for $\delta\in\{\ast,0,\gamma^{(H)}\},$ then $\dim_{\mathcal{K}_{h}}\mathcal{C}_{h}=\dim_{\mathcal{K}_{\mu(h)}}\mathcal{C}_{\mu(h)}.$ Note that for each $\delta\in\{\ast,0,\gamma^{(H)}\},$ the form $\left[\cdot,\cdot\right]_{\delta}$ is non-degenerate and Hermitian on $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)},$ i.e., \bigl{(}\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)},\left[\cdot,\cdot\right]_{\delta}\restriction_{\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}\times\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}}\bigr{)} is a Hermitian $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-space.

Proposition 3.7**.**

Let $h\in\mathfrak{M}$ and $\delta\in\{\ast,0,\gamma^{(H)}\}$ be fixed. Let $\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}$ be a $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)},$ where $\mathcal{C}_{h}$ is a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{C}_{\mu(h)}$ is a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}.$ If $\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}$ is non-degenerate, then the $\mathcal{K}_{h}$-dimension of $\mathcal{C}_{h}$ is equal to the $\mathcal{K}_{\mu(h)}$-dimension of $\mathcal{C}_{\mu(h)}.$

In order to prove this proposition, we need to define the following:

Let $\mathcal{A}_{h}$ be a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{A}_{\mu(h)}$ be a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}.$ A non-zero element $a(X)+b(X)\in\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ is said to be isotropic if $\left[a(X)+b(X),a(X)+b(X)\right]_{\delta}=0,$ otherwise $a(X)+b(X)$ is called an anisotropic element of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$ Note that all the non-zero elements of $\mathcal{A}_{h}$ and $\mathcal{A}_{\mu(h)}$ are trivially isotropic. An isotropic element $a(X)+b(X)$ of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ is said to be non-trivial if both $a(X),b(X)$ are non-zero. We say that two non-trivial isotropic elements $a(X)+b(X)$ and $c(X)+d(X)$ of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ form a hyperbolic pair if $\left[a(X)+b(X),c(X)+d(X)\right]_{\delta}=e_{h}(X)+e_{\mu(h)}(X),$ where $e_{h}(X)$ and $e_{\mu(h)}(X)$ are multiplicative identities of $\mathcal{K}_{h}$ and $\mathcal{K}_{\mu(h)}$ respectively. Throughout this section, we will denote the $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ generated by the elements $\alpha_{1}(X)+\beta_{1}(X),\alpha_{2}(X)+\beta_{2}(X),\cdots,\alpha_{m}(X)+\beta_{m}(X)\in\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ by $\langle\alpha_{1}(X)+\beta_{1}(X),\alpha_{2}(X)+\beta_{2}(X),\cdots,\alpha_{m}(X)+\beta_{m}(X)\rangle.$ Furthermore, it is easy to see that $\langle\alpha_{1}(X)+\beta_{1}(X),\alpha_{2}(X)+\beta_{2}(X),\cdots,\alpha_{m}(X)+\beta_{m}(X)\rangle^{\perp_{\delta}}=\langle\beta_{1}(X),\beta_{2}(X),\cdots,\beta_{m}(X)\rangle^{\perp_{\delta}}\oplus\langle\alpha_{1}(X),\alpha_{2}(X),\cdots,\alpha_{m}(X)\rangle^{\perp_{\delta}},$ where $\langle\beta_{1}(X),\beta_{2}(X),\cdots,\beta_{m}(X)\rangle^{\perp_{\delta}}=\{a(X)\in\mathcal{A}_{h}:\left[a(X),\beta_{j}(X)\right]_{\delta}=0\text{ for }1\leq j\leq m\}$ is a $\mathcal{K}_{h}$-subspace of $\mathcal{A}_{h}$ and $\langle\alpha_{1}(X),\alpha_{2}(X),\cdots,\alpha_{m}(X)\rangle^{\perp_{\delta}}=\{c(X)\in\mathcal{A}_{\mu(h)}:\left[c(X),\alpha_{j}(X)\right]_{\delta}\\ =0\text{ for }1\leq j\leq m\}$ is a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{A}_{\mu(h)}.$

In the following lemma, we will discuss the existence of a non-trivial isotropic element and a hyperbolic pair in a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}.$

Lemma 3.5**.**

Let $h\in\mathfrak{M}$ and $\delta\in\{\ast,0,\gamma^{(H)}\}$ be fixed. Let $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ be a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)},$ where $\mathcal{A}_{h}$ is a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ having dimension $\ell$ and $\mathcal{A}_{\mu(h)}$ is a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}$ having dimension $r.$ Then we have the following:

1. (a)

There does not exist any non-trivial isotropic element in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ when either $\ell=1$ or $r=1.$ 2. (b)

*There exists a non-trivial isotropic element in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ when both *$\ell,r\geq 2.$ 3. (c)

When both $\ell,r\geq 2,$ there exists a hyperbolic pair in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$ 4. (d)

The $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ generated by a hyperbolic pair is non-degenerate.

Proof.

1. (a)

To prove this, without any loss of generality, suppose that $r=1.$ Let $\alpha(X)+\beta(X)$ be a non-zero element of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)},$ where $\alpha(X)\in\mathcal{A}_{h}$ and $\beta(X)\in\mathcal{A}_{\mu(h)}.$ Now $\alpha(X)+\beta(X)$ is an isotropic element if and only if it satisfies $\left[\alpha(X)+\beta(X),\alpha(X)+\beta(X)\right]_{\delta}=0,$ which holds if and only if $[\alpha(X),\beta(X)]_{\delta}=0$ and $[\beta(X),\alpha(X)]_{\delta}=0,$ which is equivalent to saying that $\alpha(X)\in\langle\beta(X)\rangle^{\perp_{\delta}}$ and $\beta(X)\in\langle\alpha(X)\rangle^{\perp_{\delta}},$ where $\langle\beta(X)\rangle^{\perp_{\delta}}$ is a $\mathcal{K}_{h}$-subspace of $\mathcal{A}_{h}$ and $\langle\alpha(X)\rangle^{\perp_{\delta}}$ is a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{A}_{\mu(h)}.$ We first assert that $\text{dim}_{\mathcal{K}_{\mu(h)}}\langle\alpha(X)\rangle^{\perp_{\delta}}=0$ if $\alpha(X)\neq 0$ and $\text{dim}_{\mathcal{K}_{h}}\langle\beta(X)\rangle^{\perp_{\delta}}=\ell-1$ if $\beta(X)\neq 0.$ To prove this assertion, we define a map $\Phi_{\alpha(X)}:\mathcal{A}_{\mu(h)}\rightarrow\mathcal{K}_{\mu(h)}$ as $\Phi_{\alpha(X)}(r(X))=\left[r(X),\alpha(X)\right]_{\delta}$ for all $r(X)\in\mathcal{A}_{\mu(h)}.$ It is easy to see that $\Phi_{\alpha(X)}$ is a non-zero vector space homomorphism and $\langle\alpha(X)\rangle^{\perp_{\delta}}$ is the kernel of the map $\Phi_{\alpha(X)}.$ Now by Sylvester’s law of nullity, we have $\text{dim}_{\mathcal{K}_{\mu(h)}}\langle\alpha(X)\rangle^{\perp_{\delta}}=\text{dim}_{\mathcal{K}_{\mu(h)}}\mathcal{A}_{\mu(h)}-1=0,$ which proves the assertion. Working in a similar manner as above, one can show that if $\beta(X)\neq 0,$ then $\text{dim}_{\mathcal{K}_{h}}\langle\beta(X)\rangle^{\perp_{\delta}}=\ell-1\geq 0.$ Further, we observe that $\langle\alpha(X)\rangle^{\perp_{\delta}}=\mathcal{A}_{\mu(h)}$ if $\alpha(X)=0$ and $\langle\beta(X)\rangle^{\perp_{\delta}}=\mathcal{A}_{h}$ if $\beta(X)=0.$ From this, it follows that $\alpha(X)+\beta(X)$ is an isotropic element of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ if and only if either $\alpha(X)=0$ and $\beta(X)\in\mathcal{A}_{\mu(h)}\setminus\{0\}$ or $\alpha(X)\in\mathcal{A}_{h}\setminus\{0\}$ and $\beta(X)=0.$ This shows that there does not exist any non-trivial isotropic element in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ when $r=1.$ 2. (b)

Suppose that both $\ell,r\geq 2.$ Let $u(X)+v(X)$ be an element of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)},$ where $u(X)\in\mathcal{A}_{h}\setminus\{0\}$ and $v(X)\in\mathcal{A}_{\mu(h)}\setminus\{0\}.$ If $u(X)+v(X)$ is isotropic, then there is nothing to prove. Otherwise, we have $[u(X)+v(X),u(X)+v(X)]_{\delta}\neq 0.$ By (21), we get $[u(X)+v(X),u(X)+v(X)]_{\delta}=\left[u(X),v(X)\right]_{\delta}+\left[v(X),u(X)\right]_{\delta}.$ Now by Lemma 3.3(v) of Sharma and Kaur [16], Lemma 6(iv)-(v) of Huffman [9] and using Remark 3.2, we see that $\left[v(X),u(X)\right]_{\delta}=\tau_{1,-1}(\left[u(X),v(X)\right]_{\delta}).$ This gives \left[u(X)+v(X),u(X)+v(X)\right]_{\delta}=\left[u(X),v(X)\right]_{\delta}+\tau_{1,-1}(\left[u(X),v(X)\right]_{\delta})=d(X)+\tau_{1,-1}\big{(}d(X)\big{)}, where $d(X)\in\mathcal{K}_{h}\setminus\{0\}.$ Now consider the $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule $\langle u(X)+v(X)\rangle^{\perp_{\delta}}$ of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$ It is easy to observe that $\langle u(X)+v(X)\rangle^{\perp_{\delta}}=\langle v(X)\rangle^{\perp_{\delta}}\oplus\langle u(X)\rangle^{\perp_{\delta}}.$ Now working as in part (a), we see that $\text{dim}_{\mathcal{K}_{h}}\langle v(X)\rangle^{\perp_{\delta}}=\ell-1\geq 1$ and $\text{dim}_{\mathcal{K}_{\mu(h)}}\langle u(X)\rangle^{\perp_{\delta}}=r-1\geq 1.$ Using this, we see that there exist non-zero elements $\alpha(X)\in\langle v(X)\rangle^{\perp_{\delta}}$ and $\beta(X)\in\langle u(X)\rangle^{\perp_{\delta}},$ which implies that $\alpha(X)+\beta(X)\in\langle u(X)+v(X)\rangle^{\perp_{\delta}}.$ Now if the element $\alpha(X)+\beta(X)$ is isotropic, then we are through. Otherwise, using (21), we get $[\alpha(X)+\beta(X),\alpha(X)+\beta(X)]_{\delta}=\left[\alpha(X),\beta(X)\right]_{\delta}+\left[\beta(X),\alpha(X)\right]_{\delta}.$ Now by Sharma and Kaur [16, Lemma 3.3(v)], Huffman [9, Lemma 6(iv)] and using Remark 3.2, we see that $\left[\beta(X),\alpha(X)\right]_{\delta}=\tau_{1,-1}(\left[\alpha(X),\beta(X)\right]_{\delta}).$ This implies that \left[\alpha(X)+\beta(X),\alpha(X)+\beta(X)\right]_{\delta}=\left[\alpha(X),\beta(X)\right]_{\delta}+\tau_{1,-1}(\left[\alpha(X),\beta(X)\right]_{\delta})=\eta(X)+\tau_{1,-1}\big{(}\eta(X)\big{)}, where $\eta(X)\in\mathcal{K}_{h}\setminus\{0\}.$ Now it is easy to see that u(X)+v(X)+\bigl{\{}-d(X)+\tau_{1,-1}\bigl{(}\eta(X)^{-1}\bigr{)}\bigr{\}}\bigl{(}\alpha(X)+\beta(X)\bigr{)} is a non-trivial isotropic element of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$ 3. (c)

As both $\ell,r\geq 2,$ by part (b), we see that $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ has a non-trivial isotropic element, say $a(X)+b(X).$ Further, we have $\langle a(X)+b(X)\rangle^{\perp_{\delta}}=\langle b(X)\rangle^{\perp_{\delta}}\oplus\langle a(X)\rangle^{\perp_{\delta}}.$ Working as in part (a), we see that $\text{dim}_{\mathcal{K}_{h}}\langle b(X)\rangle^{\perp_{\delta}}=\ell-1\geq 1$ and $\text{dim}_{\mathcal{K}_{\mu(h)}}\langle a(X)\rangle^{\perp_{\delta}}=r-1\geq 1.$ Now we choose a non-zero element $c(X)\in\mathcal{A}_{h}\setminus\langle b(X)\rangle^{\perp_{\delta}}$ and $d(X)\in\mathcal{A}_{\mu(h)}\setminus\langle a(X)\rangle^{\perp_{\delta}}.$ Then $\left[a(X)+b(X),c(X)+d(X)\right]_{\delta}=\ell_{1}(X)+\ell_{2}(X),$ where $\ell_{1}(X)\in\mathcal{K}_{h}\setminus\{0\}$ and $\ell_{2}(X)\in\mathcal{K}_{\mu(h)}\setminus\{0\}.$ This gives \left[a(X)+b(X),\tau_{1,-1}\big{(}\ell_{2}(X)^{-1}\big{)}c(X)\right.\\ \left.+\tau_{1,-1}\big{(}\ell_{1}(X)^{-1}\big{)}d(X)\right]_{\delta}=e_{h}(X)+e_{\mu(h)}(X). Therefore, without any loss of generality, we assume that $c(X)+d(X)\in\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ is such that $\left[a(X)+b(X),c(X)+d(X)\right]_{\delta}=e_{h}(X)+e_{\mu(h)}(X).$

If $c(X)+d(X)$ is an isotropic element, then we are done. Otherwise, using (21), we have $[c(X)+d(X),c(X)+d(X)]_{\delta}=\left[c(X),d(X)\right]_{\delta}+\left[d(X),c(X)\right]_{\delta}.$ Now by Sharma and Kaur [16, Lemma 3.3(v)], Huffman [9, Lemma 6(iv)] and using Remark 3.2, we see that $\left[d(X),c(X)\right]_{\delta}=\tau_{1,-1}(\left[c(X),d(X)\right]_{\delta}).$ This implies that \left[c(X)+d(X),c(X)+d(X)\right]_{\delta}=\left[c(X),d(X)\right]_{\delta}+\tau_{1,-1}(\left[c(X),d(X)\right]_{\delta})=w(X)+\tau_{1,-1}\big{(}w(X)\big{)}, where $w(X)\in\mathcal{K}_{h}\setminus\{0\}.$ Now it is easy to see that c(X)+d(X)-w(X)\bigl{(}a(X)+b(X)\bigr{)}=c(X)+d(X)-w(X)a(X) is a non-trivial isotropic element in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h).}$ On the other hand, we observe that $[a(X)+b(X),c(X)+d(X)-w(X)a(X)]_{\delta}=[a(X),d(X)]_{\delta}+[b(X),c(X)-w(X)a(X)]_{\delta}=e_{h}(X)+e_{\mu(h)}(X),$ which implies that $a(X)+b(X)$ and $c(X)+d(X)-w(X)a(X)$ form a hyperbolic pair in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$ 4. (d)

Proof is trivial.

∎

The following lemma is quite useful in writing an orthogonal direct sum decomposition of a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}.$

Lemma 3.6**.**

Let $h\in\mathfrak{M}$ and $\delta\in\{\ast,0,\gamma^{(H)}\}$ be fixed. Let $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ be a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}$ and $\mathcal{W}_{h}\oplus\mathcal{W}_{\mu(h)}$ be a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)},$ where $\mathcal{A}_{j}$ is a $\mathcal{K}_{j}$-subspace of $\mathcal{J}_{j}$ and $\mathcal{W}_{j}$ is a $\mathcal{K}_{j}$-subspace of $\mathcal{A}_{j}$ for each $j\in\{h,\mu(h)\}.$ Then $(\mathcal{W}_{h}\oplus\mathcal{W}_{\mu(h)})^{\perp_{\delta}}=\{a(X)+b(X)\in\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}:\left[a(X)+b(X),c(X)+d(X)\right]_{\delta}=0\text{ for all }c(X)+d(X)\in\mathcal{W}_{h}\oplus\mathcal{W}_{\mu(h)}\}$ is a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ and

[TABLE]

Proof.

It follows from Lemma 1.1 of Knebusch et al. [13]. ∎

In the following lemma, we write an orthogonal direct sum decomposition of a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}.$

Lemma 3.7**.**

Let $h\in\mathfrak{M}$ and $\delta\in\{\ast,0,\gamma^{(H)}\}$ be fixed. Let $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ be a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)},$ where $\mathcal{A}_{h}$ is a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{A}_{\mu(h)}$ is a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}$ with $\text{dim}_{\mathcal{K}_{h}}\mathcal{A}_{h}=\ell>0$ and $\text{dim}_{\mathcal{K}_{\mu(h)}}\mathcal{A}_{\mu(h)}=r>0.$ Then the following hold:

1. (a)

When $r$ is even and $\ell\geq r,$ we have $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}=\langle m_{1}(X)+n_{1}(X),p_{1}(X)+q_{1}(X)\rangle\perp\langle m_{2}(X)+n_{2}(X),p_{2}(X)+q_{2}(X)\rangle\perp\cdots\perp\langle m_{\frac{r}{2}}(X)+n_{\frac{r}{2}}(X),p_{\frac{r}{2}}(X)+q_{\frac{r}{2}}(X)\rangle\perp\langle\theta_{1}(X)\rangle\perp\langle\theta_{2}(X)\rangle\perp\cdots\perp\langle\theta_{\ell-r}(X)\rangle,$ where \big{(}m_{j}(X)+n_{j}(X),p_{j}(X)+q_{j}(X)\big{)} is a hyperbolic pair in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ for $1\leq j\leq\frac{r}{2}$ and $\{\theta_{1}(X),\theta_{2}(X),\cdots,\theta_{\ell-r}(X)\}$ is a basis of the $(\ell-r)$-dimensional $\mathcal{K}_{h}$-subspace $\langle n_{1}(X),q_{1}(X),\cdots,n_{\frac{r}{2}}(X),\\ q_{\frac{r}{2}}(X)\rangle^{\perp_{\delta}}$ of $\mathcal{A}_{h}.$ 2. (b)

When $r$ is odd and $\ell\geq r,$ we have $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}=\langle m_{1}(X)+n_{1}(X),p_{1}(X)+q_{1}(X)\rangle\perp\langle m_{2}(X)+n_{2}(X),p_{2}(X)+q_{2}(X)\rangle\perp\cdots\perp\langle m_{\frac{r-1}{2}}(X)+n_{\frac{r-1}{2}}(X),p_{\frac{r-1}{2}}(X)+q_{\frac{r-1}{2}}(X)\rangle\perp\langle\alpha(X)+\beta(X)\rangle\perp\langle\theta_{1}(X)\rangle\perp\langle\theta_{2}(X)\rangle\perp\cdots\perp\langle\theta_{\ell-r}(X)\rangle,$ where \big{(}m_{j}(X)+n_{j}(X),p_{j}(X)+q_{j}(X)\big{)} is a hyperbolic pair in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ for $1\leq j\leq\frac{r-1}{2},$ $\alpha(X)+\beta(X)$ is an anisotropic element of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ and $\{\theta_{1}(X),\theta_{2}(X),\cdots,\theta_{\ell-r}(X)\}$ is a basis of the $(\ell-r)$-dimensional $\mathcal{K}_{h}$-subspace $\langle n_{1}(X),q_{1}(X),\cdots,\\ n_{\frac{r-1}{2}}(X),q_{\frac{r-1}{2}}(X),\beta(X)\rangle^{\perp_{\delta}}$ of $\mathcal{A}_{h}.$ 3. (c)

When $\ell$ is even and $r\geq\ell,$ we have $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}=\langle m_{1}(X)+n_{1}(X),p_{1}(X)+q_{1}(X)\rangle\perp\langle m_{2}(X)+n_{2}(X),p_{2}(X)+q_{2}(X)\rangle\perp\cdots\perp\langle m_{\frac{\ell}{2}}(X)+n_{\frac{\ell}{2}}(X),p_{\frac{\ell}{2}}(X)+q_{\frac{\ell}{2}}(X)\rangle\perp\langle\eta_{1}(X)\rangle\perp\langle\eta_{2}(X)\rangle\perp\cdots\perp\langle\eta_{r-\ell}(X)\rangle,$ where \big{(}m_{j}(X)+n_{j}(X),p_{j}(X)+q_{j}(X)\big{)} is a hyperbolic pair in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ for $1\leq j\leq\frac{\ell}{2}$ and $\{\eta_{1}(X),\eta_{2}(X),\cdots,\eta_{r-\ell}(X)\}$ is a basis of the $(r-\ell)$-dimensional $\mathcal{K}_{\mu(h)}$-subspace $\langle m_{1}(X),p_{1}(X),\cdots,\\ m_{\frac{\ell}{2}}(X),p_{\frac{\ell}{2}}(X)\rangle^{\perp_{\delta}}$ of $\mathcal{A}_{\mu(h)}.$ 4. (d)

When $\ell$ is odd and $r\geq\ell,$ we have $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}=\langle m_{1}(X)+n_{1}(X),p_{1}(X)+q_{1}(X)\rangle\perp\langle m_{2}(X)+n_{2}(X),p_{2}(X)+q_{2}(X)\rangle\perp\cdots\perp\langle m_{\frac{\ell-1}{2}}(X)+n_{\frac{\ell-1}{2}}(X),p_{\frac{\ell-1}{2}}(X)+q_{\frac{\ell-1}{2}}(X)\rangle\perp\langle\alpha(X)+\beta(X)\rangle\perp\langle\eta_{1}(X)\rangle\perp\langle\eta_{2}(X)\rangle\perp\cdots\perp\langle\eta_{r-\ell}(X)\rangle,$ where \big{(}m_{j}(X)+n_{j}(X),p_{j}(X)+q_{j}(X)\big{)} is a hyperbolic pair in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ for $1\leq j\leq\frac{\ell-1}{2},$ $\alpha(X)+\beta(X)$ is an anisotropic element of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ and $\{\eta_{1}(X),\eta_{2}(X),\cdots,\eta_{r-\ell}(X)\}$ is a basis of the $(r-\ell)$-dimensional $\mathcal{K}_{\mu(h)}$-subspace $\langle m_{1}(X),p_{1}(X),\cdots,\\ m_{\frac{\ell-1}{2}}(X),p_{\frac{\ell-1}{2}}(X),\alpha(X)\rangle^{\perp_{\delta}}$ of $\mathcal{A}_{\mu(h)}.$

Proof.

1. (a)

Suppose that $r$ is even and $\ell\geq r.$ Let us write $r=2r_{1},$ where $r_{1}\geq 1$ is an integer. In order to prove (a), we will apply induction on $r_{1}.$

First let $r_{1}=1.$ In this case, by Lemma 3.5(c) and (d), we see that $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ has a hyperbolic pair, say $(m_{1}(X)+n_{1}(X),p_{1}(X)+q_{1}(X)),$ and the $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule $\langle m_{1}(X)+n_{1}(X),p_{1}(X)+q_{1}(X)\rangle$ of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ is non-degenerate. Now by Lemma 3.6, we have $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}=\langle m_{1}(X)+n_{1}(X),p_{1}(X)+q_{1}(X)\rangle\perp\langle m_{1}(X)+n_{1}(X),p_{1}(X)+q_{1}(X)\rangle^{\perp_{\delta}},$ where $\langle m_{1}(X)+n_{1}(X),p_{1}(X)+q_{1}(X)\rangle^{\perp_{\delta}}$ is a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$ Further, it is easy to see that $\langle m_{1}(X)+n_{1}(X),p_{1}(X)+q_{1}(X)\rangle^{\perp_{\delta}}=\langle n_{1}(X),q_{1}(X)\rangle^{\perp_{\delta}}\oplus\langle m_{1}(X),p_{1}(X)\rangle^{\perp_{\delta}}.$ We next observe that $\text{dim}_{\mathcal{K}_{h}}\langle n_{1}(X),q_{1}(X)\rangle^{\perp_{\delta}}=\ell-2$ and $\text{dim}_{\mathcal{K}_{\mu(h)}}\langle m_{1}(X),p_{1}(X)\rangle^{\perp_{\delta}}=0.$ This implies that $\langle m_{1}(X)+n_{1}(X),p_{1}(X)+q_{1}(X)\rangle^{\perp_{\delta}}=\langle n_{1}(X),q_{1}(X)\rangle^{\perp_{\delta}}.$ Now let $\{\theta_{1}(X),\theta_{2}(X),\cdots,\theta_{\ell-2}(X)\}$ be a basis of the $\mathcal{K}_{h}$-subspace $\langle n_{1}(X),q_{1}(X)\rangle^{\perp_{\delta}}$ of $\mathcal{A}_{h}.$ It is clear that $\left[\theta_{j_{1}}(X),\theta_{j_{2}}(X)\right]_{\delta}=0$ for all $1\leq j_{1},j_{2}\leq\ell-2.$ From this, we obtain $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}=\langle m_{1}(X)+n_{1}(X),p_{1}(X)+q_{1}(X)\rangle\perp\langle\theta_{1}(X)\rangle\perp\langle\theta_{2}(X)\rangle\perp\cdots\perp\langle\theta_{\ell-2}(X)\rangle.$ Thus the result holds for $r_{1}=1.$

Next let $r_{1}>1$ and suppose that the result holds for every non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of the type $\mathcal{W}_{h}\oplus\mathcal{W}_{\mu(h)},$ where $\mathcal{W}_{h}$ is a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{W}_{\mu(h)}$ is a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}$ with $\text{dim}_{\mathcal{K}_{\mu(h)}}\mathcal{W}_{\mu(h)}=2(r_{1}-1)$ and $\text{dim}_{\mathcal{K}_{h}}\mathcal{W}_{h}\geq 2(r_{1}-1).$

We will now prove the result for a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)},$ where $\mathcal{A}_{h}$ is a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{A}_{\mu(h)}$ is a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}$ with $\text{dim}_{\mathcal{K}_{\mu(h)}}\mathcal{A}_{\mu(h)}=2r_{1}$ and $\text{dim}_{\mathcal{K}_{h}}\mathcal{A}_{h}=\ell\geq 2r_{1}.$ By Lemma 3.5(c) and (d), we see that $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ has a hyperbolic pair, say $(\widetilde{m}(X)+\widetilde{n}(X),\widetilde{p}(X)+\widetilde{q}(X))$ and the $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule $\langle\widetilde{m}(X)+\widetilde{n}(X),\widetilde{p}(X)+\widetilde{q}(X)\rangle$ of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ is non-degenerate. Now by Lemma 3.6, we have $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}=\langle\widetilde{m}(X)+\widetilde{n}(X),\widetilde{p}(X)+\widetilde{q}(X)\rangle\perp\langle\widetilde{m}(X)+\widetilde{n}(X),\widetilde{p}(X)+\widetilde{q}(X)\rangle^{\perp_{\delta}},$ where $\langle\widetilde{m}(X)+\widetilde{n}(X),\widetilde{p}(X)+\widetilde{q}(X)\rangle^{\perp_{\delta}}$ is a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$ Further, we see that $\langle\widetilde{m}(X)+\widetilde{n}(X),\widetilde{p}(X)+\widetilde{q}(X)\rangle^{\perp_{\delta}}=\langle\widetilde{n}(X),\widetilde{q}(X)\rangle^{\perp_{\delta}}\oplus\langle\widetilde{m}(X),\widetilde{p}(X)\rangle^{\perp_{\delta}}.$ It is easy to observe that $\text{dim}_{\mathcal{K}_{h}}\langle\widetilde{n}(X),\widetilde{q}(X)\rangle^{\perp_{\delta}}=\ell-2$ and $\text{dim}_{\mathcal{K}_{\mu(h)}}\langle\widetilde{m}(X),\widetilde{p}(X)\rangle^{\perp_{\delta}}=2r_{1}-2.$ Now by applying induction hypothesis, we have $\langle\widetilde{m}(X)+\widetilde{n}(X),\widetilde{p}(X)+\widetilde{q}(X)\rangle^{\perp_{\delta}}=\langle\widetilde{m}_{1}(X)+\widetilde{n}_{1}(X),\widetilde{p}_{1}(X)+\widetilde{q}_{1}(X)\rangle\perp\langle\widetilde{m}_{2}(X)+\widetilde{n}_{2}(X),\widetilde{p}_{2}(X)+\widetilde{q}_{2}(X)\rangle\perp\cdots\perp\langle\widetilde{m}_{r_{1}-1}(X)+\widetilde{n}_{r_{1}-1}(X),\widetilde{p}_{r_{1}-1}(X)+\widetilde{q}_{r_{1}-1}(X)\rangle\perp\langle\theta_{1}(X)\rangle\perp\langle\theta_{2}(X)\rangle\perp\cdots\perp\langle\theta_{\ell-2r_{1}}(X)\rangle,$ where \big{(}\widetilde{m}_{j}(X)+\widetilde{n}_{j}(X),\widetilde{p}_{j}(X)+\widetilde{q}_{j}(X)\big{)} is a hyperbolic pair in $\langle\widetilde{m}(X)+\widetilde{n}(X),\widetilde{p}(X)+\widetilde{q}(X)\rangle^{\perp_{\delta}}$ for $1\leq j\leq r_{1}-1$ and $\{\theta_{1}(X),\theta_{2}(X),\cdots,\theta_{\ell-2r_{1}}(X)\}$ is a basis of the $\mathcal{K}_{h}$-subspace $\langle\widetilde{n}_{1}(X),\widetilde{q}_{1}(X),\cdots,\widetilde{n}_{r_{1}-1}(X),\widetilde{q}_{r_{1}-1}(X)\rangle^{\perp_{\delta}}$ of $\langle\widetilde{n}(X),\widetilde{q}(X)\rangle^{\perp_{\delta}}.$ From this, we obtain $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}=\langle\widetilde{m}(X)+\widetilde{n}(X),\widetilde{p}(X)+\widetilde{q}(X)\rangle\perp\langle\widetilde{m}_{1}(X)+\widetilde{n}_{1}(X),\widetilde{p}_{1}(X)+\widetilde{q}_{1}(X)\rangle\perp\langle\widetilde{m}_{2}(X)+\widetilde{n}_{2}(X),\widetilde{p}_{2}(X)+\widetilde{q}_{2}(X)\rangle\perp\cdots\perp\langle\widetilde{m}_{r_{1}-1}(X)+\widetilde{n}_{r_{1}-1}(X),\widetilde{p}_{r_{1}-1}(X)+\widetilde{q}_{r_{1}-1}(X)\rangle\perp\langle\theta_{1}(X)\rangle\perp\langle\theta_{2}(X)\rangle\perp\cdots\perp\langle\theta_{\ell-2r_{1}}(X)\rangle,$ where (\widetilde{m}(X)+\widetilde{n}(X),\widetilde{p}(X)+\widetilde{q}(X)),\big{(}\widetilde{m}_{j}(X)+\widetilde{n}_{j}(X),\widetilde{p}_{j}(X)+\widetilde{q}_{j}(X)\big{)} are hyperbolic pairs in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ for $1\leq j\leq r_{1}-1$ and $\{\theta_{1}(X),\theta_{2}(X),\cdots,\theta_{\ell-2r_{1}}(X)\}$ is a basis of the $\mathcal{K}_{h}$-subspace $\langle\widetilde{n}(X),\widetilde{q}(X),\widetilde{n}_{1}(X),\widetilde{q}_{1}(X),\cdots,\widetilde{n}_{r_{1}-1}(X),\widetilde{q}_{r_{1}-1}(X)\rangle^{\perp_{\delta}}$ of $\mathcal{A}_{h}.$ This proves (a). 2. (b)

Suppose that $r$ is odd and $\ell\geq r.$ Let us write $r=2r_{1}+1,$ where $r_{1}\geq 0$ is an integer. Here we will apply induction on $r_{1}\geq 0.$

We first suppose that $r_{1}=0$ so that $\dim_{\mathcal{K}_{\mu(h)}}\mathcal{A}_{\mu(h)}=1.$ Here we assert that there exists an anisotropic element in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$ To prove this, we choose $\beta(X)\in\mathcal{A}_{\mu(h)}\setminus\{0\}.$ It is easy to see that $\dim_{\mathcal{K}_{h}}\langle\beta(X)\rangle^{\perp_{\delta}}=\ell-1.$ From this, we see that there exists $\alpha(X)\in\mathcal{A}_{h}\setminus\langle\beta(X)\rangle^{\perp_{\delta}}$ so that $\left[\alpha(X),\beta(X)\right]_{\delta}=a(X)\in\mathcal{K}_{h}\setminus\{0\}.$ This gives $\left[\alpha(X)+\beta(X),\alpha(X)+\beta(X)\right]_{\delta}=a(X)+\tau_{1,-1}(a(X)),$ i.e., $\alpha(X)+\beta(X)$ is an anisotropic element of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$ Further, it is easy to observe that $\langle\alpha(X)+\beta(X)\rangle$ is a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$ By Lemma 3.6, we get $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}=\langle\alpha(X)+\beta(X)\rangle\perp\langle\alpha(X)+\beta(X)\rangle^{\perp_{\delta}},$ where $\langle\alpha(X)+\beta(X)\rangle^{\perp_{\delta}}$ is a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$ Next we observe that $\langle\alpha(X)+\beta(X)\rangle^{\perp_{\delta}}=\langle\beta(X)\rangle^{\perp_{\delta}}\oplus\langle\alpha(X)\rangle^{\perp_{\delta}}.$ Working as in Lemma 3.5(a), we see that $\text{dim}_{\mathcal{K}_{h}}\langle\beta(X)\rangle^{\perp_{\delta}}=\ell-1$ and $\text{dim}_{\mathcal{K}_{\mu(h)}}\langle\alpha(X)\rangle^{\perp_{\delta}}=0.$ This implies that $\langle\alpha(X)+\beta(X)\rangle^{\perp_{\delta}}=\langle\beta(X)\rangle^{\perp_{\delta}}$ is a $\mathcal{K}_{h}$-subspace of $\mathcal{A}_{h}$ and suppose that $\{\theta_{1}(X),\theta_{2}(X),\cdots,\theta_{\ell-1}(X)\}$ is its basis. As $\left[\theta_{j_{1}}(X),\theta_{j_{2}}(X)\right]_{\delta}=0$ for all $1\leq j_{1},j_{2}\leq\ell-1,$ we have $\langle\beta(X)\rangle^{\perp_{\delta}}=\langle\theta_{1}(X)\rangle\perp\langle\theta_{2}(X)\rangle\perp\cdots\perp\langle\theta_{\ell-1}(X)\rangle.$ From this, we obtain $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}=\langle\alpha(X)+\beta(X)\rangle\perp\langle\theta_{1}(X)\rangle\perp\langle\theta_{2}(X)\rangle\perp\cdots\perp\langle\theta_{\ell-1}(X)\rangle.$ Thus the result holds when $r_{1}=0.$

When $r_{1}\geq 1,$ working in a similar way as in part (a) and using Lemmas 3.5 and 3.6, part (b) follows. 3. (c)

Working in a similar manner as in part (a), part (c) follows. 4. (d)

Working in a similar manner as in part (b), part (d) follows.

∎

Proof of Proposition 3.7. It follows immediately from Lemma 3.7 and using the fact that (\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)})\cap\big{(}\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}\big{)}^{\perp_{\delta}}=\{0\}. $\hfill\Box$

Remark 3.3**.**

Let $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ be a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)},$ where $\mathcal{A}_{h}$ is a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{A}_{\mu(h)}$ is a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}.$ Then in view of Proposition 3.7, we must have $\text{dim}_{\mathcal{K}_{h}}\mathcal{A}_{h}=\text{dim}_{\mathcal{K}_{\mu(h)}}\mathcal{A}_{\mu(h)}=r.$ Further, it is easy to see that $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ is a free $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}$ having rank $r.$ When $r$ is even, by Lemma 3.7(a), we see that $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ has an orthogonal direct sum decomposition of the type $\langle m_{1}(X)+n_{1}(X),p_{1}(X)+q_{1}(X)\rangle\perp\cdots\perp\langle m_{\frac{r}{2}}(X)+n_{\frac{r}{2}}(X),p_{\frac{r}{2}}(X)+q_{\frac{r}{2}}(X)\rangle,$ where $(m_{j}(X)+n_{j}(X),p_{j}(X)+q_{j}(X))$ is a hyperbolic pair in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ for $1\leq j\leq\frac{r}{2}.$ When $r$ is odd, by Lemma 3.7(b), we see that $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ has an orthogonal direct sum decomposition of the type $\langle m_{1}(X)+n_{1}(X),p_{1}(X)+q_{1}(X)\rangle\perp\cdots\perp\langle m_{\frac{r-1}{2}}(X)+n_{\frac{r-1}{2}}(X),p_{\frac{r-1}{2}}(X)+q_{\frac{r-1}{2}}(X)\rangle\perp\langle\alpha(X)+\beta(X)\rangle,$ where $(m_{j}(X)+n_{j}(X),p_{j}(X)+q_{j}(X))$ is a hyperbolic pair in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ for $1\leq j\leq\frac{r-1}{2}$ and $\alpha(X)+\beta(X)$ is an anisotropic element of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$ It is easy to see that $\mathcal{A}_{h}\cup\mathcal{A}_{\mu(h)}\setminus\{0\}$ is the set consisting of all the trivial isotropic elements in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$ From this, it follows that the number of trivial isotropic elements in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ is given by $2(q^{rd_{h}}-1).$ When $r=1,$ by Lemma 3.5(a), we see that there does not exist any non-trivial isotropic element in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$

In the following proposition, we determine the number of non-trivial isotropic elements in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ when $\text{dim}_{\mathcal{K}_{h}}\mathcal{A}_{h}=\text{dim}_{\mathcal{K}_{\mu(h)}}\mathcal{A}_{\mu(h)}=r\geq 2.$

Proposition 3.8**.**

Let $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ be a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)},$ where $\mathcal{A}_{h}$ is a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{A}_{\mu(h)}$ is a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}$ with $\text{dim}_{\mathcal{K}_{h}}\mathcal{A}_{h}=\text{dim}_{\mathcal{K}_{\mu(h)}}\mathcal{A}_{\mu(h)}=r\geq 2.$ Then the number of non-trivial isotropic elements in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ is given by $i_{r}=(q^{(r-1)d_{h}}-1)(q^{rd_{h}}-1).$

In order to prove this proposition, we need to prove the following lemma:

Lemma 3.8**.**

Let $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ be a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)},$ where $\mathcal{A}_{h}$ is a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{A}_{\mu(h)}$ is a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}$ with $\text{dim}_{\mathcal{K}_{h}}\mathcal{A}_{h}=\text{dim}_{\mathcal{K}_{\mu(h)}}\mathcal{A}_{\mu(h)}=2.$ Then the number of non-trivial isotropic elements in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ is given by $(q^{2d_{h}}-1)(q^{d_{h}}-1).$

Proof.

Since $\text{dim}_{\mathcal{K}_{h}}\mathcal{A}_{h}=\text{dim}_{\mathcal{K}_{\mu(h)}}\mathcal{A}_{\mu(h)}=2,$ by Lemma 3.7(a), we have $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}=\langle\alpha_{1}(X)+\alpha_{2}(X),\beta_{1}(X)\\ +\beta_{2}(X)\rangle,$ where $\left(\alpha_{1}(X)+\alpha_{2}(X),\beta_{1}(X)+\beta_{2}(X)\right)$ is a hyperbolic pair in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$ From this, we get $\left[\alpha_{1}(X),\beta_{2}(X)\right]_{\delta}=e_{h}(X),$ $\left[\alpha_{2}(X),\beta_{1}(X)\right]_{\delta}=e_{\mu(h)}(X),$ $\left[\alpha_{1}(X),\alpha_{2}(X)\right]_{\delta}=0$ and $\left[\beta_{1}(X),\beta_{2}(X)\right]_{\delta}=0.$ Now any element $h_{1}(X)+h_{2}(X)$ in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ with $h_{1}(X)\in\mathcal{A}_{h}\setminus\{0\}$ and $h_{2}(X)\in\mathcal{A}_{\mu(h)}\setminus\{0\}$ can be written as h_{1}(X)+h_{2}(X)=\bigl{(}k_{1}(X)+k_{2}(X)\bigr{)}\bigl{(}\alpha_{1}(X)+\alpha_{2}(X)\bigr{)}+\bigl{(}\ell_{1}(X)+\ell_{2}(X)\bigr{)}\bigl{(}\beta_{1}(X)+\beta_{2}(X)\bigr{)}, where $k_{1}(X)+k_{2}(X),\ell_{1}(X)+\ell_{2}(X)\in\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}.$ From this, we obtain $h_{1}(X)=k_{1}(X)\alpha_{1}(X)+\ell_{1}(X)\beta_{1}(X)$ and $h_{2}(X)=k_{2}(X)\alpha_{2}(X)+\ell_{2}(X)\beta_{2}(X).$ Next we observe that $h_{1}(X)=k_{1}(X)\alpha_{1}(X)+\ell_{1}(X)\beta_{1}(X)\neq 0$ if and only if $(k_{1}(X),\ell_{1}(X))\neq(0,0)$ and $h_{2}(X)=k_{2}(X)\alpha_{2}(X)+\ell_{2}(X)\beta_{2}(X)\neq 0$ if and only if $(k_{2}(X),\ell_{2}(X))\neq(0,0).$

Now in order to count all the non-trivial isotropic elements in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)},$ we will consider the following three cases separately: **I. ** $k_{1}(X)+k_{2}(X)=0$ **II. ** $\ell_{1}(X)+\ell_{2}(X)=0$ and **III. ** both $k_{1}(X)+k_{2}(X)$ and $\ell_{1}(X)+\ell_{2}(X)$ are non-zero.

Case I. Let $k_{1}(X)+k_{2}(X)=0,$ that is, $k_{1}(X)=0$ and $k_{2}(X)=0.$ Then we have $\ell_{1}(X)\neq 0$ and $\ell_{2}(X)\neq 0.$ This gives $h_{1}(X)+h_{2}(X)=(\ell_{1}(X)+\ell_{2}(X))(\beta_{1}(X)+\beta_{2}(X)),$ which is clearly a non-trivial isotropic element for all $\ell_{1}(X)\in\mathcal{K}_{h}\setminus\{0\}$ and $\ell_{2}(X)\in\mathcal{K}_{\mu(h)}\setminus\{0\}.$ So in this case, there are precisely $(q^{d_{h}}-1)^{2}$ such non-trivial isotropic elements in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$

Case II. Let $\ell_{1}(X)+\ell_{2}(X)=0.$ Then we have $k_{1}(X)\neq 0$ and $k_{2}(X)\neq 0,$ so that h_{1}(X)+h_{2}(X)=\bigl{(}k_{1}(X)+k_{2}(X)\bigr{)}\bigl{(}\alpha_{1}(X)+\alpha_{2}(X)\bigr{)}, which is clearly a non-trivial isotropic element for all $k_{1}(X)\in\mathcal{K}_{h}\setminus\{0\}$ and $k_{2}(X)\in\mathcal{K}_{\mu(h)}\setminus\{0\}.$ In this case also, there are precisely $(q^{d_{h}}-1)^{2}$ such non-trivial isotropic elements in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$

Case III. Let $k_{1}(X)+k_{2}(X)\neq 0$ and $\ell_{1}(X)+\ell_{2}(X)\neq 0.$ Then we see that $h_{1}(X)+h_{2}(X)$ is an isotropic element if and only if [\bigl{(}k_{1}(X)+k_{2}(X)\bigr{)}\bigl{(}\alpha_{1}(X)+\alpha_{2}(X)\bigr{)}+\bigl{(}\ell_{1}(X)+\ell_{2}(X)\bigr{)}\bigl{(}\beta_{1}(X)+\beta_{2}(X)\bigr{)},\bigl{(}k_{1}(X)+k_{2}(X)\bigr{)}\bigl{(}\alpha_{1}(X)+\alpha_{2}(X)\bigr{)}+\bigl{(}\ell_{1}(X)+\ell_{2}(X)\bigr{)}\bigl{(}\beta_{1}(X)+\beta_{2}(X)\bigr{)}]_{\delta}=0 if and only if \bigl{(}k_{1}(X)+k_{2}(X)\bigr{)}\tau_{1,-1}\bigl{(}\ell_{1}(X)+\ell_{2}(X)\bigr{)}+\bigl{(}\ell_{1}(X)+\ell_{2}(X)\bigr{)}\tau_{1,-1}\bigl{(}k_{1}(X)+k_{2}(X)\bigr{)}=0 if and only if

[TABLE]

Now if $k_{1}(X)=0$ and $k_{2}(X)\neq 0,$ then $\ell_{1}(X)\neq 0$ and $\ell_{2}(X)\in\mathcal{K}_{\mu(h)}.$ In this case, the condition (22) is not satisfied, which implies that $h_{1}(X)+h_{2}(X)$ is an anisotropic element of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$ Next if $k_{2}(X)=0$ and $k_{1}(X)\neq 0,$ then $\ell_{2}(X)\neq 0$ and $\ell_{1}(X)\in\mathcal{K}_{h}.$ Here also, the condition (22) is not satisfied, which implies that $h_{1}(X)+h_{2}(X)$ is an anisotropic element of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$ Working in a similar way, we see that the element $h_{1}(X)+h_{2}(X)$ is anisotropic when either $\ell_{1}(X)$ or $\ell_{2}(X)$ (but not both) is zero. Finally, we assume that $k_{1}(X),k_{2}(X),\ell_{1}(X),\ell_{2}(X)$ all are non-zero. Then by equation (22), we obtain $k_{1}(X)=-\ell_{1}(X)\tau_{1,-1}(k_{2}(X)\ell_{2}(X)^{-1}),$ which implies that the number of such non-trivial isotropic elements in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ is given by $(q^{d_{h}}-1)^{3}.$

On combining all the above cases, we see that the total number of non-trivial isotropic elements in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ is given by $2(q^{d_{h}}-1)^{2}+(q^{d_{h}}-1)^{3}=(q^{d_{h}}-1)(q^{2d_{h}}-1).$ ∎

Proof of Proposition 3.8. In order to prove this proposition, we will apply induction on $r=\text{dim}_{\mathcal{K}_{h}}\mathcal{A}_{h}=\dim_{\mathcal{K}_{\mu(h)}}\mathcal{A}_{\mu(h)}\geq 2.$ When $r=2,$ by Lemma 3.8, the result holds. Now we assume that $r>2$ and suppose that the result holds for all $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodules $\mathcal{W}_{h}\oplus\mathcal{W}_{\mu(h)}$ of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)},$ where $\mathcal{W}_{h}$ is a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{W}_{\mu(h)}$ is a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}$ with $2\leq\text{dim}_{\mathcal{K}_{h}}\mathcal{W}_{h}=\text{dim}_{\mathcal{K}_{\mu(h)}}\mathcal{W}_{\mu(h)}\leq r-1.$ Now let $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ be a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)},$ where $\mathcal{A}_{h}$ is a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{A}_{\mu(h)}$ is a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}$ with $\text{dim}_{\mathcal{K}_{h}}\mathcal{A}_{h}=\text{dim}_{\mathcal{K}_{\mu(h)}}\mathcal{A}_{\mu(h)}=r.$

First of all, we assert that there exists an anisotropic element in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$ For this, we choose $w_{2}(X)\in\mathcal{A}_{\mu(h)}\setminus\{0\}.$ Working as in Lemma 3.5(a), we see that $\dim_{\mathcal{K}_{h}}\langle w_{2}(X)\rangle^{\perp_{\delta}}=r-1.$ Thus there exists $w_{1}(X)\in\mathcal{A}_{h}\setminus\langle w_{2}(X)\rangle^{\perp_{\delta}}$ so that $\left[w_{1}(X),w_{2}(X)\right]_{\delta}=a(X)\in\mathcal{K}_{h}\setminus\{0\}.$ This gives $\left[w_{1}(X)+w_{2}(X),w_{1}(X)+w_{2}(X)\right]_{\delta}=a(X)+\tau_{1,-1}(a(X)),$ i.e., $w_{1}(X)+w_{2}(X)$ is an anisotropic element of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$ Now as $\langle w_{1}(X)+w_{2}(X)\rangle$ is a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)},$ by Lemma 3.6, we have $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}=\langle w_{1}(X)+w_{2}(X)\rangle\perp\langle w_{1}(X)+w_{2}(X)\rangle^{\perp_{\delta}}.$ We observe that $\langle w_{1}(X)+w_{2}(X)\rangle^{\perp_{\delta}}=\langle w_{2}(X)\rangle^{\perp_{\delta}}\oplus\langle w_{1}(X)\rangle^{\perp_{\delta}}.$ Working as in Lemma 3.5(a), we get $\text{dim}_{\mathcal{K}_{h}}\langle w_{2}(X)\rangle^{\perp_{\delta}}=\text{dim}_{\mathcal{K}_{\mu(h)}}\langle w_{1}(X)\rangle^{\perp_{\delta}}=r-1.$ This implies that $\langle w_{1}(X)+w_{2}(X)\rangle^{\perp_{\delta}}$ is a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ having rank $r-1.$

Now any element $h_{1}(X)+h_{2}(X)\in\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ with $h_{1}(X)\in\mathcal{A}_{h}\setminus\{0\}$ and $h_{2}(X)\in\mathcal{A}_{\mu(h)}\setminus\{0\}$ can be uniquely written as h_{1}(X)+h_{2}(X)=\bigl{(}\lambda_{1}(X)+\lambda_{2}(X)\bigr{)}\bigl{(}w_{1}(X)+w_{2}(X)\bigr{)}+\bigl{(}v_{1}(X)+v_{2}(X)\bigr{)}, where $\lambda_{1}(X)+\lambda_{2}(X)\in\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$ and $v_{1}(X)+v_{2}(X)\in\langle w_{1}(X)+w_{2}(X)\rangle^{\perp_{\delta}}$ with $v_{1}(X)\in\langle w_{2}(X)\rangle^{\perp_{\delta}}$ and $v_{2}(X)\in\langle w_{1}(X)\rangle^{\perp_{\delta}}.$ We next observe that $h_{1}(X)=\lambda_{1}(X)w_{1}(X)+v_{1}(X)\neq 0$ if and only if \bigl{(}\lambda_{1}(X),v_{1}(X)\bigr{)}\neq(0,0), while $h_{2}(X)=\lambda_{2}(X)w_{2}(X)+v_{2}(X)\neq 0$ if and only if \bigl{(}\lambda_{2}(X),v_{2}(X)\bigr{)}\neq(0,0). Now in order to count all non-trivial isotropic elements in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)},$ we shall distinguish the following two cases: **I. ** $\lambda_{1}(X)+\lambda_{2}(X)=0$ and **II. ** $\lambda_{1}(X)+\lambda_{2}(X)\neq 0.$

Case I. Let $\lambda_{1}(X)+\lambda_{2}(X)=0.$ Then we must have $v_{1}(X)\neq 0$ and $v_{2}(X)\neq 0.$ So $h_{1}(X)+h_{2}(X)=v_{1}(X)+v_{2}(X)\in\langle w_{1}(X)+w_{2}(X)\rangle^{\perp_{\delta}},$ which is a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ having rank $r-1.$ Now by applying induction hypothesis, we see that the number of such non-trivial isotropic elements in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ is given by $i_{r-1}.$

Case II. Let $\lambda_{1}(X)+\lambda_{2}(X)\neq 0.$

If $\lambda_{1}(X)=0,$ then $\lambda_{2}(X)\neq 0,$ $v_{1}(X)\neq 0$ and $v_{2}(X)\in\langle w_{1}(X)\rangle^{\perp_{\delta}}.$ Now if $v_{2}(X)=0,$ then we see that $h_{1}(X)+h_{2}(X)=\lambda_{2}(X)w_{2}(X)+v_{1}(X)$ and clearly $[\lambda_{2}(X)w_{2}(X)+v_{1}(X),\lambda_{2}(X)w_{2}(X)+v_{1}(X)]_{\delta}=0,$ which implies that $h_{1}(X)+h_{2}(X)$ is a non-trivial isotropic element for all $\lambda_{2}(X)\in\mathcal{K}_{\mu(h)}\setminus\{0\}$ and $v_{1}(X)\in\langle w_{2}(X)\rangle^{\perp_{\delta}}\setminus\{0\}.$ In this case, there are precisely $q^{(r-1)d_{h}}-1$ choices for $v_{1}(X)$ and $q^{d_{h}}-1$ choices for $\lambda_{2}(X).$ So there are precisely $(q^{(r-1)d_{h}}-1)(q^{d_{h}}-1)$ such non-trivial isotropic elements in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$ On the other hand, if $v_{2}(X)\neq 0,$ then for all $\lambda_{2}(X)\in\mathcal{K}_{\mu(h)}\setminus\{0\},$ we see that $h_{1}(X)+h_{2}(X)=v_{1}(X)+\lambda_{2}(X)w_{2}(X)+v_{2}(X)$ is an isotropic element if and only if $[v_{1}(X)+\lambda_{2}(X)w_{2}(X)+v_{2}(X),v_{1}(X)+\lambda_{2}(X)w_{2}(X)+v_{2}(X)]_{\delta}=\left[v_{1}(X)+v_{2}(X),v_{1}(X)+v_{2}(X)\right]_{\delta}=0$ if and only if $v_{1}(X)+v_{2}(X)$ is a non-trivial isotropic element in $\langle w_{1}(X)+w_{2}(X)\rangle^{\perp_{\delta}}.$ Now $\lambda_{2}(X)$ has $q^{d_{h}}-1$ choices and by the induction hypothesis, the number of choices for a non-trivial isotropic element $v_{1}(X)+v_{2}(X)\in\langle w_{1}(X)+w_{2}(X)\rangle^{\perp_{\delta}}$ is given by $i_{r-1}.$ Thus in this case, the number of non-trivial isotropic elements in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ is given by $(q^{d_{h}}-1)i_{r-1}.$

Next suppose that $\lambda_{2}(X)=0$ so that $\lambda_{1}(X)\neq 0.$ Then $v_{2}(X)\neq 0$ and $v_{1}(X)\in\langle w_{2}(X)\rangle^{\perp_{\delta}}.$ Now if $v_{1}(X)=0,$ then we see that $h_{1}(X)+h_{2}(X)=\lambda_{1}(X)w_{1}(X)+v_{2}(X),$ which is clearly a non-trivial isotropic element for all $\lambda_{1}(X)\in\mathcal{K}_{h}\setminus\{0\}$ and $v_{2}(X)\in\langle w_{1}(X)\rangle^{\perp_{\delta}}\setminus\{0\}.$ So the number of such non-trivial isotropic elements in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ is given by $(q^{(r-1)d_{h}}-1)(q^{d_{h}}-1).$ If $v_{1}(X)\neq 0,$ then $h_{1}(X)+h_{2}(X)=\lambda_{1}(X)w_{1}(X)+v_{1}(X)+v_{2}(X)$ is an isotropic element if and only if $[v_{1}(X)+\lambda_{1}(X)w_{1}(X)+v_{2}(X),v_{1}(X)+\lambda_{1}(X)w_{1}(X)+v_{2}(X)]_{\delta}=\left[v_{1}(X)+v_{2}(X),v_{1}(X)+v_{2}(X)\right]_{\delta}=0$ if and only if $v_{1}(X)+v_{2}(X)$ is a non-trivial isotropic element of $\langle w_{1}(X)+w_{2}(X)\rangle^{\perp_{\delta}}.$ Now $\lambda_{1}(X)$ has $q^{d_{h}}-1$ choices and by the induction hypothesis, there are precisely $i_{r-1}$ distinct non-trivial isotropic elements in $\langle w_{1}(X)+w_{2}(X)\rangle^{\perp_{\delta}}.$ So in this case, the number of such non-trivial isotropic elements in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ is given by $(q^{d_{h}}-1)i_{r-1}.$

Further, suppose that both $\lambda_{1}(X),\lambda_{2}(X)$ are non-zero. Then $v_{1}(X)\in\langle w_{2}(X)\rangle^{\perp_{\delta}}$ and $v_{2}(X)\in\langle w_{1}(X)\rangle^{\perp_{\delta}}.$ Now $h_{1}(X)+h_{2}(X)$ is an isotropic element if and only if [\bigl{(}\lambda_{1}(X)+\lambda_{2}(X)\bigr{)}\bigl{(}w_{1}(X)+w_{2}(X)\bigr{)}+\bigl{(}v_{1}(X)+v_{2}(X)\bigr{)},\bigl{(}\lambda_{1}(X)+\lambda_{2}(X)\bigr{)}\bigl{(}w_{1}(X)+w_{2}(X)\bigr{)}+\bigl{(}v_{1}(X)+v_{2}(X)\bigr{)}]_{\delta}=0 if and only if \bigl{(}\lambda_{1}(X)+\lambda_{2}(X)\bigr{)}\tau_{1,-1}\bigl{(}\lambda_{1}(X)+\lambda_{2}(X)\bigr{)}\bigl{(}a(X)+\tau_{1,-1}\bigl{(}a(X)\bigr{)}\bigr{)}=-\left[v_{1}(X)+v_{2}(X),v_{1}(X)+v_{2}(X)\right]_{\delta} if and only if

[TABLE]

We observe that when both $v_{1}(X),v_{2}(X)$ are zero, we have h_{1}(X)+h_{2}(X)=\bigl{(}\lambda_{1}(X)+\lambda_{2}(X)\bigr{)}\bigl{(}w_{1}(X)+w_{2}(X)\bigr{)}, which is clearly an anisotropic element. If either $v_{1}(X)$ or $v_{2}(X)$ (but not both) is non-zero, then h_{1}(X)+h_{2}(X)=\bigl{(}\lambda_{1}(X)+\lambda_{2}(X)\bigr{)}\bigl{(}w_{1}(X)+w_{2}(X)\bigr{)}+v_{j}(X) with $v_{j}(X)\neq 0$ for some $j\in\{1,2\}.$ Then by (23), we see that $h_{1}(X)+h_{2}(X)$ is an isotropic element if and only if $\lambda_{1}(X)\tau_{1,-1}(\lambda_{2}(X))a(X)=0,$ which is not possible, and hence $h_{1}(X)+h_{2}(X)$ is an anisotropic element. Now if both $v_{1}(X),v_{2}(X)$ are non-zero, then by (23), $h_{1}(X)+h_{2}(X)$ is an isotropic element if and only if $\lambda_{1}(X)\tau_{1,-1}(\lambda_{2}(X))a(X)=-\left[v_{1}(X),v_{2}(X)\right]_{\delta}.$ We observe that as $\lambda_{1}(X),\lambda_{2}(X)$ and $a(X)$ all are non-zero, we must have $\left[v_{1}(X),v_{2}(X)\right]_{\delta}\neq 0,$ which implies that $v_{1}(X)+v_{2}(X)$ is an anisotropic element of $\langle w_{1}(X)+w_{2}(X)\rangle^{\perp_{\delta}}.$ We also note that $\lambda_{2}(X)\in\mathcal{K}_{\mu(h)}\setminus\{0\}$ is determined uniquely by $\lambda_{1}(X)$ and $\left[v_{1}(X),v_{2}(X)\right]_{\delta}.$ We see that $\lambda_{1}(X)\in\mathcal{K}_{h}\setminus\{0\}$ has $q^{d_{h}}-1$ choices and by the induction hypothesis, the number of anisotropic elements in $\langle w_{1}(X)+w_{2}(X)\rangle^{\perp_{\delta}}$ is given by $(q^{2d_{h}(r-1)}-1)-2(q^{(r-1)d_{h}}-1)-i_{r-1}=(q^{2d_{h}(r-1)}-1)-2(q^{(r-1)d_{h}}-1)-(q^{(r-2)d_{h}}-1)(q^{(r-1)d_{h}}-1)=q^{(r-2)d_{h}}(q^{d_{h}}-1)(q^{(r-1)d_{h}}-1).$ Thus the number of such non-trivial isotropic elements in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ is given by $q^{(r-2)d_{h}}(q^{d_{h}}-1)^{2}(q^{(r-1)d_{h}}-1).$

On combining all the above cases, we see that the total number of non-trivial isotropic elements in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ is given by $i_{r}=i_{r-1}+2(q^{d_{h}}-1)(q^{(r-1)d_{h}}-1)+2(q^{d_{h}}-1)i_{r-1}+q^{(r-2)d_{h}}(q^{d_{h}}-1)^{2}(q^{(r-1)d_{h}}-1),$ which equals $(q^{(r-1)d_{h}}-1)(q^{rd_{h}}-1)$ after a simple computation. $\hfill\Box$

In the following proposition, we determine the number of hyperbolic pairs in a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)},$ where $\mathcal{A}_{h}$ is a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{A}_{\mu(h)}$ is a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}$ with $\text{dim}_{\mathcal{K}_{h}}\mathcal{A}_{h}=\text{dim}_{\mathcal{K}_{\mu(h)}}\mathcal{A}_{\mu(h)}=r\geq 2.$

Lemma 3.9**.**

Let $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ be a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)},$ where $\mathcal{A}_{h}$ is a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{A}_{\mu(h)}$ is a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}$ with $\text{dim}_{\mathcal{K}_{h}}\mathcal{A}_{h}=\text{dim}_{\mathcal{K}_{\mu(h)}}\mathcal{A}_{\mu(h)}=r\geq 2.$ Then the total number of hyperbolic pairs in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ is given by $\mathfrak{H}_{r}=q^{d_{h}(2r-3)}(q^{(r-1)d_{h}}-1)(q^{rd_{h}}-1).$

Proof.

Since $r\geq 2,$ by Lemma 3.5(a), $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ has a non-trivial isotropic element, say $\alpha_{1}(X)+\alpha_{2}(X).$ By Proposition 3.8, the number of choices for $\alpha_{1}(X)+\alpha_{2}(X)$ is given by $i_{r}=(q^{(r-1)d_{h}}-1)(q^{rd_{h}}-1).$ Further by Lemma 3.5(c) and (d), there exists another non-trivial isotropic element $\beta_{1}(X)+\beta_{2}(X)$ in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ such that $(\alpha_{1}(X)+\alpha_{2}(X),\beta_{1}(X)+\beta_{2}(X))$ is a hyperbolic pair in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ and the $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule $\langle\alpha_{1}(X)+\alpha_{2}(X),\beta_{1}(X)+\beta_{2}(X)\rangle$ of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ is non-degenerate. Now by Lemma 3.6, we can write $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}=\langle\alpha_{1}(X)+\alpha_{2}(X),\beta_{1}(X)+\beta_{2}(X)\rangle\perp\langle\alpha_{1}(X)+\alpha_{2}(X),\beta_{1}(X)+\beta_{2}(X)\rangle^{\perp_{\delta}},$ where $\langle\alpha_{1}(X)+\alpha_{2}(X),\beta_{1}(X)+\beta_{2}(X)\rangle^{\perp_{\delta}}$ is a non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodule of $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$ Further, we observe that $\langle\alpha_{1}(X)+\alpha_{2}(X),\beta_{1}(X)+\beta_{2}(X)\rangle^{\perp_{\delta}}=\langle\alpha_{2}(X),\beta_{2}(X)\rangle^{\perp_{\delta}}\oplus\langle\alpha_{1}(X),\beta_{1}(X)\rangle^{\perp_{\delta}},$ where $\text{dim}_{\mathcal{K}_{h}}\langle\alpha_{2}(X),\beta_{2}(X)\rangle^{\perp_{\delta}}=\text{dim}_{\mathcal{K}_{\mu(h)}}\langle\alpha_{1}(X),\beta_{1}(X)\rangle^{\perp_{\delta}}=r-2.$ So any element $h_{1}(X)+h_{2}(X)\in\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ with $h_{1}(X)\in\mathcal{A}_{h}\setminus\{0\}$ and $h_{2}(X)\in\mathcal{A}_{\mu(h)}\setminus\{0\}$ can be written as h_{1}(X)+h_{2}(X)=\bigl{(}u_{1}(X)+u_{2}(X)\bigr{)}\bigl{(}\alpha_{1}(X)+\alpha_{2}(X)\bigr{)}+\bigl{(}v_{1}(X)+v_{2}(X)\bigr{)}\bigl{(}\beta_{1}(X)+\beta_{2}(X)\bigr{)}+\bigl{(}\eta_{1}(X)+\eta_{2}(X)\bigr{)}, where $u_{1}(X)+u_{2}(X),$ $v_{1}(X)+v_{2}(X)\in\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$ and $\eta_{1}(X)+\eta_{2}(X)\in\langle\alpha_{1}(X)+\alpha_{2}(X),\beta_{1}(X)+\beta_{2}(X)\rangle^{\perp_{\delta}}.$ Note that \bigl{(}h_{1}(X)+h_{2}(X),\alpha_{1}(X)+\alpha_{2}(X)\bigr{)} is a hyperbolic pair if and only if $\left[h_{1}(X)+h_{2}(X),\alpha_{1}(X)+\alpha_{2}(X)\right]_{\delta}=e_{h}(X)+e_{\mu(h)}(X)$ and $[h_{1}(X)+h_{2}(X),h_{1}(X)+h_{2}(X)]_{\delta}=0.$ Now $\left[h_{1}(X)+h_{2}(X),\alpha_{1}(X)+\alpha_{2}(X)\right]_{\delta}=e_{h}(X)+e_{\mu(h)}(X)$ if and only if $v_{1}(X)=e_{h}(X)$ and $v_{2}(X)=e_{\mu(h)}(X),$ which gives h_{1}(X)+h_{2}(X)=\bigl{(}u_{1}(X)+u_{2}(X)\bigr{)}\bigl{(}\alpha_{1}(X)+\alpha_{2}(X)\bigr{)}+\bigl{(}\beta_{1}(X)+\beta_{2}(X)\bigr{)}+\bigl{(}\eta_{1}(X)+\eta_{2}(X)\bigr{)}. We also observe that both $h_{1}(X)=u_{1}(X)\alpha_{1}(X)+\beta_{1}(X)+\eta_{1}(X)$ and $h_{2}(X)=u_{2}(X)\alpha_{2}(X)+\beta_{2}(X)+\eta_{2}(X)$ are non-zero. Further, we see that $\left[h_{1}(X)+h_{2}(X),h_{1}(X)+h_{2}(X)\right]_{\delta}=0$ if and only if u_{1}(X)+u_{2}(X)+\tau_{1,-1}\bigl{(}u_{1}(X)+u_{2}(X)\bigr{)}+\left[\eta_{1}(X)+\eta_{2}(X),\eta_{1}(X)+\eta_{2}(X)\right]_{\delta}=0 if and only if u_{1}(X)+\tau_{1,-1}\bigl{(}u_{2}(X)\bigr{)}=-\left[\eta_{1}(X),\eta_{2}(X)\right]_{\delta}. Now we observe that $u_{1}(X)$ is determined uniquely by $u_{2}(X)$ and $\eta_{1}(X)+\eta_{2}(X)\in\langle\alpha_{1}(X)+\alpha_{2}(X),\beta_{1}(X)+\beta_{2}(X)\rangle^{\perp_{\delta}}.$ Further, it is easy to see that each of $\eta_{1}(X)$ and $\eta_{2}(X)$ has $q^{(r-2)d_{h}}$ choices and $u_{2}(X)$ has $q^{d_{h}}$ choices. From this, it follows that the element $h_{1}(X)+h_{2}(X)$ has $q^{(2r-3)d_{h}}$ choices for $(\alpha_{1}(X)+\alpha_{2}(X),h_{1}(X)+h_{2}(X))$ to be a hyperbolic pair in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}.$ Therefore the total number of hyperbolic pairs in $\mathcal{A}_{h}\oplus\mathcal{A}_{\mu(h)}$ is given by $\mathfrak{H}_{r}=q^{d_{h}(2r-3)}i_{r},$ which equals $q^{d_{h}(2r-3)}(q^{(r-1)d_{h}}-1)(q^{rd_{h}}-1).$ ∎

Proof of Proposition 3.6. In view of Remark 3.2, we need to consider $\delta\in\{\ast,0,\gamma^{(H)}\}$ so that \big{(}\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)},\left[\cdot,\cdot\right]_{\delta}\restriction_{\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}\times\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}}\big{)} is a Hermitian $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-space having rank $t.$

Now to prove this proposition, for each integer $k~{}(0\leq k\leq t),$ let $N_{h,k}$ denote the number of pairs $\left(\mathcal{C}_{h},\mathcal{C}_{\mu(h)}\right)$ with $\mathcal{C}_{h}$ as a $k$-dimensional $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{C}_{\mu(h)}$ as a $k$-dimensional $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}$ satisfying $\mathcal{C}_{h}\cap\mathcal{C}_{h}^{(\delta)}=\{0\}$ and $\mathcal{C}_{\mu(h)}\cap\mathcal{C}_{\mu(h)}^{(\delta)}=\{0\}.$ Then we have $N_{h}=\sum\limits_{k=0}^{t}N_{h,k}.$

To begin with, we note, by Lemma 3.4(c), that for $0\leq k\leq t,$ the number $N_{h,k}$ is equal to the number of $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodules $\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}$ of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}$ satisfying $\text{dim}_{\mathcal{K}_{h}}\mathcal{C}_{h}=\text{dim}_{\mathcal{K}_{\mu(h)}}\mathcal{C}_{\mu(h)}=k$ and (\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)})\cap\bigl{(}\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}\bigr{)}^{\perp_{\delta}}=\{0\}, that is, $N_{h,k}$ is equal to the number of non-degenerate $\mathcal{K}_{h}\oplus\mathcal{K}_{\mu(h)}$-submodules $\mathcal{C}_{h}\oplus\mathcal{C}_{\mu(h)}$ of $\mathcal{J}_{h}\oplus\mathcal{J}_{\mu(h)}$ satisfying $\text{dim}_{\mathcal{K}_{h}}\mathcal{C}_{h}=\text{dim}_{\mathcal{K}_{\mu(h)}}\mathcal{C}_{\mu(h)}=k,$ where $\mathcal{C}_{h}$ is a $\mathcal{K}_{h}$-subspace of $\mathcal{J}_{h}$ and $\mathcal{C}_{\mu(h)}$ is a $\mathcal{K}_{\mu(h)}$-subspace of $\mathcal{J}_{\mu(h)}.$

Here by Lemma 3.3, we see that $N_{h,0}=N_{h,t}=1.$ Further, for any integer $k$ satisfying $1\leq k\leq t-1,$ using Remark 3.3, Proposition 3.8 and Lemma 3.9 and working in a similar manner as in Proposition 3.2, we see that $N_{h,k}=\displaystyle q^{kd_{h}(t-k)}{t\brack k}_{q^{d_{h}}}$ for $1\leq k\leq t-1.$ From this, the desired result follows. $\hfill\Box$

Proof of Theorem 3.1. It follows immediately from Propositions 3.1-3.6. $\hfill\Box$