Discriminants of a Class of Self-inversive Polynomials and Real Binary Forms

TL;DR

This paper explores the relationships between certain classes of self-inversive polynomials and real binary forms, providing explicit discriminant formulas via determinants of coefficient-dependent matrices.

Contribution

It establishes bijections between classes of self-inversive polynomials and real binary forms, and derives explicit discriminant expressions using polynomial coefficient matrices.

Findings

Bijections between self-inversive polynomials and real binary forms

Discriminants expressed as determinants of matrices

Matrices have entries as polynomials in polynomial coefficients

Abstract

A class of self-inversive polynomials includes all the self-reciprocal polynomials. Let A denote the set of all self-reciprocal polynomials with n+1 coefficients. Let B denote the set of certain self-inversive and non self-reciprocal polynomials with n+1 coefficients for odd n. Let C denote the set of real binary n-ic forms. Then there exist a bijection between A and C and another bijection between B and C. Let f be a monic polynomial in A and g be the corresponding polynomial in C. If the reading coefficient of g is not zero, then the discriminant of g is expressed by the determinant of a matrix of type (n, n). Any element of the matrix is a polynomial in the coefficients of f with integer coefficients. The same holds for monic polynomials in B.

Equations

$$P(z)=a_0{z}^n+a_1{z}^{n-1}+\cdot \cdot \cdot +a_n$$

$$P^{\ast }(z)=z^n\overline{P(1/\bar{z})}={\bar{a}}_n{z}^n+{\bar{a}}_{n-1}{z}^{n-1}+\cdot \cdot \cdot +{\bar{a}}_n,$$

$$a_0\ne 0\text{mbox}and{a}_k=\mu {\bar{a}}_{n-k}\text{mbox}forevery0\le k\le n.$$

$$g_2(x,y)=a{x}^3+b{x}^{2}y+cx{y}^2+d{y}^3,(a,b,c,d\in \mathbb{R})$$

$$g_2(x,y)=Re(\alpha z^3+\beta z^2\bar{z}),\alpha ,\beta ,\in \mathbb{C}.$$

$$g_3(x,y)=a{x}^4+b{x}^{3}y+c{x}^3{y}^2+dx{y}^3+e{y}^4$$

$$g_3(x,y)=Re(\alpha z^4+\beta z^3\bar{z}+\gamma z^2{\bar{z}}^2),\alpha ,\beta \in \mathbb{C},\gamma \in \mathbb{R}.$$

$$\chi (X)=X^4-\tau X^3+\sigma X^2-\bar{\tau }X+1,\tau \in \mathbb{C},\sigma \in \mathbb{R}.$$

$$P(z,w)=\sum _{j=0}^{n+1}{\zeta }_j{z}^{n+1-j}{w}^j.$$

$${\mathcal{A}}_n:=\{P(z,w):{\zeta }_j=(-1{)}^{n+1}{\overline{\zeta }}_{n+1-j},\text{mbox}forj=0,1,\dots ,n+1\}.$$

$${\mathcal{B}}_n:=\{P(z,w):{\zeta }_j=(-1{)}^n{\overline{\zeta }}_{n+1-j},\text{mbox}forj=0,1,\dots ,n+1\}.$$

$$g(X,Y)=\sum _{j=0}^{n+1}{a}_j{X}^{n+1-j}{Y}^j,a_0,a_1,\dots ,a_{n+1}\in \mathbb{R}.$$

$$f_n(T,U)={\zeta }_0{T}^{n+1}-{\zeta }_1{T}^{n}U+{\zeta }_2{T}^{n-1}{U}^2+\cdot \cdot \cdot +(-1{)}^n{\zeta }_{n}T{U}^n+(-1{)}^{n+1}{\zeta }_{n+1}{U}^{n+1}.$$

$${\zeta }_k\in \mathbb{C},{\zeta }_k={\bar{\zeta }}_{n+1-k},k=0,\cdot \cdot \cdot ,n+1.$$

$$\begin{array}{r}{g}_n(X,Y)=a_0{X}^{n+1}+a_1{X}^{n}Y+\cdot \cdot \cdot +a_{n}X{Y}^n+a_{n+1}{Y}^{n+1},\\ \text{mbox}with{a}_0,a_1,\cdot \cdot \cdot ,a_{n+1}\in \mathbb{R}.\end{array}$$

$$T=i(X+iY)\text{mbox}andU=i(X-iY).$$

$$a_{2h+1}=(-1{)}^{n/2+h+1}(\genfrac{}{}{0ex}{}{n+1}{2h+1})({\zeta }_0+{\zeta }_{n+1})$$

$$+(-1)^{n/2+1}\sum_{k=1}^{n/2}\left[(-1)^{k}\Bigg{\{}\sum_{j=0}^{h}(-1)^{j}{{k}\choose{h-j}}{{n+1-2k}\choose{2j+1}}\Bigg{\}}(\zeta_{k}+\zeta_{n+1-k})\right],$$

$$a_{2h}=(-1{)}^{n/2+h}(\genfrac{}{}{0ex}{}{n+1}{2h})({\zeta }_0-{\zeta }_{n+1})\sqrt{-1}$$

$$+(-1)^{n/2}\sum_{k=1}^{n/2}\left[(-1)^{k}\Bigg{\{}\sum_{j=0}^{h}(-1)^{j}{{k}\choose{h-j}}{{n+1-2k}\choose{2j}}\Bigg{\}}(\zeta_{k}-\zeta_{n+1-k})\sqrt{-1}\right],$$

$$\text{mbox}whereh=0,1,\dots ,n/2.$$

$$a_{2h+1}=(-1{)}^{(n+1)/2+h}(\genfrac{}{}{0ex}{}{n+1}{2h+1})({\zeta }_0-{\zeta }_{n+1})\sqrt{-1}$$

$$+(-1)^{(n+1)/2}\sum_{k=1}^{(n-1)/2}\left[(-1)^{k}\Bigg{\{}\sum_{j=0}^{h}(-1)^{j}{{k}\choose{h-j}}{{n+1-2k}\choose{2j+1}}\Bigg{\}}(\zeta_{k}-\zeta_{n+1-k})\sqrt{-1}\right],$$

$$\text{mbox}whereh=0,1,\dots ,(n-1)/2,$$

$$a_{2h}=(-1{)}^{(n+1)/2+h}(\genfrac{}{}{0ex}{}{n+1}{2h})({\zeta }_0+{\zeta }_{n+1})+(\genfrac{}{}{0ex}{}{(n+1)/2}{h}){\zeta }_{(n+1)/2}$$

$$+(-1)^{(n+1)/2}\sum_{k=1}^{(n-1)/2}\left[(-1)^{k}\Bigg{\{}\sum_{j=0}^{h}(-1)^{j}{{k}\choose{h-j}}{{n+1-2k}\choose{2j}}\Bigg{\}}(\zeta_{k}+\zeta_{n+1-k})\right],$$

$$\text{mbox}whereh=0,1,\dots ,(n+1)/2.$$

$$\left(\genfrac{}{}{0ex}{}{k}{0}\right)=1,\text{mbox}and\left(\genfrac{}{}{0ex}{}{k}{j}\right)=0,\text{mbox}ifj>k\ge 0.$$

$$Q_k:=i^{n+1}(-1{)}^k(X^2+Y^2{)}^k\{{\zeta }_k(X+iY{)}^{n+1-2k}-{\zeta }_{n+1-k}(X-iY{)}^{n+1-2k}\},$$

$$k=1,\dots ,[n/2].$$

Full text

1. Introduction

1.1. Self-inversive polynomials

A polynomial

[TABLE]

is said to be a self-inversive polynomial of degree $n$ if it satisfies $a_{0}\neq 0$ and $P(z)=\mu P^{*}(z)$ where $\mid\mu\mid=1$ and

[TABLE]

or equivalently,

[TABLE]

See e.g. Marden [7].

In particular, if $P(z)=P^{*}(z),\quad P(z)$ is called be self-reciprocal. In the literature [10] such a polynomial is called a self-inversive polynomial.

The zeros of a self-inversive polynomial are on or symmetric in the unit circle $C:\mid z\mid=1$.

In 1922, Cohn [2] proved that a polynomial $P(z)$ has all of its zeros on $C$ if and only if it is self-inversive and its derivative $P^{\prime}(z)$ has all its zeros in the closed unit disk $\mid z\mid\leq 1$.

Chen [1] found another necessary and sufficient condition for which all zeros of a self-inversive polynomial lie on $C$.

1.2. Real binary forms in terms of the complex variable

Zeeman [14] described a real binary cubic form

[TABLE]

in terms of the complex variable $z=x+iy$ and showed that $g_{2}$ can be expressed uniquely as

[TABLE]

Poston and Stewart [9] applied this method to the quartic forms. A real binary quartic form

[TABLE]

can be expressed uniquely as

[TABLE]

In this paper we will study relation between real binary forms and certain self-inversive polynomials via the terms of the complex variable. Such self-inversive polynomials appear in several branches of mathematics.

1.3. Monic self-inversive polynomials

Gongopadhyay and Parker [5] and Gongopadhyay, Parker and Parsad [6] classified the dynamical action in SU(p,q) using the coefficients of their characteristic polynomial. In the case $p+q=4$, the characteristic polynomial is

[TABLE]

The locus where the resultant $R(\chi,\chi^{\prime})=0$ was studied by Poston and Stewart [9]. The locus was named the holy grail. The characteristic polynomial $\chi(X)$ is a monic self-reciprocal polynomial.

Uchimura [13] studied the dynamics of holomorphic endomorphisms $P_{A_{3}}^{d}$ on ${\mathbb{P}}^{3}({\mathbb{C}})$ which is related to a complex Lie algebra of type $A_{3}$. The set of the critical values of $P_{A_{3}}^{d}$ restricted to a real three-dimensional subspace of ${\mathbb{C}}^{3}$ was proved to be equal to the holy grail. Uchimura [13] studied the relation between real binary quartic forms and monic self-reciprocal polynomials of degree four. In [12], monic self-inversive polynomials of degree three were studied.

Marden [7] stated that in a linear difference equation with constant coefficients, the requirement for a stable solution is that all the zeros of the characteristic polynomial lie in the unit circle. On the other hand, the map $P_{A_{3}}^{d}$ causes chaos in the region $W_{3}$ that corresponds to the case that all the zeros of its self-inversive polynomial lie on the unit circle.

Peterson and Sinclair [8] and Sinclair and Vaaler [11] studied monic self-reciprocal polynomials. Those polynomials have appeared in the study of random polynomials, random matrix theory and number theory.

1.4. Summary of results

We consider complex binary forms expressed by

[TABLE]

Let ${\mathcal{A}}_{n}$ denote the set of the self-inversive forms $P(z,w)$ satisfying that ($n$ is odd and $\mu=1$) or ($n$ is even and $\mu=-1$) , where $\mu$ is the constant in (1.1). That is,

[TABLE]

Let ${\mathcal{B}}_{n}$ denote the set of the self-inversive forms $P(z,w)$ satisfying that ($n$ is odd and $\mu=-1$) or ($n$ is even and $\mu=1$) . That is,

[TABLE]

We also consider real binary forms

[TABLE]

Let ${\mathcal{F}}_{n}$ denote the set of the real binary forms $g(X,Y)$.

We will show that there exist an isomorphism from ${\mathcal{A}}_{n}$ to ${\mathcal{F}}_{n}$ and another isomorphism from ${\mathcal{B}}_{n}$ to ${\mathcal{F}}_{n}$.

Let ${\mathcal{MA}}_{n}$ denote the set of monic self-reciprocal forms in ${\mathcal{A}}_{n}$ satisfying $\zeta_{0}=1$. Let ${\mathcal{MF}}_{n}$ denote the set of the corresponding forms in ${\mathcal{F}}_{n}$. We will show that for any form $g(X,Y)$ in ${\mathcal{MF}}_{n}$ satisfying $a_{0}\neq 0$, the discriminant of the polynomial $g(X,1)$ is expressed by the determinant of a $(n+1)\times(n+1)$ matrix. Any element of the matrix is a polynomial in $\zeta_{1},\zeta_{2},\dots,\zeta_{n}$ with integer coefficients. The matrix is a slightly modified version of the matrix introduced by [3]. Their matrices are related to generalized Chebyshev polynomials of the first kind in several complex variables.

Let ${\mathcal{MB}}_{n}$ denote the set of monic self-reciprocal forms in ${\mathcal{B}}_{n}$ for even $n$. The same result holds for ${\mathcal{MB}}_{n}$.

2. Results and proofs

2.1. The spaces ${\mathcal{A}}_{n}$ and ${\mathcal{F}}_{n}$

We consider self-inversive forms in the form

[TABLE]

where

[TABLE]

Let ${\mathcal{A}}_{n}$ denote the space of these self-inversive forms. Note that if $n$ is odd, then $f_{n}(T,U)$ is self-reciprocal and that if $n$ is even, then $f_{n}(T,-U)$ is self-reciprocal. The space ${\mathcal{A}}_{n}$ is related to a complex Lie algebra of type $A_{n}$ . See [12] and [13].

We also consider binary forms in the forms

[TABLE]

Let ${\mathcal{F}}_{n}$ denote the space of these forms. We do not assume that $a_{0}\neq 0$. We show that ${\mathcal{A}}_{n}$ is isomorphic to ${\mathcal{F}}_{n}$.

Proposition 2.1**.**

There exists an isomorphism $\Phi_{n}$ from ${\mathcal{A}}_{n}$ onto ${\mathcal{F}}_{n}$.

We define the map $\Phi_{n}$ as follows. Clearly $\overline{f_{n}(T,-{\overline{T}})}=f_{n}(T,-{\overline{T}}).$ Then we set

[TABLE]

Hence $f_{n}(i(X+iY),i(X-iY))$ is a polynomial in variables $X$ and $Y$ with real coefficients. So $f_{n}(i(X+iY),i(X-iY))$ is written as $g_{n}(X,Y)$ in (2.3). The image of a self-inversive form (2.1) under the map $\Phi_{n}$ is the binary form $g_{n}(X,Y)$. The coefficients $a_{k}$ are given by the following proposition.

Proposition 2.2**.**

(1) If $n$ is even, then

[TABLE]

[TABLE]

and

[TABLE]

[TABLE]

[TABLE]

(2) If $n$ is odd, then

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

[TABLE]

[TABLE]

Here we use conventions

[TABLE]

Proof of Proposition 2.2..

Case 1 :  $n$ is even.

We consider pairs of terms of $f_{n}(i(X+iY),i(X+iY))$. Set

[TABLE]

[TABLE]

Easy calculations reveal that the coefficient of $X^{n-2h}Y^{2h+1}$ in $Q_{k}$ is

[TABLE]

and the coefficient of $X^{n+1-2h}Y^{2h}$ in $Q_{k}$ is

[TABLE]

To get $a_{2h+1}$ and $a_{2h}$, we sum the coefficients of $X^{n-2h}Y^{2h+1}$ and $X^{n+1-2h}Y^{2h}$  in $Q_{k}$ over $k=1,\dots,n/2$  and those coefficients of the pair of terms  $\zeta_{0}(X+iY)^{n+1}-\zeta_{n+1}(X-iY)^{n+1}$, respectively.

Case 2 :  $n$ is odd. By the similar method, we can prove the assertion. ∎

Conversely we consider the inverse of the map $\Phi_{n}$. From (2.4), we have

[TABLE]

Then we need to show

[TABLE]

We set

[TABLE]

Then it suffices to prove the equality

[TABLE]

where, for a form

[TABLE]

we define the form $\hat{f}(T,U)$ by

[TABLE]

To prove this we note that

[TABLE]

[TABLE]

[TABLE]

Then we have (2.6) and so $\Phi_{n}$ is a bijection. Since $\Phi_{n}$ is a linear map, we have

[TABLE]

Hence $\Phi_{n}$ is an isomorphism.

Next we give an alternative proof of Proposition 2.1. This is useful in the sections below.

Set $Z=X+iY$. Let

[TABLE]

Then $f_{n}(Z)$ is written as follows.

Lemma 2.3**.**

If $n$ is even, then

[TABLE]

If $n$ is odd, then

[TABLE]

Proof.

Case 1 :  $n$ is even. Set

[TABLE]

Then

[TABLE]

Setting $T=iZ$,  we have (2.7).

Case 2 :  $n$ is odd. The proof is similar. ∎

Let

[TABLE]

We regard $w_{j}$ as a polynomial in ${\mathbb{C}}[X,Y]$.

Set

[TABLE]

Probably, the following lemma may have been known. But, here we give a proof of it.

Lemma 2.4**.**

*(1) The family ${\mathcal{S}}_{1}$ is linearly independent over ${\mathbb{C}}$.

(2) We define real polynomials $R_{j}(X,Y)$ and $I_{j}(X,Y)$ by*

[TABLE]

(a) If $n$ is even, then

[TABLE]

If $n$ is odd, then

[TABLE]

[TABLE]

(b) Set

[TABLE]

*Then the family ${\mathcal{S}}_{2}$ is linearly independent over ${\mathbb{C}}$.

Proof.

(1)  We define a Hermitian product on ${\mathcal{S}}_{1}$ by

[TABLE]

Then elements $w_{j}(X,Y)$ are orthogonal on the unit disk. Hence the family ${\mathcal{S}}_{1}$ is linearly independent over ${\mathbb{C}}$.

(2) The assertion (a) is trivial.

(b) Case 1 :  $n$ is even. Note that

[TABLE]

Since the family $\{w_{j}\}$ is linearly independent, the family ${\mathcal{S}}_{2}$ is linearly independent.

Case 2 :  $n$ is odd. By the similar method we can prove the assertion (b). ∎

We recall $f_{n}(T,U)$ and the definition of the map $\Phi_{n}$.

[TABLE]

Substituting

[TABLE]

in $f_{n}(T,U)$ we obtain a real polynomial

[TABLE]

Let

[TABLE]

Note that if $n$ is odd, then $\eta_{(n+1)/2}=0.$

We denote by $v_{1}$ the transpose of the row vector $(a_{0},a_{1},\dots,a_{n+1}).$ We denote by $v_{2}$ the transpose of the row vector $(\xi_{0},\eta_{0},\xi_{1},\eta_{1},\dots,\xi_{n/2},\eta_{n/2})$ if $n$ is even, or the row vector $(\xi_{0},\eta_{0},,\dots,\xi_{(n-1)/2},\eta_{(n-1)/2},\xi_{(n+1)/2})$ if $n$ is odd.

Lemma 2.5**.**

There exists an invertible matrix $M$ satisfying

[TABLE]

Proof.

Case 1 :  $n$ is even. From Lemma 2.3, we know that

[TABLE]

We denote by $V$ the vector space over ${\mathbb{C}}$ generated by homogeneous polynomials $X^{n+1},X^{n}Y,\dots.$ and $Y^{n+1}$. We see from Lemma 2.4 2(b) that $\{R_{0},I_{0},\dots,R_{n/2},I_{n/2}\}$ is another base of $V$. Then there exists an invertible matrix $M$ satisfying

[TABLE]

Case 2 :  $n$ is odd. The proof is similar. ∎

Proof of Proposition 2.1..

From Lemma 2.5, we conclude that the map $\Phi_{n}$ is bijective. ∎

Next we consider the relation between the roots of $f_{n}(T,U)$ and those of the corresponding form $g_{n}(X,Y)$ under $\Phi_{n}$. Let

[TABLE]

We recall (2.5) :

[TABLE]

We define a map $\varphi$ from ${\mathbb{P}}^{1}({\mathbb{C}})$ to ${\mathbb{P}}^{1}({\mathbb{C}})$ by

[TABLE]

Then with a suitable numbering

[TABLE]

The map $\varphi$ is a M$\ddot{o}$bius transformation on ${\mathbb{P}}^{1}$.

Clearly  $y_{j}=0$ if and only if  $t_{j}=u_{j}$.

If  $y_{j}\neq 0$, we may set  $y_{j}=u_{j}=1$. Then if  $t_{j}\neq 1,$

[TABLE]

If  $y_{j}=0$, then  $\varphi^{-1}(1:0)=(1:1)$.

The point (1 : 0) is the point at infinity and the point (1 : 1) lies on the unit circle $C$. The real axis of ${\mathbb{P}}^{1}$ maps into the unit circle under the M$\ddot{o}$bius transformation $\varphi^{-1}$. Then we have the following proposition.

Proposition 2.6**.**

*We suppose  $y_{j}=u_{j}=1$  and  $t_{j}\neq 1$ .

(1) $x_{j}$ is real if and only if  $\mid t_{j}\mid=1$.

(2) $x_{j},x_{j+1}\notin{\mathbb{R}}$  and  $\overline{x_{j}}=x_{j+1}$

if and only if  $t_{j}=re^{i\theta}$  and  $t_{j+1}=\frac{1}{r}e^{i\theta}$  with  $r\neq 1$.*

The pair $(t_{j}:1)$ and $(t_{j+1}:1)$ is said to be symmetric in the unit circle C if it satisfies that  $t_{j}=re^{i\theta}$  and  $t_{j+1}=\frac{1}{r}e^{i\theta}$  with  $r\neq 1$ .

2.2. The spaces ${\mathcal{B}}_{n}$ and ${\mathcal{F}}_{n}$

In this section we will give an isomorphism $\Psi_{n}$ from the space ${\mathcal{B}}_{n}$ onto ${\mathcal{F}}_{n}$ . We divide ${\mathcal{B}}_{n}$ into two spaces ${\mathcal{B}}(e,1)$ and ${\mathcal{B}}(o,-1)$. Let ${\mathcal{B}}(e,1)$ denote the space of the self-reciprocal forms :

[TABLE]

where $n$ is even and

[TABLE]

Let ${\mathcal{B}}(o,-1)$ denote the space of the self-inversive forms :

[TABLE]

where $n$ is odd and

[TABLE]

Since $\zeta_{(n+1)/2}=-{\bar{\zeta}}_{(n+1)/2},$  we may set

[TABLE]

In the first place we study the space ${\mathcal{B}}(e,1)$ . Clearly

[TABLE]

Then we set

[TABLE]

We denote the real polynomial $p_{n}(X+iY,X-iY)$ by

[TABLE]

We define the map $\Psi_{n}$ by $\Psi_{n}(p_{n})=q_{n}.$ Set $Z=X+iY.$ Then as in Lemma 2.3 we have

[TABLE]

Next we consider the space ${\mathcal{B}}(o,-1)$ . Set $\epsilon=e^{\pi i/(2(n+1))}.$ Clearly

[TABLE]

Then we set

[TABLE]

We denote the real polynomial $p_{n}(\epsilon(X+iY),\epsilon(X-iY))$ by

[TABLE]

We define the map $\Psi_{n}$ by $\Psi_{n}(p_{n})=q_{n}.$ Set $Z=X+iY.$ Then

[TABLE]

Hence in both cases, by the same proof as the alternative proof of Proposition 2.1, we can show the following proposition.

Proposition 2.7**.**

There exists an isomorphism $\Psi_{n}$ from ${\mathcal{B}}_{n}$ onto ${\mathcal{F}}_{n}$.

Next we consider the relation between the roots of $p_{n}(T,U)$ and those of the corresponding form $q_{n}(X,Y)$ under $\Psi_{n}$. Let

[TABLE]

By (2.11) and (2.12), in both cases we may define a map $\psi$ from ${\mathbb{P}}^{1}({\mathbb{C}})$ to ${\mathbb{P}}^{1}({\mathbb{C}})$ by

[TABLE]

We assume that  $y^{\prime}_{j}=u^{\prime}_{j}=1$  and  $t^{\prime}_{j}\neq 1$. Then

[TABLE]

This is the same as that in (2.9).

2.3. Discriminants of monic self-inversive polynomials in ${\mathcal{A}}_{n}$

We consider monic self-inversive polynomials in ${\mathcal{A}}_{n}$ and their discriminant. Let ${\mathcal{MA}}_{n}$ denote self-inversive forms that are written in (2.1) satisfying $\zeta_{0}=\zeta_{n+1}=1.$ The image of ${\mathcal{MA}}_{n}$ under $\Phi_{n}$ is denoted by $\mathcal{MF}_{n}$. Any element of ${\mathcal{MF}}_{n}$ is written as

[TABLE]

In general, the general linear group $GL(2,{\mathbb{R}})$ acts on the space of binary forms ${\mathcal{F}}_{n}$ via linear substitution :

[TABLE]

[TABLE]

[TABLE]

Lemma 2.8**.**

Suppose an element $g_{n}$ in ${\mathcal{F}}_{n}$ satisfies that $\Phi_{n}^{-1}(g_{n})$ is written as

[TABLE]

*with  $\zeta_{0}\neq 0.$

Then there exists a matrix $M$ in $GL(2,{\mathbb{R}})$ satisfying*

[TABLE]

Proof.

From Lemma 2.3, we know that

[TABLE]

Let

[TABLE]

To replace  $\zeta_{0}Z^{n+1}$ by $(Z^{\prime})^{n+1}$, we set

[TABLE]

where  $Z=X+iY,\quad Z^{\prime}=X^{\prime}+iY^{\prime}.$ Then we have new coefficients  $\zeta^{\prime}_{0}=1,\zeta^{\prime}_{1},\dots,\zeta^{\prime}_{n/2}$ . Hence by (2.2) we get new coefficients $\zeta^{\prime}_{0},\zeta^{\prime}_{1},\dots,\zeta^{\prime}_{n+1}$. The same folds when $n$ is odd. ∎

Note that this transformation is not unique.

Next we define the discriminant of $\tilde{g}_{n}(X,1)$ in (2.15). To define the discriminant of $\tilde{g}_{n}(X,1)$ we assume that $c_{0}\neq 0$. We define the discriminant by

[TABLE]

where $x_{1},x_{2},\dots,x_{n+1}$ are roots of $\tilde{g}_{n}(X,1)$. Though Gelfand, Kapranov and Zelevinsky [4] defined $Dis(\tilde{g}_{n}(X,1))$ to be $(-1)^{n(n+1)/2}$ times the right-hand side of (2.16), we adopt the definition in (2.16). In our definition, if all the roots of  $\tilde{g}_{n}(X,1)$ are real, then we have  $Dis(\tilde{g}_{n}(X,1))\geq 0$ . This is convenient to Corollary 2.10 below.

In this section, we consider monic self-inversive polynomials in ${\mathcal{A}}_{n}$. We denote the image of $\tilde{g}_{n}(X,Y)$ in (2.15) under the map $\Phi_{n}^{-1}$ by

[TABLE]

We consider the roots of $\tilde{f}_{n}$ and $\tilde{g}_{n}$. We use the notation in (2.8). In this case we may set $u_{j}=y_{j}=1$ for $j=1,\dots,n+1$. Hence $x_{1},x_{2},\dots,x_{n+1}$ are roots of $\tilde{g}_{n}(X,1)$ and $t_{1},t_{2},\dots,t_{n+1}$ are roots of $\tilde{f}_{n}(T,1)$.

We will define a $(n+1)\times(n+1)$ matrix $H_{n}$ that is a slightly modified version of the matrix introduced by Eier and Lidl [3]. Let $t_{1},1\leq j\leq n+1,$ be elements in ${\mathbb{C}}$ satisfying $t_{1}t_{2}\dots t_{n+1}=1$. The coefficient $\zeta_{j}$ is the $j$-th elementary symmetric function in $t_{1},\dots,t_{n+1}$. We define the $m$-th power $h_{m}(m\in{\mathbb{Z}})$ by

[TABLE]

Since  $t_{1}t_{2}\dots t_{n+1}=1$,  we may regard $h_{m}$ as a polynomial in  $\zeta_{1},\zeta_{2},\dots,\zeta_{n}$ even when  $m<0$.

We define a $(n+1)\times(n+1)$ matrix $H_{n}$ by

[TABLE]

In our case

[TABLE]

Eier and Lidl [3] considered only the case that any $t_{j}$ lies on the unit circle $C$.

The determinant $det\enskip H_{n}$ is a polynomials in $\zeta_{1},\dots,\zeta_{n}$. The relation between the discriminant $Dis(\tilde{g}_{n}(X,1))$ in (2.16) and the determinant $det\enskip H_{n}$ is shown in the following theorem.

Theorem 2.9**.**

Under the above notations we have

[TABLE]

Proof.

Case 1 :  $n$ is even. Using the formula for the Vandermonde matrix we see

[TABLE]

[TABLE]

[TABLE]

[TABLE]

On the other hand by (2.16),

[TABLE]

From (2.9) it follows that

[TABLE]

Hence

[TABLE]

Clearly

[TABLE]

Replacing $(T,U)=(1,1)$ by $(X,Y)$ under the transformation (2.5), we see that $X=-i$ and $Y=0.$ Hence we have

[TABLE]

Substituting this into (2.18), we have

[TABLE]

Case 2 :  $n$ is odd. The proof is similar. ∎

Since $\tilde{g}_{n}(X,1)$ is a real polynomial, its discriminant $Dis(\tilde{g}_{n}(X,1))$ is real. Thus $det\enskip H_{n}$ is also real. We may regard the space

[TABLE]

as the space $\mathbb{R}^{n}$. The set $\{det\enskip H_{n}=0\}$ may be seen as a hypersurface in $R_{n}$. We define a subset $W_{n}$ of $R_{n}$ by

[TABLE]

[TABLE]

Any element of $W_{n}$ corresponds to the $\tilde{f}_{n}(T,1)$ all the roots of which are on the unit circle $C$ and distinct.

We consider the n-simplex $S_{n}$ defined by

[TABLE]

[TABLE]

We define a map $\phi$ from $S_{n}$ to $R_{n}$ by

[TABLE]

where  $\zeta_{j}$ is the j-th elementary symmetric function in $e^{i\theta_{1}},\dots,e^{i\theta_{n+1}},\enskip\\ (j=1,2,\dots,n).$  By [3], we see that $\phi$ is a diffeomorphism from int$(S_{n})$ onto $W_{n}$ and $\partial S_{n}$ is mapped into the set  $\{det\enskip H_{n}=0\}$ .

Then the set $W_{n}$ is a connected component in $R_{n}\setminus\{det\enskip H_{n}=0\}$ that contains the origin  $(\zeta_{1}=\zeta_{2}=\dots=\zeta_{n}=0)$.

We consider the case that ${\tilde{f}}_{n}(T,1)$ has exactly $k$ roots on the unit circle $C$ . Note that $n+1-k$ is even.

Corollary 2.10**.**

We assume that ${\tilde{f}}_{n}(T,1)$ has exactly $k$ roots on $C$ and that all the roots of ${\tilde{f}}_{n}(T,1)$ are distinct. Then

[TABLE]

where  $sgn\enskip det\enskip H_{n}$ denotes the signature of  $det\enskip H_{n}.$

Proof.

The assumption implies that ${\tilde{g}}_{n}(X,1)$ has $(n+1-k)/2$ pairs of complex conjugate roots and $k$ real roots . Then the corollary follows from (2.16) and Theorem 2.9. ∎

In theorem 2.9, we assume that $c_{0}\neq 0$. Here we consider this condition. If $c_{0}=0,$ the positive integer $k$ is uniquely determined by the condition that $c_{k}\neq 0$ and $c_{j}=0$ for $0\leq j<k.$ We may consider $\tilde{g}_{n}(X,Y)/Y^{k}.$ We study this case through the corresponding self- inversive forms. Let $\tilde{f}_{n}(T,U)$ be the corresponding self-inversive form of $\tilde{g}_{n}(X,Y)$  under $\Phi_{n}^{-1}$. Then it is equivalent to consider $\tilde{f}_{n}(T,U)/(T-U)^{k}$. We will see that $\tilde{f}_{n}(T,U)/(T-U)^{k}\in{\mathcal{MA}}_{n-k}$ by the following proposition.

Proposition 2.11**.**

Assume that $\tilde{f}_{n}(T,U)$ is an element of ${\mathcal{MA}}_{n}$ and has a root $T=U$. Then $\tilde{f}_{n}(T,U)/(T-U)$ is an element of ${\mathcal{MA}}_{n-1}$.

Proof.

We may assume that

[TABLE]

and $t_{1}t_{2}\dots t_{n+1}=1$. It is known e.g. in [7] that a polynomial whose roots are on or symmetric in the unit circle $C$ is a self-inversive polynomial and that the converse is true. Hence all the roots of $\tilde{f}_{n}(T,U)$ are on or symmetric in $C$. Since $\tilde{f}_{n}(T,U)$ has a root $T=U$, we may assume $t_{1}=1$. Then the roots of $\tilde{f}_{n}(T,U)/(T-U)$ are on or symmetric in $C$ and satisfy $t_{2}t_{3}\dots t_{n+1}=1$. So $\tilde{f}_{n}(T,U)/(T-U)$ is self-inversive and the coefficients of $T^{n}$ and $U^{n}$ are 1 and $(-1)^{n}$, respectively. Hence $\tilde{f}_{n}(T,U)/(T-U)\in{\mathcal{MA}}_{n-1}$. ∎

We note a remark. In the proof of Lemma 2.8, we replace $\zeta_{0}Z^{n+1}$ by $(Z^{\prime})^{n+1}$ to make $\zeta_{0}=1$. If we use another replacement $(Z^{\prime\prime})^{n+1}$ of $\zeta_{0}Z^{n+1}$, then  $Z^{\prime\prime}=\omega Z^{\prime}$ where  $\omega^{n+1}=1$. The determinant $det\enskip H_{n}$ does not change under this new replacement.

2.4. Discriminants of monic self-reciprocal polynomials in ${\mathcal{B}}_{n}$

In the section we assume that $n$ is even. We consider monic self-reciprocal polynomials in ${\mathcal{B}}_{n}$ and their discriminants. Let ${\mathcal{MB}}_{n}$ denote self-reciprocal forms that are written in (2.10) satisfying $\zeta_{0}=\zeta_{n+1}=1.$ The image of ${\mathcal{MB}}_{n}$ under $\Psi_{n}$ is denoted by $\mathcal{MF^{\prime}}_{n}$. Any element of ${\mathcal{MF^{\prime}}}_{n}$ is written as

[TABLE]

To define the discriminant of $\tilde{q}_{n}(X,1)$ we assume that $d_{0}\neq 0$. We define the discriminant by

[TABLE]

where $x^{\prime}_{1},x^{\prime}_{2},\dots,x^{\prime}_{n+1}$ are roots of $\tilde{q}_{n}(X,1)$.

We consider monic self-reciprocal polynomials in ${\mathcal{B}}_{n}$. We denote the image of $\tilde{q}_{n}(X,Y)$ in (2.20) under the map $\Psi_{n}^{-1}$ by

[TABLE]

We consider the roots of $\tilde{p}_{n}$ and $\tilde{q}_{n}$. We use the notation in (2.13). In this case we may set $u^{\prime}_{j}=y^{\prime}_{j}=1$ for $j=1,\dots,n+1$. Hence $x^{\prime}_{1},x^{\prime}_{2},\dots,x^{\prime}_{n+1}$ are roots of $\tilde{q}_{n}(X,1)$ and $t^{\prime}_{1},t^{\prime}_{2},\dots,t^{\prime}_{n+1}$ are roots of $\tilde{p}_{n}(T,1)$.

The coefficient $\zeta_{j}$ is $(-1)^{j}$ times the $j$-th elementary symmetric function in $t^{\prime}_{1},\dots,t^{\prime}_{n+1}$. Clearly $t^{\prime}_{1},\dots,t^{\prime}_{n+1}=-1$. We define the $m$-th power $k_{m}(m\in{\mathbb{Z}})$ by

[TABLE]

Clearly

[TABLE]

We define a $(n+1)\times(n+1)$ matrix $K_{n}$ by

[TABLE]

In our case

[TABLE]

Theorem 2.12**.**

Under the above notations we have

[TABLE]

Proof.

The proof is essentially the same as that of Theorem 2.9. For any monic self-reciprocal polynomial $\tilde{p}_{n}$ in (2.22) , we have $\tilde{p}_{n}(1,1)=d_{0}$, by (2.11) . This corresponds to $\tilde{f}_{n}(1,1)=c_{0}(-i)^{n+1}$ in (2.19). Since $n$ is even, it follows that  $(t^{\prime}_{1},\dots,t^{\prime}_{n+1})^{n}=1$.

Then we have

[TABLE]

Clearly

[TABLE]

Hence by (2.8) and (2.13) we have with a suitable numbering

[TABLE]

Then

[TABLE]

Therefore by the proof of Theorem 2.9 we have

[TABLE]

∎