Graded Thread Modules over the Positive Part of the Witt (Virasoro) Algebra

TL;DR

This paper classifies and introduces new examples of graded thread modules over the positive part of the Witt (Virasoro) algebra, expanding understanding of their structure and representations.

Contribution

It provides a classification of high-dimensional graded thread modules over the Witt algebra's positive part, including new examples beyond known families.

Findings

Classified (n+1)-dimensional graded thread modules for large n.

Identified new examples of graded thread modules not related to known families.

Extended the understanding of module structures over the Witt algebra.

Abstract

We study ${\mathbb Z}$-graded thread $W^+$-modules $$V=\oplus_i V_i, \; \dim{V_i}=1, -\infty \le k< i < N\le +\infty, \; \dim{V_i}=0, \; {\rm \; otherwise},$$ over the positive part $W^+$ of the Witt (Virasoro) algebra $W$. There is well-known example of infinite-dimensional ($k=-\infty, N=\infty$) two-parametric family $V_{\lambda, \mu}$ of $W^+$-modules induced by the twisted $W$-action on tensor densities $P(x)x^{\mu}(dx)^{-\lambda}, \mu, \lambda \in {\mathbb K}, P(x) \in {\mathbb K}[t]$. Another family $C_{\alpha, \beta}$ of $W^+$-modules is defined by the action of two multiplicative generators $e_1, e_2$ of $W^+$ as $e_1f_i=\alpha f_{i+1}$ and $e_2f_j=\beta f_{j+2}$ for $i,j \in {\mathbb Z}$ and $\alpha, \beta$ are two arbitrary constants ($e_if_j=0, i \ge 3$). We classify $(n+1)$-dimensional graded thread $W^+$-modules for $n$ sufficiently large $n$ of three important types. New examples of graded thread $W^+$-modules different from finite-dimensional quotients of $V_{\lambda, \mu}$ and $C_{\alpha, \beta}$ were found.

Tables

Table 1. Table 1. Graded thread W + superscript 𝑊 W^{+} -modules of the type ( 1 , 1 , … , 1 ) , dim V = n + 1 ≥ 10 1 1 … 1 dimension 𝑉 𝑛 1 10 (1,1,\dots,1),\dim{V}=n+1\geq 10 .

module	$b_1$	$b_2$	$\dots$	$b_i$	$\dots$	$b_{n-2}$	$b_{n-1}$
$\begin{array}{c}{V}_{\lambda ,\mu }\left(n+1\right),\\ u=\mu -3\lambda ,\\ v=\mu -2\lambda ,\\ v\ne -1,\dots ,-n,\end{array}$	$\frac{6\left(u+1\right)}{\left(v+1\right)\left(v+2\right)}$	$\frac{6\left(u+2\right)}{\left(v+2\right)\left(v+3\right)}$	$\dots$	$\frac{6\left(u+i\right)}{\left(v+i\right)\left(v+i+1\right)}$	$\dots$	$\frac{6\left(u+n-2\right)}{\left(v+n-2\right)\left(v+n-1\right)}$	$\frac{6\left(u+n-1\right)}{\left(v+n-1\right)\left(v+n\right)}$
$C_{1,x}\left(n+1\right)$	$x$	$x$	$\dots$	$x$	$\dots$	$x$	$x$
$\begin{array}{c}{V}_{-2,-3}^t\left(n+1\right),\\ t\ne 4\end{array}$	$t$	$\frac{5}{2}$	$\dots$	$\frac{6\left(i+3\right)}{\left(i+1\right)\left(i+2\right)}$	$\dots$	$\frac{6\left(n+1\right)}{\left(n-1\right)n}$	$\frac{6\left(n+2\right)}{n\left(n+1\right)}$
$\begin{array}{c}{V}_{1,3-n}^t\left(n+1\right),\\ t\ne 4\end{array}$	$-\frac{6\left(n+2\right)}{n\left(n+1\right)}$	$-\frac{6\left(n+1\right)}{\left(n-1\right)n}$	$\dots$	$-\frac{6\left(n-i\right)}{\left(n-i-2\right)\left(n-i-1\right)}$	$\dots$	$-\frac{5}{2}$	$-t$
$\begin{array}{c}{V}_{0,-1}^t\left(n+1\right),\\ t\ne 6\end{array}$	$t$	$3$	$\dots$	$\frac{6}{i}$	$\dots$	$\frac{6}{\left(n-2\right)}$	$\frac{6}{\left(n-1\right)}$
$\begin{array}{c}{V}_{-1,-2-n}^t\left(n+1\right),\\ t\ne 6\end{array}$	$-\frac{6}{\left(n-1\right)}$	$-\frac{6}{\left(n-2\right)}$	$\dots$	$-\frac{6}{\left(n-i\right)}$	$\dots$	$-3$	$-t$

Table 2. Table 2. Graded thread W + superscript 𝑊 W^{+} -modules of the type ( 1 , … , 1 , 0 , 1 ​ … , 1 ) , dim V = n + 1 ≥ 16 1 … 1 0 1 … 1 dimension 𝑉 𝑛 1 16 (1,\dots,1,0,1\dots,1),\dim{V}=n+1\geq 16 .

module	$b_1$	$b_2$	$\dots$	$b_{k-2}$	$b_{k-1}$	$b_k$	$b_{k+1}$	$\dots$	$b_{n-1}$
	$\frac{6\left(-\lambda -k+1\right)}{\left(k-1\right)\left(k-2\right)}$	$\frac{6\left(-\lambda -k+2\right)}{\left(k-2\right)\left(k-3\right)}$	$\dots$	$\frac{6\left(-\lambda -2\right)}{2\cdot 1}$	$6\left(-\lambda -1\right)$	$-6\lambda$	$\frac{6\left(-\lambda +1\right)}{1\cdot 2}$	$\dots$	$\frac{6\left(-\lambda +n-1-k\right)}{\left(n-k-1\right)\left(n-k\right)}$
$\begin{array}{c}{V}_{\lambda ,2\lambda -2}\left(n+1\right),\\ \lambda \ne -1,k=2\end{array}$	$6\left(\lambda +1\right)$	$-6\lambda$	$\dots$	$\dots$	$\frac{6\left(-\lambda +k-3\right)}{\left(k-3\right)\left(k-2\right)}$	$\dots$	$\frac{6\left(-\lambda +k-1\right)}{\left(k-1\right)k}$	$\dots$	$\frac{6\left(-\lambda +n-3\right)}{\left(n-3\right)\left(n-2\right)}$
$\begin{array}{c}{V}_{\lambda ,2\lambda -1}\left(n+1\right),\\ \lambda \ne 0,-1,k=1\end{array}$	$-6\lambda$	$\frac{6\left(-\lambda +1\right)}{1\cdot 2}$	$\dots$	$\dots$	$\frac{6\left(-\lambda +k-2\right)}{\left(k-2\right)\left(k-1\right)}$	$\dots$	$\frac{6\left(-\lambda +k\right)}{k\left(k+1\right)}$	$\dots$	$\frac{6\left(-\lambda +n-2\right)}{\left(n-2\right)\left(n-1\right)}$
$\begin{array}{c}{V}_{\lambda ,2\lambda -n}\left(n+1\right),\\ \lambda \ne 0,-1,k=n\end{array}$	$\frac{6\left(-\lambda -n+1\right)}{\left(n-2\right)\left(n-1\right)}$	$\frac{6\left(-\lambda -n+2\right)}{\left(n-2\right)\left(n-3\right)}$	$\dots$	$\dots$	$\frac{6\left(-\lambda -n+k-1\right)}{\left(n-k+2\right)\left(n-k+1\right)}$	$\dots$	$\frac{6\left(-\lambda -n+k\right)}{\left(n-k\right)\left(n-k-1\right)}$	$\dots$	$6\left(-\lambda -1\right)$
$\begin{array}{c}{V}_{\lambda ,2\lambda -n+1}\left(n+1\right),\\ \lambda \ne 0,k=n-1\end{array}$	$\frac{6\left(-\lambda -n+2\right)}{\left(n-2\right)\left(n-3\right)}$	$\frac{6\left(-\lambda -n+3\right)}{\left(n-3\right)\left(n-4\right)}$	$\dots$	$\dots$	$\frac{6\left(-\lambda -n+k\right)}{\left(n-k\right)\left(n-k-1\right)}$	$\dots$	$\frac{6\left(-\lambda -n+k+2\right)}{\left(n-k-2\right)\left(n-k-3\right)}$	$\dots$	$-6\lambda$
	$-\frac{6}{k-1}$	$-\frac{6}{k-2}$	$\dots$	$-\frac{6}{2}$	$0$	$6$	$t$	$\dots$	$\frac{6}{n-k-1}$
$\begin{array}{c}{V}_{-1,-4}^t\left(n+1\right),\\ k=2\end{array}$	$0$	$6$	$\dots$	$\frac{6}{k-4}$	$\frac{6}{k-3}$	$\frac{6}{k-2}$	$\frac{6}{k-1}$	$\dots$	$\frac{6}{n-3}$
$\begin{array}{c}{V}_{-1,-3}^t\left(n+1\right),\\ k=1\end{array}$	$6$	$t$	$\dots$	$\frac{6}{k-3}$	$\frac{6}{k-2}$	$\frac{6}{k-1}$	$\frac{6}{k}$	$\dots$	$\frac{6}{n-2}$
	$-\frac{6}{\left(k-2\right)}$	$-\frac{6}{\left(k-3\right)}$	$\dots$	$t$	$-6$	$0$	$\frac{6}{2}$	$\dots$	$\frac{6}{n-k}$
$\begin{array}{c}{V}_{0,-n+1}^t\left(n+1\right),\\ k=n-1\end{array}$	$-\frac{6}{n-3}$	$-\frac{6}{n-2}$	$\dots$	$-\frac{6}{n-k}$	$-\frac{6}{n-k-1}$	$-\frac{6}{n-k-2}$	$-\frac{6}{n-k-3}$	$\dots$	$0$
$\begin{array}{c}{V}_{0,-n}^t\left(n+1\right),\\ k=n\end{array}$	$-\frac{6}{n-2}$	$-\frac{6}{n-1}$	$\dots$	$-\frac{6}{n-k+1}$	$-\frac{6}{n-k}$	$-\frac{6}{n-k-1}$	$-\frac{6}{n-k-2}$	$\dots$	$-6$
$\begin{array}{c}{\tilde{V}}_{0,-1}\left(n+1\right),\\ k=1\end{array}$	$6$	$\frac{6}{2}$	$\dots$	$\frac{6}{k-2}$	$\frac{6}{k-1}$	$\frac{6}{k}$	$\frac{6}{k+1}$	$\dots$	$\frac{6}{n-1}$
$\begin{array}{c}{\tilde{V}}_{-1,-2-n}\left(n+1\right),\\ k=n\end{array}$	$-\frac{6}{n-1}$	$-\frac{6}{n-2}$	$\dots$	$-\frac{6}{n-k+2}$	$-\frac{6}{n-k+1}$	$\frac{6}{n-k}$	$\frac{6}{n-k-1}$	$\dots$	$-6$
$\begin{array}{c}{\tilde{V}}_{-2,-3}\left(n+1\right),\\ k=1\end{array}$	$6$	$\frac{5}{2}$	$\dots$	$\frac{6\left(1+k\right)}{\left(k-1\right)k}$	$\frac{6\left(2+k\right)}{k\left(k+1\right)}$	$\frac{6\left(3+k\right)}{\left(k+1\right)\left(k+2\right)}$	$\frac{6\left(4+k\right)}{\left(k+2\right)\left(k+3\right)}$	$\dots$	$\frac{6\left(n+2\right)}{n\left(n+1\right)}$
$\begin{array}{c}{\tilde{V}}_{1,-n}\left(n+1\right),\\ k=n\end{array}$	$-\frac{6\left(n+2\right)}{n\left(n+1\right)}$	$-\frac{6\left(n+1\right)}{\left(n-1\right)n}$	$\dots$	$\dots$	$-\frac{6\left(n+4-k\right)}{\left(n+3-k\right)\left(n+2-k\right)}$	$\dots$	$-\frac{6\left(n+2-k\right)}{\left(n+1-k\right)\left(n-k\right)}$	$\dots$	$-6$

Equations

$$V={\oplus }_i{V}_i,\dim V_i=1,-\infty \le k<i<N\le +\infty ,\dim V_i=0,\mathrm{otherwise},$$

$$[e_i,e_j]=(j-i)e_{i+j},i,j\in \mathbb{Z}.$$

$$[e_i,z]=0,\forall i\in \mathbb{Z},[e_i,e_j]=(j-i)e_{i+j}+\frac{j^3-j}{12}{\delta }_{-i,j}z.$$

$$e_{i}V_{j}\subset V_{i{+}j},i\in{\mathbb{N}},j\in{\mathbb{Z}},\;\;\dim V_{i}=\left\{\begin{array}[]{cr}1,&k<i<N,\\ 0,&{\rm otherwise}\\ \end{array}\right.,$$

$$f_i,i\in \mathbb{Z},k<i<N,⟨f_i⟩=V_i,$$

$${\alpha }_i,{\beta }_j,i,j\in \mathbb{Z},k<i<N-2,k<j<N-3,$$

$$e_1{f}_i={\alpha }_i{f}_{i+1},e_2{f}_j={\beta }_j{f}_{j+2}.$$

$$L_{\xi }P(x)x^{\mu }(dx{)}^{-\lambda }=\left(f(x)(P(x)x^{\mu }{)}^{\prime }-\lambda P(x)x^{\mu }{f}^{\prime }(x)\right)(dx{)}^{-\lambda }.$$

$$e_k{f}_j=(j+\mu -\lambda (k+1))f_{k+j}.$$

$${\alpha }_i=i+\mu -2\lambda ,{\beta }_j=j+\mu -3\lambda .$$

$$V(k,N)={\oplus }_{i>k}^{\infty }{V}_i/{\oplus }_{j>N-1}^{\infty }{V}_j,V(k,N)=⟨f_{k+1},\dots ,f_{N-1}⟩.$$

$$f_1^{\prime }=f^{n+1},f_2^{\prime }=f^n,\dots ,f_n^{\prime }=f^2,f_{n+1}^{\prime }=f^1,$$

$$\{-{\alpha }_n,\dots ,-{\alpha }_1,-{\beta }_{n-1},\dots ,-{\beta }_1\}$$

$$\begin{array}[]{l}e_{1}f^{\prime}_{j}=e_{1}f^{n{+}2{-}j}={-}\alpha_{n{+}1{-}j}f^{n{+}1{-}j}={-}\alpha_{n{+}1{-}j}f^{\prime}_{j{+}1},\;j=1,2,\dots,n;\\ e_{2}f^{\prime}_{k}=e_{2}f^{n{+}2{-}k}={-}\beta_{n{-}k}f^{n{-}k}={-}\beta_{n{-}k}f^{\prime}_{k{+}2},\;k=1,2,\dots,n-1.\end{array}$$

$$\{{\alpha }_1,\dots ,{\alpha }_m,0,{\alpha }_{m+1},\dots ,{\alpha }_n,{\beta }_1,\dots ,{\beta }_{m-1},0,0,{\beta }_{m+2},\dots ,{\beta }_{n-1}\},0\le m<n$$

$$V=V_1\oplus V_2,V_1=⟨f_1,\dots ,f_{m+1}⟩,V_2=⟨f_{m+2},\dots ,f_{n+1}⟩.$$

$$\{{\alpha }_1,\dots ,{\alpha }_m,{\beta }_1,\dots ,{\beta }_{m-1}\},\{{\alpha }_{m+1},\dots ,{\alpha }_n,{\beta }_{m+2},\dots ,{\beta }_{n-1}\},$$

$$\exists k,p,1\le k<k+p\le n+1,p\ge 2,e_1{f}_k=e_1{f}_{k+p}=0.$$

$$(p-1)e_{p+1}{f}_k=\left(e_1{e}_p{f}_k-e_p{e}_1{f}_k\right)=0.$$

$$p{e}_{p+2}{f}_k=\left(e_1{e}_{p+1}{f}_k-e_{p+1}{e}_1{f}_k\right)=0,p{e}_{p+2}{f}_{k-1}=\left(e_1{e}_{p+1}{f}_{k-1}-e_{p+1}{e}_1{f}_{k-1}\right)=0.$$

$$(p+s-1)e_{p+s+1}{f}_k=0,\dots ,(p+s-1)e_{p+s+1}{f}_{k-s}=\left(e_1{e}_{p+s}{f}_{k-s}-e_{p+s}{e}_1{f}_{k-s}\right)=0.$$

$$(k+p-1)e_{k+p+1}{f}_1=(e_1{e}_{k+p}{f}_1-e_{k+p}{e}_1{f}_1)=\cdots =(k+p-1)e_{k+p+1}{f}_k=0.$$

$${\tilde{e}}_1=e_1,{\tilde{e}}_i=6(i-2)!e_i.$$

$$[{\tilde{e}}_1,{\tilde{e}}_i]={\tilde{e}}_{i+1},[{\tilde{e}}_2,{\tilde{e}}_3]={\tilde{e}}_5,[{\tilde{e}}_2,{\tilde{e}}_5]=\frac{9}{10}{\tilde{e}}_7.$$

$$\left[{\tilde{e}}_2,[{\tilde{e}}_1,{\tilde{e}}_2]\right]=\left[{\tilde{e}}_1,[{\tilde{e}}_1,[{\tilde{e}}_1,{\tilde{e}}_2]]\right],\left[{\tilde{e}}_2,\left[{\tilde{e}}_1,[{\tilde{e}}_1,[{\tilde{e}}_1,{\tilde{e}}_2]]\right]\right]=\frac{9}{10}\left[{\tilde{e}}_1,\left[{\tilde{e}}_1,\left[{\tilde{e}}_1,[{\tilde{e}}_1,[{\tilde{e}}_1,{\tilde{e}}_2]]\right]\right]\right]$$

$$R_i^5:([{\tilde{e}}_2,{\tilde{e}}_3]-{\tilde{e}}_5)f_i=0,i=1,\dots ,n-5,$$

$$R_j^7:([{\tilde{e}}_2,{\tilde{e}}_5]-\frac{9}{10}{\tilde{e}}_7)f_i=0,i=1,\dots ,n-7.$$

$${\tilde{e}}_1{f}_i=f_{i+1},i=1,\dots ,n,{\tilde{e}}_2{f}_j=b_j{f}_{j+2},j=1,\dots ,n-1.$$

$$\begin{array}{r}{R}_i^5:b_{i+3}(b_i-b_{i+1})-b_i(b_{i+2}-b_{i+3})=b_i-3b_{i+1}+3b_{i+2}-b_{i+3},\\ R_j^7:b_{j+5}(b_j-3b_{j+1}+3b_{j+2}-b_{j+3})-b_j(b_{j+2}-3b_{j+3}+3b_{j+4}-b_{j+5})=\\ =\frac{9}{10}(b_j-5b_{j+1}+10b_{j+2}-10b_{j+3}+5b_{j+4}-b_{j+5}),\\ i=1,\dots ,n,j=1,\dots ,n-1.\end{array}$$

$$b_i=\frac{6{\beta }_i}{{\alpha }_i{\alpha }_{i+1}},i=1,\dots ,i-2.$$

$$u=\mu -3\lambda ,v=\mu -2\lambda .$$

Full text

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over the positive part $W^{+}$ of the Witt (Virasoro) algebra $W$. There is well-known example of infinite-dimensional ($k=-\infty,N=\infty$) two-parametric family $V_{\lambda,\mu}$ of $W^{+}$-modules induced by the twisted $W$-action on tensor densities $P(x)x^{\mu}(dx)^{-\lambda},\mu,\lambda\in{\mathbb{K}},P(x)\in{\mathbb{K}}[t]$. Another family $C_{\alpha,\beta}$ of $W^{+}$-modules is defined by the action of two multiplicative generators $e_{1},e_{2}$ of $W^{+}$ as $e_{1}f_{i}=\alpha f_{i+1}$ and $e_{2}f_{j}=\beta f_{j+2}$ for $i,j\in{\mathbb{Z}}$ and $\alpha,\beta$ are two arbitrary constants ($e_{i}f_{j}=0,i\geq 3$).

We classify $(n+1)$-dimensional graded thread $W^{+}$-modules for $n$ sufficiently large $n$ of three important types. New examples of graded thread $W^{+}$-modules different from finite-dimensional quotients of $V_{\lambda,\mu}$ and $C_{\alpha,\beta}$ were found.

Supported by the RFBR grant N 16-51-55017 (Russian Foundation for Basic Research)

1. Positive part of the Witt algebra and its finite-dimensional graded modules

The representation theory of the Virasoro algebra $Vir$ was intensively studied in the 80s of the last century, now one may consider it as a classical section of the representation theory of infinite-dimensional Lie algebras. For instance, the structure of Verma modules over Virasoro algebra and Fock modules over $Vir$ were completely determined by B. Feigin and D. Fuchs. O. Mathieu [11] proved V. Kac’s conjecture which says that any simple ${\mathbb{Z}}$-graded $Vir$-module with finite-dimensional homogeneous components is either a highest weight module, a lowest weight module, or the module of type $V_{\lambda,\mu}$. However two particular cases of the theorem were already proved: the classification of Harish-Chandra modules for which all the multiplicities of weights are $1$ (by I. Kaplansky and J. Santharoubane [9]) and the classification of unitarizable Harish-Chandra modules (by V. Chari and A. Pressley [2]). Some partial results on Kac’s conjecture were obtained in [10].

It should be noted that a great number of well-known mathematicians and mathematical physicists contributed to the development of the theory of representations of the Virasoro algebra, for complete survey we recommend the monograph [8].

In 1992 Benoist answering negatively to Milnor’s question [14] on left-invariant affine structures on nilpotent Lie groups presented examples of compact $11$-dimensional nilmanifolds that carry no complete affine structure. For that he constructed examples of $11$-dimensional nilpotent Lie algebras with no faithfull linear representations of dimension $12$. In his proof [3] he classified ${\mathbb{N}}$-graded Lie algebras ${\mathfrak{a}}_{r}$ defined by two generators $e_{1}$ and $e_{2}$ of degrees $1$ and $2$ respectively and two relations $[e_{2},e_{3}]=e_{5}$ and $[e_{2},e_{5}]=re_{7}$, where $r$ is an arbirary scalar and $e_{i{+}1}=[e_{1},e_{i}]$ for all $i\geq 2$.

Lemma 1.1** (Benoist, [3]).**

If $r{\neq}\frac{9}{10},1$, then ${\mathfrak{a}}_{r}$ is a finite-dimensional Lie algebra.

1) Let $r=\frac{9}{10}$, then ${\mathfrak{a}}_{r}\cong W^{+}$, positive part of the Witt (Virasoro) algebra, it is infinite-dimensional Lie algebra with the base $\tilde{e}_{i},i\geq 1$ and the relations $[\tilde{e}_{i},\tilde{e}_{j}]=(j{-}i)\tilde{e}_{i{+}j}$, where $\tilde{e}_{i}{=}e_{i}/(i{-}2)!$.

2) Let $r=1$, then ${\mathfrak{a}}_{r}\cong{\mathfrak{m}}_{2}$.

3) Let $r{\neq}0,\frac{9}{10},1,2,3$, then ${\mathfrak{a}}_{r}$ is a $11$-dimensional filiform Lie algebra.

We recall that a finite-dimensional nilpotent Lie algebra ${\mathfrak{g}}$ is called filiform if it has the maximal possible (for its dimension) value of nil-index $s{(\mathfrak{g}})=\dim{\mathfrak{g}}{-}1$. In its turn the nil-index $s{(\mathfrak{g}})$ is the length of the descending central series of ${\mathfrak{g}}$.

The idea of Benoist’s construction was the following. Benoist considered the algebra ${\mathfrak{a}}_{-2}$ of the family ${\mathfrak{a}}_{r}$. It is a $11$-dimensional filiform Lie algebra. Itself ${\mathfrak{a}}_{-2}$ admits a complete affine structure because it is positively graded. Benoist considered its filtered deformation ${\mathfrak{a}}_{-2,s,t}$ that it is not positively graded and he proved that ${\mathfrak{a}}_{-2,s,t}$ does not admit any faithfull $12$-dimensional representation. That is an abstruction for existence of a complete affine structure on the corresponding nilmanifold. The proof of non-existence of faithfull $12$-representation was based in particular on the classification of graded faithfull ${\mathfrak{a}}_{-2}$-modules.

Despite the fact that the positive part $W^{+}$ of the Witt algebra was not used in his proof, Benoist suggested that special deformations of finite-dimensional factors of $W^{+}$ can also be used for counterexamples to Milnor’s conjecture in higher dimensions $n>11$. Hence the classification of graded thread $W^{+}$-modules is quite necessary for the possible proof.

The aim of this paper is a classification of $(n+1)$-dimensional graded thread $W^{+}$-modules with additional structure restriction $e_{n}f_{1}\neq 0$ which means that the corresponding representation of the $n$-dimensional quotient $W^{+}/\langle e_{n+1},e_{n+2},\dots,\rangle$ is faithfull.

Also finite-dimensional graded thread $W^{+}$-modules played an essential role in explicite constructions of singular Virasoro vectors in [1] and [13]. Besides this graded thread $W^{+}$-modules were used for the construction of trivial Massey products in the cohomology $H^{*}(W^{+},{\mathbb{K}})$ in [5], answering V. Buchstaber’s conjecture, that the cohomology $H^{*}(W^{+},{\mathbb{K}})$ is generated by non-trivial Massey products of one-dimensional cohomology classes. Finaly V. Buchstaber’s conjecture was proved for non-trivial Massey products in [12].

The paper is organized as follows. In the Section 1 we recall necessary definitions and facts on positive part of the Witt (Virasoro) algebra and its graded modules. In particular we introduce the important class of $(n{+}1)$-dimensional graded thread $W^{+}$-modules by a property $e_{n}f_{1}\neq 0$ mentioned above. We prove the key Lemma 2.2 and its Corollary stating that we have only three types of $(n+1)$-dimensional graded thread $W^{+}$-modules defined by basis $f_{1},\dots,f_{n+1}$:

a) no zeroes of $e_{1}$, i.e. $e_{1}f_{i}\neq 0,\;i=1,\dots,n$;

b) one zero of $e_{1}$, i.e. $\exists!k,\;1\leq k\leq n,\;e_{1}f_{k}{=}0$;

c) two neighboring zeroes of $e_{1}$, i.e. $\exists!k,1{\leq}k{\leq}n{-}1,e_{1}f_{k}{=}e_{1}f_{k{+}1}{=}0$.

We classify the modules of the type a) (we call them graded thread $W^{+}$-modules of the type $(1,1,\dots,1)$) in Section 3. $W^{+}$-modules with one zero, subcase b), so called modules of the type $(1,\dots,1,0,1,\dots,1)$ are classified in Section 4. We remark that $W^{+}$-modules of the family $V_{\lambda,\mu}$ are generic modules for both types.

Section 5 is devoted to the most interesting case of graded $W^{+}$-modules with two neighboring zeroes of $e_{1}$. Modules of this type were applied in [12] for the proof of V. Buchstaber’s conjecture on Massey products in Lie algebra cohomology $H^{*}(W^{+},{\mathbb{K}})$ as we mentioned above.

2. Definitions and examples

The Witt algebra $W$ can be defined by its infinite basis $e_{i}=t^{i{+}1}\frac{d}{dt},i\in{\mathbb{Z}}$ and the Lie bracket is given by

[TABLE]

The Virasoro algebra Vir is infinite-dimensional Lie algebra, defined by its basis $\{z,e_{i},i\in{\mathbb{Z}}\}$ and commutation relations:

[TABLE]

Vir is a one-dimensional central extension of the Witt algebra $W$ (the one-dimensional center is spaned by $z$).

There is a subalgebra $W^{+}\subset W$ spanned by $e_{1},e_{2},e_{3},\dots,$ basis vectors with positive subscripts that we cal the positive part of the Witt algebra.

Definition 2.1** ([5] ).**

A $W^{+}$-module $V$ is called graded thread $W^{+}$-module if there exist two integers $k,N,k<N$, or $k=-\infty,N=+\infty$ and a decompostion $V=\oplus_{{-}\infty}^{{+}\infty}V_{j},j\in{\mathbb{Z}},$ such that:

[TABLE]

where $e_{1},e_{2},\dots,e_{k},\dots$ is a graded basis of the positive part $W^{+}$ of the Witt (Virasoro) algebra.

Remark*.*

A graded thread $W^{+}$-module $V=\oplus_{{-}\infty}^{{+}\infty}V_{j}$ can be defined by its basis

[TABLE]

and by the set of its structure constants

[TABLE]

such that

[TABLE]

Certainly the constants $\alpha_{i},\beta_{j}$ can not be arbitrary. They must satisfy certain algebraic relations that we are going to discuss later.

To begin with, we present two infinite-dimensional examples.

• •

One can take $\alpha_{i}=\alpha,\;\beta_{j}=\beta$. It is a $W^{+}$-module $C_{\alpha,\beta}$ where all other basic elements $e_{i},i\geq 3$ act trivially.

• •

We have defined basic vectors $e_{i}$ of $W^{+}$ as differential operators $e_{i}=x^{i+1}\frac{d}{dx}$ on the real (complex) line. One can consider the space $V_{\lambda,\mu}$ of tensor densities of the form $P(x)x^{\mu}(dx)^{-\lambda}$, where $P(x)$ is some polynomial on $x$ and the parameters $\lambda,\mu$ are arbitrary real (complex) numbers. Operator $\xi=f(x)\frac{d}{dx}$ acts on $F_{\lambda,\mu}$ by means of the Lie derivative $L_{\xi}$:

[TABLE]

Taking the infinite basis $f_{j}=x^{j+\mu}(dx)^{-\lambda}$ of $F_{\lambda,\mu}$ we have the following $W^{+}$-action [6]:

[TABLE]

In other words $V_{\lambda,\mu}$ is a twist of the natural action of $W$ on ${\mathbb{C}}[t,t^{-1}]$. The defining set for a $W^{+}$-module $V_{\lambda,\mu}$ is

[TABLE]

Remark*.*

The vector space $V_{\lambda,\mu}$ can be regarded as a $W$-module over the entire Witt algebra $W$, or, that is equivalent, as a zero-energy Virasoro representation [7]. As $W$-module $V_{\lambda,\mu}$ is reducible if $\lambda\in{\mathbb{Z}}$ and $\beta{=}0$ or $\lambda\in{\mathbb{Z}}$ and $\beta{=}1$ otherwise it is irreducible [7]. The infinite-dimensional $W$-modules $V_{\lambda,\mu}$ and $V_{\lambda,\mu{+}m},m\in{\mathbb{Z}}$ are isomorphic.

Having an infinite-dimensional graded thread $W^{+}$-module $V=\langle f_{i},i\in{\mathbb{Z}}\rangle$ one can construct a $(N-k-1)$-dimensional $W^{+}$-module $V(k,N)$ taking a subquotient of $V$:

[TABLE]

From now we deal with a $(n+1)$-dimensional graded thread $W^{+}$-module $V$ defined by its basis $f_{1},f_{2},\dots,f_{n{+}1}$ and a finite set of constants $\left\{\alpha_{1},\dots,\alpha_{n},\beta_{1},\dots,\beta_{n-1}\right\}$.

The dual module $V^{*}$ of a finite-dimensional graded thread $W^{+}$-module $V=\langle f_{1},f_{2},\dots,f_{n{+}1}\rangle$ in its turn has the structure of a graded thread $W^{+}$-module with respect to the basis

[TABLE]

where $f^{1},\dots,f^{n},f^{n{+}1}$ is the dual basis in $V^{*}$ with respect to the basis $f_{1},\dots,f_{n},f_{n{+}1}$ of $V$, $f^{i}(f_{j})=\delta^{i}_{j}$. Let $\{\alpha_{1},\dots,\alpha_{n},\beta_{1},\dots,\beta_{n-1}\}$ be the set of structure constants of the $W^{+}$-module $V$ with respect to the basis $f_{1},\dots,f_{n{+}1}$. Then the set

[TABLE]

defines the structure of the dual $W^{+}$-module $V^{*}$:

[TABLE]

We recall that the dual $W^{+}$-action on $V^{*}$ is defined by $(g\cdot f)(x):=-f(gx)$, where $g\in W^{+},x\in V,f\in V^{*}$.

For obvious reasons there is no sense in discussing the irreducibility of graded thread $W^{+}$-modules: they are all reducible. Instead of irreducibility one has to discuss indecomposability. Let the defining sets of structure constants of a $W^{+}$-module $V$ be of the following type

[TABLE]

Then $V$ is a direct sum of two $W^{+}$-modules $V_{1}$ and $V_{2}$

[TABLE]

where $V_{1}$ and $V_{2}$ have the following defining sets of structure constants

[TABLE]

respectively. The converse is also true. One can consider a direct sum $V_{1}\oplus V_{2}$ of two finite-dimensional graded thread $W^{+}$-modules $V_{1}$ and $V_{2}$, may be after possible renumbering of the basis vectors from $V_{2}$.

Lemma 2.2**.**

Let $V=\langle f_{1},f_{2},\dots,f_{n+1}\rangle$ be a $(n+1)$-dimensional graded thread $W^{+}$-module such that

[TABLE]

Then $e_{n}f_{1}=0$.

Proof.

The equalities $e_{1}f_{k}=e_{1}f_{k{+}p}=0$ imply that

[TABLE]

On the next step we have

[TABLE]

One can suppose by an inductive assumption that $e_{p+s}f_{k}=\dots=e_{p+s}f_{k-s+1}=0$ for some $s,1\leq s\leq k-1$. Then it follows that

[TABLE]

Hence $e_{k+p}f_{1}=\dots=e_{k+p}f_{k}=0$ and

[TABLE]

Continuing these calculations we will have that $e_{i}f_{1}=\dots=e_{i}f_{k}=0$ for all $i\geq k{+}p$. In particular $e_{n}f_{1}=0$. ∎∎

Corollary 2.3**.**

Let $V=\langle f_{1},f_{2},\dots,f_{n{+}1}\rangle$ be a $(n{+}1)$-dimensional graded thread $W^{+}$-module such that $e_{n}f_{1}\neq 0$. Then for its defining set of constants $\alpha_{i},i=1,\dots,n$, we have three possibilities:

• a)

no zeroes, $\alpha_{i}\neq 0,\;i=1,\dots,n$;

• b)

the only one zero, $\exists!k,\;1\leq k\leq n,\;\alpha_{k}{=}0$;

• c)

two neighboring zeroes, $\exists!k,1{\leq}k{\leq}n{-}1,\alpha_{k}{=}\alpha_{k{+}1}{=}0$.

Following [3] we will consider a new basis of $W^{+}$:

[TABLE]

Now we have in particular that

[TABLE]

It was proved in [3] that the Lie algebra generated by two elements $\tilde{e}_{1},\tilde{e}_{2}$ with the following two relations on them

[TABLE]

is isomorphic to $W^{+}$.

Hence the defining relations (2) will give us the following set of equations

[TABLE]

It is possible to write out the explicit expressions for $R^{5}_{i}$ and $R^{7}_{j}$ in terms of $\alpha_{i},\beta_{j}$. However rescaling $f_{i}\to\gamma_{i}f_{i}$ we can make the constants $\alpha_{i}$ equal to one or to zero.

3. Graded thread $W^{+}$-modules of the type $(1,1,\dots,1)$.

Let us consider the case when all constants $\alpha_{i}$ of the defining set for a graded thread $W^{+}$-module are non-trivial and hence we may assume (after a suitable rescaling of the basis vectors $f_{1},\dots,f_{n+1}$) that

[TABLE]

Then the equations $R^{5}_{i},R^{7}_{j}$ are read as

[TABLE]

There is the relation between $b_{i}$ and the initial structure constants $\alpha_{i},\beta_{j}$ of our graded $W^{+}$-module $V$:

[TABLE]

Consider a module $F_{\lambda,\mu}$. Recall that it has the definig set with $\alpha_{i}=i+\mu-2\lambda$ and $\beta_{j}=j+\mu-3\lambda$. We introduce new parameters

[TABLE]

Suppose that $v=\mu-2\lambda\neq{-}1,{-}2,\dots,{-}n.$ Then $F_{\lambda,\mu}$ is of the type $(1,1,\dots,1)$ and its coordinates $(b_{1},b_{2},\dots,b_{n-1})$ are

[TABLE]

Definition 3.1**.**

An affine (projective) variety defined by the system of algebraic equations (1) in ${\mathbb{K}}^{n-1}({\mathbb{P}\mathbb{K}}^{n-2})$ is called the affine (projective) variety of $(n+1)$-dimensional graded thread $W^{+}$-modules of the type $(1,1,\dots,1)$.

Theorem 3.2**.**

Let $V$ be a $(n{+}1)$-dimensional graded thread $W^{+}$-module of the type $(1,1,\dots,1)$ and $n\geq 9$, i.e. $W^{+}$-module defined by its basis and defining set of relations

[TABLE]

Then $V$ is isomorphic to the one and only one $W^{+}$-module from the list below.

• •

$V_{\lambda,\mu}(n{+}1),\quad\mu-2\lambda\neq{-}1,{-}2,\dots,{-}n.$**

• •

$C_{1,x}(n+1),\quad b_{1}=b_{2}=\dots=b_{n-1}=x;$**

• •

$V_{-2,-3}^{t}(n+1),\;t\neq 4,\quad b_{1}=t,b_{i}=\frac{6(i+3)}{(i+1)(i+2)},i=2,\dots,n-1;$**

• •

$V_{1,3{-}n}^{t}(n+1),\;t\neq 4,\quad b_{i}=-\frac{6(n-i)}{(n-i-2)(n-i-1)},i=1,\dots,n-2,b_{n-1}=-t;$**

• •

$V_{0,-1}^{t}(n+1),\;t\neq 6,\quad b_{1}=t,b_{i}=\frac{6}{i},i=2,\dots,n-1;$**

• •

$V_{{-}1,{-}2{-}n}^{t}(n+1),\;t\neq 6,\quad b_{i}=-\frac{6}{(n-i)},i=1,\dots,n-2,b_{n-1}=t;$**

Remark*.*

1. A one-parametric family $V_{-2,-3}^{t}(n+1)$ of graded thread $W^{+}$-modules is a linear deformation of $V_{-2,-3}(n+1)$. Moreover $V_{-2,-3}^{4}(n+1)=V_{-2,-3}(n+1)$. A module $V_{1,3{-}n}^{t}(n+1)$ is dual to $V_{-2,-3}^{t}(n+1)$.

2. Family $V_{0,-1}^{t}(n+1)$ is one-parametric linear deformation of $V_{0,-1}(n+1)$ and $V_{0,-1}^{6}(n+1)=V_{0,-1}(n+1)$. $V_{1,{-}2{-}n}^{t}(n+1)$ is the dual module to $V_{0,-1}^{t}(n+1)$.

3. $C_{1,x}^{*}(n+1)=C_{1,-x}(n+1)$;

Proof.

We prove the Theorem by induction on dimension $\dim{V}$. The equation $R_{1}^{5}$ appears first time for a $5$-dimensional graded thread module. In dimension $6$ we have two relations $R_{1}^{5},R_{2}^{5}$ and coordinates $(b_{1},b_{2},\dots,b_{5})$ of an arbitrary $7$-dimensional graded thread $W^{+}$-module $V$ of the type $(1,1,\dots,1)$ satisfies three relations $R_{1}^{5},R_{2}^{5},R_{1}^{7}$ respectively.

We start with the classification of $8$-dimensional graded thread $W^{+}$-modules.

Lemma 3.3**.**

Consider an affine variety $M$ of $8$-dimensional graded thread $W^{+}$-modules of the type $(1,1,\dots,1)$ defined by the following system of quadratic equations in ${\mathbb{K}}^{6}$:

[TABLE]

then the variety $M$ can be decomposed as the union of the following two- and one-parametric algebraic subsets:

• $M_{1}:$

$b_{i}=6\frac{(u+i)}{(v+i)(v+i+1)},\;i=1,2,\dots,6,$*  $u\neq v,u\neq v{+}1,\;v\neq{-}1,{-}2,\dots,{-}7$;*

• $M_{1}^{0}:$

$b_{i}=\frac{6}{v{+}i},\;i=1,2,\dots,6,$*  $v\neq{-}1,{-}2,\dots,{-}6$;*

• $M_{2}:$

\begin{array}[]{l}b_{1}=\frac{5xy-17x+10y+2}{5y-9},\;b_{2}=x,\;b_{3}=y,\;b_{4}=y-\frac{2}{5},\;b_{5}=\frac{1}{5}\frac{5xy+3x-6}{2x-y+1},\\ b_{6}=\frac{5xy^{2}-2xy-22y+10y^{2}+21x-12}{(2x-y+1)(5y+7)},\;y\neq\frac{9}{5},{-}\frac{7}{5},2x-y+1\neq 0\end{array};

• $M_{3}:$

$b_{1}=b_{2}=b_{3}=b_{4}=b_{5}=b_{6}=t$;

• $M_{4}^{\pm}:$

\begin{array}[]{l}b_{1}=12\pm 3\sqrt{19},b_{2}=-\frac{2}{5}\pm\frac{1}{5}\sqrt{19},b_{3}=\frac{1}{5}\pm\frac{2}{5}\sqrt{19},b_{4}=-\frac{1}{5}\pm\frac{2}{5}\sqrt{19},\\ b_{5}=\frac{2}{5}\pm\frac{1}{5}\sqrt{19}+t,b_{6}=-12\pm 3\sqrt{19}+t(\pm\frac{4}{3}\sqrt{19}-\frac{13}{3}).\end{array};

• $M_{5}^{-}:$

$b_{1}=-\frac{27}{28},b_{2}=-\frac{8}{7},b_{3}=-\frac{7}{5},b_{4}=-\frac{9}{5},b_{5}=-\frac{5}{2},b_{6}=t$;

• $M_{5}^{+}:$

$b_{1}=t,b_{2}=\frac{5}{2},b_{3}=\frac{9}{5},b_{4}=\frac{7}{5},b_{5}=\frac{8}{7},b_{6}=\frac{27}{28}$;

• $M_{6}^{+}:$

$b_{1}=t,b_{2}=\frac{6}{2},b_{3}=\frac{6}{3},b_{4}=\frac{6}{4},b_{5}=\frac{6}{5},b_{6}=\frac{6}{6}$;

• $M_{6}^{-}:$

$b_{1}=-\frac{6}{6},b_{2}=-\frac{6}{5},b_{3}=-\frac{6}{4},b_{4}=-\frac{6}{3},b_{5}=-\frac{6}{2},b_{6}=-t$;

Proof.

Denote $b_{2}=x,b_{3}=y,b_{4}=z$ and rewrite after that first two equations of the system (2)

[TABLE]

Then we multiply the third equation of (2) by $(2x-y+1)$ and exclude $b_{5}$

[TABLE]

Finally we multiply the last equation of (2) by $(2z{-}y{-}1)(2x{-}y{+}1)(2y{-}z{+}1)$ and exclude $b_{1},b_{5},b_{6}$. We will get the following fifth-order equation of unknowns $x,y,z$:

[TABLE]

where

[TABLE]

First of all we are going to study an algebraic variety $M_{F}\subset{\mathbb{K}}^{3}$ defined by the equation $F(x,y,z)=0$. Consider mapping $f:{\mathbb{K}}^{2}\backslash\{(u,v),v=0,{-}1,{-}2,{-}3\}\to{\mathbb{K}}^{3}$:

[TABLE]

Proposition 3.4**.**

The variety $M_{F}\subset{\mathbb{K}}^{3}$ is the union of $Imf$ and three lines $l_{1},l_{2},l_{3}$ defined by

[TABLE]

Proof.

Let a point $(x,y,z)\in Imf$. It means that for some $(u,v),v\neq 0,{-}1,{-}2,{-}3$ we have

[TABLE]

Consider (5) as a system of equations with respect to unknowns $u,v$. For $z\neq x$ it is equivalent to

[TABLE]

1. Let $(3y{-}x)(z{-}x)-(5z{-}x)(y{-}x)=2(2xz-y(x+z))\neq 0$, then substituting

[TABLE]

in the second equation of (6) we get

[TABLE]

where the polynomial $F(x,y,z)$ is defined by (4). Hence a point $(x,y,z)$ of the surface $M_{F}$ with $x\neq z$ and $y(x{+}z)\neq 2xz$ is in the image $Imf$ and the corresponding parameters $u,v$ are determined uniquely.

2. Consider a point $(x,y,z)\in M_{F}$ such that

[TABLE]

This system can be rewritten in a following way

[TABLE]

Substitute $y=\frac{2xz}{x+z}$ ($x+z=0$ implies $x=z=0$) in the first equation. We obtain $(z{-}x)(xz{+}3z{-}3x){=}0$, hence if $z\neq x$ then

[TABLE]

Changing parameter $z=\frac{6}{t+2}$ we see that the set of solutions of (7) coincides with the curve

[TABLE]

Parameters $u,v$ for a point $\left(\frac{6}{t},\frac{6}{t+1},\frac{6}{t+2}\right)$ of $\gamma$ are determined not uniquely: $u=v=t$ or $u=v+1=t+2$.

Now let us consider the case $x=z$. Then the system (6) is equivalent to the following one

[TABLE]

Substituting $v$ by $\frac{6-3x}{2x}$ in the second equation of (9), we’ll get

[TABLE]

It follows from the last equation that $y=\frac{36x+3x^{3}}{36-x^{2}},x\neq 6$.

On the another hand, the square equation (with respect to $y$, we assume also that $x\neq\pm 6$)

[TABLE]

has two roots $y=x$ and $y=\frac{36x+3x^{3}}{36-x^{2}}$. One has to remark also that if $x=z=\pm 6$, then $y=\pm 6$.

Hence an arbitrary point $P=(x,y,z)$ of the surface $M_{F}$ with $x=z\neq y$ also belongs to the image $Imf$. We conclude with a remark that the system (5) never has the solutions with $v=-1,-2$, but it has the solutions with $v=0$ and $v=-3$. More precisely, if $v=0$ then it follows from the first equation of (5) that $u=0$ and this is the case for $y=3,z=2$ and arbitrary $x$. We get the line $l_{2}$.

Analogously the case $v=-3$ implies $u=-2$ and $y=-3,x=-2$. The corresponding set of solutions is $l_{3}$.

∎

Remark*.*

The polynomial $F(x,y,z)$ has the degree two with respect to each variable $x,y,z$ (thinking other two are parameters). One can verify directly that the curve $\gamma(t)$ defined by (8) coincides with the set of singular points of $M_{F}$ (it follows from the proof of our proposition). Also one can see that $M_{F}$ has an involution $\sigma_{F}:M_{F}\to M_{F}$:

[TABLE]

More precisely

[TABLE]

Now we came to the following natural question.

Does an arbitrary point $(x,y,z)\in M_{F}$ correspond to some solution $(b_{1},x,y,z,b_{5},b_{6})$ of the initial system (2)?

Before answering this question, we state a few preliminary remarks

Proposition 3.5**.**

It exists the unique solution $(b_{1},b_{2},\dots,b_{6})$ of the system (2) with $b_{2}=b_{3}=b_{4}=t$ and it is $b_{1}=b_{2}=b_{3}=b_{4}=b_{5}=b_{6}=t$.

Proof.

If $b_{2}=b_{3}=b_{4}=t$ then the first two equations of (2) can be rewritten as

[TABLE]

If $t\neq\pm 1$ then $b_{1}=b_{5}=t$ and $b_{6}=\frac{t^{3}+2t^{2}+t}{(t+1)^{2}}=t$.

If $t=1$ then $b_{5}=b_{6}=1$ and the fourth equation of (2) will look in the following way:

[TABLE]

Hence $b_{1}=1$. The case $t=-1$ is studied analogously.

∎

Proposition 3.6**.**

There are no solutions $(b_{1},b_{2},\dots,b_{6})$ of the system (2) such that:

• a)

$b_{3}=3,b_{4}=2$;

• b)

$b_{3}=-3,b_{2}=-2$.

Proof.

Solving first three equations of (2) with $b_{3}=3,b_{4}=2$ one get

[TABLE]

But the fourth equation of (2) looks as $0\cdot b_{6}=-\frac{339}{10}$.

If $b_{3}=-3$ and $b_{2}=-2$ we obtain $b_{1}=-\frac{3}{2},b_{4}=-7,b_{6}=-9$ and the last equation of (2) will be inconsistent with respect to $b_{5}$.

∎

Now we are going to study the same question for the points of $Imf$. Namely let suppose that for some choice of $u,v,v\neq 0,-1,-2,-3,$ we have

[TABLE]

Then the first three equations of the system (2) we rewrite as

[TABLE]

First of all let study a generic case.

Proposition 3.7**.**

Let $v\neq 0,-1,-2,-3$ and moreover

[TABLE]

Then there exists the only one solution $(b_{1},b_{5},b_{6})$ of the system (10)

[TABLE]

Proof.

Direct verification. ∎

Proposition 3.8**.**

Let $v\neq 0,-1,-2,-3$. Then if one between of the three given expressions

[TABLE]

is equal to zero then the rest two of them are non trivial.

Now we assume that $v\neq 1,0,-1,-2,-3,-4,-5$.

1. $6u-v^{2}-7v=0$. Then from the first and the third equations of (2) we have

[TABLE]

The fourth equation of (2) will look as

[TABLE]

If $v\neq 2,3$ then $b_{1}=\frac{v^{2}{+}7v{-}6}{v(v{-}1)}=\frac{6(u{-}1)}{v(v{-}1)}$ like in generic situation.

a) Now let $v=2$ then $u=\frac{v^{2}+7v}{6}=3$ and we have

[TABLE]

The component $b_{1}$ can take an arbitrary value $t$.

b) In the case $v=3$ analogously we obtain $u=5$ and

[TABLE]

We obtain another one-parametric family of solutions for (2):

[TABLE]

2. $6u+v^{2}-v=0$. Then $b_{1}=\frac{6(u{-}1)}{(v{-}1)v}$ and the third equation of (2) will look as

[TABLE]

Expressing $b_{6}$ and rewriting the last equation of (2) we will have

[TABLE]

If $v^{2}+8v-3\neq 0$ then $b_{5}$ is determined uniquely and hence we have

[TABLE]

If $v^{2}+8v-3=0$, i.e. $v=-4\pm\sqrt{19}$, then the component $b_{5}$ can take an arbitrary values and then we have $u=-\frac{13}{2}\pm\frac{3}{2}\sqrt{19}$ and

[TABLE]

After a parameter change $t\to t+\frac{2}{5}\pm\frac{1}{5}\sqrt{19}$ we will obtain the final version.

3. $6u+v^{2}+v+6=0$. Then $b_{1}=\frac{6(u{-}1)}{(v{-}1)v}$ and $b_{5}=\frac{6(u{+}3)}{(v{+}3)(v{+}4)}$. The fourth equation of the system (2) will look as

[TABLE]

If $v\neq-6,-7$ then $b_{6}=\frac{6(u{+}4)}{(v{+}4)(v{+}5)}$.

For $v=-6,u=-6$ we get an one-parametric family

[TABLE]

The case $v=-7,u=-8$ corresponds to another one line of solutions:

[TABLE]

4. Let $v=-4$ then the first equation implies that $u=-3$ or $u=-\frac{10}{3}$.

a) Let $v=-4,u=-3$, then $b_{1}=-\frac{6}{5},b_{2}=-\frac{3}{2},b_{3}=-2,b_{4}=-3$. The third equation gives $b_{5}=-7$ but the fourth equation becomes inconsistent with respect to $b_{6}$.

b) Let $v=-4,u=-\frac{10}{3}$, then third and fourth equations of (2) look as

[TABLE]

and this system is inconsistent with respect to $b_{5},b_{6}$.

5. $v=1$, the the second equation of (2) implies that $u=1$ or $u=\frac{4}{3}$.

a) The case $v=u=1$ corresponds to the family

[TABLE]

b) If $v=1,u=\frac{4}{3}$ the fourth equation of (2) degenerates to $-\frac{1}{900}=0$ and hence the system (2) is inconsistent.

6. Let $v=5$, the first equation of (2) gives us $b_{5}=3(u{+}3)$, the third equation after substitution $v=-5,b_{5}=3(u{+}3)$ will look like

[TABLE]

It has the roots $u=-\frac{13}{3},-4$.

a) The case $v=-5,u=-4$ corresponds to the family that we already obtained:

[TABLE]

b) If $v=-5,u=-\frac{13}{3}$ then $b_{1}=-\frac{16}{15},b_{2}=-\frac{13}{10},b_{3}=-\frac{5}{3},b_{4}=-\frac{7}{3},b_{5}=-4$ and the fourth equation is inconsistent with respect to $b_{6}$.

We studied the solutions of the main system (2) that correspond to the points of algebraic variety $M_{F}=\{F(x,y,z)=0\}$. Now we will consider the case

[TABLE]

Then one can rewrite the first three equations of (2) as

[TABLE]

Proposition 3.9**.**

There are no solutions of (11) with $y=\frac{9}{5},-\frac{7}{5}$.

Proof.

Direct calculations. ∎

Proposition 3.10**.**

Let $2x-y+1=0$ then the system (11) is consistent if and only if

[TABLE]

Proof.

Indeed $2x-y+1=0$ implies $xy+\frac{3}{5}x-\frac{6}{5}=0$ that is equivalent to $2x^{2}+\frac{8}{5}x-\frac{6}{5}=0$. The roots of this square equation will give us the values given above. Now one can express $b_{1}=12\pm 3\sqrt{19}$, take $b_{5}=t$, then express $b_{6}$ in terms of $b_{5}$ taking into account third equation. Finally we obtain

[TABLE]

That corresponds to the family of solutions that we have already obtained.

∎

Now it is easy to see that in generic situation ($2x-y+1\neq 0$) we have

[TABLE]

The last statement concludes the proof of the theorem. ∎

Remark*.*

We shifted the arguments $u,v$ in our answer for $M_{1}$, for instance we considered $b_{2}=\frac{6u}{v(v{+}1)}$ in our proof and now $b_{2}=\frac{6(u{+}2)}{(v{+}2)(v{+}3)}$. One has to point out some properties of the subsets $M_{i}$:

• •

$M_{3}$ does not intersect other subsets $M_{i}$;

• •

$P(u,v){\in}M_{1}$ belongs to $M_{2}$ if and only if $u{=}\frac{1}{30}(v{+}3)(v{+}4)(v{+}5){+}\frac{1}{2}(v{-}3)$;

• •

$M_{1}^{0}$ intersects $M_{2}$ at $v=\frac{{-}7{\pm}\sqrt{61}}{2}$;

• •

$M_{4}^{\pm}$ intersects $M_{1}$ at $t{=}{-}\frac{16}{15}\pm\frac{68}{285}\sqrt{19}$ ($P_{4}^{\pm}{=}P({-}\frac{9}{2}{\pm}\frac{3}{2}\sqrt{19},-2{\pm}\sqrt{19})$ and does not intersect other subsets $M_{i}$;

• •

$M_{5}^{\pm}$ intersects $M_{1}$ at $t=\pm 4$ (the points $P_{5}^{+}{=}P(3,1)$ and $P_{5}^{-}{=}P({-}10,{-}9)$ respectively);

• •

$M_{5}^{\pm}$ intersects $M_{2}$ at $t=0$;

• •

$M_{6}^{\pm}$ intersects $M_{1}^{0}$ at $t{=}{\pm}6$ and $v=0$ and does not intersect other subsets;

One can also remark that there exists an involution $\sigma:M\to M$

[TABLE]

with the properties:

• •

$M_{1}$ is invariant with respect to $\sigma$: $\sigma(P(u,v))=P({-}u{-}7,{-}v{-}8)$;

• •

$\sigma(M_{1}^{0})=M_{1}^{0}$;

• •

$\sigma(M_{2}{\cup}M_{4}^{\pm})=M_{2}{\cup}M_{4}^{\pm}$;

• •

$\sigma(M_{3})=M_{3}$;

• •

$\sigma(M_{5}^{-})=M_{5}^{+}$;

• •

$\sigma(M_{6}^{-})=M_{6}^{+}$;

Corollary 3.11**.**

The affine variety of $9$-dimensional graded thread $W^{+}$-modules

[TABLE]

can be parametrized by means of two- and one-parametric algebraic subsets

• $\tilde{M}_{1}:$

$b_{i}=6\frac{(u+i)}{(v+i)(v+i+1)},\;i=1,2,\dots,6,7,$*  $u\neq v,u\neq v{+}1,\;v\neq{-}1,{-}2,\dots,{-}7,{-}8$;*

• $\tilde{M}_{1}^{0}:$

$b_{i}=\frac{6}{v{+}i},\;i=1,2,\dots,6,7,$*  $v\neq{-}1,{-}2,\dots,{-}6,{-}7$;*

• $\tilde{M}_{2}:$

\begin{array}[]{c}b_{1}{=}\frac{(y{-}\frac{3}{5})(y{+}\frac{3}{5})(y{-}2)}{(y-\frac{9}{5})(y-\frac{7}{5})},b_{2}{=}\frac{(y{-}\frac{8}{5})(y{+}\frac{3}{5})}{(y-\frac{9}{5})},b_{3}{=}y{+}\frac{2}{5},b_{4}{=}y,b_{5}{=}y{-}\frac{2}{5},\\ b_{6}{=}\frac{(y{+}\frac{8}{5})(y{-}\frac{3}{5})}{(y{+}\frac{9}{5})},b_{7}{=}\frac{(y{-}\frac{3}{5})(y{+}\frac{3}{5})(y{+}2)}{(y{+}\frac{9}{5})(y{+}\frac{7}{5})}\end{array};

• $\tilde{M}_{3}:$

$\;b_{1}=b_{2}=b_{3}=b_{4}=b_{5}=b_{6}=b_{7}=t$;

• $\tilde{M}_{5}^{-}:$

$b_{1}=-\frac{5}{6},b_{2}=-\frac{27}{28},b_{3}=-\frac{8}{7},b_{4}=-\frac{7}{5},b_{5}=-\frac{9}{5},b_{6}=-\frac{5}{2},b_{7}=t$;

• $\tilde{M}_{5}^{+}:$

$b_{1}=t,b_{2}=\frac{5}{2},b_{3}=\frac{9}{5},b_{4}=\frac{7}{5},b_{5}=\frac{8}{7},b_{6}=\frac{27}{28},b_{7}=\frac{5}{6}$;

• $\tilde{M}_{6}^{+}:$

$b_{1}=t,b_{2}=\frac{6}{2},b_{3}=\frac{6}{3},b_{4}=\frac{6}{4},b_{5}=\frac{6}{5},b_{6}=\frac{6}{6},b_{7}=\frac{6}{7}$;

• $\tilde{M}_{6}^{-}:$

$b_{1}=-\frac{6}{7},b_{2}=-\frac{6}{6},b_{3}=-\frac{6}{5},b_{4}=-\frac{6}{4},b_{5}=-\frac{6}{3},b_{6}=-\frac{6}{2},b_{7}=-t$.

Proof.

The families $M_{4}^{\pm}$ do not survive to the dimension $9$, $\tilde{M}_{2}$ became an one-parametric family instead of two-parametric $M_{2}$. $\tilde{M}_{2}$ intersects $\tilde{M}_{1}$ at $y={\pm}\frac{2}{5}\sqrt{21}$.

∎

We already classified $9$-dimensional graded thread $W^{+}$-modules. Let $V$ be a $10$-dimensional $W^{+}$-module with the basis $f_{1},\dots,f_{10}$. Consider its quotient module $\hat{V}=V/\langle f_{10}\rangle$ and its submodule $\tilde{V}=\langle f_{2},\dots,f_{10}\rangle$. Both of them are $9$-dimensional graded thread modules of the type $(1,1,\dots,1)$ with the defining sets $b_{1},\dots,b_{7}$ and $b_{2},\dots,b_{8}$ respectively.

Only two points $y={\pm}\frac{2}{5}\sqrt{21}$ from $\tilde{M}_{2}$ survive to dimension $10$. These points are in the intersection $\tilde{M}_{1}\cap\tilde{M}_{2}$.

Let $(b_{1},b_{2},\dots,b_{n-2},b_{n-1})$ be coordinates (defining set) of a $(n+1)$-dimensional graded thread $W^{+}$-module $V=\langle f_{1},f_{2},\dots,f_{n},f_{n+1}\rangle$ of the type $(1,1,\dots,1)$ then its subsets $(b_{1},b_{2},\dots,b_{n-2})$ and $(b_{2},\dots,b_{n-2},b_{n-1})$ are coordinates of $n$-dimensional quotient $V/\langle f_{n+1}\rangle$ and submodule $\langle f_{2},\dots,f_{n},f_{n+1}\rangle$ respectively. Both of them are $n$-dimensional graded thread $W^{+}$-modules of the type $(1,1,\dots,1)$ and we can apply the induction hypothesis.

For instance let $(b_{2},b_{3},\dots,b_{n-2},b_{n-1})=(t,3,\dots,\frac{6}{n-2},\frac{6}{n-1})$ then $(b_{1},t,3,\dots,\frac{6}{n-2})$ have also to be present in the Table below and it is impossible. On the another hand if $(b_{1},b_{2},\dots,b_{n-2})=(t,3,\dots,\frac{6}{n-2})$ then we have to find the set $(b_{2},\dots,b_{n-1})=(\frac{6}{2},\dots,\frac{6}{n-2},b_{n-1})$ in the Table. It can occure if and only if $b_{n-1}=\frac{6}{n-1}$.

∎

4. Graded thread modules of the type $(1,\dots,1,0,1,\dots,1)$.

Now we consider the case, when one and only one of the defining constants $\alpha_{i}$ vanishes.

[TABLE]

A graded module of this type is called ”décousu” in [3].

Theorem 4.1**.**

Let $V$ be a $(n{+}1)$-dimensional, $n\geq 16$, indecomposable graded thread $W^{+}$-module of the type $(1,\dots,1,0,1,\dots,1)$, i.e. there exists a basis $f_{i},i=1,\dots,n{+}1,$ of $V$ and $\exists!k,1\leq k\leq n$, such that

[TABLE]

Then $V$ is isomorphic to one and only one module from the following list

A) $(n+1)$-dimensional quotients of $V_{\lambda,2\lambda-k},\lambda\neq 0,{-}1$,

• •

$V_{\lambda,2\lambda-k}(n+1),\lambda\neq 0,{-}1,\;1\leq k\leq n;$**

B) linear deformations of $(n+1)$-dimensional quotients $V_{{-}1,{-}k{-}2}$ and $V_{0,-k}$

• •

$V_{{-}1,{-}k{-}2}^{t}(n+1),\;1\leq k<n-1,\;t\in{\mathbb{K}};$* defined by*

[TABLE]

• •

$V_{0,-k}^{t}(n+1),2<k\leq n,\;t\in{\mathbb{K}};$* it is dual to the $W^{+}$-module $V_{{-}1,{-}k{-}2}^{-t}(n+1)$*

[TABLE]

C) degenerate cases for $k=1,2,n{-}1,n$.

• •

$V_{0,{-}2}(n+1),\;k=2;$**

• •

$V_{{-}1,{-}4}(n+1)=V_{0,{-}2}^{*}(n+1),\;k=n{-}1;$**

• •

$\tilde{V}_{0,{-}1}(n+1),\;k=1,$* defined by*

[TABLE]

• •

$\tilde{V}_{{-}1,{-}2{-}n}(n+1)=\tilde{V}_{0,{-}1}^{*}(n+1),\;k=n;$**

• •

$\tilde{V}_{{-}2,{-}3}(n+1),\;k=1,$* defined by*

[TABLE]

• •

$\tilde{V}_{1,{-}n}(n+1)=\tilde{V}_{{-}2,{-}3}^{*}(n+1),\;k=n.$**

Remark*.*

1. There is another relation of duality

[TABLE]

2. $\tilde{V}_{0,{-}1}(n+1)=\tilde{V}_{{-}1,{-}2}(n+1)$, where $\tilde{V}_{{-}1,{-}2}(n+1)$ is defined by

[TABLE]

Undeformed modules are non-isomorphic $V_{0,{-}1}(n+1)\neq V_{{-}1,{-}2}(n+1)$. The module $V_{0,{-}1}(n+1)$ is decomposable $V_{0,{-}1}(n+1)=\langle f_{1}\rangle\oplus\langle f_{2},\dots,f_{n{+}1}\rangle$ and $V_{{-}1,{-}2}(n+1)$ is not. However their $n$-dimensional submodules $\langle f_{2},\dots,f_{n{+}1}\rangle$ are isomorphic.

Proof.

The first example $V_{\lambda,2\lambda-k}(n+1)$ in the list of graded thread $W^{+}$-modules from the Theorem is absolutely ovbious, we have for all $j,1\leq j\leq n$

[TABLE]

What other graded thread $W^{+}$-modules exist of the type $(1,\dots,1,0,1,\dots,1)$?

Lemma 4.2**.**

*Let $V$ be a $(n+1)$-dimensional graded thread $W^{+}$-module of the type $(1,\dots,1,0,1,\dots,1)$ with $\alpha_{k}=0,1\leq k\leq n$, i.e. $e_{1}f_{k}=0$.

The corresponding graded $W^{+}$-module $V$ is decomposable in a direct sum of two graded $W^{+}$-modules:*

[TABLE]

if and only if $\beta_{k}=\beta_{k{-}1}=0$, i.e. $e_{2}f_{k-1}=0,e_{2}f_{k}=0$. We denoted by $f_{1},\dots,f_{n{+}1}$ the graded basis of $V$.

Proof.

It is evident that the subspace $V_{2}$ is invariant. On the another hand $e_{1}f_{k}=e_{2}f_{k}=e_{2}f_{k{-}1}=0$. Hence the subspace $V_{1}$ is invariant with respect to $e_{1},e_{2}$ and therefore it is invariant with respect to the entire $W^{+}$-action.

∎

Now we have to rewrite the basic equations (1) for a module with $\alpha_{k}=0$. It is easy to see that one have to substitute $b_{i}$ by $\frac{\beta_{i}}{\alpha_{i}\alpha_{i{+}1}}$ in (1) and then multiply $R^{5}_{i}$ by the product $\alpha_{i}\alpha_{i+1}\alpha_{i+2}\alpha_{i+3}$ and $R^{7}_{i}$ by $\alpha_{i}\alpha_{i+1}\dots\alpha_{i+5}$ respectively. We suppose that $\alpha_{i}{=}1,\;i\neq k,\;\alpha_{k}{=}0$. One can meet $\alpha_{k}$ only in the denominators of $b_{k}$ and $b_{k{-}1}$. Hence the new equation that involves $b_{k}$ is obtained from the old one by a very simple procedure: we keep summands only of the form $b_{k}b_{j}$ or $b_{k-1}b_{l}$, for instance if $1<k<n-2$ we have

[TABLE]

It follows from Lemma 4.2 that for an indecomposable module $V$ with $\alpha_{k}=0$ the constants $b_{k},b_{k-1}$ can not vanish simultaneously.

Lemma 4.3**.**

Let $V$ be a $9$-dimensional graded thread $W^{+}$-module defined by its basis $f_{k},f_{k{+}1},\dots,f_{k{+}8},$ and the defining set of relations:

[TABLE]

Then $V$ is either decomposable as a direct sum of $W^{+}$-modules $\langle f_{k}\rangle\oplus\langle f_{k{+}1},\dots,f_{k{+}8}\rangle$ ($b_{k}=0$) or it is indecomposable ($b_{k}\neq 0$) and $V$ isomorphic to the one and only one graded thread $W^{+}$-module with the defining set $(b_{k},\dots,b_{k+6})$ from the table below

[TABLE]

Proof.

Consider two equations with $b_{k}$.

[TABLE]

Remark that if $b_{k}=0$ then for the first basis vector $f_{k}$ we have

[TABLE]

and hence the one-dimensional subspace $\langle f_{k}\rangle$ is invariant with respect to the entire $W^{+}$-action.

If $b_{k}\neq 0$ we have two linear equations (18) on $b_{k{+}2},b_{k{+}3},b_{k{+}4},b_{k{+}5}$.

Now we can apply the description of $8$-dimensional graded thread modules from the Theorem 3.3. We consider the submodule $\langle f_{k{+}1},\dots,f_{k{+}8}\rangle$ of $V$ as a $8$-dimensional graded thread module of the type $(1,1,\dots,1)$.

1. Coordinates $b_{k{+}i}=\frac{6(u{+}i)}{(v{+}i)(v{+}i{+}1)},i=1,\dots,6,$ of a point $P(u,v)\in M_{1}$ satisfy both equations $R^{5}_{k}$ and $R^{7}_{k}$ if and only if

• •

$u{=}4,v{=}2$, i.e. $b_{k{+}1}{=}\frac{5}{2},b_{k{+}2}{=}\frac{9}{5},b_{k{+}3}{=}\frac{7}{5},b_{k{+}4}{=}\frac{8}{7},b_{k{+}5}{=}\frac{27}{28},b_{k{+}6}{=}\frac{5}{6}$;

• •

$v{=}0$, i.e. $b_{k{+}i}{=}\frac{6(u{+}i)}{i(i{+}1)},\;i=1,\dots,6,\;u\neq 0,1$.

2. Coordinates $b_{k{+}i}{=}\frac{6}{(v{+}i)}$ of a point $P(v)\in M_{1}^{0}$ satisfy $R^{5}_{k}$ and $R^{7}_{k}$ also in two cases

• •

$v=0$, i.e. $b_{k{+}1}{=}6,b_{k{+}2}{=}3,b_{k{+}3}{=}2,b_{k{+}4}{=}\frac{3}{2},b_{k{+}5}{=}\frac{6}{5},b_{k{+}6}{=}1$;

• •

$v=1$, i.e. $b_{k{+}1}{=}3,b_{k{+}2}{=}2,b_{k{+}3}{=}\frac{3}{2},b_{k{+}4}{=}\frac{6}{5},b_{k{+}5}{=}1,b_{k{+}6}{=}\frac{6}{7}$.

Hence we have to remove the restriction $u\neq 0,1$ in the first line of our table.

3. There are only two points in $M_{2}$ satisfying $R^{5}_{k}$ and $R^{7}_{k}$:

• •

$x{=}\frac{5}{2},y{=}\frac{7}{4}$, i.e. $b_{k{+}1}{=}\frac{9}{2},b_{k{+}2}{=}\frac{5}{2},b_{k{+}3}{=}\frac{7}{4},b_{k{+}4}{=}\frac{27}{20},b_{k{+}5}{=}\frac{11}{20},b_{k{+}6}{=}\frac{13}{14}$;

• •

$x{=}\frac{9}{5},y{=}\frac{7}{5}$, i.e. $b_{k{+}1}{=}1,b_{k{+}2}{=}\frac{9}{5},b_{k{+}3}{=}\frac{7}{5},b_{k{+}4}{=}1,b_{k{+}5}{=}\frac{3}{4},b_{k{+}6}{=}\frac{17}{28}$.

But the point with parameters $x{=}\frac{5}{2},y{=}\frac{7}{4}$ coincides with the point $P(\frac{1}{2},0)\in M_{1}$ with $u{=}\frac{1}{2},v{=}0$ that we have already considered above.

4. Coordinates $(b_{k{+}1},b_{k{+}2},\dots,b_{k{+}6})$ of a point $P(t)\in M_{3}$ satisfies $R^{5}_{k}$ only if $t=1$. However $P(1)$ does not satisfy $R^{7}_{k}$.

5. It is easy to verify directly that there is no point in the subsets $M_{4}^{\pm}$, $M_{5}^{\pm}$,$M_{6}^{-}$ that satisfies the equation $R^{5}_{k}$.

6. All points from $M_{6}^{+}$ satisfy both equations (18). However it does not hold for points from $M_{6}^{-}$.

∎

Corollary 4.4**.**

Let $V$ be a idecomposable $(n+1)$-dimensional graded thread $W^{+}$-module defined by its basis $f_{1},f_{2},\dots,f_{n{+}1},n+1\geq 10$ and the defining set of relations:

[TABLE]

then $V$ is isomorphic to the one and only one $W^{+}$-module from the table below

[TABLE]

Proof.

We rescaled the first vector $f_{1}$ in order to fix the value of $b_{1}$ (we recall that $b_{1}$ is not determined by equations $R_{1}^{5}$ and $R^{7}_{1}$). It is possible because $e_{1}f_{1}=0$.

We can conclude that if $b_{k}\neq 0$ then $b_{k{+}1},\dots,b_{k{+}6}$ have the values prescribed by Lemma 18. It follows from Theorem 3.2 that $b_{k{+}i},1\leq i\leq s$ if $7\leq s\leq n-9$ is determined uniquely for all three subcases of (17)

[TABLE]

However for the subcase

[TABLE]

the defining set can not be extended to a system $\{b_{k+1},\dots,b_{k{+}7}\}$. It follows from the fact that 4) corresponds to the point $\left(1,\frac{9}{5},\frac{7}{5},1,\frac{3}{4},\frac{17}{28}\right)$ of the subset $M_{2}$ defined in Theorem 3.2 by parameters $x{=}y{+}\frac{2}{5}{=}\frac{9}{5},y{=}\frac{7}{5}$. This set can not be extended to $\left(1,\frac{9}{5},\frac{7}{5},1,\frac{3}{4},\frac{17}{28},b_{k+7}\right)$. One can also very it directly considering the equations $R^{5}_{k{+}7},R^{7}_{k{+}7}$.

1. For convenience of notations we denote by $V_{{-}1,{-}3}^{t}$ a linear deformation of $V_{{-}1,{-}3}$ ($t$ is a parameter). It is defined by the formulas

[TABLE]

2. The $W^{+}$-module $\tilde{V}_{0,{-}1}$ is also deformed $W^{+}$-module $V_{0,{-}1}$. It is defined by

[TABLE]

3. The $W^{+}$-module $\tilde{V}_{{-}2,{-}3}$ is defined by

[TABLE]

We also substituted for convinience $u=-\lambda$.

∎

Lemma 4.5**.**

Let $V$ be a $k+8$-dimensional graded thread $W^{+}$-module defined by its basis $f_{1},\dots,f_{k},f_{k{+}1},\dots,f_{k{+}8},2\leq k\leq 8,$ and the defining set of relations:

[TABLE]

Then $V$ is either decomposable as a direct sum of $W^{+}$-modules $\langle f_{1},\dots,f_{k}\rangle\oplus\langle f_{k{+}1},\dots,f_{k{+}8}\rangle$ ($b_{k}=0$) or it is indecomposable ($b_{k}\neq 0$) and $V$ isomorphic to the one and only one graded thread $W^{+}$-module with the defining set $(b_{1},\dots,b_{k+6})$ from the table below

[TABLE]

Proof.

1. Consider equations $R^{5}_{k{-}1}$ and $R^{7}_{k{-}1}$

[TABLE]

In the subcase 1) from the Table (17) both equations are equivalent to

[TABLE]

Hence one can take $b_{k}=6u\gamma,\;b_{k{-}1}=-6(u{-}1)\gamma,\;u\neq 0,1,\gamma\neq 0$. If $u=0$ then $b_{k}=0$, if $u=1$ then $b_{k-1}=0$. After (if necessary) rescaling of $f_{k}$ we may assume that $\gamma=1$.

In the subcase 2) we have $b_{k}\neq 0,b_{k{+}1}{=}t,b_{k{+}2}{=}\frac{6}{2},b_{k{+}3}{=}\frac{6}{3},b_{k{+}4}{=}\frac{6}{4}$. Then $R^{5}_{k{-}1}$ and $R^{7}_{k{-}1}$ are

[TABLE]

That implies $b_{k{-}1}=0$. We set $b_{k}=6$.

For the subcase 3) in (17) the system $R^{5}_{k{-}1}$ and $R^{7}_{k{-}1}$ look

[TABLE]

and it is inconsistent. Hence the case 3) is not extendable to the left.

The subcase 4) also leeds to a inconsistent system on unknowns $b_{k-1}$ and $b_{k}$ and also is not extendable.

Now we have to study the case $b_{k}{=}0$. As $b_{k{-}1}\neq 0$ we remark that (20) is equivalent to the following linear system

[TABLE]

One can remark that the equations (21) can be obtained from (18) just by shifting the index $k\to k-1$. The mimic of the proof of the Lemma 4.3 will give us the following answer ($b_{k{-}1}$ can take arbitrary values but after rescaling (if necessary) we may assume that $b_{k{-}1}{=}{-}6$).

[TABLE]

We summarize our results by means of the following table:

[TABLE]

2. Suppose now that $k\geq 3$. Hence we may consider equations $R^{5}_{k{-}2}$ and $R^{7}_{k{-}2}$

[TABLE]

For the module 1) in (22) the equations (23) are equivalent to one equation

[TABLE]

Hence we take $b_{k{-}2}=3(u{-}2)=\frac{6(u{-}2)}{1\cdot 2}$ if $u\neq 0$.

For the module 2) in (22) our equations imply $b_{k{-}2}={-}3$.

In the subcase 3) taking into account $b_{k}{=}0$ we conclude that all values of $b_{k{-}2}$ are valid and we set $b_{k{-}2}=t$.

3. Suppose now that $k\geq 4$. The equations on $b_{k{-}3}$ are the following ones

[TABLE]

For the first module 1) they are equivalent to

[TABLE]

Hence $b_{k{-}3}=u{-}3$. For the second module 2) both equations are equivalent to $b_{k{-}3}={-}2$. For the module 3) we have $b_{k{-}3}={-}3$.

4. Let $k\geq 5$. The equations on $b_{k{-}4}$ are

[TABLE]

they will give us for 1), 2), 3) respectively

[TABLE]

5. The remaining cases ($k\geq 6,7$) are treated similarly to the previous ones.

∎

Now considering a general case of $(n+1)$-dimensional graded thread $W^{+}$-module $V$, $n+1\geq 16$, with $e_{1}f_{k}=0$ we may assume that $k+8\leq n$. If $k>n-8\geq 7$ we take the dual $W^{+}$-module $V^{*}$ instead of $V$. We have for $V^{*}$

[TABLE]

It means that $\alpha^{*}_{n+1-k}=-\alpha_{k}=0$ and $(n+1-k)+7\leq 15\leq n$.

Assuming $k+8\leq n$ we can define a $9$-dimensional subquotient $\tilde{V}$ of $V$

[TABLE]

and apply Lemmas 4.3, 4.5 and Corollary 4.4.

Every graded thread $W^{+}$-module $\tilde{V}$ (the set $(b_{k},\dots,b_{k+7})$) presented in the classification lists of Lemmas 4.3, 4.5 and Corollary 4.4 can be uniquely extended to the graded thread $W^{+}$-module $V$ ($(b_{1},\dots,b_{k},\dots,b_{k+8},\dots,b_{n-1})$. We remove the restriction $k+8\leq n$ considering their dual modules.

The results of this classification are presented in the Table 2.

∎

5. Graded thread modules of the type $(1,\dots,1,0,0,1,\dots,1)$.

Now we consider modules with vanishing two consecutive $\alpha_{i},\alpha_{i+1}$, i.e.

[TABLE]

Modules of this type are called ”reprisé” in [3]

In [12] an infinite-dimensional graded thread $W^{+}$-module $\tilde{V}_{gr}$ was constructed, it was defined by its basis $\{f_{j},j\in{\mathbb{Z}}\}$ and the relations

[TABLE]

It holds for this module $e_{1}f_{-1}=e_{1}f_{0}=0.$ This module and its finite-dimensional subquotients played the crucial role in the proof of Buchstaber’s conjecture on Massey products in Lie algebra cohomology $H^{*}(W^{+},{\mathbb{K}})$ [12]. It has interesting nature, it is not a module of $V_{\lambda,\mu}$ family or its degeneration or deformation, in some sense it is the result of gluing together of two modules: the quotient of $V_{{-}1,1}$ with a submodule of $V_{0,0}$ and it is unique infinite-dimensional module with the property $\exists!\;k,\alpha_{k}=\alpha_{k{+}1}=0$ [12] .

Theorem 5.1**.**

Let $V$ be a $(n{+}1)$-dimensional, $n{+}1\geq 11$, indecomposable graded thread $W^{+}$-module of the type $(1,\dots,1,0,0,1,\dots,1)$, i.e. there exists a basis $f_{1},\dots,f_{n+1}$ of $V$ and $\exists!k,1\leq k\leq n{-}1$ such that:

[TABLE]

then if $k\neq 1,n{-}1,$ the module $V$ is isomorphic to one and only one module from the list

• •

$R_{k},1\leq k\leq n-1,$* defined by its basis $f_{1},\dots,f_{n+1}$ and relations*

[TABLE]

• •

its dual module $R_{k}^{*},1\leq k\leq n-1$.

Proof.

The equations that involves $b_{k},b_{k{+}1},b_{k{+}2}$ will be obtained from the standard ones by a very simple procedure: we will keep the summands only of the form $b_{k}b_{j}$, $b_{k+1}b_{l}$, $b_{k+2}b_{l}$.

1. Let consider the case $k=1$ (and hence the dual module with $k=n{-}2$). The first four equations $R^{5}_{1},R^{5}_{2},R^{5}_{3},R^{7}_{1}$ are:

[TABLE]

If $b_{1}=0$ then there is a decomposition $V=\langle f_{1}\rangle\oplus\langle f_{2},\dots,f_{n}\rangle$ in the sum of two submodules, if $b_{2}=0$ then $V=\langle f_{2}\rangle\oplus\langle f_{1},f_{3},\dots,f_{n}\rangle$ is also the sum of its submodules. Hence we may assume that $b_{1}=b_{2}=1$.

The system (29) with $b_{1}=b_{2}=1$ has two solutions

[TABLE]

It follows from the Proposition 4.3 that the module $\langle f_{2},f_{3},\dots,f_{7}\rangle$ corresponding to the second solution can not be extended to $\langle f_{2},f_{3},\dots,f_{7},\dots,f_{n}\rangle$ with $n\geq 10$. On the another hand one can check out that we have the only one module with

[TABLE]

that corresponds to the first solution.

2. Let us suppose now that $2\leq k\leq n{-}6$. Then the equations $R^{5}_{k{-}1},R^{5}_{k},R^{5}_{k{+}1},R^{7}_{k{-}1}$ will have the following form:

[TABLE]

Proposition 5.2**.**

Let $b_{k}=0$ or $b_{k{-}1}=b_{k{+}1}=0$ then the module $V$ is decomposable.

Proof.

Indeed it follows from the first equation of the system above that if $b_{k}=0$ then $b_{k{-}1}b_{k{+}1}=0$. In the case of $b_{k{-}1}=b_{k}=0$ we have the invariant decomposition:

[TABLE]

If $b_{k}=b_{k{+}1}=0$ then $V$ is decomposed in another way:

[TABLE]

If both $b_{k{-}1}=b_{k{+}1}=0$ then $V$ is also decomposable:

[TABLE]

∎

Now after some rescaling of the basic vectors we have to consider two possibilities:

$b_{k{-}1}=b_{k}=1$;

• 2)

$b_{k{-}1}=0,\;b_{k}=b_{k{+}1}=1$.

In the first case the system (30) has the only one solution

[TABLE]

In the second case we also have the unique solution $b_{k{+}2}{=}3,b_{k{+}3}{=}2,b_{k{+}4}{=}\frac{3}{2}$. The components $b_{i},i>k{+}4$ are also determined uniquely as it follows from the Proposition 4.3. If we want to find $b_{k{-}2}$ we have to suppose that $k\geq 3$ and to consider two new equations:

[TABLE]

Evidently in both cases ($b_{k{+}1}=0$ or $b_{k{+}1}=1$) we have the same answer $b_{k{-}2}={-}3$.

Now supposing that $k\geq 4$ we have two new additional equations:

[TABLE]

Again it follows that $b_{k{-}3}={-}2$ in both situations.

Let $k\geq 5$. We have two equations on $b_{k{-}4}$:

[TABLE]

In both cases we have $b_{k{-}4}={-}\frac{3}{2}$.

The last case $k\geq 6$ can be treated absolutely similarly. In fact we have obtained two graded modules:

[TABLE]

The cases $n{-}5\leq k\leq n{-}3$ follow from the previous considerations: one have to take the corresponding dual module instead of initial one.

∎