Super homologies associated with low dimensional Lie algebras

Kentaro Mikami Akita University Tadayoshi Mizutani Professor Emeritus, Saitama University

1 Introduction

A Poisson structure $\pi$ on a manifold $M$ is a 2-vector field characterized as $[\pi,\pi]_{\text{\footnotesize S}}=0$ , where $[\cdot,\cdot]_{\text{\footnotesize S}}$ is the Schouten bracket. The graded algebra $\sum\Lambda^{p}\mathrm{T}(M)$ with the Schouten bracket is a prototype of Lie superalgebra (cf. [2]). The Poisson condition $[\pi,\pi]_{\text{\footnotesize S}}=0$ means that a 2-chain $\pi\wedge\pi$ is a cycle. Thus, studying the second super homology group of Lie superalgebra of tangent bundle of $M$ is an activity of Poisson geometry (cf. [3]). Given a $\mathbb{Z}$ -graded Lie superalgebra, the 0-graded subspace is a Lie algebra. In this note, using the DGA $\sum\Lambda^{p}\mathrm{T}(M)$ with the Schouten bracket as a model, we start from an abstract Lie algebra, construct non-trivial Lie superalgebra by Schouten-like bracket. Then it may be natural to ask how the core Lie algebra control the Lie superalgebra. One trial here is to investigate the Betti numbers of the super homology groups. For abelian Lie algebras, the boundary operator is trivial, and the Betti number is equal to the dimension of chain space for a given weight. So, we study their super homology groups for low dimensional Lie algebras of dimension smaller than 4.

For non-abelian Lie algebras of dimension 2, we have

$\begin{array}[]{c | *{5}{c}}w>0&C_{w}^{[w]}&C_{w+1}^{[w]}&C_{w+2}^{[w]}\\ \hline\cr\text{SpaceDim}&1&2&1\\ \hline\cr\text{KerDim}&1&1&0\\ \hline\cr\text{Betti}&0&0&0\\ \hline\cr\end{array}$

We classify non-abelian Lie algebras $\mathfrak{g}$ of dimension 3 into four cases:

(1) $\textrm{dim}[\mathfrak{g},\mathfrak{g}]=1$ and $\textrm{dim}[\mathfrak{g},\mathfrak{g}]\subset Z(\mathfrak{g})$ , (2) $\textrm{dim}[\mathfrak{g},\mathfrak{g}]=1$ and $\textrm{dim}[\mathfrak{g},\mathfrak{g}]\not\subset Z(\mathfrak{g})$ ,

(3) $\textrm{dim}[\mathfrak{g},\mathfrak{g}]=2$ , (4) $\textrm{dim}[\mathfrak{g},\mathfrak{g}]=3$ .

Even though the case (4) has two kinds of Lie algebras when the base field is $\mathbb{R}$ , we made the same treatment and got the same results.

When $w=0$ the homology groups are the same with usual Lie algebra homology groups, so we show the Betti numbers for super homology groups when $w>0$ .

$\begin{array}[]{c|*{5}{c}}\text{weight}=w>0&w-1&w&w+1&w+2&w+3\\ \hline\cr\text{SpaceDim}&\binom{w}{2}&3\binom{w}{2}+\binom{w+2}{2}&3\binom{w}{2}+3\binom{w+2}{2}&\binom{w}{2}+3\binom{w+2}{2}&\binom{w+2}{2}\\ \hline\cr\hline\cr\text{(1)'s Betti}&0&w&3w+1&3w+2&w+1\\ \hline\cr\text{(2)'s Betti}&0&1&2&1&0\\ \hline\cr\text{(3)'s Betti}&0&0&\kappa&2\kappa&\kappa\\ \hline\cr\text{(4)'s Betti}&0&0&0&0&0\end{array}$

where $\kappa$ in (3) is defined by $\kappa=\begin{cases}1&\text{if\quad}\alpha=-1\\ 0&\text{if\quad}\alpha\neq-1\end{cases}$ , and $\alpha$ is a non-zero parameter of the Lie bracket relations $[z_{1},z_{2}]=0$ , $[z_{1},z_{3}]=z_{1}$ , $[z_{2},z_{3}]=\alpha z_{2}$ .

2 Preliminaries, Notations and Basic Facts

Definition 2.1 (Lie superalgebra).

Suppose a real vector space $\mathfrak{g}$ is graded by $\displaystyle\mathbb{Z}$ as $\displaystyle\mathfrak{g}=\sum_{j\in\mathbb{Z}}\mathfrak{g}_{j}$ and has a bilinear operation $[\cdot,\cdot]$ satisfying

(2.1)		$\displaystyle[\mathfrak{g}_{i},\mathfrak{g}_{j}]\subset\mathfrak{g}_{i+j}$
(2.2)		$\displaystyle[X,Y]+(-1)^{xy}[Y,X]=0\quad\text{ where }X\in\mathfrak{g}_{x}\text{ and }Y\in\mathfrak{g}_{y}$
(2.3)		$\displaystyle(-1)^{xz}[[X,Y],Z]+(-1)^{yx}[[Y,Z],X]+(-1)^{zy}[[Z,X],Y]=0\;.$

Then we call $\mathfrak{g}$ a $\mathbb{Z}$ -graded (or pre) Lie superalgebra.

Remark 2.1.

Super Jacobi identity (2.3) above is equivalent to the one of the following.

(2.4)		$\displaystyle[[X,Y],Z]$	$\displaystyle=[X,[Y,Z]]+(-1)^{yz}[[X,Z],Y]$
(2.5)		$\displaystyle[X,[Y,Z]]$	$\displaystyle=[[X,Y],Z]+(-1)^{xy}[Y,[X,Z]]$

As in a usual Lie algebra homology theory, $m$ -th chain space is the exterior product $\displaystyle\Lambda^{m}\mathfrak{g}$ of $\mathfrak{g}$ and the boundary operator essentially comes from the operation $\displaystyle X\wedge Y\mapsto[X,Y]$ , in the case of pre Lie superalgebras, ”exterior algebra” is defined as the quotient of the tensor algebra $\displaystyle\otimes^{m}\mathfrak{g}$ of $\mathfrak{g}$ by the two-sided ideal generated by

(2.6)

\displaystyle X\otimes Y+(-1)^{xy}Y\otimes X\quad\text{where }\quad X\in\mathfrak{g}_{x},Y\in\mathfrak{g}_{y}\;,

and we denote the equivalence class of $\displaystyle X\otimes Y$ by $\displaystyle X\bigtriangleup Y$ . Since $\displaystyle X_{\text{odd}}\bigtriangleup Y_{\text{odd}}=Y_{\text{odd}}\bigtriangleup X_{\text{odd}}$ and $\displaystyle X_{\text{even}}\bigtriangleup Y_{\text{any}}=-Y_{\text{any}}\bigtriangleup X_{\text{even}}$ hold, $\bigtriangleup^{m}\mathfrak{g}_{k}$ has a symmetric property for odd $k$ and has a skew-symmetric property for even $k$ with respect to $\bigtriangleup$ .

Suppose we have an exterior product $\displaystyle Y_{1}\bigtriangleup\cdots\bigtriangleup Y_{m}$ of $\displaystyle Y_{1},\dots,Y_{m}$ . Omitting $i$ -th element, we have $\displaystyle Y_{1}\bigtriangleup\cdots\bigtriangleup Y_{i-1}\bigtriangleup Y_{i+1}\bigtriangleup\cdots\bigtriangleup Y_{m}$ , which is often denoted as $\displaystyle Y_{1}\bigtriangleup\cdots\widehat{Y_{i}}\cdots\bigtriangleup Y_{m}$ .

The boundary operator $\displaystyle\partial:\text{C}_{m}\to\text{C}_{m-1}$ is defined by

(2.7)

\displaystyle\partial(Y_{1}\bigtriangleup\cdots\bigtriangleup Y_{m})

\displaystyle=\sum_{i<j}(-1)^{i-1+y_{i}\mathop{\sum}_{i<s<j}y_{s}}Y_{1}\bigtriangleup\cdots\widehat{Y_{i}}\cdots\bigtriangleup\overbrace{[Y_{i},Y_{j}]}^{j}\bigtriangleup\cdots\bigtriangleup Y_{m}

where $\displaystyle y_{i}$ is the homogeneous degree of $Y_{i}$ , i.e., $\displaystyle Y_{i}\in\mathfrak{g}_{y_{i}}$ . $\displaystyle\partial\circ\partial=0$ holds and we have the homology groups

\displaystyle\textrm{H}_{m}(\mathfrak{g},\mathbb{R})=\ker(\partial:\text{C}_{m}\rightarrow\text{C}_{m-1})/\partial(\text{C}_{m+1})\;.

	If all $Y_{i}$ are even (i.e., $y_{i}$ are even) in (2.7), then
(2.8)		$\displaystyle\partial(Y_{1}\bigtriangleup\cdots\bigtriangleup Y_{m})$	$\displaystyle=-\sum_{i<j}(-1)^{i+j}[Y_{i},Y_{j}]\bigtriangleup Y_{1}\bigtriangleup\cdots\widehat{Y_{i}}\cdots\widehat{Y_{j}}\cdots\bigtriangleup Y_{m}\;.$
	If all $Y_{i}$ are odd (i.e., $y_{i}$ are odd) in (2.7), then
(2.9)		$\displaystyle\partial(Y_{1}\bigtriangleup\cdots\bigtriangleup Y_{m})$	$\displaystyle=\sum_{i<j}[Y_{i},Y_{j}]\bigtriangleup Y_{1}\bigtriangleup\cdots\widehat{Y_{i}}\cdots\widehat{Y_{j}}\cdots\bigtriangleup Y_{m}\;.$

Define a binary operation $[A,B]_{res}$ by
(2.11)		$\displaystyle[A,B]_{res}$	$\displaystyle=\partial(A\bigtriangleup B)-(\partial A)\bigtriangleup B-(-1)^{\bar{a}}A\bigtriangleup\partial B$
	If all $A_{i}$ are even and all $B_{j}$ are odd, then
(2.12)		$\displaystyle[A,B]_{res}$	$\displaystyle=\sum_{i,j}(-1)^{i+1}A_{1}\bigtriangleup\cdots\widehat{A_{i}}\cdots A_{\bar{a}}\bigtriangleup[A_{i},B_{j}]_{\text{\footnotesize S}}\bigtriangleup B_{1}\bigtriangleup\cdots\widehat{B_{j}}\cdots\bigtriangleup B_{\bar{b}}\;.$

3 Superalgebras associated with Lie algebras

Definition 1 (Schouten-like bracket).

Let $\mathfrak{g}$ be a Lie algebra. Then we have a $\mathbb{Z}$ -graded Lie superalgebra $\overline{\mathfrak{g}}$ given by

\overline{\mathfrak{g}}=\sum_{i}\Lambda^{i}\mathfrak{g}\;,\;\text{ where, the grade of the element in }\Lambda^{i}\;\text{ is }\;i-1\;,\;

and the bracket of $a={a}_{1}\wedge\cdots\wedge{a}_{\bar{a}}\in\Lambda^{\bar{a}}\mathfrak{g}$ and $b={b}_{1}\wedge\cdots\wedge{b}_{\bar{b}}\in\Lambda^{\bar{a}}\mathfrak{g}$ is given by

(3.1)	$\displaystyle[a,b]_{\text{\footnotesize S}}$	$\displaystyle=\sum_{i,j}(-1)^{i+j}[a_{i},b_{j}]\wedge a[i]\wedge b[j]$
(3.2)	$\displaystyle a[i]$	$\displaystyle={a}_{1}\wedge\cdots\wedge{a}_{i-1}\wedge{a}_{i+1}\wedge\cdots\wedge{a}_{\bar{a}}\in\Lambda^{\bar{a}-1}\mathfrak{g}\;\text{ for }i=1,\ldots,\bar{a}\;$
(3.3)	$\displaystyle b[j]$	$\displaystyle={b}_{1}\wedge\cdots\wedge{b}_{j-1}\wedge{b}_{j+1}\wedge\cdots\wedge{b}_{\bar{b}}\in\Lambda^{\bar{b}-1}\mathfrak{g}\;\text{for}j=1,\ldots,\bar{b}\;.$

Proposition 3.1.

(3.4)

[b,a]_{\text{\footnotesize S}}=-(-1)^{(\bar{b}-1)(\bar{a}-1)}[a,b]_{\text{\footnotesize S}}

(3.5)		$\displaystyle[a,b\wedge c]_{\text{\footnotesize S}}=$	$\displaystyle[a,b]_{\text{\footnotesize S}}\wedge c+(-1)^{(\bar{a}-1)\bar{b}}b\wedge[a,c]_{\text{\footnotesize S}}$
(3.6)		$\displaystyle[a\wedge b,c]_{\text{\footnotesize S}}=$	$\displaystyle a\wedge[b,c]_{\text{\footnotesize S}}+(-1)^{\bar{b}(\bar{c}-1)}[a,c]_{\text{\footnotesize S}}\wedge b$

(3.7)

\mathop{\mathfrak{S}}_{a,b,c}(-1)^{(\bar{a}-1)(\bar{c}-1)}[[a,b]_{\text{\footnotesize S}},c]_{\text{\footnotesize S}}=0

4 Homology groups of superalgebras associated with Lie algebras

Let $\mathfrak{g}$ be a finite dimensional Lie algebra with the Lie bracket $[\cdot,\cdot]$ . Then we have chain complex consisting of $\text{C}_{i}=\Lambda^{i}\mathfrak{g}$ and homology groups.

$\displaystyle\overline{\mathfrak{g}}=\mathfrak{\oplus}_{i=0}^{-1+\textrm{dim}\mathfrak{g}}\mathfrak{g}_{i}$ becomes a $\mathbb{Z}$ -graded Lie superalgebra (where $\mathfrak{g}_{i}=\text{C}_{i+1}$ ) with the Schouten bracket $[\cdot,\cdot]_{\text{\footnotesize S}}$ . From the definition, $w$ -weighted $m$ -th chain space $C_{m}^{[w]}$ is defined as follows:

	$\displaystyle C_{m}^{[w]}$	$\displaystyle=\sum\bigtriangleup^{p_{0}}\mathfrak{g}_{0}\bigtriangleup^{p_{1}}\mathfrak{g}_{1}\bigtriangleup^{p_{2}}\mathfrak{g}_{2}\bigtriangleup\cdots\bigtriangleup^{p_{s}}\mathfrak{g}_{s}\text{ where }s=-1+\textrm{dim}\mathfrak{g}$
	$\displaystyle m$	$\displaystyle=p_{0}+p_{1}+\cdots+p_{s}\;,\quad w=0p_{0}+1p_{1}+\cdots+sp_{s}\;,$
(4.1)		$\displaystyle p_{2i}$	$\displaystyle\leqq\binom{\textrm{dim}\mathfrak{g}}{2i+1}\text{ , which is the dimensional restriction for }\mathfrak{g}_{2i}\;.$
(4.2)		$\displaystyle C_{m}^{[0]}$	$\displaystyle=\bigtriangleup^{m}\mathfrak{g}_{0}\;\text{ for }\;m=0,\ldots,\textrm{dim}\mathfrak{g}\;.$

By the same discussion we already developed, we see

w+m=1p_{0}+2p_{1}+\cdots+({\textrm{dim}\mathfrak{g}})p_{s}

and we deal with the Young diagrams of the area $w+m$ with the length $m$ and pick up those satisfy the dimensional condition (4.1). Since $p_{0}$ does not contribute for the weight $w$ , we assume $p_{0}=0$ , and for given $w$ we define the subspace $\displaystyle\widetilde{C}_{m}^{[w]}$ of $\displaystyle C_{m}^{[w]}$ by

\widetilde{C}_{m}^{[w]}=\sum\bigtriangleup^{p_{1}}\mathfrak{g}_{1}\bigtriangleup^{p_{2}}\mathfrak{g}_{2}\bigtriangleup\cdots\bigtriangleup^{p_{s}}\mathfrak{g}_{s}\text{ where }\sum_{i}ip_{i}=w\text{ , }\sum_{i}p_{i}=m\;,\text{ and satisfy \eqref{dim:cond}.}

In order to control $\displaystyle\widetilde{C}_{m}^{[w]}$ , we define the subspace $\displaystyle\widehat{C}_{m}^{[w]}$ of $\displaystyle C_{m}^{[w]}$ by $p_{0}=p_{1}=0$ . Thus,

\widehat{C}_{m}^{[w]}=\sum\bigtriangleup^{p_{2}}\mathfrak{g}_{2}\bigtriangleup\cdots\bigtriangleup^{p_{s}}\mathfrak{g}_{s}\text{ where }\sum_{i>1}ip_{i}=w\text{ , }\sum_{i>1}p_{i}=m\;,\text{ and satisfy \eqref{dim:cond}.}

We define subspaces $\displaystyle\widetilde{C}_{\bullet}^{[w]}=\sum_{m}\widetilde{C}_{m}^{[w]}$ and $\displaystyle\widehat{C}_{\bullet}^{[w]}=\sum_{m}\widehat{C}_{m}^{[w]}$ . Then we have

(4.3)		$\displaystyle C_{\bullet}^{[w]}$	$\displaystyle=C_{\bullet}^{[0]}\bigtriangleup\widetilde{C}_{\bullet}^{[w]}\;,\quad\text{ precisely }C_{m}^{[w]}=\sum_{a+b=m}\bigtriangleup^{a}\mathfrak{g}_{0}\bigtriangleup\widetilde{C}_{b}^{[w]}\;,$
(4.4)		$\displaystyle\widetilde{C}_{\bullet}^{[w]}$	$\displaystyle=\widehat{C}_{\bullet}^{[w]}+\mathfrak{g}_{1}\bigtriangleup\widetilde{C}_{\bullet}^{[w-1]}\;,\quad\widehat{C}_{\bullet}^{[1]}=0\;,\;\widetilde{C}_{\bullet}^{[0]}=1\;,\;\widetilde{C}_{\bullet}^{[1]}=\mathfrak{g}_{1}\;.$

(4.3) implies the next theorem:

Theorem 4.1.

The Euler number of $w$ -weighted homology groups of superalgebra induced from finite dimensional Lie algebra is 0.

Proof:

	The Euler num	$\displaystyle=\sum_{m}(-1)^{m}\textrm{dim}C_{m}^{[w]}=\sum_{a+b=m}(-1)^{a+b}\textrm{dim}(\bigtriangleup^{a}\mathfrak{g}_{0})\textrm{dim}\widetilde{C}_{b}^{[w]}$
		$\displaystyle=\sum_{a}(-1)^{a}\textrm{dim}(\bigtriangleup^{a}\mathfrak{g}_{0})\sum_{b}(-1)^{b}\textrm{dim}\widetilde{C}_{b}^{[w]}$
		$\displaystyle=\sum_{a}(-1)^{a}\binom{\textrm{dim}\mathfrak{g}_{0}}{a}\sum_{b}(-1)^{b}\textrm{dim}\widetilde{C}_{b}^{[w]}=0\;.$

$\blacksquare$

The equation (4.4) shows the space $\widetilde{C}_{\bullet}^{[w]}$ is determined by $\widehat{C}_{\bullet}^{[w]}$ and $\widetilde{C}_{\bullet}^{[w-1]}$ , and suggests discussions by induction on the weight $w$ .

5 Super homologies of 2-dimensional Lie algebras

Since our discussion is trivial for abelian Lie algebras, we consider a Lie algebra $\mathfrak{g}$ generated by $z_{1},z_{2}$ with $[z_{1},z_{2}]=z_{1}$ . We have only 1-dimensional 2-vectors, so we take $u=z_{1}\wedge z_{2}$ . For a given weight $w$ , the chain complex becomes as follows. Using $\partial(z_{1}\bigtriangleup z_{2})=z_{1}$ , $[z_{1},u]_{res}=0$ and $[z_{2},u]_{res}=-u$ , we see $[z_{1},u^{w}]_{res}=0$ and $[z_{2},u^{w}]_{res}=-wu^{w}$ . We have $\partial u^{w}=0$ , $\partial(z_{1}\bigtriangleup u^{w})=[z_{1},u^{w}]_{res}=0$ , $\partial(z_{2}\bigtriangleup u^{w})=[z_{2},u^{w}]_{res}=-wu^{w}$ , and $\partial(z_{1}\bigtriangleup z_{2}\bigtriangleup u^{w})=(1+w)z_{1}\bigtriangleup u^{w}$ . Using those, we complete the table below.

$\begin{array}[]{c | *{5}{c}}&C_{0}^{[0]}&C_{1}^{[0]}&C_{2}^{[0]}\\ \hline\cr\hline\cr\text{generators}&1&z_{i}&z_{1}\bigtriangleup z_{2}\\ \hline\cr\text{SpaceDim}&1&2&1\\ \hline\cr\text{KerDim}&1&2&0\\ \hline\cr\text{Betti}&1&1&0\\ \hline\cr\end{array}$ $\begin{array}[]{c | *{5}{c}}w>0&C_{w}^{[w]}&C_{w+1}^{[w]}&C_{w+2}^{[w]}\\ \hline\cr\hline\cr\text{generators}&u^{w}&z_{i}\bigtriangleup u^{w}&z_{1}\bigtriangleup z_{2}\bigtriangleup u^{w}\\ \hline\cr\text{SpaceDim}&1&2&1\\ \hline\cr\text{KerDim}&1&1&0\\ \hline\cr\text{Betti}&0&0&0\\ \hline\cr\end{array}$

6 Super homologies of 3-dimensional Lie algebras

We know the all Lie algebras of lower dimensional cases (cf. Lie algebras by Jacobson), in this section we try to get super homology tables for 3-dimensional Lie algebras. For that purpose, we prepare notations here. Let $z_{1},z_{2},z_{3}$ be a basis of $\mathfrak{g}$ , $u_{1},\;u_{2},\;u_{3}$ be a basis of $\Lambda^{2}\mathfrak{g}$ , and $z_{4}=V=z_{1}\wedge z_{2}\wedge z_{3}$ be a basis of $\Lambda^{3}\mathfrak{g}$ . Since $\{u_{i}\}$ have symmetric property, we put $U^{a,b,c}=u_{1}^{a}\bigtriangleup u_{2}^{b}\bigtriangleup u_{3}^{c}$ .

Denote the 3-vector $(a,b,c)$ by $A$ and also $\mathfrak{e}_{j}$ means 3-vector $(\delta^{k}_{j})_{k=1..3}$ , and $A-\mathfrak{e}_{j}$ means the difference of 3-vectors. To handle even basis at once, we introduce an odd notation:

(6.1)

W^{\epsilon_{1},\epsilon_{2},\epsilon_{3},\epsilon_{4}}=z_{1}^{\epsilon_{1}}\bigtriangleup z_{2}^{\epsilon_{2}}\bigtriangleup z_{3}^{\epsilon_{3}}\bigtriangleup z_{4}^{\epsilon_{4}}

where $\epsilon_{1},\epsilon_{2},\epsilon_{3},\epsilon_{4}$ are 0 or 1, and if $\epsilon_{1}=1$ then $z_{1}^{\epsilon_{1}}=z_{1}$ as expected otherwise we require $z_{1}^{\epsilon_{1}}$ disappear or becomes nothing/“empty”. For instance, $W^{1,0,1,0}=z_{1}\bigtriangleup z_{3}$ . A generic chain of the chain complex of superalgebra is

(6.2)

W^{\epsilon_{1},\epsilon_{2},\epsilon_{3},\epsilon_{4}}\bigtriangleup{U}^{a,b,c}_{\ell}

The degree(length) is $\sum_{i=1}^{4}\epsilon_{i}+a+b+c$ and the weight is $2\epsilon_{4}+a+b+c$ for the chain (6.2).

For a given weight $w$ , the chain complex becomes as follows:

$\begin{array}[]{c | *{5}{c}}&C_{w-1}^{[w]}&C_{w}^{[w]}&C_{w+1}^{[w]}&C_{w+2}^{[w]}&C_{w+3}^{[w]}\\ \hline\cr\hline\cr W^{\mathcal{E}}\bigtriangleup{U}^{A}_{w}&\text{none}&\begin{tabular}[]{c}$\epsilon\epsilon=0$\\ $\epsilon_{4}=0$\end{tabular}&\begin{tabular}[]{c}$\epsilon\epsilon=1$\\ $\epsilon_{4}=0$\end{tabular}&\begin{tabular}[]{c}$\epsilon\epsilon=2$\\ $\epsilon_{4}=0$\end{tabular}&\begin{tabular}[]{c}$\epsilon\epsilon=3$\\ $\epsilon_{4}=0$\end{tabular}\\ \hline\cr W^{\mathcal{E}}\bigtriangleup{U}^{P}_{w-2}&\begin{tabular}[]{c}$\epsilon\epsilon=0$\\ $\epsilon_{4}=1$\end{tabular}&\begin{tabular}[]{c}$\epsilon\epsilon=1$\\ $\epsilon_{4}=1$\end{tabular}&\begin{tabular}[]{c}$\epsilon\epsilon=2$\\ $\epsilon_{4}=1$\end{tabular}&\begin{tabular}[]{c}$\epsilon\epsilon=3$\\ $\epsilon_{4}=1$\end{tabular}&\text{none}\\ \hline\cr\hline\cr W^{\mathcal{E}}\bigtriangleup{U}^{A}_{w}&\text{none}&{U}^{A}_{w}&z_{i}\bigtriangleup{U}^{A}_{w}&z_{i}\bigtriangleup z_{j}\bigtriangleup{U}^{A}_{w}&W^{1110}\bigtriangleup{U}^{A}_{w}\\ \hline\cr W^{\mathcal{E}}\bigtriangleup{U}^{P}_{w-2}&z_{4}\bigtriangleup{U}^{P}_{w-2}&z_{i}\bigtriangleup z_{4}\bigtriangleup{U}^{P}_{w-2}&W^{\epsilon_{i}=0}\bigtriangleup{U}^{P}_{w-2}&W^{1111}\bigtriangleup{U}^{P}_{w-2}&\text{none}\\ \hline\cr\hline\cr\text{SpaceDim}&\binom{w}{2}&3\binom{w}{2}+\binom{w+2}{2}&3\binom{w}{2}+3\binom{w+2}{2}&\binom{w}{2}+3\binom{w+2}{2}&\binom{w+2}{2}\\ \hline\cr\end{array}$

where $\mathcal{E}=(\epsilon_{1},\epsilon_{2},\epsilon_{3},\epsilon_{4})$ and $\epsilon\epsilon=\sum_{i=1}^{3}\epsilon_{i}$ . In general, the boundary operator works as below:

(6.3)	$\displaystyle\partial(W^{\mathcal{E}}\wedge U^{A})$	$\displaystyle=\partial(W^{\mathcal{E}})\wedge U^{A}+(-1)^{\|\mathcal{E}\|}W^{\mathcal{E}}\wedge\partial U^{A}+[W^{\mathcal{E}},U^{A}]_{res}$
(6.4)	$\displaystyle\partial W^{\mathcal{E}}$	$\displaystyle=-\sum_{i<j}\tilde{\epsilon}_{i}\tilde{\epsilon}_{j}[z_{i},z_{j}]_{\text{\footnotesize S}}\bigtriangleup W^{\mathcal{E}-\mathfrak{e}_{i}-\mathfrak{e}_{j}}\;,\quad\text{ here }\tilde{\epsilon}_{i}=\epsilon_{i}(-1)^{\sum_{k\leqq i}\epsilon_{k}}$
(6.5)	$\displaystyle[W^{\mathcal{E}},U^{A}]_{res}$	$\displaystyle=-\sum_{i\leqq 3}\tilde{\epsilon}_{i}W^{\epsilon_{i}=0}\bigtriangleup[z_{i},U^{A}]_{res}$
(6.6)	$\displaystyle\partial U^{A}$	$\displaystyle=\sum_{i}\binom{a_{i}}{2}[u_{i},u_{i}]_{\text{\footnotesize S}}\bigtriangleup U^{A-2\mathfrak{e}_{i}}+\sum_{i<j}{a_{i}}{a_{j}}[u_{i},u_{j}]_{\text{\footnotesize S}}\bigtriangleup U^{A-\mathfrak{e}_{i}-\mathfrak{e}_{j}}$

The boundary operator on $C_{w-1}^{[w]}$ is trivial:

A general chain is given by $M=\sum_{P}{\mu}_{P}z_{4}\bigtriangleup{U}^{P}_{w-2}$ and $\partial M=-\sum_{P}{\mu}_{P}z_{4}\bigtriangleup\partial{U}^{P}_{w-2}+\sum_{P}{\mu}_{P}[z_{4},{U}^{P}_{w-2}]_{res}=0$ because $\partial{U}^{P}_{w-2}=z_{4}\bigtriangleup\text{"some"}$ and $[z_{4},u_{i}]_{\text{\footnotesize S}}=0$ . Thus, we have

Proposition 6.1.

Even though the boundary operator is really depending on the Schouten bracket, the boundary operator on the lowest chain space is trivial.

The boundary operator on $C_{w}^{[w]}$ :

	$\displaystyle\quad\partial(\sum\lambda_{A}{U}^{A}_{w}+\sum{\mu}^{j}_{P}z_{i}\bigtriangleup z_{4}\bigtriangleup{U}^{P}_{w-2})$
	$\displaystyle=\sum\lambda_{A}\partial{U}^{A}_{w}+\sum{\mu}^{i}_{P}([z_{i},z_{4}]_{\text{\footnotesize S}}\bigtriangleup{U}^{P}_{w-2}+z_{i}\bigtriangleup z_{4}\bigtriangleup\partial{U}^{P}_{w-2}+[z_{i}\bigtriangleup z_{4},{U}^{P}_{w-2}]_{res})$
	$\displaystyle=\sum\lambda_{A}\partial{U}^{A}_{w}+\sum{\mu}^{i}_{P}[z_{i},z_{4}]_{\text{\footnotesize S}}\bigtriangleup{U}^{P}_{w-2}+z_{4}\bigtriangleup\sum{\mu}^{i}_{P}[z_{i},{U}^{P}_{w-2}]_{res}$

The boundary operator on $C_{w+1}^{[w]}$ :

	$\displaystyle\quad\partial(\sum\lambda^{i}_{A}z_{i}\bigtriangleup{U}^{A}_{w}+\sum{\mu}^{j}_{P}W^{\epsilon_{j}=0}\bigtriangleup{U}^{P}_{w-2})$
	$\displaystyle=\sum\lambda^{i}_{A}(-z_{i}\bigtriangleup\partial{U}^{A}_{w}+[z_{i},{U}^{A}_{w}]_{res})+\sum{\mu}^{j}_{P}(\partial W^{\epsilon_{j}=0})\bigtriangleup{U}^{P}_{w-2}+\sum{\mu}^{j}_{P}[W^{\epsilon_{j}=0},{U}^{P}_{w-2}]_{res}$

The boundary operator on $C_{w+2}^{[w]}$ :

	$\displaystyle\quad\partial(\sum\lambda^{i}_{A}W^{\epsilon_{i}=\epsilon_{4}=0}\bigtriangleup{U}^{A}_{w}+\sum{\mu}_{P}W^{1111}\bigtriangleup{U}^{P}_{w-2})$
	$\displaystyle=\sum\lambda^{i}_{A}(\partial W^{\epsilon_{i}=\epsilon_{4}=0})\bigtriangleup{U}^{A}_{w}+\sum\lambda^{i}_{A}W^{\epsilon_{i}=\epsilon_{4}=0}\bigtriangleup\partial{U}^{A}_{w}+\sum\lambda^{i}_{A}[W^{\epsilon_{i}=\epsilon_{4}=0},{U}^{A}_{w}]_{res}$
	$\displaystyle\quad+\sum{\mu}_{P}\partial(W^{1111})\bigtriangleup{U}^{P}_{w-2}+\sum{\mu}_{P}[W^{1111},{U}^{P}_{w-2}]_{res}$

The boundary operator on $C_{w+3}^{[w]}$ :

	$\displaystyle\quad\partial\sum\lambda_{A}W^{\epsilon_{4}=0}\bigtriangleup{U}^{A}_{w}$
	$\displaystyle=\sum\lambda^{i}_{A}(\partial W^{\epsilon_{4}=0})\bigtriangleup{U}^{A}_{w}-\sum\lambda^{i}_{A}W^{\epsilon_{4}=0}\bigtriangleup\partial{U}^{A}_{w}+\sum\lambda^{i}_{A}[W^{\epsilon_{4}=0},{U}^{A}_{w}]_{res}$
	$\displaystyle=([z_{1},z_{2}]_{\text{\footnotesize S}}\bigtriangleup z_{3}-[z_{1},z_{3}]_{\text{\footnotesize S}}\bigtriangleup z_{2}+[z_{2},z_{3}]_{\text{\footnotesize S}}\bigtriangleup z_{1})\bigtriangleup\sum\lambda^{i}_{A}{U}^{A}_{w}-\sum W^{\epsilon_{4}=0}\bigtriangleup\lambda^{i}_{A}\partial{U}^{A}_{w}$
	$\displaystyle\quad+W^{0110}\bigtriangleup\lambda_{A}[z_{1},{U}^{A}_{w}]_{res}-W^{1010}\bigtriangleup\lambda_{A}[z_{2},{U}^{A}_{w}]_{res}+W^{1100}\bigtriangleup\lambda_{A}[z_{3},{U}^{A}_{w}]_{res}$

6.1 $\textrm{dim}\mathfrak{g}=3$ , $\textrm{dim}[\mathfrak{g},\mathfrak{g}]=1$ and $[\mathfrak{g},\mathfrak{g}]\subset Z(\mathfrak{g})$

Consider a Lie algebra $\mathfrak{g}$ where $[\mathfrak{g},\mathfrak{g}]$ is 1-dimensional and is in of the center of $\mathfrak{g}$ . Then we find a basis $z_{1},z_{2},z_{3}$ of $\mathfrak{g}$ so that $[z_{1},z_{2}]=-[z_{2},z_{1}]=z_{3}$ and the other brackets are zero. We take $u_{1}=z_{2}\wedge z_{3},\;u_{2}=z_{3}\wedge z_{1},\;u_{3}=z_{1}\wedge z_{2}$ as a basis of $\Lambda^{2}\mathfrak{g}$ . It should be noted that this definition does not necessarily follow a natural order. Take $V=z_{4}=z_{1}\wedge z_{2}\wedge z_{3}$ as a basis of $\Lambda^{3}\mathfrak{g}$ . Now we have the multiplication (by the Schouten bracket) tables:

$\begin{array}[t]{c|*{3}{c}|c|*{3}{c}|}&z_{1}&z_{2}&z_{3}&z_{4}&u_{1}&u_{2}&u_{3}\\ \hline\cr z_{1}&0&z_{3}&0&0&0&0&-u_{2}\\ z_{2}&-z_{3}&0&0&0&0&0&u_{1}\\ z_{3}&0&0&0&0&0&0&0\end{array}$ $\begin{array}[t]{c|*{3}{c}|c}&u_{1}&u_{2}&u_{3}&z_{4}\\ \hline\cr u_{1}&0&0&0&0\\ u_{2}&0&0&0&0\\ u_{3}&0&0&2z_{4}&0\\ \end{array}$

Then we have

	$\displaystyle\partial W^{1111}$	$\displaystyle=0\;,\quad\partial W^{1110}=0\;,\quad\partial W^{1101}=z_{3}\bigtriangleup z_{4}\;,\quad\partial W^{1011}=0\;,\quad\partial W^{0111}=0\;,\quad$
	$\displaystyle\partial U^{a,b,c}$	$\displaystyle=z_{4}\bigtriangleup 2\tbinom{c}{2}U^{a,b,c-2}\;.\quad\sum_{A}\lambda_{A}\partial U^{a,b,c}=z_{4}\bigtriangleup\sum U^{p,q,r}(r+2)(r+1)\lambda_{p,q,r+2}\;.$
	$\displaystyle[z_{1},U^{A}]_{res}$	$\displaystyle=-cU^{a,b+1,c-1}\;,\quad[z_{2},U^{A}]_{res}=cU^{a+1,b,c-1}\;,\quad[z_{3},U^{A}]_{res}=0$
	$\displaystyle\sum_{A}\lambda^{i}_{A}[z_{1},U^{A}]_{res}$	$\displaystyle=\sum_{A}(-(c+1)\lambda^{i}_{a,b-1,c+1})U^{A}\;,\quad\sum_{A}\lambda^{i}_{A}[z_{2},U^{A}]_{res}=\sum_{A}((c+1)\lambda^{i}_{a-1,b,c+1})U^{A}\;.$

Kernel of $\displaystyle C_{w+3}^{[w]}\mathop{\to}^{\partial}C_{w+2}^{[w]}$

A basis of $C_{w+3}^{[w]}$ , we have $W^{1,1,1,0}\bigtriangleup U^{a,b,c}$ with $a+b+c=w$ and take a linear combination $X=\sum_{a+b+c=w}\lambda_{a,b,c}W^{1,1,1,0}\bigtriangleup U^{a,b,c}$ by unknown scalars $\lambda_{a,b,c}$ . Then

	$\displaystyle\partial X=$	$\displaystyle\sum_{a+b+c=w}\lambda_{a,b,c}\left(cW^{0,1,1,0}\bigtriangleup U^{a,b+1,c-1}+cW^{1,0,1,0}\bigtriangleup U^{a+1,b,c-1}\right.$
		$\displaystyle\qquad\qquad\qquad\left.-c(c-1)W^{1,1,1,1}\bigtriangleup U^{a,b,c-2}\right)$
	$\displaystyle=$	$\displaystyle W^{0,1,1,0}\sum_{a+b+c=w}\lambda_{a,b,c}c\bigtriangleup U^{a,b+1,c-1}+W^{1,0,1,0}\sum_{a+b+c=w}\lambda_{a,b,c}c\bigtriangleup U^{a+1,b,c-1}$
		$\displaystyle\quad-W^{1,1,1,1}\sum_{a+b+c=w}\lambda_{a,b,c}c(c-1)\bigtriangleup U^{a,b,c-2}$

$\partial X=0$ is generated by


(6.7a)			$\displaystyle\lambda_{a,b-1,c+1}(c+1)$
(6.7b)			$\displaystyle\lambda_{a-1,b,c+1}(c+1)$
(6.7c)			$\displaystyle\lambda_{p,q,r+2}(r+2)(r+1)$

Thus, the rank is $\binom{w+2}{2}-\#\{\lambda_{a,b,0}\mid a+b=w\}=\binom{w+2}{2}-(w+1)$ from (6.7a) $\sim$ (6.7c), and the kernel dimension is $w+1$ .

Kernel of $\displaystyle C_{w+2}^{[w]}\mathop{\to}^{\partial}C_{w+1}^{[w]}$

As a basis of $C_{w+2}^{[w]}$ , we have $W^{\epsilon_{1},\epsilon_{2},\epsilon_{3},0}\bigtriangleup{U}^{a,b,c}_{w}$ with $\epsilon_{1}+\epsilon_{2}+\epsilon_{3}=2$ and $W^{1,1,1,1}\bigtriangleup{U}^{p,q,r}_{w-2}$ .

Take a linear combination

X=\sum_{a+b+c=w}\lambda^{k}_{a,b,c}W^{\epsilon_{1},\epsilon_{2},\epsilon_{3},0}\bigtriangleup{U}^{a,b,c}_{w}+\sum_{p+q+r=w-2}{\mu}_{p,q,r}W^{1,1,1,1}\bigtriangleup{U}^{p,q,r}_{w-2}

by unknown scalars $\lambda^{k}_{a,b,c},\mu_{p,q,r}$ with $\epsilon_{1}+\epsilon_{2}+\epsilon_{3}=2$ and $\epsilon_{k}=0$ .

	$\displaystyle\partial X=$	$\displaystyle\sum\lambda^{1}_{a,b,c}(cW^{0,0,1,0}{U}^{a+1,b,c-1}_{w}+2\binom{c}{2}W^{0,1,1,1}{U}^{a,b,c-2}_{w-2})$
		$\displaystyle+\sum\lambda^{2}_{a,b,c}(-cW^{0,0,1,0}{U}^{a,b+1,c-1}_{w}+2\binom{c}{2}W^{1,0,1,1}{U}^{a,b,c-2}_{w-2})$
		$\displaystyle+\sum\lambda^{3}_{a,b,c}(W^{0,0,1,0}{U}^{a,b,c}_{w}-cW^{0,1,0,0}{U}^{a,b+1,c-1}_{w}-cW^{1,0,0,0}{U}^{a+1,b,c-1}_{w}+2\binom{c}{2}W^{1,1,0,1}{U}^{a,b,c-2}_{w-2})$
		$\displaystyle-\sum{\mu}_{p,q,r}(rW^{0,1,1,1}{U}^{p,q+1,r-1}_{w-2}+rW^{1,0,1,1}{U}^{p+1,q,r-1}_{w-2})$
	$\displaystyle=$	$\displaystyle W^{0,0,1,0}(\sum\lambda^{1}_{a,b,c}c{U}^{a+1,b,c-1}_{w}-\sum\lambda^{2}_{a,b,c}c{U}^{a,b+1,c-1}_{w}+\sum\lambda^{3}_{a,b,c}{U}^{a,b,c}_{w})$
		$\displaystyle-W^{0,1,0,0}\sum\lambda^{3}_{a,b,c}c{U}^{a,b+1,c-1}_{w}-W^{1,0,0,0}\sum\lambda^{3}_{a,b,c}c{U}^{a+1,b,c-1}_{w}$
		$\displaystyle+W^{0,1,1,1}(\sum\lambda^{1}_{a,b,c}2\binom{c}{2}{U}^{a,b,c-2}_{w-2}-\sum{\mu}_{p,q,r}r{U}^{p,q+1,r-1}_{w-2})$
		$\displaystyle+W^{1,0,1,1}(\sum\lambda^{2}_{a,b,c}2\binom{c}{2}{U}^{a,b,c-2}_{w-2}-\sum{\mu}_{p,q,r}r{U}^{p+1,q,r-1}_{w-2})+W^{1,1,0,1}\sum\lambda^{3}_{a,b,c}2\binom{c}{2}{U}^{a,b,c-2}_{w-2}$

Thus, $\partial X=0$ is determined by


(6.8a)			$\displaystyle\lambda^{1}_{a-1,b,c+1}(c+1)-\lambda^{2}_{a,b-1,c+1}(c+1)+\lambda^{3}_{a,b,c}$
(6.8b)			$\displaystyle\lambda^{3}_{a,b-1,c+1}(c+1)$
(6.8c)			$\displaystyle\lambda^{3}_{a-1,b,c+1}(c+1)$
(6.8d)			$\displaystyle\lambda^{1}_{p,q,r+2}2\binom{r+2}{2}-{\mu}_{p,q-1,r+1}(r+1)$
(6.8e)			$\displaystyle\lambda^{2}_{p,q,r+2}2\binom{r+2}{2}-{\mu}_{p-1,q,r+1}(r+1)$
(6.8f)			$\displaystyle\lambda^{3}_{p,q,r+2}2\binom{r+2}{2}\;.$

(6.8b), (6.8c), and (6.8f) tell that $\lambda^{3}_{a,b,c}$ are single generators for $c>0$ . With $c=0$ for (6.8a), $\lambda^{3}_{A}$ are linear independent generators. When $c>0$ for (6.8a), then we have

\lambda^{1}_{a-1,b,c+1}-\lambda^{2}_{a,b-1,c+1}

Those are obtained from (6.8d) and (6.8e). Thus, the rank is $\binom{w+2}{2}+2\binom{w}{2}$ , the kernel dimension is $(\binom{w}{2}+3\binom{w+2}{2})-(\binom{w+2}{2}+2\binom{w}{2})=-\binom{w}{2}+2\binom{w+2}{2}$ .

Kernel of $\displaystyle C_{w+1}^{[w]}\mathop{\to}^{\partial}C_{w}^{[w]}$

As a basis of $C_{w+1}^{[w]}$ , we have $z_{i}\bigtriangleup{U}^{a,b,c}_{w}$ with $i=1,2,3$ , and $W^{\epsilon_{1},\epsilon_{2},\epsilon_{3},1}\bigtriangleup{U}^{p,q,r}_{w-2}$ with $\epsilon_{1}+\epsilon_{2}+\epsilon_{3}=2$ . Take a linear combination

X=\sum\lambda^{i}_{a,b,c}z_{i}\bigtriangleup{U}^{a,b,c}_{w}+\sum_{\epsilon_{j}=0,\epsilon_{1}+\epsilon_{2}+\epsilon_{3}=2}{\mu}^{j}_{p,q,r}W^{\epsilon_{1},\epsilon_{2},\epsilon_{3},1}\bigtriangleup{U}^{p,q,r}_{w-2}

for unknown scalars $\lambda^{i}_{a,b,c}$ and ${\mu}^{j}_{p,q,r}$ .

	$\displaystyle\partial X=$	$\displaystyle\sum\lambda^{1}_{a,b,c}\left(c{U}^{a,b+1,c-1}_{w}-2\binom{c}{2}z_{1}\bigtriangleup V\bigtriangleup{U}^{a,b,c-2}_{w-2}\right)-\sum\lambda^{2}_{a,b,c}\left(c{U}^{a+1,b,c-1}_{w}+2\binom{c}{2}z_{2}\bigtriangleup V\bigtriangleup{U}^{a,b,c-2}_{w-2}\right)$
		$\displaystyle-\sum\lambda^{3}_{a,b,c}\left(2\binom{c}{2}z_{3}\bigtriangleup V\bigtriangleup{U}^{a,b,c-2}_{w-2}\right)-\sum{\mu}^{1}_{p,q,r}\left(rz_{3}\bigtriangleup V\bigtriangleup{U}^{p+1,q,r-1}_{w-2}\right)$
		$\displaystyle+\sum{\mu}^{2}_{p,q,r}\left(rz_{3}\bigtriangleup V\bigtriangleup{U}^{p,q+1,r-1}_{w-2}\right)+\sum{\mu}^{3}_{p,q,r}0\quad\text{(remember that ${\mu}^{3}_{p,q,r}$ disappeared.) }$
	$\displaystyle=$	$\displaystyle\sum\lambda^{1}_{a,b,c}c{U}^{a,b+1,c-1}_{w}-\sum\lambda^{2}_{a,b,c}c{U}^{a+1,b,c-1}_{w}-z_{1}\bigtriangleup V\bigtriangleup\sum\lambda^{1}_{a,b,c}2\binom{c}{2}{U}^{a,b,c-2}_{w-2}$
		$\displaystyle-z_{2}\bigtriangleup V\bigtriangleup\sum\lambda^{2}_{a,b,c}2\binom{c}{2}{U}^{a,b,c-2}_{w-2}$
		$\displaystyle-z_{3}\bigtriangleup V\bigtriangleup\left(\sum{\mu}^{1}_{p,q,r}r{U}^{p+1,q,r-1}_{w-2}-\sum{\mu}^{2}_{p,q}r{U}^{p,q+1,r-1}_{w-2}+\sum\lambda^{3}_{a,b,c}2\binom{c}{2}{U}^{a,b,c-2}_{w-2}\right)$

Thus, we have linear equations

(F1)		$\displaystyle\sum\lambda^{1}_{a,b,c}c{U}^{a,b+1,c-1}_{w}-\sum\lambda^{2}_{a,b,c}c{U}^{a+1,b,c-1}_{w}=0$
(F2)		$\displaystyle\sum\lambda^{1}_{a,b,c}2\binom{c}{2}{U}^{a,b,c-2}_{w-2}=0$
(F3)		$\displaystyle\sum\lambda^{2}_{a,b,c}2\binom{c}{2}{U}^{a,b,c-2}_{w-2}=0$
(FF)		$\displaystyle\sum{\mu}^{1}_{p,q,r}r{U}^{p+1,q,r-1}_{w-2}-\sum{\mu}^{2}_{p,q,r}r{U}^{p,q+1,r-1}_{w-2}+\sum\lambda^{3}_{a,b,c}2\binom{c}{2}{U}^{a,b,c-2}_{w-2}=0$

(F1) $\sim$ (FF) yield the generators

	$\displaystyle\lambda^{1}_{a,b-1,c+1}(c+1)-\lambda^{2}_{a-1,b,c+1}(c+1)$
	$\displaystyle\lambda^{1}_{p,q,r+2}2\tbinom{r+2}{2}$
	$\displaystyle\lambda^{2}_{p,q,r+2}2\tbinom{r+2}{2}$
	$\displaystyle{\mu}^{1}_{p-1,q,r+1}(r+1)-{\mu}^{2}_{p,q-1,r+1}(r+1)+\lambda^{3}_{p,q,r+2}2\tbinom{r+2}{2}$

The last three are linearly independent and the first type give $w+1$ independent equations. Thus the rank is $3\binom{w}{2}+(w+1)$ and the kernel dimension is $3\binom{w}{2}+3\binom{w+2}{2}-3\binom{w}{2}-(w+1)=\binom{w}{2}+2\binom{w+2}{2}+w$ .

Kernel of $\displaystyle C_{w}^{[w]}\mathop{\to}^{\partial}C_{w-1}^{[w]}$

As a basis of $C_{w}^{[w]}$ , we have ${U}^{a,b,c}_{w}$ with $a+b+c=w$ and $z_{k}\bigtriangleup z_{4}\bigtriangleup{U}^{p,q,r}_{w-2}$ with $k=1,2,3$ and $p+q+r=w-2$ , take a linear combination

X=\sum\lambda_{a,b,c}\bigtriangleup{U}^{a,b,c}_{w}+\sum_{k=1}^{3}\sum{\mu}^{k}_{p,q,r}z_{k}\bigtriangleup z_{4}\bigtriangleup{U}^{p,q,r}_{w-2}

by unknown scalars $\lambda_{a,b,c},{\mu}^{k}_{p,q,r}$ .

\displaystyle\partial X=

\displaystyle\sum\lambda_{a,b,c}c(c-1)z_{4}\bigtriangleup{U}^{a,b,c-2}_{w-2}+\sum{\mu}^{1}_{p,q,r}(-r)z_{4}\bigtriangleup{U}^{p,q+1,r-1}_{w-2}+\sum{\mu}^{2}_{p,q,r}rz_{4}\bigtriangleup{U}^{p+1,q,r-1}_{w-2}\;,

$\partial X=0$ is generated by

\lambda_{p,q,r+2}(r+2)(r+1)-{\mu}^{1}_{p,q-1,r+1}(r+1)+{\mu}^{2}_{p-1,q,r+1}(r+1)\;,\quad p+q+r=w-2\;.

Thus the rank is $\binom{w}{2}$ and the kernel dimension is $3\binom{w}{2}+\binom{w+2}{2}-\binom{w}{2}=2\binom{w}{2}+\binom{w+2}{2}$ .

Final table of chain complex

Theorem 6.1.

$\begin{array}[]{c|*{5}{c}}\text{weight}=w&w-1&w&w+1&w+2&w+3\\ \hline\cr\text{SpaceDim}&\binom{w}{2}&3\binom{w}{2}+\binom{w+2}{2}&3\binom{w}{2}+3\binom{w+2}{2}&\binom{w}{2}+3\binom{w+2}{2}&\binom{w+2}{2}\\ \ker\textrm{dim}&\binom{w}{2}&2\binom{w}{2}+\binom{w+2}{2}&\binom{w}{2}+2\binom{w+2}{2}+w&-\binom{w}{2}+2\binom{w+2}{2}&w+1\\ \hline\cr\text{Betti}&0&w&3w+1&3w+2&w+1\end{array}$

6.2 $\textrm{dim}\mathfrak{g}=3$ , $\textrm{dim}[\mathfrak{g},\mathfrak{g}]=1$ and $[\mathfrak{g},\mathfrak{g}]\not\subset Z(\mathfrak{g})$

Consider a Lie algebra $\mathfrak{g}$ where $[\mathfrak{g},\mathfrak{g}]$ is 1-dimensional and not in of the center of $\mathfrak{g}$ . Then we find a basis $z_{1},z_{2},z_{3}$ of $\mathfrak{g}$ so that $[z_{1},z_{2}]=-[z_{2},z_{1}]=z_{2}$ and the other brackets are zero. We take $u_{1}=z_{1}\wedge z_{2},\;u_{2}=z_{2}\wedge z_{3},\;u_{3}=z_{3}\wedge z_{1}$ as a basis of $\Lambda^{2}\mathfrak{g}$ , and $V=z_{4}=z_{1}\wedge z_{2}\wedge z_{3}$ as a basis of $\Lambda^{3}\mathfrak{g}$ . Now we have the multiplication (by the Schouten bracket) tables:

$\begin{array}[t]{c|*{3}{c}|c|*{3}{c}|}&z_{1}&z_{2}&z_{3}&z_{4}&u_{1}&u_{2}&u_{3}\\ \hline\cr z_{1}&0&z_{2}&0&z_{4}&u_{1}&u_{2}&0\\ z_{2}&-z_{2}&0&0&0&0&0&u_{2}\\ z_{3}&0&0&0&0&0&0&0\end{array}$ $\begin{array}[t]{c|*{3}{c}|c}&u_{1}&u_{2}&u_{3}&z_{4}\\ \hline\cr u_{1}&0&0&z_{4}&0\\ u_{2}&0&0&0&0\\ u_{3}&z_{4}&0&0&0\\ \end{array}$

Then we have

	$\displaystyle\partial W^{1111}$	$\displaystyle=2W^{0111}\;,\quad\partial W^{1110}=z_{2}\bigtriangleup z_{3}\;,\quad$
	$\displaystyle\partial W^{1101}$	$\displaystyle=2z_{2}\bigtriangleup z_{4}\;,\quad\partial W^{1011}=z_{3}\bigtriangleup z_{4}\;,\quad\partial W^{0111}=0\;,\quad$
	$\displaystyle\sum_{A}\lambda_{A}\partial U^{a,b,c}$	$\displaystyle=z_{4}\bigtriangleup\sum U^{p,q,r}(p+1)(r+1)\lambda_{p+1,q,r+1}\;.$
	$\displaystyle[z_{1},U^{A}]_{res}$	$\displaystyle=bU^{a,b-1,c+1}-\beta cU^{a,b+1,c-1}$
	$\displaystyle[z_{2},U^{A}]_{res}$	$\displaystyle=-aU^{a-1,b,c+1}+\alpha cU^{a+1,b,c-1}$
	$\displaystyle[z_{3},U^{A}]_{res}$	$\displaystyle=\beta aU^{a-1,b+1,c}-\alpha bU^{a+1,b-1,c}$
	$\displaystyle\sum_{A}\lambda^{i}_{A}[z_{1},U^{A}]_{res}$	$\displaystyle=\sum_{A}(a+b)\lambda^{i}_{a,b,c}U^{a,b,c}\;,$
	$\displaystyle\sum_{A}\lambda^{i}_{A}[z_{2},U^{A}]_{res}$	$\displaystyle=\sum_{A}(c+1)\lambda^{i}_{a,b-1,c+1}U^{a,b,c}\;,$
	$\displaystyle[z_{3},U^{A}]_{res}$	$\displaystyle=0\;.$

Kernel of $\displaystyle C_{w+3}^{[w]}\mathop{\to}^{\partial}C_{w+2}^{[w]}$

A basis of $C_{w+3}^{[w]}$ , we have $W^{1,1,1,0}\bigtriangleup U^{a,b,c}$ with $a+b+c=w$ . Take a linear combination $X=\sum_{a+b+c=w}\lambda_{a,b,c}W^{1,1,1,0}\bigtriangleup U^{a,b,c}$ by unknown scalars $\lambda_{a,b,c}$ . Then

	$\displaystyle\partial X=$	$\displaystyle\sum_{a+b+c=w}\lambda_{a,b,c}\left((a+b+1)W^{0,1,1,0}\bigtriangleup U^{a,b,c}-cW^{1,0,1,0}\bigtriangleup U^{a,b+1,c-1}\right.$
		$\displaystyle\qquad\qquad\qquad\left.-acW^{1,1,1,1}\bigtriangleup U^{a-1,b,c-1}\right)$
	$\displaystyle=$	$\displaystyle W^{0,1,1,0}\sum_{a+b+c=w}\lambda_{a,b,c}(a+b+1)\bigtriangleup U^{a,b,c}-W^{1,0,1,0}\sum_{a+b+c=w}\lambda_{a,b,c}c\bigtriangleup U^{a,b+1,c-1}$
		$\displaystyle\quad-W^{1,1,1,1}\sum_{a+b+c=w}\lambda_{a,b,c}ac\bigtriangleup U^{a-1,b,c-1}$

Suppose $\partial X=0$ . Then $\lambda_{a,b,c}(a+b+1)=0$ for $a+b+c=w$ and all $\lambda_{a,b,c}=0$ . Thus, the kernel space is 0-dimensional. Or, the linear equations is generated by

	$\displaystyle\lambda_{p+1,q,r+1}(p+1)(r+1)\;,\quad(p+q+r=w-2)$
	$\displaystyle\lambda_{a,b,c}(1+a+b)\;,\quad\lambda_{a,b-1,c+1}(c+1)\;,\quad(a+b+c=w)\;,$

and the rank is $\binom{w+2}{2}$ and the kernel dimension is 0.

Kernel of $\displaystyle C_{w+2}^{[w]}\mathop{\to}^{\partial}C_{w+1}^{[w]}$

Take a linear combination

X=\sum_{a+b+c=w}\lambda^{k}_{a,b,c}W^{\epsilon_{1},\epsilon_{2},\epsilon_{3},0}\bigtriangleup{U}^{a,b,c}_{w}+\sum_{p+q+r=w-2}\mu_{p,q,r}W^{1,1,1,1}\bigtriangleup{U}^{p,q,r}_{w-2}

by unknown scalars $\lambda^{k}_{a,b,c},\mu_{p,q,r}$ with $\epsilon_{1}+\epsilon_{2}+\epsilon_{3}=2$ and $\epsilon_{k}=0$ . Again,

	$\displaystyle\partial X=$	$\displaystyle\sum\lambda^{1}_{a,b,c}(cW^{0,0,1,0}{U}^{a,b+1,c-1}_{w}+acW^{0,1,1,1}{U}^{a-1,b,c-1}_{w-2})$
		$\displaystyle+\sum\lambda^{2}_{a,b,c}((a+b)W^{0,0,1,0}{U}^{a,b,c}_{w}+acW^{1,0,1,1}{U}^{a-1,b,c-1}_{w-2})$
		$\displaystyle+\sum\lambda^{3}_{a,b,c}((a+b+1)W^{0,1,0,0}{U}^{a,b,c}_{w}-cW^{1,0,0,0}{U}^{a,b+1,c-1}_{w}+acW^{1,1,0,1}{U}^{a-1,b,c-1}_{w-2})$
		$\displaystyle+\sum\mu_{p,q,r}((p+q+2)W^{0,1,1,1}{U}^{p,q,r}_{w-2}-rW^{1,0,1,1}{U}^{p,q+1,r-1}_{w-2})$
	$\displaystyle=$	$\displaystyle\sum\lambda^{1}_{a,b,c}cW^{0,0,1,0}{U}^{a,b+1,c-1}_{w}+\sum\lambda^{2}_{a,b,c}(a+b)W^{0,0,1,0}{U}^{a,b,c}_{w}$
		$\displaystyle+\sum\lambda^{3}_{a,b,c}(a+b+1)W^{0,1,0,0}{U}^{a,b,c}_{w}-\sum\lambda^{3}_{a,b,c}(w-a-b)W^{1,0,0,0}{U}^{a,b+1}_{w}$
		$\displaystyle+\sum\lambda^{1}_{a,b,c}a(w-a-b)W^{0,1,1,1}{U}^{a-1,b,c-1}_{w-2}+\sum\lambda^{2}_{a,b,c}a(w-a-b)W^{1,0,1,1}{U}^{a-1,b,c-1}_{w-2}$
		$\displaystyle+\sum\lambda^{3}_{a,b,c}a(w-a-b)W^{1,1,0,1}{U}^{a-1,b,c-1}_{w-2}+\sum\mu_{p,q,r}(p+q+2)W^{0,1,1,1}{U}^{p,q,r}_{w-2}$
		$\displaystyle-\sum\mu_{p,q,r}rW^{1,0,1,1}{U}^{p,q+1,r-1}_{w-2}$
	$\displaystyle=$	$\displaystyle W^{0,0,1,0}(\sum\lambda^{1}_{a,b,c}c{U}^{a,b+1,c-1}_{w}+\sum\lambda^{2}_{a,b,c}(a+b){U}^{a,b,c}_{w})$
		$\displaystyle+W^{0,1,0,0}\sum\lambda^{3}_{a,b,c}(a+b+1){U}^{a,b,c}_{w}-W^{1,0,0,0}\sum\lambda^{3}_{a,b,c}c{U}^{a,b+1,c-1}_{w}$
		$\displaystyle+W^{0,1,1,1}(\sum\lambda^{1}_{a,b,c}ac{U}^{a-1,b,c-1}_{w-2}+\sum\mu_{p,q,r}(p+q+2){U}^{p,q,r}_{w-2})$
		$\displaystyle+W^{1,0,1,1}(\sum\lambda^{2}_{a,b,c}ac{U}^{a-1,b,c-1}_{w-2}-\sum\mu_{p,q,r}r{U}^{p,q+1,r-1}_{w-2})$
		$\displaystyle+W^{1,1,0,1}\sum\lambda^{3}_{a,b,c}ac{U}^{a-1,b,c-1}_{w-2}$

Suppose $\partial X=0$ . Then we have

(6.9)		$\displaystyle\sum\lambda^{1}_{a,b,c}(w-a-b){U}^{a,b+1,c-1}_{w}+\sum\lambda^{2}_{a,b,c}(a+b){U}^{a,b,c}_{w}=0$
(6.10)		$\displaystyle\sum\lambda^{3}_{a,b,c}(a+b+1){U}^{a,b,c}_{w}=0$
(6.11)		$\displaystyle\sum\lambda^{3}_{a,b,c}(w-a-b){U}^{a,b+1,c-1}_{w}=0$
(6.12)		$\displaystyle\sum\lambda^{1}_{a,b,c}a(w-a-b){U}^{a-1,b,c-1}_{w-2}+\sum\mu_{a,b,c}(a+b+2){U}^{a,b,c}_{w-2}=0$
(6.13)		$\displaystyle\sum\lambda^{2}_{a,b,c}a(w-a-b){U}^{a-1,b,c-1}_{w-2}-\sum\mu_{p,q,r}r{U}^{p,q+1,r-1}_{w-2}=0$
(6.14)		$\displaystyle\sum\lambda^{3}_{a,b,c}ac{U}^{a-1,b,c-1}_{w-2}=0$

The linear equations are generated by

(6.15a)		$\displaystyle\lambda^{3}_{a,b-1,c+1}(c+1)$
(6.15b)		$\displaystyle\lambda^{3}_{a,b,c}(a+b+1)$
(6.15c)		$\displaystyle\lambda^{1}_{a,b-1,c+1}(c+1)+\lambda^{2}_{a,b,c}(a+b)$
(6.15d)		$\displaystyle\lambda^{1}_{p+1,q,r+1}(p+1)(r+1)+(2+p+q){\mu}_{p,q,r}$
(6.15e)		$\displaystyle\lambda^{2}_{p+1,q,r+1}(p+1)-{\mu}_{p,q-1,r+1}$
(6.15f)		$\displaystyle\lambda^{3}_{p+1,q,r+1}(p+1)(r+1)$

From (6.15a), (6.15b) and (6.15f), $\lambda^{3}_{A}$ are linearly independent monomials.

We take linearly independent generators (6.15c) with leading term $\lambda^{2}_{A}$ whose number is $\binom{w+2}{2}-1$ , where $a+b=0$ is the exception. Finally, We take linearly independent generators (6.15d) with leading term ${\mu}_{P}$ whose number is $\binom{w}{2}$ . (6.15e) are linearly dependent on (6.15c) and (6.15d). Thus, the rank is $2\binom{w+2}{2}-1+\binom{w}{2}$ and the kernel dimension is $\binom{w+2}{2}+1$ .

Kernel of $\displaystyle C_{w+1}^{[w]}\mathop{\to}^{\partial}C_{w}^{[w]}$

X=\sum\lambda^{i}_{a,b,c}z_{i}\bigtriangleup{U}^{a,b,c}_{w}+\sum_{\epsilon_{j}=0,\epsilon_{1}+\epsilon_{2}+\epsilon_{3}=2}{\mu}^{j}_{p,q,r}W^{\epsilon_{1},\epsilon_{2},\epsilon_{3},1}\bigtriangleup{U}^{p,q,r}_{w-2}

for unknown scalars $\lambda^{i}_{a,b,c}$ and ${\mu}^{j}_{p,q,r}$ .

	$\displaystyle\partial X=$	$\displaystyle\sum\lambda^{1}_{a,b,c}\left((a+b){U}^{a,b,c}_{w}-acz_{1}\bigtriangleup V\bigtriangleup{U}^{a-1,b,c-1}_{w-2}\right)$
		$\displaystyle+\sum\lambda^{2}_{a,b,c}\left(c{U}^{a,b+1,c-1}_{w}-acz_{2}\bigtriangleup V\bigtriangleup{U}^{a-1,b,c-1}_{w-2}\right)$
		$\displaystyle-\sum\lambda^{3}_{a,b,c}\left(acz_{3}\bigtriangleup V\bigtriangleup{U}^{a-1,b,c-1}_{w-2}\right)$
		$\displaystyle+\sum{\mu}^{1}_{p,q,r}\left(rz_{3}\bigtriangleup V\bigtriangleup{U}^{p,q+1,r-1}_{w-2}\right)$
		$\displaystyle+\sum{\mu}^{2}_{p,q,r}\left((p+q-1)z_{3}\bigtriangleup V\bigtriangleup{U}^{p,q,r}_{w-2}\right)$
		$\displaystyle+\sum{\mu}^{3}_{p,q,r}\left((p+q)z_{2}\bigtriangleup V\bigtriangleup{U}^{p,q,r}_{w-2}-rz_{1}\bigtriangleup V\bigtriangleup{U}^{p,q+1,r-1}_{w-2}\right)$
	$\displaystyle=$	$\displaystyle\sum\lambda^{1}_{a,b,c}(a+b){U}^{a,b,c}_{w}-\sum\lambda^{1}_{a,b,c}acz_{1}\bigtriangleup V\bigtriangleup{U}^{a-1,b,c-1}_{w-2}$
		$\displaystyle+\sum\lambda^{2}_{a,b,c}c{U}^{a,b+1,c-1}_{w}-\sum\lambda^{2}_{a,b,c}acz_{2}\bigtriangleup V\bigtriangleup{U}^{a-1,b,c-1}_{w-2}$
		$\displaystyle-\sum\lambda^{3}_{a,b,c}acz_{3}\bigtriangleup V\bigtriangleup{U}^{a-1,b,c-1}_{w-2}$
		$\displaystyle+\sum{\mu}^{1}_{p,q,r}rz_{3}\bigtriangleup V\bigtriangleup{U}^{p,q+1,r-1}_{w-2}$
		$\displaystyle+\sum{\mu}^{2}_{p,q,r}(p+q-1)z_{3}\bigtriangleup V\bigtriangleup{U}^{p,q,r}_{w-2}$
		$\displaystyle+\sum{\mu}^{3}_{p,q,r}(p+q)z_{2}\bigtriangleup V\bigtriangleup{U}^{p,q,r}_{w-2}-\sum{\mu}^{3}_{p,q,r}rz_{1}\bigtriangleup V\bigtriangleup{U}^{p,q+1,r-1}_{w-2}$
	$\displaystyle=$	$\displaystyle\sum\lambda^{1}_{a,b,c}(a+b){U}^{a,b,c}_{w}+\sum\lambda^{2}_{a,b,c}c{U}^{a,b+1,c-1}_{w}$
		$\displaystyle-z_{1}\bigtriangleup V\bigtriangleup(\sum\lambda^{1}_{a,b,c}ac{U}^{a-1,b,c-1}_{w-2}+\sum{\mu}^{3}_{p,q,r}r{U}^{p,q+1,r-1}_{w-2})$
		$\displaystyle+z_{2}\bigtriangleup V\bigtriangleup(\sum{\mu}^{3}_{p,q,r}(p+q){U}^{p,q,r}_{w-2}-\sum\lambda^{2}_{a,b,c}ac{U}^{a-1,b,c-1}_{w-2})$
		$\displaystyle+z_{3}\bigtriangleup V\bigtriangleup(-\sum\lambda^{3}_{a,b,c}ac{U}^{a-1,b,c-1}_{w-2}+\sum{\mu}^{1}_{p,q,r}r){U}^{p,q+1,r-1}_{w-2}+\sum{\mu}^{2}_{p,q,r}(p+q-1){U}^{p,q,r}_{w-2})$

Thus, we have linear equations

(E1)		$\displaystyle\sum\lambda^{1}_{a,b,c}(a+b){U}^{a,b,c}_{w}+\sum\lambda^{2}_{a,b,c}c{U}^{a,b+1,c-1}_{w}=0$
(E2)		$\displaystyle\sum\lambda^{1}_{a,b,c}ac{U}^{a-1,b,c-1}_{w-2}+\sum{\mu}^{3}_{p,q,r}r{U}^{p,q+1,r-1}_{w-2}=0$
(E3)		$\displaystyle\sum{\mu}^{3}_{p,q,r}(p+q){U}^{p,q,r}_{w-2}-\sum\lambda^{2}_{a,b,c}ac{U}^{a-1,b,c-1}_{w-2}=0$
(EF)		$\displaystyle\sum\lambda^{3}_{a,b,c}ac{U}^{a-1,b,c-1}_{w-2}-\sum{\mu}^{1}_{p,q,r}r{U}^{p,q+1,r-1}_{w-2}-\sum{\mu}^{2}_{p,q,r}(p+q-1){U}^{p,q,r}_{w-2}=0$

From (E1) $\sim$ (EF), we have generators

(6.16a)		$\displaystyle\lambda^{1}_{a,b,c}(a+b)+\lambda^{2}_{a,b-1,c+1}(c+1)$
(6.16b)		$\displaystyle\lambda^{1}_{p+1,q,r+1}(p+1)(r+1)+{\mu}^{3}_{p,q-1,r+1}(r+1)$
(6.16c)		$\displaystyle\lambda^{2}_{p+1,q,r+1}(p+1)(r+1)-{\mu}^{3}_{p,q,r}(2+p+q)$
(6.16d)		$\displaystyle\lambda^{3}_{p+1,q,r+1}(p+1)(r+1)-{\mu}^{1}_{p,q-1,r+1}(r+1)-{\mu}^{2}_{p,q,r}(p+q+1)$

We take linearly independent generators (6.16a) with leading term $\lambda^{1}_{a,b,c}(a+b)$ , (6.16c) with leading term ${\mu}^{3}_{p,q,r}$ , and (6.16d) with leading term ${\mu}^{2}_{p,q,r}$ . Then (6.16b) are linearly dependent on those of 3 types. Thus, the rank is $\binom{w+2}{2}-1+2\binom{w}{2}$ and the kernel dimension is $1+\binom{w}{2}+2\binom{w+2}{2}$ .

Kernel of $\displaystyle C_{w}^{[w]}\mathop{\to}^{\partial}C_{w-1}^{[w]}$

As a basis of $C_{w}^{[w]}$ , we have $U^{a,b,c}$ with $a+b+c=w$ and $z_{k}\bigtriangleup V\bigtriangleup U^{p,q,r}$ with $k=1,2,3$ and $p+q+r=w-2$ . A basis of $C_{w}^{[w]}$ is ${U}^{a,b,c}_{w}$ and $z_{k}\bigtriangleup V\bigtriangleup{U}^{p,q,r}_{w-2}$ . Consider a linear combination

X=\sum_{a+b+c=w}\lambda_{a,b,c}{U}^{a,b,c}_{w}+\sum_{k=1}^{3}\sum_{p+q+r=w-2}{\mu}^{k}_{p,q,r}z_{k}\bigtriangleup V\bigtriangleup{U}^{p,q,r}_{w-2}

by unknown scalars $\lambda_{a,b,c},{\mu}^{k}_{p,q,r}$ . Then

	$\displaystyle\partial X=$	$\displaystyle\sum\lambda_{a,b,c}acV\bigtriangleup{U}^{a-1,b,c-1}_{w-2}+\sum{\mu}^{1}_{p,q,r}(p+q+1)V\bigtriangleup{U}^{p,q,r}_{w-2}$
		$\displaystyle+\sum{\mu}^{2}_{p,q,r}rV\bigtriangleup{U}^{p,q+1,r-1}_{w-2}\;.$

Since $\partial X=V\bigtriangleup\sum{U}^{p,q,r}_{w-2}(\lambda_{p+1,q,r+1}(p+1)(r+1)+(1+p+q){\mu}^{1}_{p,q,r}+(1+r){\mu}^{2}_{p,q-1,r+1})$ , the rank is $\binom{w}{2}$ because the leading terms are ${\mu}^{1}_{p,q,r}$ , thus the kernel dimension is $(3\binom{w}{2}+\binom{w+2}{2})-\binom{w}{2}=2\binom{w}{2}+\binom{w+2}{2}$ .

Final table of chain complex

Theorem 6.2.

$\begin{array}[]{c|*{5}{c}}\text{weight}=w&w-1&w&w+1&w+2&w+3\\ \hline\cr\text{SpaceDim}&\binom{w}{2}&3\binom{w}{2}+\binom{w+2}{2}&3\binom{w}{2}+3\binom{w+2}{2}&\binom{w}{2}+3\binom{w+2}{2}&\binom{w+2}{2}\\ \ker\textrm{dim}&\binom{w}{2}&2\binom{w}{2}+\binom{w+2}{2}&1+\binom{w}{2}+2\binom{w+2}{2}&1+\binom{w+2}{2}&0\\ \hline\cr\text{Betti}&0&1&2&1&0\end{array}$

6.3 $\textrm{dim}[\mathfrak{g},\mathfrak{g}]=2$

Take a 3-dimensional Lie algebra whose derived subalgebra is 2-dimensional. Then we have a basis satisfying

[z_{1},z_{2}]=0,\ [z_{1},z_{3}]=z_{1},\ [z_{2},z_{3}]=\alpha z_{2}\quad(\alpha\neq 0)\;.

Now we have the multiplication (by the Schouten bracket) tables with notations $u_{1}=z_{1}\wedge z_{2}\;,\;u_{2}=z_{1}\wedge z_{3}\;,\;u_{3}=z_{2}\wedge z_{3}\;,\;z_{4}=V=z_{1}\wedge z_{2}\wedge z_{3}$ .

$\begin{array}[t]{c|*{3}{c}|c|*{3}{c}|}&z_{1}&z_{2}&z_{3}&z_{4}&u_{1}&u_{2}&u_{3}\\ \hline\cr z_{1}&0&0&z_{1}&0&0&0&-u_{1}\\ z_{2}&0&0&\alpha z_{2}&0&0&\alpha u_{1}&0\\ z_{3}&-z_{1}&-\alpha z_{2}&0&-(1+\alpha)V&-(1+\alpha)u_{1}&-u_{2}&-\alpha u_{3}\end{array}$ $\begin{array}[t]{c|*{3}{c}|c}&u_{1}&u_{2}&u_{3}&z_{4}\\ \hline\cr u_{1}&0&0&0&0\\ u_{2}&0&0&(1-\alpha)z_{4}&0\\ u_{3}&0&(1-\alpha)z_{4}&0&0\\ \end{array}$

Then we have

	$\displaystyle\partial W^{1111}$	$\displaystyle=-2(1+\alpha)z_{1}\bigtriangleup z_{2}\bigtriangleup z_{4}\;,\quad\partial W^{1110}=-(1+\alpha)z_{1}\bigtriangleup z_{2}\;,\quad$
	$\displaystyle\partial W^{1101}$	$\displaystyle=0\;,\quad\partial W^{1011}=(2+\alpha)z_{1}\bigtriangleup z_{4}\;,\quad\partial W^{0111}=(2\alpha+1)z_{2}\bigtriangleup z_{4}\;,\quad$
	$\displaystyle\partial U^{a,b,c}$	$\displaystyle=z_{4}\bigtriangleup(bcU^{a,b-1,c-1})\;,\quad\sum_{A}\lambda^{i}_{A}\partial U^{a,b,c}=z_{4}\bigtriangleup\sum_{i}U^{p,q,r}((q+1)(r+1)\lambda^{i}_{p,q+1,r+1})\;.$
	$\displaystyle[z_{1},U^{A}]_{res}$	$\displaystyle=-cU^{a+1,b,c-1}\;,\quad\sum_{A}\lambda^{i}_{A}[z_{1},U^{A}]_{res}=\sum_{A}(-(c+1)\lambda^{i}_{a-1,b,c+1})U^{A}\;,$
	$\displaystyle[z_{2},U^{A}]_{res}$	$\displaystyle=\alpha bU^{a+1,b-1,c}\;,\quad\sum_{A}\lambda^{i}_{A}[z_{2},U^{A}]_{res}=\sum_{A}(\alpha(b+1)\lambda^{i}_{a-1,b+1,c})U^{A}\;,$
	$\displaystyle[z_{3},U^{A}]_{res}$	$\displaystyle=(-(1+\alpha)a-b-\alpha c)U^{a,b,c}\;.$

Remark 6.1.

When the weight is 0, the usual Lie algebra homology groups are obtained as below:

$\begin{array}[]{c|*{4}{c}}&\Lambda^{0}\mathfrak{g}&\Lambda^{1}\mathfrak{g}&\Lambda^{2}\mathfrak{g}&\Lambda^{3}\mathfrak{g}\\ \hline\cr\text{SpaceDim}&1&3&3&1\\ \ker\textrm{dim}&1&3&1&\kappa\\ \hline\cr\text{Betti}&1&1&\kappa&\kappa\\ \hline\cr\end{array}$

Since $\partial(z_{1}\wedge z_{2}\wedge z_{3})=-(1+\alpha)z_{1}\wedge z_{2}$ , we have $\kappa=\begin{cases}1&\text{if\quad}\alpha=-1\\ 0&\text{if\quad}\alpha\neq-1\end{cases}$ . Encountering with this phenomena, we are interested in how the super homology groups depend on non-zero $\alpha$ , which is a parameter of the Lie algebra structures.

The space of cycles in $C_{w+3}^{[w]}$ :

We study the space of cycles in $C_{w+3}^{[w]}$ . Take $W^{1,1,1,0}\bigtriangleup U^{a,b,c}$ with $a+b+c=w$ as a basis of the chain space. Consider a general chain $\sum_{a+b+c=w}\lambda_{a,b,c}W^{1,1,1,0}\bigtriangleup U^{a,b,c}$ with unknown scalars $\lambda_{a,b,c}$ .

		$\displaystyle\partial(\sum_{a+b+c=w}\lambda_{a,b,c}W^{1,1,1,0}\bigtriangleup U^{a,b,c})$
	$\displaystyle=$	$\displaystyle W^{0,1,1,0}\bigtriangleup\sum_{a+b+c=w}\lambda_{a,b,c}cU^{a+1,b,c-1}-W^{1,0,1,0}\bigtriangleup\sum_{a+b+c=w}\lambda_{a,b,c}\alpha bU^{a+1,b-1,c}$
		$\displaystyle-W^{1,1,0,0}\bigtriangleup\sum_{a+b+c=w}\lambda_{a,b,c}((1+\alpha)(a+1)+b+\alpha c)U^{a,b,c}$
		$\displaystyle-W^{1,1,1,1}\bigtriangleup\sum_{a+b+c=w}\lambda_{a,b,c}bc(1-\alpha)U^{a,b-1,c-1}$

Now assume that the above is zero. Then we have


(6.17a)		$\displaystyle\sum_{a+b+c=w}\lambda_{a,b,c}cU^{a+1,b,c-1}=0$
(6.17b)		$\displaystyle\sum_{a+b+c=w}\lambda_{a,b,c}bU^{a+1,b-1,c}=0$
(6.17c)		$\displaystyle\sum_{a+b+c=w}\lambda_{a,b,c}((1+\alpha)(1+a)+b+\alpha c)U^{a,b,c}=0$
(6.17d)		$\displaystyle\sum_{a+b+c=w}\lambda_{a,b,c}bc(1-\alpha)U^{a,b-1,c-1}=0$
(6.17a) $\sim$ (6.17d) yield generators of linear equation system as follows:

(6.17e)		$\displaystyle\lambda_{a-1,b,c+1}(c+1)$
(6.17f)		$\displaystyle\lambda_{a-1,b+1,c}(b+1)$
(6.17g)		$\displaystyle\lambda_{a,b,c}((1+\alpha)(1+a)+b+\alpha c)$
(6.17h)		$\displaystyle\lambda_{a,b+1,c+1}(b+1)(c+1)(1-\alpha)$
If $\alpha=1$ , then we see directly $\lambda_{A}$ are the linearly independent generators by (6.17g), and so the rank is $\binom{w+2}{2}$ .

If $\alpha\neq 1$ , (6.17h) provide $\lambda_{a,b,c}$ with $bc>0$ , (6.17e) provide $\lambda_{a,0,c}$ with $c>0$ , (6.17f) provide $\lambda_{a,b,0}$ with $b>0$ , and (6.17g) provide $\lambda_{a,0,0}(1+\alpha)(1+a)$ . Thus, the rank is $\binom{w+2}{2}-1$ if $\alpha+1=0$ or the rank is $\binom{w+2}{2}$ if $\alpha+1\neq 0$ .

The space of cycles in $C_{w+2}^{[w]}$ :

We study the space of cycles in $C_{w+2}^{[w]}$ . Take $W[\epsilon_{i}=0,\epsilon_{4}=0]\bigtriangleup{U}^{a,b,c}_{w}$ for $i=1,2,3$ and $W^{1,1,1,1}\bigtriangleup{U}^{p,q,r}_{w-2}$ as a basis of the chain space, and consider a general chain $\sum\lambda^{i}_{a,b,c}W[{\epsilon_{i}=0,\epsilon_{4}=0}]\bigtriangleup{U}^{a,b,c}_{w}+\sum{\mu}_{p,q,r}W^{1,1,1,1}\bigtriangleup{U}^{p,q,r}_{w-2}$ with unknown scalars $\lambda^{i}_{a,b,c},{\mu}_{p,q,r}$ .

		$\displaystyle\ \partial(\sum\lambda^{i}_{a,b,c}W[{\epsilon_{i}=0,\epsilon_{4}=0}]\bigtriangleup{U}^{a,b,c}_{w}+\sum{\mu}_{p,q,r}W^{1,1,1,1}\bigtriangleup{U}^{p,q,r}_{w-2})$
	$\displaystyle=$	$\displaystyle\ +W^{1,0,0,0}\bigtriangleup\sum(\lambda^{2}_{a,b,c}((1+\alpha)a+b+\alpha c+1){U}^{a,b,c}_{w}-\lambda^{3}_{a,b,c}\alpha b{U}^{a+1,b-1,c}_{w})$
		$\displaystyle+W^{0,1,0,0}\bigtriangleup\sum(\lambda^{1}_{a,b,c}((1+\alpha)a+b+\alpha c+\alpha){U}^{a,b,c}_{w}-\lambda^{3}_{a,b,c}c{U}^{a+1,b,c-1}_{w})$
		$\displaystyle+W^{0,0,1,0}\bigtriangleup\sum(\alpha\lambda^{1}_{a,b,c}b{U}^{a+1,b-1,c}_{w}-\lambda^{2}_{a,b,c}c{U}^{a+1,b,c-1}_{w})$
		$\displaystyle+W^{0,1,1,1}\bigtriangleup\left(-\sum{\mu}_{p,q,r}r{U}^{p+1,q,r-1}_{w-2}+\sum\lambda^{1}_{a,b,c}(1-\alpha)bc{U}^{a,b-1,c-1}_{w-2}\right)$
		$\displaystyle+W^{1,0,1,1}\bigtriangleup\left(\sum{\mu}_{p,q,r}(-\alpha q){U}^{p+1,q-1,r}_{w-2}+\sum\lambda^{2}_{a,b,c}(1-\alpha)bc{U}^{a,b-1,c-1}_{w-2}\right)$
		$\displaystyle+W^{1,1,0,1}\bigtriangleup\left(\sum{\mu}_{p,q,r}(-(1+\alpha)(p+2)+q+\alpha r){U}^{p,q,r}_{w-2}+\sum\lambda^{3}_{a,b,c}(1-\alpha)bc{U}^{a,b-1,c-1}_{w-2}\right)$

We pick up the coefficient of $W^{\mathcal{E}}\bigtriangleup U^{a,b,c}$ :


(6.18a)	$\displaystyle W^{1,0,0,0}$	$\displaystyle:\quad\lambda^{2}_{a,b,c}((1+\alpha)a+b+\alpha c+1)-\alpha\lambda^{3}_{a-1,b+1,c}(b+1)$
(6.18b)	$\displaystyle W^{0,1,0,0}$	$\displaystyle:\quad\lambda^{1}_{a,b,c}((1+\alpha)a+b+\alpha c+\alpha)-\lambda^{3}_{a-1,b,c+1}(c+1)$
(6.18c)	$\displaystyle W^{0,0,1,0}$	$\displaystyle:\quad\alpha\lambda^{1}_{a-1,b+1,c}(b+1)-\lambda^{2}_{a-1,b,c+1}(c+1)$
(6.18d)	$\displaystyle W^{0,1,1,1}$	$\displaystyle:\quad-{\mu}_{p-1,q,r+1}(r+1)+(1-\alpha)\lambda^{1}_{p,q+1,r+1}(q+1)(r+1)$
(6.18e)	$\displaystyle W^{1,0,1,1}$	$\displaystyle:\quad-\alpha{\mu}_{p-1,q+1,r}(q+1)+(1-\alpha)\lambda^{2}_{p,q+1,r+1}(q+1)(r+1)$
(6.18f)	$\displaystyle W^{1,1,0,1}$	$\displaystyle:\quad-{\mu}_{p,q,r}((1+\alpha)(p+2)+q+\alpha r)+(1-\alpha)\lambda^{3}_{p,q+1,r+1}(q+1)(r+1)$

If $\alpha=1$ , then (6.18d) $\sim$ (6.18f) yield independent generators ${\mu}_{p,q,r}$ with $p+q+r=w-2$ . (6.18a) and (6.18b) are independent generators which express (6.18c). Thus, the rank is $\binom{w}{2}+2\binom{w+2}{2}$ .

If $\alpha\neq 1$ , (6.18d) $\sim$ (6.18f) yield independent generators. Be careful that (6.18f) with $q=r=0$ is $-{\mu}_{p,0,0}(1+\alpha)(p+2)+(1-\alpha)\lambda^{3}_{p,1,1}$ . We need more generators from (6.18a) $\sim$ (6.18c) with the condition $bc=0$ . (6.18a) with $b=0,c=1$ is $-\lambda^{2}_{w-1,0,1}(1+\alpha)w-\alpha\lambda^{3}_{w-2,1,1}$ . (6.18b) with $b=1,c=0$ is $\lambda^{1}_{w-1,1,0}(1+\alpha)w-\lambda^{3}_{w-2,1,1}$ .

If $\alpha+1\neq 0$ , then (6.18c) are expressed by (6.18a) and (6.18b), the rank is $3\binom{w}{2}+2(2w+1)=\binom{w}{2}+2\binom{w+2}{2}$ .

If $\alpha+1=0$ , as described above, we have 3 cases which mean the same generator $\lambda^{3}_{w-2,1,1}$ . Depending to those cases, we need relation $\alpha\lambda^{1}_{w-1,1,0}-\lambda^{2}_{w-1,0,1}$ from (6.18c). Thus, the rank is $3\binom{w}{2}+2(w+1)-2+1=\binom{w}{2}+2\binom{w+2}{2}-1$ .

The space of cycles in $C_{w+1}^{[w]}$ :

When $w=0$ , then $C_{w+1}^{[w]}=C_{1}^{[0]}=\mathfrak{g}$ and the boundary operator is trivial, and so the kernel dimension is $\textrm{dim}\mathfrak{g}$ . Thus, we study the space of cycles in $C_{w+1}^{[w]}$ for $w>0$ . Take $z_{i}\bigtriangleup{U}^{a,b,c}_{w}$ and $z_{i}\bigtriangleup z_{j}\bigtriangleup z_{4}\bigtriangleup{U}^{a,b,c}_{w-2}$ ( $1\leqq i<j\leqq 3$ ) as a basis of the chain space, and consider a general chain $L+M$ where $L=\sum\lambda^{i}_{a,b,c}z_{i}\bigtriangleup{U}^{a,b,c}_{w}$ and $M=\sum{\mu}^{i}_{p,q,r}W[\epsilon_{i}=0,\epsilon_{4}=1]\bigtriangleup{U}^{p,q,r}_{w-2}$ with unknown scalars $\lambda^{i}_{a,b,c},{\mu}^{i}_{p,q,r}$ .

	$\displaystyle\partial L=$	$\displaystyle-z_{1}\bigtriangleup\sum\lambda^{1}_{a,b,c}\partial{U}^{a,b,c}_{w}-\sum\lambda^{1}_{a,b,c}c{U}^{a+1,b,c-1}_{w}$
		$\displaystyle-z_{2}\bigtriangleup\sum\lambda^{2}_{a,b,c}\partial{U}^{a,b,c}_{w}+\sum\lambda^{2}_{a,b,c}\alpha b{U}^{a+1,b-1,c}_{w}$
		$\displaystyle-z_{3}\bigtriangleup\sum\lambda^{3}_{a,b,c}\partial{U}^{a,b,c}_{w}-\sum\lambda^{3}_{a,b,c}((1+\alpha)a+b+\alpha c){U}^{a,b,c}_{w}$

Using the fact $[z_{i}\bigtriangleup z_{j}\bigtriangleup z_{4},U^{a,b,c}]_{res}=z_{j}\bigtriangleup z_{4}\bigtriangleup[z_{i},U^{a,b,c}]_{res}-z_{i}\bigtriangleup z_{4}\bigtriangleup[z_{j},U^{a,b,c}]_{res}$ we have

	$\displaystyle\partial M=$	$\displaystyle+W^{0,1,0,1}\bigtriangleup\sum{\mu}^{1}_{P}(1+2\alpha+(1+\alpha)p+q+\alpha r)U^{P}+W^{0,0,1,1}\bigtriangleup\sum{\mu}^{1}_{P}\alpha qU^{p+1,q-1,r}$
		$\displaystyle+W^{1,0,0,1}\bigtriangleup\sum{\mu}^{2}_{P}(2+\alpha+(1+\alpha)p+q+\alpha r)U^{P}+W^{0,0,1,1}\bigtriangleup\sum{\mu}^{2}_{P}(-r)U^{p+1,q,r-1}$
		$\displaystyle+W^{1,0,0,1}\bigtriangleup\sum{\mu}^{3}_{P}(-\alpha rU^{p,q+1,r-1})+W^{0,1,0,1}\bigtriangleup\sum{\mu}^{3}_{P}(-r)U^{p+1,q,r-1}$

Assume that $\partial(L+M)=0$ . Knowing $\partial U^{a,b,c}=z_{4}\bigtriangleup(1-\alpha)bcU^{a,b-1,c-1}$ ,


	$\displaystyle 0$	$\displaystyle=\sum\lambda^{1}_{a,b,c}(1-\alpha)bc{U}^{a,b-1,c-1}_{w}-\sum{\mu}^{2}_{P}(2+\alpha+(1+\alpha)p+q+\alpha r)U^{P}$
		$\displaystyle\qquad+\sum{\mu}^{3}_{P}(\alpha r)U^{p,q+1,r-1}$
	$\displaystyle 0$	$\displaystyle=\sum\lambda^{2}_{a,b,c}(1-\alpha)bc{U}^{a,b-1,c-1}_{w}-\sum{\mu}^{1}_{P}(1+2\alpha+(1+\alpha)p+q+\alpha r)U^{P}$
		$\displaystyle\qquad-\sum{\mu}^{3}_{P}(-r)U^{p+1,q,r-1}$
	$\displaystyle 0$	$\displaystyle=\sum\lambda^{3}_{a,b,c}(1-\alpha)bc{U}^{a,b-1,c-1}_{w}-\sum{\mu}^{1}_{P}\alpha qU^{p+1,q-1,r}-\sum{\mu}^{2}_{P}(-r)U^{p+1,q,r-1}$
	$\displaystyle 0$	$\displaystyle=-\sum\lambda^{1}_{a,b,c}c{U}^{a+1,b,c-1}_{w}+\sum\lambda^{2}_{a,b,c}\alpha b{U}^{a+1,b-1,c}_{w}$
		$\displaystyle\qquad-\sum\lambda^{3}_{a,b,c}((1+\alpha)a+b+\alpha c){U}^{a,b,c}_{w}$
Taking coefficient of ${U}^{a,b,c}$ , we have

(6.19a)			$\displaystyle(1-\alpha)\lambda^{1}_{p,q+1,r+1}(q+1)(r+1)-{\mu}^{2}_{p,q,r}(2+\alpha+(1+\alpha)p+q+\alpha r)$
		$\displaystyle\qquad+\alpha{\mu}^{3}_{p,q-1,r+1}(r+1)$
(6.19b)			$\displaystyle(1-\alpha)\lambda^{2}_{p,q+1,r+1}(q+1)(r+1)-{\mu}^{1}_{p,q,r}(1+2\alpha+(1+\alpha)p+q+\alpha r)$
		$\displaystyle\qquad+{\mu}^{3}_{p-1,q,r+1}(r+1)$
(6.19c)			$\displaystyle(1-\alpha)\lambda^{3}_{p,q+1,r+1}(q+1)(r+1)-\alpha{\mu}^{1}_{p-1,q+1,r}(q+1)+{\mu}^{2}_{p-1,q,r+1}(r+1)$
(6.19d)			$\displaystyle-\lambda^{1}_{a-1,b,c+1}(c+1)+\alpha\lambda^{2}_{a-1,b+1,c}(b+1)-\lambda^{3}_{a,b,c}((1+\alpha)a+b+\alpha c)$

Assume $1-\alpha=0$ .

We have

(6.20a)		$\displaystyle-{\mu}^{2}_{p,q,r}(1+w+p)+{\mu}^{3}_{p,q-1,r+1}(r+1)$
(6.20b)		$\displaystyle-{\mu}^{1}_{p,q,r}(1+w+p)+{\mu}^{3}_{p-1,q,r+1}(r+1)$
(6.20c)		$\displaystyle-{\mu}^{1}_{p-1,q+1,r}(q+1)+{\mu}^{2}_{p-1,q,r+1}(r+1)$
(6.20d)		$\displaystyle-\lambda^{1}_{a-1,b,c+1}(c+1)+\lambda^{2}_{a-1,b+1,c}(b+1)-\lambda^{3}_{a,b,c}(w+a)$

When $\alpha=1$ , from (6.19a) and (6.19b), we have $2\binom{w}{2}$ generators and (6.19c) are controlled by (6.19a) and (6.19b). (6.19d) yield $\binom{w+2}{2}$ generators. Thus, the rank is $2\binom{w}{2}+\binom{w+2}{2}$ .

If $\alpha\neq 1$ , from (6.19a) $\sim$ (6.19c), we have $3(\binom{w+2}{2}-(2w+1))$ generators, and (6.19d) yield generators depending on $(a,b,c)$ with $bc=0$ . Thus, the rank is $3(\binom{w+2}{2}-(2w+1))+(2w+1)=3\binom{w+2}{2}-2(2w+1)=\binom{w+2}{2}+2\binom{w}{2}$ .

The space of cycles in $C_{w}^{[w]}$ :

When $w=0$ , then $C_{w}^{[w]}=\Lambda^{0}\mathfrak{g}$ and the boundary operator is trivial, and so the kernel dimension is 1. Thus, we study the space of cycles in $C_{w}^{[w]}$ ) when $w>0$ . Take ${U}^{a,b,c}_{w}$ and $z_{i}\bigtriangleup z_{4}\bigtriangleup{U}^{p,q,r}_{w-2}$ ( $i=1,2,3$ ) as a basis of the chain space, and consider a general chain $L+M$ where $L=\sum\lambda_{a,b,c}{U}^{a,b,c}_{w}$ and $M=\sum{\mu}^{i}_{p,q,r}z_{i}\bigtriangleup z_{4}\bigtriangleup{U}^{p,q,r}_{w-2}$ with unknown scalars $\lambda_{a,b,c},{\mu}^{i}_{p,q,r}$ .

\displaystyle\partial L=

\displaystyle(1-\alpha)z_{4}\bigtriangleup\sum\lambda_{a,b,c}bc{U}^{a,b-1,c-1}_{w-2}=(1-\alpha)z_{4}\bigtriangleup\sum\lambda_{a,b+1,c+1}(b+1)(c+1){U}^{a,b,c}_{w-2}

Using the facts $z_{4}\bigtriangleup\partial U^{A}=0$ and $[z_{i}\bigtriangleup z_{4},U^{A}]_{res}=z_{4}\bigtriangleup[z_{i},U^{A}]_{res}$ , we have

	$\displaystyle\partial M=$	$\displaystyle-z_{4}\bigtriangleup\sum{\mu}^{1}_{p,q,r}c{U}^{p+1,q,r-1}_{w-2}+z_{4}\bigtriangleup\sum{\mu}^{2}_{p,q,r}\alpha q{U}^{p+1,q-1,r}_{w-2}$
		$\displaystyle-z_{4}\bigtriangleup\sum{\mu}^{3}_{p,q,r}(1+\alpha){U}^{p,q,r}_{w-2}-z_{4}\bigtriangleup\sum{\mu}^{3}_{p,q,r}((1+\alpha)p+q+\alpha r){U}^{p,q,r}_{w-2}$
	$\displaystyle=$	$\displaystyle z_{4}\bigtriangleup\sum{U}^{p,q,r}_{w-2}(-{\mu}^{1}_{p-1,q,r+1}(r+1)+{\mu}^{2}_{p-1,q+1,r}\alpha(q+1)-{\mu}^{3}_{p,q,r}((1+\alpha)(p+1)+q+\alpha r))$

The linear equation $\partial(L+M)=0$ is generated by

(6.21)			$\displaystyle\quad(\alpha-1)\lambda_{p,q+1,r+1}(q+1)(r+1)$
		$\displaystyle+{\mu}^{1}_{p-1,q,r+1}(r+1)-{\mu}^{2}_{p-1,q+1,r}\alpha(q+1)+{\mu}^{3}_{p,q,r}((1+\alpha)(p+1)+q+\alpha r)$

If $w<2$ , then generators are $(\alpha-1)\lambda_{p,q+1,r+1}(q+1)(r+1)$ , and we get none. Thus, the rank is 0.

Assume that $w\geqq 2$ . If $\alpha\neq 1$ , then we have $\binom{w}{2}$ linearly independent generators with leading term $\lambda_{p,q+1,r+1}$ , and the rank is $\binom{w}{2}$ . If $\alpha=1$ , then generators are ${\mu}^{1}_{p-1,q,r+1}(r+1)-{\mu}^{2}_{p-1,q+1,r}\alpha(q+1)+{\mu}^{3}_{p,q,r}(p+w)$ with leading term ${\mu}^{3}_{p,q,r}$ , and the rank is $\binom{w}{2}$ and the kernel dimension is $2\binom{w}{2}+\binom{w+2}{2}$ for two cases.

The space of cycles in $C_{w-1}^{[w]}$ :

Generators of this chain space are $z_{4}\bigtriangleup{U}^{A}_{w-2}$ .

Since $\partial(z_{4}\bigtriangleup{U}^{A}_{w-2})=-z_{4}\bigtriangleup\partial({U}^{A}_{w-2})+[z_{4},{U}^{A}_{w-2}]_{res}=-z_{4}\bigtriangleup z_{4}\wedge(1-\alpha)bc{U}^{a,b-1,c-1}_{w-2}+0=0$ , the boundary operator is trivial and the kernel dimension is $\binom{w}{2}$ .

Final table of chain complex

Theorem 6.3.

$\begin{array}[]{c|*{5}{c}}\text{weight}=w>0&w-1&w&w+1&w+2&w+3\\ \hline\cr\text{SpaceDim}&\binom{w}{2}&3\binom{w}{2}+\binom{w+2}{2}&3\binom{w}{2}+3\binom{w+2}{2}&\binom{w}{2}+3\binom{w+2}{2}&\binom{w+2}{2}\\ \ker\textrm{dim}&\binom{w}{2}&2\binom{w}{2}+\binom{w+2}{2}&\binom{w}{2}+2\binom{w+2}{2}&\binom{w+2}{2}+\kappa&\kappa\\ \hline\cr\text{Betti}&0&0&\kappa&2\kappa&\kappa\end{array}$

where $\kappa=\begin{cases}1&\text{if\quad}\alpha=-1\\ 0&\text{if\quad}\alpha\neq-1\end{cases}$ .

6.4 $\textrm{dim}[\mathfrak{g},\mathfrak{g}]=3$

3-dimensional Lie algebras with $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$ are completely classified in [1]. By an equivalence, called multiplicativity cogredience, the bracket relations are given by $[z_{1},z_{2}]=z_{3},[z_{1},z_{3}]=-\beta z_{2},[z_{2},z_{3}]=\alpha z_{1}$ . [1] says that if the base field of $\mathfrak{g}$ is $\mathbb{R}$ , then we may choose $\alpha=\beta=1$ or $-\alpha=\beta=1$ . Those include $A_{1}$ type by Dynkin diagram. Here, we care about two parameters $\alpha,\beta$ are just non-zero.

We choose a basis of 2-vectors by

u_{1}=\frac{1}{\alpha}z_{2}\wedge z_{3}\;,\;u_{2}=\frac{-1}{\beta}z_{1}\wedge z_{3}\;,\;u_{3}=z_{1}\wedge z_{2}

and 3-vector by $z_{4}=V=z_{1}\wedge z_{2}\wedge z_{3}$ . The multiplication (by the Schouten bracket) tables are the followings:

$\begin{array}[t]{c|*{3}{c}|c|*{3}{c} }&z_{1}&z_{2}&z_{3}&z_{4}&u_{1}&u_{2}&u_{3}\\ \hline\cr z_{1}&0&z_{3}&-\beta z_{2}&0&0&u_{3}&-\beta u_{2}\\ z_{2}&-z_{3}&0&\alpha z_{1}&0&-u_{3}&0&\alpha u_{1}\\ z_{3}&\beta z_{2}&-\alpha z_{1}&0&0&\beta u_{2}&-\alpha u_{1}&0\end{array}$ $\begin{array}[t]{c|*{3}{c}|c}&u_{1}&u_{2}&u_{3}&z_{4}\\ \hline\cr u_{1}&\frac{2}{\alpha}z_{4}&0&0&0\\ u_{2}&0&\frac{2}{\beta}z_{4}&0&0\\ u_{3}&0&0&2z_{4}&0\\ \end{array}$

From the tables, we have

	$\displaystyle\partial W^{1111}$	$\displaystyle=0\;,\quad\partial W^{1110}=0\;,\quad$
	$\displaystyle\partial W^{1101}$	$\displaystyle=z_{3}\bigtriangleup z_{4}\;,\quad\partial W^{1011}=-\beta z_{2}\bigtriangleup z_{4}\;,\quad\partial W^{0111}=\alpha z_{1}\bigtriangleup z_{4}\;,\quad$
	$\displaystyle\partial U^{a,b,c}$	$\displaystyle=z_{4}\bigtriangleup(\frac{2}{\alpha}\tbinom{a}{2}U^{a-2,b,c}+\frac{2}{\beta}\tbinom{b}{2}U^{a,b-2,c}+2\tbinom{c}{2}U^{a,b,c-2})\;.$
	$\displaystyle\sum_{A}\lambda_{A}\partial U^{a,b,c}$	$\displaystyle=z_{4}\bigtriangleup\sum U^{p,q,r}(\frac{2}{\alpha}\tbinom{p+2}{2}\lambda_{p+2,q,r}+\frac{2}{\beta}\tbinom{q+2}{2}\lambda_{p,q+2,r}+2\tbinom{r+2}{2}\lambda_{p,q,r+2})\;.$
	$\displaystyle[z_{1},U^{A}]_{res}$	$\displaystyle=bU^{a,b-1,c+1}-\beta cU^{a,b+1,c-1}$
	$\displaystyle[z_{2},U^{A}]_{res}$	$\displaystyle=-aU^{a-1,b,c+1}+\alpha cU^{a+1,b,c-1}$
	$\displaystyle[z_{3},U^{A}]_{res}$	$\displaystyle=\beta aU^{a-1,b+1,c}-\alpha bU^{a+1,b-1,c}$
	$\displaystyle\sum_{A}\lambda^{i}_{A}[z_{1},U^{A}]_{res}$	$\displaystyle=\sum_{A}((b+1)\lambda^{i}_{a,b+1,c-1}-\beta(c+1)\lambda^{i}_{a,b-1,c+1})U^{A}\;,$
	$\displaystyle\sum_{A}\lambda^{i}_{A}[z_{2},U^{A}]_{res}$	$\displaystyle=\sum_{A}(-(a+1)\lambda^{i}_{a+1,b,c-1}+\alpha(c+1)\lambda^{i}_{a-1,b,c+1})U^{A}\;,$
	$\displaystyle\sum_{A}\lambda^{i}_{A}[z_{3},U^{A}]_{res}$	$\displaystyle=\sum_{A}(\beta(a+1)\lambda^{i}_{a+1,b-1,c}-\alpha(b+1)\lambda^{i}_{a-1,b+1,c})U^{A}\;.$

The space of cycles in $C_{w+3}^{[w]}$ :

Take a general chain $L=\sum_{a+b+c=w}\lambda_{a,b,c}W^{1,1,1,0}\bigtriangleup U^{a,b,c}$ with unknown scalars $\lambda_{a,b,c}$ .

	$\displaystyle\partial L=$	$\displaystyle\partial(\sum_{A}\lambda_{A}W^{1110}\bigtriangleup U^{A})$
	$\displaystyle=$	$\displaystyle\sum\lambda_{a,b,c}(\partial W^{1,1,1,0})\bigtriangleup U^{A}-W^{1110}\sum\lambda_{A}\bigtriangleup\partial U^{A}+\sum_{i=1}^{3}(-1)^{i+1}W^{\epsilon_{i}=\epsilon_{4}=0}\sum\lambda_{A}[z_{i},U^{A}]_{res}$
	$\displaystyle=$	$\displaystyle-W^{1111}\bigtriangleup\sum\lambda_{A}(\frac{2}{\alpha}\tbinom{a}{2}U^{a-2,b,c}+\frac{2}{\beta}\tbinom{b}{2}U^{a,b-2,c}+2\tbinom{c}{2}U^{a,b,c-2})$
		$\displaystyle+z_{2}\bigtriangleup z_{3}\bigtriangleup\sum\lambda_{A}(bU^{a,b-1,c+1}-\beta cU^{a,b+1,c-1})$
		$\displaystyle-z_{1}\bigtriangleup z_{3}\bigtriangleup\sum\lambda_{A}(-aU^{a-1,b,c+1}+\alpha cU^{a+1,b,c-1})$
		$\displaystyle+z_{1}\bigtriangleup z_{2}\bigtriangleup\sum\lambda_{A}(\beta aU^{a-1,b+1,c}-\alpha bU^{a+1,b-1,c})$

The linear equation $\partial L=0$ is generated by the followings:


(6.22a)		$\displaystyle\frac{(p+2)(p+1)}{\alpha}\lambda_{p+2,q,r}+\frac{(q+2)(q+1)}{\beta}\lambda_{p,q+2,r}+(r+2)(r+1)\lambda_{p,q,r+2}$
(6.22b)		$\displaystyle(b+1)\lambda_{a,b+1,c-1}-\beta(c+1)\lambda_{a,b-1,c+1}$
(6.22c)		$\displaystyle(a+1)\lambda_{a+1,b,c-1}-\alpha(c+1)\lambda_{a-1,b,c+1}$
(6.22d)		$\displaystyle\beta(a+1)\lambda_{a+1,b-1,c}-\alpha(b+1)\lambda_{a-1,b+1,c}$

When $w=0$ , then the unknown parameter is only one $\lambda_{0,0,0}$ and the above equations give no information. This means $\lambda_{0,0,0}$ is free, and so the kernel dimension is 1. In fact, $\partial(z_{1}\bigtriangleup z_{2}\bigtriangleup z_{3})=0$ .

When $w=1$ , the unknown parameters are $\lambda_{a,b,c}$ with $a+b+c=1$ . (6.22a) gives no information, but (6.22d) tells $\lambda_{1,0,0}=0$ , $\lambda_{0,1,0}=0$ . $\lambda_{0,0,1}=0$ comes from (6.22b) by $(a,b,c)=(1,0,0)$ . Thus, the kernel dimension is 0.

Here we assume $w\geqq 2$ . From (6.22a) $\sim$ (6.22c), we have generators of the linear equations. Here we denote them by

(6.23)	$\displaystyle G_{0}(p,q,r)$	$\displaystyle=\frac{2}{\alpha}\tbinom{p+2}{2}\lambda_{p+2,q,r}+\frac{2}{\beta}\tbinom{q+2}{2}\lambda_{p,q+2,r}+(r+2)(r+1)\lambda_{p,q,r+2}\;,$
(6.24)	$\displaystyle G_{1}(a,b,c)$	$\displaystyle=(b+1)\lambda_{a,b+1,c-1}-\beta(c+1)\lambda_{a,b-1,c+1}\;,$
(6.25)	$\displaystyle G_{2}(a,b,c)$	$\displaystyle=(a+1)\lambda_{a+1,b,c-1}-\alpha(c+1)\lambda_{a-1,b,c+1}\;,$
(6.26)	$\displaystyle G_{3}(a,b,c)$	$\displaystyle=\beta(a+1)\lambda_{a+1,b-1,c}-\alpha(b+1)\lambda_{a-1,b+1,c}\;.$

The above pictures suggest properties of $G_{1},G_{2},G_{3}$ are vertical, horizontal, or slant relationship, where $\Box$ means the barycenter. We call them twins and call $G_{0}$ triple stars. When $\Box$ move around the area $T=\{(a,b,c)\mid a+b+c=w\}$ , some $G_{i}(a,b,c)$ becomes single star near the border of $T$ . For instance, $G_{1}(a,0,c)=\lambda_{a,1,c-1}$ . By this single star and vertical relations, we see that $\lambda_{a,2b+1,c}$ are single star. By the same idea, we see $\lambda_{2a+1,b,c}$ and $\lambda_{a,b,2c+1}$ are single star. Take the particular $\lambda_{0,w-1,1}$ which is single star. If $w$ is odd, then $w-1$ which is the 2nd entry of the single star, and by applying vertical relation, $\lambda_{0,2b,c}$ are single star. Now applying horizontal relation, we conclude $\lambda_{a,b,c}$ are all single star when $w$ is odd. So the rank is $\binom{w+2}{2}$ .

Now assume $w=2k$ . Since $\lambda_{a,b,c}$ is single star if one of $a,b,c$ is odd, we care about $\lambda_{a,b,c}$ of even entries $a,b,c$ . Take

(6.27)	$\displaystyle H_{0}=$	$\displaystyle G_{0}(2p,2q,2r)=(2p+2)(2p+1)\lambda_{2p+2,2q,2r}+(2q+2)(2q+1)\beta\lambda_{2p,2q+2,2r}$
	$\displaystyle+(2r+2)(2r+1)\alpha\lambda_{2p,2q,2r+2}$
(6.28)	$\displaystyle H_{1}=$	$\displaystyle G_{1}(2p,1+2q,1+2r)=(2+2q)\lambda_{2p,2q+2,2r}-\beta(2+2r)\lambda_{2p,2q,2r+2}$
(6.29)	$\displaystyle H_{2}=$	$\displaystyle G_{2}(1+2p,2q,1+2r)=(2+2p)\lambda_{2p+2,2q,2r}-\alpha(2+2r)\lambda_{2p,2q,2r+2}$
(6.30)	$\displaystyle H_{3}=$	$\displaystyle G_{3}(1+2p,1+2q,2r)=\beta(2+2p)\lambda_{2p+2,2q,2r}-\alpha(2+2q)\lambda_{2p,2q,2r+2}$

We see that $\alpha H_{1}-\beta H_{2}+H_{3}=0$ and

(6.31)	$\displaystyle H_{0}=$	$\displaystyle\frac{2p+1}{\alpha}\left(H_{2}+\alpha(2+2r)\lambda_{2p,2q,2r+2}\right)+\frac{2q+1}{\beta}\left(H_{1}+\beta(2+2r)\lambda_{2p,2q,2r+2}\right)$
(6.32)		$\displaystyle+(2r+2)(2r+1)\alpha\lambda_{2p,2q,2r+2}$
(6.33)	$\displaystyle=$	$\displaystyle\frac{2p+1}{\alpha}H_{2}+\frac{2q+1}{\beta}H_{1}+\alpha(2w-1)(2r+2)\lambda_{2p,2q,2r+2}$

This shows $\lambda_{2p,2q,2r+2}$ is a single star. We also see that $\lambda_{2p+2,2q,2r}$ is a single star by (6.29) and $\lambda_{2p,2q+2,2r}$ is a single star by (6.30). Thus, every $\lambda_{a,b,c}$ is single star and the rank is $\binom{w+2}{2}$ .

The space of cycles in $C_{w+2}^{[w]}$ :

We study the space of cycles in $C_{w+2}^{[w]}$ . If $w=0$ , $z_{i}\bigtriangleup z_{j}$ ( $i<j\leqq 3$ ) are a basis and since $\partial(z_{1}\bigtriangleup z_{2})=z_{3}$ , $\partial(z_{1}\bigtriangleup z_{3})=-\beta z_{2}$ , and $\partial(z_{2}\bigtriangleup z_{3})=\alpha z_{1}$ , the rank is 3 and the kernel dimension is 0. If $w=1$ , $W[\epsilon_{i}=\epsilon_{4}=0]\bigtriangleup u_{j}$ are a basis of the chain space. Take a general chain $X=\sum_{i,j}\lambda^{i}_{j}W[\epsilon_{i}=\epsilon_{4}=0]\bigtriangleup u_{j}$ .

	$\displaystyle\partial X=$	$\displaystyle\sum_{j}\lambda^{1}_{j}(\alpha z_{1}\bigtriangleup u_{j}+z_{3}\bigtriangleup[z_{2},u_{j}]-z_{2}\bigtriangleup[z_{3},u_{j}])$
		$\displaystyle+\sum_{j}\lambda^{2}_{j}(-\beta z_{2}\bigtriangleup u_{j}+z_{3}\bigtriangleup[z_{1},u_{j}]-z_{1}\bigtriangleup[z_{3},u_{j}])$
		$\displaystyle+\sum_{j}\lambda^{3}_{j}(z_{3}\bigtriangleup u_{j}+z_{2}\bigtriangleup[z_{1},u_{j}]-z_{1}\bigtriangleup[z_{2},u_{j}])$
	$\displaystyle=$	$\displaystyle+\lambda^{1}_{1}(\alpha z_{1}\bigtriangleup u_{1}+z_{3}\bigtriangleup(-{u_{3}})-z_{2}\bigtriangleup(\beta{u_{2}})$
		$\displaystyle+\lambda^{1}_{2}(\alpha z_{1}\bigtriangleup u_{2}+z_{3}\bigtriangleup 0-z_{2}\bigtriangleup(-\alpha{u_{1}})+\lambda^{1}_{3}(\alpha z_{1}\bigtriangleup u_{3}+z_{3}\bigtriangleup\alpha{u_{1}}-z_{2}\bigtriangleup 0)$
		$\displaystyle+\lambda^{2}_{1}(-\beta z_{2}\bigtriangleup u_{1}+z_{3}\bigtriangleup 0-z_{1}\bigtriangleup\beta{u_{2}})$
		$\displaystyle+\lambda^{2}_{2}(-\beta z_{2}\bigtriangleup u_{2}+z_{3}\bigtriangleup{u_{3}}-z_{1}\bigtriangleup(-\alpha{u_{1}}))+\lambda^{2}_{3}(-\beta z_{2}\bigtriangleup u_{3}+z_{3}\bigtriangleup(-\beta{u_{2}})-z_{1}\bigtriangleup 0)$
		$\displaystyle+\lambda^{3}_{1}(z_{3}\bigtriangleup u_{1}+z_{2}\bigtriangleup 0-z_{1}\bigtriangleup(-{u_{3}}))$
		$\displaystyle+\lambda^{3}_{2}(z_{3}\bigtriangleup u_{2}+z_{2}\bigtriangleup{u_{3}}-z_{1}\bigtriangleup 0)+\lambda^{3}_{3}(z_{3}\bigtriangleup u_{3}+z_{2}\bigtriangleup(-\beta{u_{2}})-z_{1}\bigtriangleup\alpha{u_{1}})$

$\partial X=0$ is defined by 6 linearly independent polynomials $\lambda^{1}_{1}+\lambda^{2}_{2}-\lambda^{3}_{3}$ , $\lambda^{1}_{1}+\lambda^{2}_{2}+\lambda^{3}_{3}$ , $-\lambda^{1}_{1}+\lambda^{2}_{2}+\lambda^{3}_{3}$ , $\alpha\lambda^{1}_{2}-\beta\lambda^{2}_{1}$ , $\alpha\lambda^{1}_{3}+\lambda^{3}_{1}$ , $-\beta\lambda^{2}_{3}+\lambda^{3}_{2}$ . Thus the rank is 6 and the kernel dimension -s 3.

When $w>1$ , take a general chain $L+M$ with $L=\sum\lambda^{i}_{a,b,c}W[{\epsilon_{i}=\epsilon_{4}=0}]\bigtriangleup{U}^{a,b,c}_{w}$ and $M=\sum{\mu}_{p,q,r}W^{1111}\bigtriangleup{U}^{p,q,r}_{w-2}$ with unknown scalars $\lambda^{i}_{a,b,c},{\mu}_{p,q,r}$ .

Then

		$\displaystyle\partial(L+M)$
	$\displaystyle=$	$\displaystyle\sum\lambda^{i}_{A}(\partial(W[{\epsilon_{i}=\epsilon_{4}=0}])\bigtriangleup U^{A}+W[{\epsilon_{i}=\epsilon_{4}=0}]\bigtriangleup\partial U^{A})+[W[{\epsilon_{i}=\epsilon_{4}=0}],U^{A}]_{res}$
		$\displaystyle+0+0+\sum{\mu}_{P}[W^{1111},U^{P}]_{res}$
	$\displaystyle=$	$\displaystyle\alpha z_{1}\bigtriangleup\sum\lambda^{1}_{A}U^{A}-\beta z_{2}\bigtriangleup\sum\lambda^{2}_{A}U^{A}+z_{3}\bigtriangleup\sum\lambda^{3}_{A}U^{A}$
		$\displaystyle+W^{0111}\bigtriangleup\sum\lambda^{1}_{A}(\frac{2}{\alpha}\tbinom{a}{2}U^{a-2,b,c}+\frac{2}{\beta}\tbinom{b}{2}U^{a,b-2,c}+2\tbinom{c}{2}U^{a,b,c-2})$
		$\displaystyle+W^{1011}\bigtriangleup\sum\lambda^{2}_{A}(\frac{2}{\alpha}\tbinom{a}{2}U^{a-2,b,c}+\frac{2}{\beta}\tbinom{b}{2}U^{a,b-2,c}+2\tbinom{c}{2}U^{a,b,c-2})$
		$\displaystyle+W^{1101}\bigtriangleup\sum\lambda^{3}_{A}(\frac{2}{\alpha}\tbinom{a}{2}U^{a-2,b,c}+\frac{2}{\beta}\tbinom{b}{2}U^{a,b-2,c}+2\tbinom{c}{2}U^{a,b,c-2})$
		$\displaystyle+\sum\lambda^{1}_{A}(-z_{2}[z_{3},U^{A}]_{res}+z_{3}[z_{2},U^{A}]_{res})+\sum\lambda^{2}_{A}(-z_{1}[z_{3},U^{A}]_{res}+z_{3}[z_{1},U^{A}]_{res})$
		$\displaystyle+\sum\lambda^{3}_{A}(-z_{1}[z_{2},U^{A}]_{res}+z_{2}[z_{1},U^{A}]_{res})$
		$\displaystyle+W^{0111}\bigtriangleup{\mu}_{P}[z_{1},U^{P}]_{res}-W^{1011}\bigtriangleup{\mu}_{P}[z_{2},U^{P}]_{res}+W^{1101}\bigtriangleup{\mu}_{P}[z_{3},U^{P}]_{res}$
	$\displaystyle=$	$\displaystyle+z_{1}\bigtriangleup\sum(\alpha\lambda^{1}_{A}U^{A}-\lambda^{2}_{A}[z_{3},U^{A}]_{res}-\lambda^{3}_{A}[z_{2},U^{A}]_{res})$
		$\displaystyle+z_{2}\bigtriangleup\sum(-\beta\lambda^{2}_{A}U^{A}-\lambda^{1}_{A}[z_{3},U^{A}]_{res}+\lambda^{3}_{A}[z_{1},U^{A}]_{res})$
		$\displaystyle+z_{3}\bigtriangleup\sum(\lambda^{3}_{A}U^{A}+\lambda^{1}_{A}[z_{2},U^{A}]_{res}+\lambda^{2}_{A}[z_{1},U^{A}]_{res})$
		$\displaystyle+W^{0111}\bigtriangleup(\sum{\mu}_{P}[z_{1},U^{P}]_{res}+\sum\operatorname{L}_{1}{p,q,r}U^{p,q,r})$
		$\displaystyle+W^{1011}\bigtriangleup\sum(-{\mu}_{P}[z_{2},U^{P}]_{res}+\sum\operatorname{L}_{2}{p,q,r}U^{p,q,r})$
		$\displaystyle+W^{1101}\bigtriangleup\sum({\mu}_{P}[z_{3},U^{P}]_{res}+\sum\operatorname{L}_{3}{p,q,r}U^{p,q,r})$
where
	$\displaystyle\operatorname{L}_{i}{p,q,r}$	$\displaystyle=\frac{(p+2)(p+1)}{\alpha}\lambda^{i}_{p+2,q,r}+\frac{(q+2)(q+1)}{\beta}\lambda^{i}_{p,q+2,r}+(r+2)(r+1)\lambda^{i}_{p,q,r+2}\;.$

Then $\partial(L+M)=0$ is defined by the following polynomials of unknown $\lambda^{i}_{A}$ and ${\mu}_{P}$ :

(6.34)	$\displaystyle E^{1}_{a,b,c}=$	$\displaystyle\alpha\lambda^{1}_{a,b,c}-\beta(a+1)\lambda^{2}_{a+1,b-1,c}+\alpha(b+1)\lambda^{2}_{a-1,b+1,c}$
	$\displaystyle\quad+(a+1)\lambda^{3}_{a+1,b,c-1}-\alpha(c+1)\lambda^{3}_{a-1,b,c+1}$
(6.35)	$\displaystyle E^{2}_{a,b,c}=$	$\displaystyle-\beta\lambda^{2}_{a,b,c}-\beta(a+1)\lambda^{1}_{a+1,b-1,c}+\alpha(b+1)\lambda^{1}_{a-1,b+1,c}$
	$\displaystyle\quad+(b+1)\lambda^{3}_{a,b+1,c-1}-\beta(c+1)\lambda^{3}_{a,b-1,c+1}$
(6.36)	$\displaystyle E^{3}_{a,b,c}=$	$\displaystyle\lambda^{3}_{a,b,c}-(a+1)\lambda^{1}_{a+1,b,c-1}+\alpha(c+1)\lambda^{1}_{a-1,b,c+1}$
	$\displaystyle\quad+(b+1)\lambda^{2}_{a,b+1,c-1}-\beta(c+1)\lambda^{2}_{a,b-1,c+1}$
(6.37)	$\displaystyle E^{4}_{p,q,r}=$	$\displaystyle(q+1){\mu}_{p,q+1,r-1}-\beta(r+1){\mu}_{p,q-1,r+1}+\operatorname{L}_{1}{p,q,r}$
(6.38)	$\displaystyle E^{5}_{p,q,r}=$	$\displaystyle(p+1){\mu}_{p+1,q,r-1}-\alpha(r+1){\mu}_{p-1,q,r+1}+\operatorname{L}_{2}{p,q,r}$
(6.39)	$\displaystyle E^{6}_{p,q,r}=$	$\displaystyle\beta(p+1){\mu}_{p+1,q-1,r}-\alpha(q+1){\mu}_{p-1,q+1,r}+\operatorname{L}_{3}{p,q,r}$

6.4.1 Rank of ${\mu}_{p,q,r}$

We see that $E^{6}_{p,q,r}-\beta E^{5}_{p,q-1,r+1}+\alpha E^{4}_{p-1,q,r+1}$ does not have ${\mu}_{P}$ as follows:

(6.40)		$\displaystyle+p(p+1)\lambda^{1}_{p+1,q,r+1}+\alpha(r+3)(r+2)\lambda^{1}_{p-1,q,r+3}+\alpha(q+2)(q+1)/\beta\lambda^{1}_{p-1,q+2,r+1}$
	$\displaystyle-q(q+1)\lambda^{2}_{p,q+1,r+1}-\beta(r+3)(r+2)\lambda^{2}_{p,q-1,r+3}-\beta(p+2)(p+1)/\alpha\lambda^{2}_{p+2,q-1,r+1}$
	$\displaystyle+(r+2)(r+1)\lambda^{3}_{p,q,r+2}+(p+2)(p+1)/\alpha\lambda^{3}_{p+2,q,r}+(q+2)(q+1)/\beta\lambda^{3}_{p,q+2,r}\;.$

In order to know the rank concerning to ${\mu}_{p,q,r}$ , we handle one of three, $\operatorname{F}^{6}=E^{6}\negmedspace\mod\lambda^{i}_{A}$ , i.e., $\operatorname{F}^{6}_{p,q,r}=\beta(p+1){\mu}_{p+1,q-1,r}-\alpha(q+1){\mu}_{p-1,q+1,r}$ . Among $\operatorname{F}^{6}_{p,q,r}$ with $p+q+r=w-2$ , first fix $r$ and put $H=w-2-r$ and study of linearly independence of $\mathfrak{S}^{[w-2]}_{r}=\{\operatorname{F}^{6}_{p,q,r}\mid p+q=H=w-2-r\}$ as polynomials of ${\mu}_{P}$ . Basically, $\operatorname{F}^{6}_{p,q,r}$ looks like slant double stars with distance 2, we define

	$\displaystyle OD^{[r]}_{2i+1}$	$\displaystyle=\operatorname{F}^{6}_{1,H-1,r}\wedge\operatorname{F}^{6}_{3,H-3,r}\wedge\cdots\wedge\operatorname{F}^{6}_{2i+1,H-1-2i,r}\quad\mathtt{,where}\quad 2i+1\leqq H\;,$
		$\displaystyle=\mathop{\wedge}_{i=0}^{2i+1\leqq H}(-\alpha(H-2i){\mu}_{2i,H-2i,r}+\beta(2+2i){\mu}_{2i+2,H-2i-2,r})$
and
	$\displaystyle EV^{[r]}_{2i}$	$\displaystyle=\operatorname{F}^{6}_{0,H,r}\wedge\operatorname{F}^{6}_{2,H-2,r}\wedge\cdots\wedge\operatorname{F}^{6}_{2i,H-2i,r}\quad\mathtt{where}\quad 2i\leqq H\;.$
For instance,
	$\displaystyle EV^{[r]}_{0}$	$\displaystyle=\operatorname{F}^{6}_{0,H,r}=\beta{\mu}_{1,H-1,r}-\alpha(H+1){\mu}_{-1,H+1,r}=\beta{\mu}_{1,H-1,r}\;,$
	$\displaystyle EV^{[r]}_{2}$	$\displaystyle=EV^{[r]}_{0}\wedge\operatorname{F}^{6}_{2,H-2,r}=\beta{\mu}_{1,H-1,r}\wedge(\beta 3{\mu}_{3,H-3,r}-\alpha(q+1){\mu}_{1,H-1,r})$
		$\displaystyle=3\beta^{2}{\mu}_{1,H-1,r}\wedge{\mu}_{3,H-3,r}$
		$\displaystyle\vdots$
	$\displaystyle EV^{[r]}_{2k}$	$\displaystyle=(2k+1)!!\beta^{k+1}{\mu}_{1,H-1,r}\wedge{\mu}_{3,H-3,r}\wedge\cdots\wedge{\mu}_{2k+1,H-2k-1,r}$
If $H=2h$ even, then
	$\displaystyle EV^{[r]}_{2h}$	$\displaystyle=EV^{[r]}_{2h-2}\wedge\operatorname{F}^{6}_{2h,0,r}$
		$\displaystyle=(2h-1)!!\beta^{h}{\mu}_{1,H-1,r}\wedge\wedge\cdots\wedge{\mu}_{2h-1,1,r}\wedge(0-\alpha{\mu}_{2h-1,1,r})$
		$\displaystyle=0\;.$

Thus, we have the next lemma.

Lemma 6.1.

If $H=w-2-r=$ odd, then the set $\{\operatorname{F}^{6}_{p,q,r}\mid p+q=H=w-2-r\}$ consists of linearly independent polynomials.

If $H=w-2-r=$ even, then the set $\{\operatorname{F}^{6}_{p,q,r}\mid p+q=H=w-2-r\}$ is not linearly independent, but the set $\{\operatorname{F}^{6}_{p,q,r}\mid p+q=H=w-2-r\}\setminus\{\operatorname{F}^{6}_{H,0,r}\}$ consists of linearly independent polynomials.

As a direct corollary, we see that

Corollary 6.2.

The dimension of the space spanned by $E^{4},E^{5},E^{6}$ with leading term ${\mu}_{P}$ is $\binom{w}{2}$ if $w$ is odd, and is $\binom{w}{2}-1$ when $w$ is even.

6.4.2 Rank of $\lambda^{i}_{a,b,c}$

We define $\widehat{E}^{1}_{a,b,c}$ symbolically by $E^{1}_{a,b,c}/E^{3}_{A}$ and $\widehat{E}^{2}_{a,b,c}$ by $E^{2}_{a,b,c}/E^{3}_{A}$ because all $\lambda^{3}_{a^{\prime},b^{\prime},c^{\prime}}$ are completely determined by $\lambda^{1}_{A},\lambda^{2}_{A}$ in (6.36). $\widehat{E}^{1},\widehat{E}^{2}$ are linear polynomials of only $\lambda^{1}_{A},\lambda^{2}_{A}$ as follows:

(6.41)	$\displaystyle\widehat{E}^{1}_{a,b,c}$	$\displaystyle=+\alpha^{2}(c+2)(c+1)\lambda^{1}_{a-2,b,c+2}-\alpha(2ac+a+c-1)\lambda^{1}_{a,b,c}+(a+2)(a+1)\lambda^{1}_{a+2,b,c-2}$
	$\displaystyle+\alpha(c+2)(b+1)\lambda^{2}_{a-1,b+1,c}-(b+1)(a+1)\lambda^{2}_{a+1,b+1,c-2}$
	$\displaystyle-\alpha\beta(c+2)(c+1)\lambda^{2}_{a-1,b-1,c+2}+\beta(c-1)(a+1)\lambda^{2}_{a+1,b-1,c}$
(6.42)	$\displaystyle\widehat{E}^{2}_{a,b,c}$	$\displaystyle=-\alpha(c-1)(b+1)\lambda^{1}_{a-1,b+1,c}+(a+1)(b+1)\lambda^{1}_{a+1,b+1,c-2}$
	$\displaystyle+\alpha\beta(c+2)(c+1)\lambda^{1}_{a-1,b-1,c+2}-\beta(a+1)(c+2)\lambda^{1}_{a+1,b-1,c}$
	$\displaystyle-(b+2)(b+1)\lambda^{2}_{a,b+2,c-2}+\beta(2bc+b+c-1)\lambda^{2}_{a,b,c}-\beta^{2}(c+2)(c+1)\lambda^{2}_{a,b-2,c+2}$

In (6.41) or (6.42), almost all coefficients of each term are non-zero, but we see $c-1$ , $2ac+a+c-1$ or $2bc+b+c-1$ . We get special cases as below:

(6.43)		$\displaystyle\frac{1}{\alpha}\widehat{E}^{1}_{a,b,1}$	$\displaystyle=\frac{1}{\beta}\widehat{E}^{2}_{a-1,b+1,1}\;,$
(6.44)		$\displaystyle\widehat{E}^{1}_{0,w-1,1}$	$\displaystyle=0\;,\quad\widehat{E}^{2}_{w-1,0,1}=0\;.$

We saw $E^{6}_{p,q,r}-\beta E^{5}_{p,q-1,r+1}+\alpha E^{4}_{p-1,q,r+1}$ , (say $G$ ) does not have ${\mu}_{P}$ in (6.40). We define $\widetilde{G}=G/E^{3}_{A}$ . Then we have

	$\displaystyle\widetilde{G}=$	$\displaystyle+(p+3)(p+2)(p+1)/\alpha\lambda^{1}_{p+3,q,r-1}+(q+2)(q+1)(p+1)/\beta\lambda^{1}_{p+1,q+2,r-1}$
		$\displaystyle-r(p+1)(p-r-1)\lambda^{1}_{p+1,q,r+1}-\alpha r(r+3)(r+2)\lambda^{1}_{p-1,q,r+3}$
		$\displaystyle-\alpha r(q+2)(q+1)/\beta\lambda^{1}_{p-1,q+2,r+1}+r(q+1)(q-r-1)\lambda^{2}_{p,q+1,r+1}$
		$\displaystyle+\beta r(r+3)(r+2)\lambda^{2}_{p,q-1,r+3}+\beta r(p+2)(p+1)/\alpha\lambda^{2}_{p+2,q-1,r+1}$
		$\displaystyle-(p+2)(p+1)(q+1)/\alpha\lambda^{2}_{p+2,q+1,r-1}-(q+3)(q+2)(q+1)/\beta\lambda^{2}_{p,q+3,r-1}$

Now we define $\operatorname{F}^{1}_{a,b,c}$ symbolically by $\widehat{E}^{1}_{a,b,c}/\lambda^{2}_{A}$ . Since $\operatorname{F}^{1}_{a,b,c}$ looks like horizontal 3 stars with 2 step distance in its shape, we study of linear independence of the subset $\{\operatorname{F}^{1}_{a,b,c}\mid a+c=w-b\}$ and $\{\widehat{E}^{1}_{a,b,c}\mid a+c=w-b\}$ for each $b$ ( $0\leqq b\leqq w$ ) by the same discussion in the previous subsection, there are slant double stars with 2 step distance in its shape.

When $a+b+c=w$ , we fix $b$ and abbreviate $w-b$ as $H$ , $c$ is determined by $c=H-a$ , we use the following notations

\operatorname{F}^{1}_{a,b,c}=\operatorname{F}^{1,w}_{a,b,c}=\operatorname{F}^{1,w,b}_{a}=\operatorname{F}^{1,w}_{a}=\operatorname{F}^{1}_{a}\;,\;\lambda^{1}_{a,b,c}={L}^{w}_{a,b,c}={L}^{w,b}_{a}={L}^{w}_{a}={L}_{a}

unless there is confusion. The following are matrices of $\operatorname{F}^{1}$ with respect $\lambda^{1}$ in which the coefficients of simultaneous linear equations are arranged. Since the size is large, we write them separately with an even number of subscripts. They are tri-diagonal matrices.

Case of $H=w-b=1+2h=$ odd :

$\begin{array}[]{c | *{14}{c} }\hline\cr&{L}^{w,b}_{0}&{L}^{w,b}_{2}&{L}^{w,b}_{4}&&&&&{L}^{w,b}_{2h-4}&{L}^{w,b}_{2h-2}&{L}^{w,b}_{2h}\\ \hline\cr\operatorname{F}^{1,w,b}_{0}&-\alpha(H-1)&2&0\\ \operatorname{F}^{1,w,b}_{2}&\alpha^{2}H(H-1)&*&3\cdot 4\\ \\ \operatorname{F}^{1,w,b}_{a}&&&&\alpha^{2}(1+c)(2+c)&*&(1+a)(2+a)\\ \\ \operatorname{F}^{1,w,b}_{2h-2}&&&&&&&&20\alpha^{2}&*&(H-1)(H-2)\\ \operatorname{F}^{1,w,b}_{2h}&&&&&&&&0&6\alpha^{2}&-3\alpha(H-1)\\ \hline\cr\end{array}$

$\begin{array}[]{c | *{14}{c} }\hline\cr&{L}^{w,b}_{1}&{L}^{w,b}_{3}&{L}^{w,b}_{5}&&&&&{L}^{w,b}_{2h-3}&{L}^{w,b}_{2h-1}&{L}^{w,b}_{2h+1}\\ \hline\cr\operatorname{F}^{1,w,b}_{1}&-3\alpha(H-1)&2\cdot 3&0\\ \operatorname{F}^{1,w,b}_{3}&\alpha^{2}(H-1)(H-2)&*&4\cdot 5\\ \\ \operatorname{F}^{1,w,b}_{a}&&&&\alpha^{2}(1+c)(2+c)&*&(1+a)(2+a)\\ \\ \operatorname{F}^{1,w,b}_{2h-1}&&&&&&&&12\alpha^{2}&*&H(H-1)\\ \operatorname{F}^{1,w,b}_{2h+1}&&&&&&&&0&2\alpha^{2}&-\alpha(H-1)\\ \hline\cr\end{array}$

where $*=\alpha\left(1-(a+c+2ac)\right)$ .

From the shape of matrices, their rank is $h$ or $h+1$ , and we want to show the last line is not linearly independent.

Method of deforming triple stars to twins: This discussion works for any parity $H$ . Assume that we have triple stars $\widetilde{\text{F}}^{1,w,b}_{a}=K_{a}{L}^{w,b}_{a}+(a+1)(a+2){L}^{w,b}_{a+2}$ with $K_{a}\neq 0$ (from twins $\operatorname{F}^{1,w,b}_{a}$ ). We already have $\widetilde{\text{F}}^{1,w,b}_{0}=\operatorname{F}^{1,w,b}_{0}$ and $\widetilde{\text{F}}^{1,w,b}_{1}=\operatorname{F}^{1,w,b}_{1}$ with $K_{0}=-\alpha(H-1)$ , $K_{1}=-3\alpha(H-1)$ .

Define a twins by

(6.45)	$\displaystyle\widetilde{\text{F}}^{1,w,b}_{a+2}$	$\displaystyle=\operatorname{F}^{1,w,b}_{a+2}-\frac{\alpha^{2}(H-a)(H-1-a)}{K_{a}}\widetilde{\text{F}}^{1,w,b}_{a}$
(6.46)		$\displaystyle=K_{a+2}{L}^{w,b}_{a+2}+(a+3)(a+4){L}^{w,b}_{a+4}$
(6.47)	$\displaystyle K_{a+2}$	$\displaystyle=-\alpha(2Ha+5H-2a^{2}-8a-9)-\frac{\alpha^{2}(H-a)(H-1-a)}{K_{a}}(a+1)(a+2)$
Inductively, starting from 0 by step 2, we have
(6.48)	$\displaystyle K_{2a}$	$\displaystyle=-\alpha(2a+1)(H-1-2a)$
starting from 1 by step2 similarly, we have
(6.49)	$\displaystyle K_{2a+1}$	$\displaystyle=-\alpha(2+(2a+1))(H-(2a+1))$

$\begin{array}[]{c | *{14}{c} }\hline\cr&{L}^{w,b}_{0}&{L}^{w,b}_{2}&{L}^{w,b}_{4}&&&&&{L}^{w,b}_{2h-4}&{L}^{w,b}_{2h-2}&{L}^{w,b}_{2h}\\ \hline\cr\widetilde{\text{F}}^{1,w,b}_{0}&-\alpha(H-1)&2&0\\ \widetilde{\text{F}}^{1,w,b}_{2}&0&K_{2}&3\cdot 4\\ \\ \widetilde{\text{F}}^{1,w,b}_{a}&&&&0&K_{a}&(1+a)(2+a)\\ \\ \widetilde{\text{F}}^{1,w,b}_{2h-2}&&&&&&&&0&K_{2h-2}&(H-1)(H-2)\\ \widetilde{\text{F}}^{1,w,b}_{2h}&&&&&&&&0&6\alpha^{2}&-3\alpha(H-1)\\ \hline\cr\end{array}$

$\begin{array}[]{c | *{14}{c} }\hline\cr&{L}^{w,b}_{1}&{L}^{w,b}_{3}&{L}^{w,b}_{5}&&&&&{L}^{w,b}_{2h-3}&{L}^{w,b}_{2h-1}&{L}^{w,b}_{2h+1}\\ \hline\cr\widetilde{\text{F}}^{1,w,b}_{1}&-3\alpha(H-1)&2\cdot 3&0\\ \widetilde{\text{F}}^{1,w,b}_{3}&0&K_{3}&4\cdot 5\\ \\ \widetilde{\text{F}}^{1,w,b}_{a}&&&&0&K_{a}&(1+a)(2+a)\\ \\ \widetilde{\text{F}}^{1,w,b}_{2h-1}&&&&&&&&0&K_{2h-1}&H(H-1)\\ \widetilde{\text{F}}^{1,w,b}_{2h+1}&&&&&&&&0&2\alpha^{2}&-\alpha(H-1)\\ \hline\cr\end{array}$

$H=w-b=2h+1$

K_{2h-2}=-2(2h-1)\alpha\;,\quad K_{2h-1}=-2(2h+1)\alpha\;,\quad\begin{vmatrix}K_{2h-2}&(H-1)(H-2)\\ 6\alpha^{2}&-3\alpha(H-1)\end{vmatrix}=\begin{vmatrix}-2(2h-1)\alpha&2h(2h-1)\\ 6\alpha^{2}&-3\alpha 2h\end{vmatrix}=0

\begin{vmatrix}K_{2h-1}&H(H-1)\\ 2\alpha^{2}&-\alpha(H-1)\end{vmatrix}=\begin{vmatrix}-2(2h+1)\alpha&2h(2h+1)\\ 2\alpha^{2}&-\alpha 2h\end{vmatrix}=0

Thus,

	$\displaystyle\{\widetilde{\text{F}}^{1}_{2a,b,H-2a}\mid 0\leqq 2a\leqq H-3\}\quad\text{are linearly independent, and }$
	$\displaystyle\{\widetilde{\text{F}}^{1}_{2a,b,H-2a}\mid 0\leqq 2a\leqq H-1\}\quad\text{are not linearly independent. }$
	$\displaystyle\{\widetilde{\text{F}}^{1}_{2a+1,b,H-1-2a}\mid 1\leqq 2a+1\leqq H-2\}\quad\text{are linearly independent, and }$
	$\displaystyle\{\widetilde{\text{F}}^{1}_{2a+1,b,H-1-2a}\mid 1\leqq 2a+1\leqq H\}\quad\text{are not linearly independent. }$

Case of $H=w-b=2h=$ even : $\#\{\operatorname{F}^{1,w,b}_{2a}\}=\#\{{L}^{w,b}_{2a}\}=h+1\neq h=\#\{\operatorname{F}^{1,w,b}_{2a+1}\}=\#\{{L}^{w,b}_{2a+1}\}$ . This is different from the case of $H=$ odd.

$\begin{array}[]{c | *{14}{c} }\hline\cr&{L}^{w,b}_{0}&{L}^{w,b}_{2}&{L}^{w,b}_{4}&&&&&{L}^{w,b}_{2h-4}&{L}^{w,b}_{2h-2}&{L}^{w,b}_{2h}\\ \hline\cr\operatorname{F}^{1,w,b}_{0}&-\alpha(H-1)&2&0\\ \operatorname{F}^{1,w,b}_{2}&\alpha^{2}H(H-1)&*&12\\ \\ \operatorname{F}^{1,w,b}_{a}&&&&\alpha^{2}(1+c)(2+c)&*&(1+a)(2+a)\\ \\ \operatorname{F}^{1,w,b}_{2h-2}&&&&&&&&12\alpha^{2}&*&H(H-1)\\ \operatorname{F}^{1,w,b}_{2h}&&&&&&&&0&2\alpha^{2}&-\alpha(H-1)\\ \hline\cr\end{array}$

$\begin{array}[]{c | *{9}{c} }\hline\cr&{L}^{w,b}_{1}&{L}^{w,b}_{3}&{L}^{w,b}_{5}&&&&{L}^{w,b}_{2h-5}&{L}^{w,b}_{2h-3}&{L}^{w,b}_{2h-1}\\ \hline\cr\operatorname{F}^{1,w,b}_{1}&-3\alpha(H-1)&6&0\\ \operatorname{F}^{1,w,b}_{3}&\alpha^{2}(H-1)(H-2)&*&20\\ \\ \operatorname{F}^{1,w,b}_{a}&&&&\alpha^{2}(1+c)(2+c)&*&(1+a)(2+a)\\ \\ \operatorname{F}^{1,w,b}_{2h-3}&&&&&&&20\alpha^{2}&*&(H-1)(H-2)\\ \operatorname{F}^{1,w,b}_{2h-1}&&&&&&&0&6\alpha^{2}&-3\alpha(H-1)\\ \hline\cr\end{array}$

$\begin{array}[]{c | *{14}{c} }\hline\cr&{L}^{w,b}_{0}&{L}^{w,b}_{2}&{L}^{w,b}_{4}&&&&&{L}^{w,b}_{2h-4}&{L}^{w,b}_{2h-2}&{L}^{w,b}_{2h}\\ \hline\cr\widetilde{\text{F}}^{1,w,b}_{0}&-\alpha(H-1)&2&0\\ \widetilde{\text{F}}^{1,w,b}_{2}&0&K_{2}&12\\ \\ \widetilde{\text{F}}^{1,w,b}_{a}&&&&0&K_{a}&(1+a)(2+a)\\ \\ \widetilde{\text{F}}^{1,w,b}_{2h-2}&&&&&&&&0&K_{2h-2}&H(H-1)\\ \widetilde{\text{F}}^{1,w,b}_{2h}&&&&&&&&0&2\alpha^{2}&-\alpha(H-1)\\ \hline\cr\end{array}$

$\begin{array}[]{c | *{9}{c} }\hline\cr&{L}^{w,b}_{1}&{L}^{w,b}_{3}&{L}^{w,b}_{5}&&&&{L}^{w,b}_{2h-5}&{L}^{w,b}_{2h-3}&{L}^{w,b}_{2h-1}\\ \hline\cr\widetilde{\text{F}}^{1,w,b}_{1}&-3\alpha(H-1)&6&0\\ \widetilde{\text{F}}^{1,w,b}_{3}&0&K_{3}&20\\ \\ \widetilde{\text{F}}^{1,w,b}_{a}&&&&0&K_{a}&(1+a)(2+a)\\ \\ \widetilde{\text{F}}^{1,w,b}_{2h-3}&&&&&&&0&K_{2h-3}&(H-1)(H-2)\\ \widetilde{\text{F}}^{1,w,b}_{2h-1}&&&&&&&0&6\alpha^{2}&-3\alpha(H-1)\\ \hline\cr\end{array}$

$H=w-b=2h$

K_{2h-2}=-(2h-1)\alpha\;,\quad K_{2h-3}=-3(2h-1)\alpha

\begin{vmatrix}K_{2h-2}&H(H-1)\\ 2\alpha^{2}&-\alpha(H-1)\end{vmatrix}=\begin{vmatrix}-2(2h-1)\alpha&2h(2h-1)\\ 2\alpha^{2}&-\alpha(2h-1)\end{vmatrix}=-\alpha^{2}(2h-1)(2h+1)

\begin{vmatrix}K_{2h-3}&(H-1)(H-2)\\ 6\alpha^{2}&-3\alpha(H-1)\end{vmatrix}=\begin{vmatrix}-3(2h-1)\alpha&(2h-1)(2h-2)\\ 6\alpha^{2}&-3\alpha(2h-1)\end{vmatrix}=3\alpha^{2}(2h-1)(2h+1)

Thus,

	$\displaystyle\{\widetilde{\text{F}}^{1}_{2a,b,H-2a}\mid 0\leqq 2a\leqq H\}\quad\text{are linearly independent, and }$
	$\displaystyle\{\widetilde{\text{F}}^{1}_{2a+1,b,H-1-2a}\mid 1\leqq 2a+1\leqq H\}\quad\text{are linearly independent, and so }$
	$\displaystyle\{\widetilde{\text{F}}^{1}_{a,b,2h-a}\mid 0\leqq a\leqq 2h\}\quad\text{are linearly independent. }$
	$\displaystyle\{\widehat{E}^{1}_{a,b,2h-a}\mid 0\leqq a\leqq 2h\}\quad\text{are linearly independent. }$

By the above discussions, we have a result about $\textrm{dim}\{\widehat{E}^{1}_{a,b,c}\mid a+c=w-b\}$ , which is similar to Lemma 6.1. We denote $\mathfrak{S}^{[w]}_{b}=\{\widehat{E}^{1}_{a,b,c}\mid a+c=w-b\}$ .

Lemma 6.3.

If $w-b=\text{ even}$ , $\textrm{dim}\mathfrak{S}^{[w]}_{b}={}^{\#}\mathfrak{S}^{[w]}_{b}=w-b+1$ , (i.e., full rank).

If $w-b=\text{ odd}$ , then $\textrm{dim}\mathfrak{S}^{[w]}_{b}={}^{\#}\mathfrak{S}^{[w]}_{b}-1=w-b$ and $\{\widehat{E}^{1}_{a,b,c}\in\mathfrak{S}^{[w]}_{b}\mid c\neq 1\}$ are a base, by which $\widehat{E}^{1}_{w-b-1,b,1}$ are expressed.

As a direct result, we have

Corollary 6.4.

$\displaystyle\textrm{dim}\{\widehat{E}^{1}_{a,b,c}\mid a+b+c=w\}=\binom{w+2}{2}-\frac{w+1-\epsilon}{2}$ , where $\epsilon=1$ if $w$ is even, $\epsilon=0$ otherwise.

Proof: From corollary above,

	$\displaystyle\textrm{dim}\{\widehat{E}^{1}_{a,b,c}\mid a+b+c=w\}$	$\displaystyle=\sum_{b=0}^{w}\mathfrak{S}^{[w]}_{b}=\sum_{b=0}^{w}(w-b)+\#\{w-b=\text{ even }\mid 0\leqq b\leqq w\}$
		$\displaystyle=w(w+1)-\frac{w(w+1)}{2}+\#\{2b\mid 0\leqq b\leqq w\}$
		$\displaystyle=\frac{w(w+1)}{2}+\frac{w+1+\epsilon}{2}=\frac{(w+2)(w+1)}{2}-\frac{w+1-\epsilon}{2}\;.$

The parity of $w$ still has some effect. $\epsilon=2[w/2]-w+1$ holds if we use the Gauss symbol. $\blacksquare$

So far, we got $\binom{w}{2}-\epsilon,\binom{w+2}{2}$ by ${\mu}_{P}$ or $\lambda^{3}_{A}$ , and $\binom{w+2}{2}-\frac{w+1-\epsilon}{2}$ by Lemma 6.4 generators.

The last thing to do is to find which $\widehat{E}^{2}_{a,b,c}$ contributes to enlarging the linearly independent system. By bare hands, or by an experiment of symbol calculus, we know that $\{\widehat{E}^{2}_{2a,w-2a,0}\mid 2a\leqq w\}$ contribute as linear independent members, where

	$\displaystyle\widehat{E}^{2}_{2a,w-2a,0}=$	$\displaystyle 2\alpha\beta\lambda^{1}_{2a-1,w-2a-1,2}-\alpha(2a-w-1)\lambda^{1}_{2a-1,w-2a+1,0}-2\beta(2a+1)\lambda^{1}_{2a+1,w-2a-1,0}$
		$\displaystyle-2\beta^{2}\lambda^{2}_{2a,w-2a-2,2}-\beta(2a-w+1)\lambda^{2}_{2a,w-2a,0}\;.$

The cardinality is $\frac{w+1+\epsilon}{2}$ . This means the rank $\text{Rk}_{[w+2]}$ should be

	$\displaystyle\text{Rk}_{[w+2]}$	$\displaystyle\geqq(\binom{w}{2}-\epsilon)+\binom{w+2}{2}+(\binom{w+2}{2}-\frac{w+1-\epsilon}{2})+\frac{w+1+\epsilon}{2}$
		$\displaystyle=\binom{w}{2}+2\binom{w+2}{2}$

Since the Betti number is given by $\text{Bet}_{[w+2]}=\textrm{dim}C_{w+2}^{[w]}-\text{Rk}_{[w+2]}-\text{Rk}_{[w+3]}$ and it is non-negative, $\text{Rk}_{[w+2]}\leqq\textrm{dim}C_{w+2}^{[w]}-\text{Rk}_{[w+3]}=\binom{w}{2}+2\binom{w+2}{2}$ holds in our case. Thus, we conclude the rank is $\text{Rk}_{[w+2]}=\binom{w}{2}+2\binom{w+2}{2}$ and so the kernel dimension of $C_{w+3}^{[w]}$ is $\binom{w+2}{2}$ .

The space of cycles in $C_{w+1}^{[w]}$ :

When $w=0$ , then $C_{w+1}^{[w]}=C_{1}^{[0]}=\mathfrak{g}$ and the boundary operator is trivial, and so the kernel dimension is $\textrm{dim}\mathfrak{g}$ . We then study the space of cycles for $w>0$ .

Take a general chain $L+M$ where $L=\sum\lambda^{i}_{a,b,c}z_{i}\bigtriangleup{U}^{a,b,c}_{w}$ and $M=\sum{\mu}^{i}_{p,q,r}W[\epsilon_{i}=0]\bigtriangleup{U}^{p,q,r}_{w-2}$ with unknown scalars $\lambda^{i}_{a,b,c},{\mu}^{i}_{p,q,r}$ .

	$\displaystyle\partial(L+M)=$	$\displaystyle\sum\lambda^{i}_{A}(-z_{i}\bigtriangleup\partial{U}^{A}_{w}+[z_{i},{U}^{A}_{w}]_{res})$
		$\displaystyle+\sum{\mu}^{j}_{P}((\partial W[\epsilon_{j}=0])\bigtriangleup{U}^{P}_{w-2}+[W[\epsilon_{j}=0],{U}^{P}_{w-2}]_{res})$
	$\displaystyle=$	$\displaystyle-\sum z_{i}\bigtriangleup z_{4}\bigtriangleup U^{p,q,r}(\frac{2}{\alpha}\tbinom{p+2}{2}\lambda^{i}_{p+2,q,r}+\frac{2}{\beta}\tbinom{q+2}{2}\lambda^{i}_{p,q+2,r}+2\tbinom{r+2}{2}\lambda^{i}_{p,q,r+2})+\sum\lambda^{i}_{A}[z_{i},U^{A}]_{res}$
		$\displaystyle+z_{1}\bigtriangleup z_{4}\bigtriangleup\sum\alpha{\mu}^{1}_{P}U^{P}+z_{2}\bigtriangleup z_{4}\bigtriangleup\sum(-\beta){\mu}^{2}_{P}U^{P}+z_{3}\bigtriangleup z_{4}\bigtriangleup\sum{\mu}^{3}_{P}U^{P}$
		$\displaystyle+\sum{\mu}^{1}_{P}(z_{3}\bigtriangleup z_{4}\bigtriangleup[z_{2},U^{P}]_{res}-z_{2}\bigtriangleup z_{4}\bigtriangleup[z_{3},U^{P}]_{res})$
		$\displaystyle+\sum{\mu}^{2}_{P}(z_{3}\bigtriangleup z_{4}\bigtriangleup[z_{1},U^{P}]_{res}-z_{1}\bigtriangleup z_{4}\bigtriangleup[z_{3},U^{P}]_{res})$
		$\displaystyle+\sum{\mu}^{3}_{P}(z_{2}\bigtriangleup z_{4}\bigtriangleup[z_{1},U^{P}]_{res}-z_{1}\bigtriangleup z_{4}\bigtriangleup[z_{2},U^{P}]_{res})$
	$\displaystyle=$	$\displaystyle+\sum\lambda^{i}_{A}[z_{i},U^{A}]_{res}$
		$\displaystyle+z_{1}\bigtriangleup z_{4}\bigtriangleup(\sum\alpha{\mu}^{1}_{P}U^{P}-\sum\lambda^{1}_{A}(\partial U^{A})/z_{4}-\sum{\mu}^{2}_{P}[z_{3},U^{P}]_{res}-\sum{\mu}^{3}_{P}[z_{2},U^{P}]_{res})$
		$\displaystyle+z_{2}\bigtriangleup z_{4}\bigtriangleup(\sum-\beta{\mu}^{2}_{P}U^{P}-\sum\lambda^{2}_{A}(\partial U^{A})/z_{4}-\sum{\mu}^{1}_{P}[z_{3},U^{P}]_{res}+\sum{\mu}^{3}_{P}[z_{1},U^{P}]_{res})$
		$\displaystyle+z_{3}\bigtriangleup z_{4}\bigtriangleup(\sum{\mu}^{3}_{P}U^{P}-\sum\lambda^{3}_{A}(\partial U^{A})/z_{4}+\sum{\mu}^{1}_{P}[z_{2},U^{P}]_{res}+\sum{\mu}^{2}_{P}[z_{1},U^{P}]_{res})$

Thus, the linear equation $\partial(L+M)=0$ is defined by


(6.50a)	$\displaystyle G_{0}=$	$\displaystyle(b+1)\lambda^{1}_{a,b+1,c-1}-\beta(c+1)\lambda^{1}_{a,b-1,c+1}-(a+1)\lambda^{2}_{a+1,b,c-1}+\alpha(c+1)\lambda^{2}_{a-1,b,c+1}$
	$\displaystyle\quad+\beta(a+1)\lambda^{3}_{a+1,b-1,c}-\alpha(b+1)\lambda^{3}_{a-1,b+1,c}$
(6.50b)	$\displaystyle G_{1}=$	$\displaystyle\alpha{\mu}^{1}_{p,q,r}-\frac{1}{\alpha}2\tbinom{p+2}{2}\lambda^{1}_{p+2,q,r}-\frac{1}{\beta}2\tbinom{q+2}{2}\lambda^{1}_{p,q+2,r}-2\tbinom{r+2}{2}\lambda^{1}_{p,q,r+2}$
	$\displaystyle-\beta(p+1){\mu}^{2}_{p+1,q-1,r}+\alpha(q+1){\mu}^{2}_{p-1,q+1,r}$
	$\displaystyle+(p+1){\mu}^{3}_{p+1,q,r-1}-\alpha(r+1){\mu}^{3}_{p-1,q,r+1}$
(6.50c)	$\displaystyle G_{2}=$	$\displaystyle-\beta{\mu}^{2}_{p,q,r}-\frac{1}{\alpha}2\tbinom{p+2}{2}\lambda^{2}_{p+2,q,r}-\frac{1}{\beta}2\tbinom{q+2}{2}\lambda^{2}_{p,q+2,r}-2\tbinom{r+2}{2}\lambda^{2}_{p,q,r+2}$
	$\displaystyle-\beta(p+1){\mu}^{1}_{p+1,q-1,r}+\alpha(q+1){\mu}^{1}_{p-1,q+1,r}$
	$\displaystyle+(q+1){\mu}^{3}_{p,q+1,r-1}-\beta(r+1){\mu}^{3}_{p,q-1,r+1}$
(6.50d)	$\displaystyle G_{3}=$	$\displaystyle{\mu}^{3}_{p,q,r}-\frac{1}{\alpha}2\tbinom{p+2}{2}\lambda^{3}_{p+2,q,r}-\frac{1}{\beta}2\tbinom{q+2}{2}\lambda^{3}_{p,q+2,r}-2\tbinom{r+2}{2}\lambda^{3}_{p,q,r+2}$
	$\displaystyle-(p+1){\mu}^{1}_{p+1,q,r-1}+\alpha(r+1){\mu}^{1}_{p-1,q,r+1}$
	$\displaystyle+(q+1){\mu}^{2}_{p,q+1,r-1}-\beta(r+1){\mu}^{2}_{p,q-1,r+1}$

If we focus on ${\mu}^{3}_{p,q,r}$ in (6.50d) and substitute them to (6.50b) and (6.50c), then we have next equations without ${\mu}^{1}_{P}$ and ${\mu}^{3}_{P}$ :

(6.51)	$\displaystyle H_{1}(p,q,r)$	$\displaystyle=-(r+2)(r+1)\lambda^{1}_{p,q,r+2}-(p+2)(p+1)\frac{1}{\alpha}\lambda^{1}_{p+2,q,r}-(q+2)(q+1)\frac{1}{\beta}\lambda^{1}_{p,q+2,r}$
	$\displaystyle+(p+3)(p+2)(p+1)\frac{1}{\alpha}\lambda^{3}_{p+3,q,r-1}+(q+2)(q+1)(p+1)\frac{1}{\beta}\lambda^{3}_{p+1,q+2,r-1}$
	$\displaystyle-(r+1)(p+1)(p-r)\lambda^{3}_{p+1,q,r+1}-\alpha(r+3)(r+2)(r+1)\lambda^{3}_{p-1,q,r+3}$
	$\displaystyle-\alpha(q+2)(q+1)(r+1)\frac{1}{\beta}\lambda^{3}_{p-1,q+2,r+1}+\alpha(r+2)(q+1){\mu}^{2}_{p-1,q+1,r}$
	$\displaystyle+\beta(r-1)(p+1){\mu}^{2}_{p+1,q-1,r}-(q+1)(p+1){\mu}^{2}_{p+1,q+1,r-2}$
	$\displaystyle-\beta\alpha(r+2)(r+1){\mu}^{2}_{p-1,q-1,r+2}\;,$
(6.52)	$\displaystyle H_{2}(p,q,r)$	$\displaystyle=-(r+2)(r+1)\lambda^{2}_{p,q,r+2}-(p+2)(p+1)\frac{1}{\alpha}\lambda^{2}_{p+2,q,r}-(q+2)(q+1)\frac{1}{\beta}\lambda^{2}_{p,q+2,r}$
	$\displaystyle+(p+2)(p+1)(q+1)\frac{1}{\alpha}\lambda^{3}_{p+2,q+1,r-1}+(q+3)(q+2)(q+1)\frac{1}{\beta}\lambda^{3}_{p,q+3,r-1}$
	$\displaystyle-(r+1)(q+1)(q-r)\lambda^{3}_{p,q+1,r+1}-\beta(r+3)(r+2)(r+1)\lambda^{3}_{p,q-1,r+3}$
	$\displaystyle-\beta(p+2)(p+1)(r+1)\frac{1}{\alpha}\lambda^{3}_{p+2,q-1,r+1}+\beta(2qr+q+r-1){\mu}^{2}_{p,q,r}$
	$\displaystyle-(q+2)(q+1){\mu}^{2}_{p,q+2,r-2}-\beta^{2}(r+2)(r+1){\mu}^{2}_{p,q-2,r+2}\;.$

Now we look for the dimension of the space generated by $G_{0}$ , $H_{1}$ and $H_{0}$ , and we get $\binom{w+2}{2}+\binom{w}{2}$ . Thus, the rank is $2\binom{w}{2}+\binom{w+2}{2}$ and the kernel dimension is $\binom{w}{2}+2\binom{w+2}{2}$ .

The space of cycles in $C_{w}^{[w]}$ :

When $w=0$ , then $C_{w}^{[w]}=\Lambda^{0}\mathfrak{g}$ and the boundary operator is trivial, and so the kernel dimension is 1. We study the space of cycles in $C_{w}^{[w]}$ for $w>0$ . Take a general chain $L+M$ where $L=\sum\lambda_{A}{U}^{A}_{w}$ and $M=\sum{\mu}^{i}_{P}z_{i}\bigtriangleup z_{4}\bigtriangleup{U}^{P}_{w-2}$ with unknown scalars $\lambda_{A},{\mu}^{i}_{P}$ . Since

\displaystyle\partial(L+M)=

\displaystyle\sum\lambda_{A}\partial U^{A}+0+0+\sum{\mu}^{j}_{P}z_{4}\bigtriangleup[z_{j},U^{P}]_{res}

the linear equations $\partial(L+M)=0$ is defined by

(6.53)		$\displaystyle\frac{(p+2)(p+1)}{\alpha}\lambda_{p+2,q,r}+\frac{(q+2)(q+1)}{\beta}\lambda_{p,q+2,r}+(r+2)(r+1)\lambda_{p,q,r+2}$
	$\displaystyle+(q+1){\mu}^{1}_{p,q+1,r-1}-\beta(r+1){\mu}^{1}_{p,q-1,r+1}-(p+1){\mu}^{2}_{p+1,q,r-1}+\alpha(r+1){\mu}^{2}_{p-1,q,r+1}$
	$\displaystyle+\beta(p+1){\mu}^{3}_{p+1,q-1,r}-\alpha(q+1){\mu}^{3}_{p-1,q+1,r}$

Thus, the rank is $\binom{w}{2}$ , and the kernel dimension is $2\binom{w}{2}+\binom{w+2}{2}$ .

Final table of chain complex

Theorem 6.4.

$\begin{array}[]{c|*{5}{c}}\text{weight}=w>0&w-1&w&w+1&w+2&w+3\\ \hline\cr\text{SpaceDim}&\binom{w}{2}&3\binom{w}{2}+\binom{w+2}{2}&3\binom{w}{2}+3\binom{w+2}{2}&\binom{w}{2}+3\binom{w+2}{2}&\binom{w+2}{2}\\ \ker\textrm{dim}&\binom{w}{2}&2\binom{w}{2}+\binom{w+2}{2}&\binom{w}{2}+2\binom{w+2}{2}&\binom{w+2}{2}&0\\ \hline\cr\text{Betti}&0&0&0&0&0\end{array}$

7 Further questions

So far, we discussed super homology groups associated with Lie algebras of dimension 2 or 3. We observed that if $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$ then the Betti numbers are all zero, and it may be interesting question if this is true when the dimension is larger than 3.

As an example of 4-dimensional Lie algebra, we take $\mathfrak{gl}(2,\mathbb{R})$ of $2\times 2$ matrices, which is an extension of $\mathfrak{sl}(2,\mathbb{R})$ . By experiments for lower weights, we have the following information about Betti numbers (we do not note dimension or rank for each chain space here).

$\begin{array}[]{ c | *{10}{r|} }m\text{-th chain}&0&1&2&3&4&5&6&7&8&9\\ \hline\cr w=0&1&1&0&1&1\\ \hline\cr w=1&&0&0&0&0&0\\ \hline\cr w=2&&0&2&2&0&2&2\\ \hline\cr w=3&&0&1&1&0&1&1&0\\ \hline\cr w=4&&&0&0&2&2&0&2&2\\ \hline\cr w=5&&&0&1&2&1&1&2&1&0\\ \hline\cr\end{array}$

Looking at this table, we expect some “rule”’ and expect the rule comes from that the Lie algebra is an extension of $\mathfrak{sl}(2,\mathbb{R})$ .

References

[1] Nathan Jacobson. Lie algebras. Dover Publications, Inc., New York, 1979. Republication of the 1962 original.
[2] Kentaro Mikami and Tadayoshi Mizutani. Euler number and Betti numbers of homology groups of pre Lie superalgebra. arXiv:1809.08028v2, December 2018.
[3] Kentaro Mikami and Tadayoshi Mizutani. The second Betti number of doubly weighted homology groups of some pre Lie superalgebra. arXiv:1902.09137, February 2019.

Super homologies associated with low dimensional Lie algebras

1 Introduction

2 Preliminaries, Notations and Basic Facts

Definition 2.1 (Lie superalgebra).

Remark 2.1.

3 Superalgebras associated with Lie algebras

Definition 1 (Schouten-like bracket).

Proposition 3.1.

4 Homology groups of superalgebras associated with Lie algebras

Theorem 4.1.

5 Super homologies of 2-dimensional Lie algebras

6 Super homologies of 3-dimensional Lie algebras

The boundary operator on Cw−1[w]C_{w-1}^{[w]} is trivial:

Proposition 6.1.

The boundary operator on Cw[w]C_{w}^{[w]}:

The boundary operator on Cw+1[w]C_{w+1}^{[w]}:

The boundary operator on Cw+2[w]C_{w+2}^{[w]}:

The boundary operator on Cw+3[w]C_{w+3}^{[w]}:

6.1 dim​𝔤=3\textrm{dim}\mathfrak{g}=3, dim​[𝔤,𝔤]=1\textrm{dim}[\mathfrak{g},\mathfrak{g}]=1 and [𝔤,𝔤]⊂Z​(𝔤)[\mathfrak{g},\mathfrak{g}]\subset Z(\mathfrak{g})

Kernel of Cw+3[w]​→∂Cw+2[w]\displaystyle C_{w+3}^{[w]}\mathop{\to}^{\partial}C_{w+2}^{[w]}

Kernel of Cw+2[w]​→∂Cw+1[w]\displaystyle C_{w+2}^{[w]}\mathop{\to}^{\partial}C_{w+1}^{[w]}

Kernel of Cw+1[w]​→∂Cw[w]\displaystyle C_{w+1}^{[w]}\mathop{\to}^{\partial}C_{w}^{[w]}

Kernel of Cw[w]​→∂Cw−1[w]\displaystyle C_{w}^{[w]}\mathop{\to}^{\partial}C_{w-1}^{[w]}

Final table of chain complex

Theorem 6.1.

6.2 dim​𝔤=3\textrm{dim}\mathfrak{g}=3, dim​[𝔤,𝔤]=1\textrm{dim}[\mathfrak{g},\mathfrak{g}]=1 and [𝔤,𝔤]⊄Z​(𝔤)[\mathfrak{g},\mathfrak{g}]\not\subset Z(\mathfrak{g})

Kernel of Cw+3[w]​→∂Cw+2[w]\displaystyle C_{w+3}^{[w]}\mathop{\to}^{\partial}C_{w+2}^{[w]}

Kernel of Cw+2[w]​→∂Cw+1[w]\displaystyle C_{w+2}^{[w]}\mathop{\to}^{\partial}C_{w+1}^{[w]}

Kernel of Cw+1[w]​→∂Cw[w]\displaystyle C_{w+1}^{[w]}\mathop{\to}^{\partial}C_{w}^{[w]}

Kernel of Cw[w]​→∂Cw−1[w]\displaystyle C_{w}^{[w]}\mathop{\to}^{\partial}C_{w-1}^{[w]}

Final table of chain complex

Theorem 6.2.

6.3 dim​[𝔤,𝔤]=2\textrm{dim}[\mathfrak{g},\mathfrak{g}]=2

Remark 6.1.

The space of cycles in Cw+3[w]C_{w+3}^{[w]}:

The space of cycles in Cw+2[w]C_{w+2}^{[w]}:

The space of cycles in Cw+1[w]C_{w+1}^{[w]}:

The space of cycles in Cw[w]C_{w}^{[w]}:

The space of cycles in Cw−1[w]C_{w-1}^{[w]}:

Final table of chain complex

Theorem 6.3.

6.4 dim​[𝔤,𝔤]=3\textrm{dim}[\mathfrak{g},\mathfrak{g}]=3

The space of cycles in Cw+3[w]C_{w+3}^{[w]}:

The space of cycles in Cw+2[w]C_{w+2}^{[w]}:

6.4.1 Rank of μp,q,r{\mu}_{p,q,r}

Lemma 6.1.

Corollary 6.2.

6.4.2 Rank of λa,b,ci\lambda^{i}_{a,b,c}

Lemma 6.3.

Corollary 6.4.

The space of cycles in Cw+1[w]C_{w+1}^{[w]}:

The space of cycles in Cw[w]C_{w}^{[w]}:

Final table of chain complex

Theorem 6.4.

7 Further questions

References

The boundary operator on $C_{w-1}^{[w]}$ is trivial:

The boundary operator on $C_{w}^{[w]}$ :

The boundary operator on $C_{w+1}^{[w]}$ :

The boundary operator on $C_{w+2}^{[w]}$ :

The boundary operator on $C_{w+3}^{[w]}$ :

6.1 $\textrm{dim}\mathfrak{g}=3$ , $\textrm{dim}[\mathfrak{g},\mathfrak{g}]=1$ and $[\mathfrak{g},\mathfrak{g}]\subset Z(\mathfrak{g})$

Kernel of $\displaystyle C_{w+3}^{[w]}\mathop{\to}^{\partial}C_{w+2}^{[w]}$

Kernel of $\displaystyle C_{w+2}^{[w]}\mathop{\to}^{\partial}C_{w+1}^{[w]}$

Kernel of $\displaystyle C_{w+1}^{[w]}\mathop{\to}^{\partial}C_{w}^{[w]}$

Kernel of $\displaystyle C_{w}^{[w]}\mathop{\to}^{\partial}C_{w-1}^{[w]}$

6.2 $\textrm{dim}\mathfrak{g}=3$ , $\textrm{dim}[\mathfrak{g},\mathfrak{g}]=1$ and $[\mathfrak{g},\mathfrak{g}]\not\subset Z(\mathfrak{g})$

Kernel of $\displaystyle C_{w+3}^{[w]}\mathop{\to}^{\partial}C_{w+2}^{[w]}$

Kernel of $\displaystyle C_{w+2}^{[w]}\mathop{\to}^{\partial}C_{w+1}^{[w]}$

Kernel of $\displaystyle C_{w+1}^{[w]}\mathop{\to}^{\partial}C_{w}^{[w]}$

Kernel of $\displaystyle C_{w}^{[w]}\mathop{\to}^{\partial}C_{w-1}^{[w]}$

6.3 $\textrm{dim}[\mathfrak{g},\mathfrak{g}]=2$

The space of cycles in $C_{w+3}^{[w]}$ :

The space of cycles in $C_{w+2}^{[w]}$ :

The space of cycles in $C_{w+1}^{[w]}$ :

The space of cycles in $C_{w}^{[w]}$ :

The space of cycles in $C_{w-1}^{[w]}$ :

6.4 $\textrm{dim}[\mathfrak{g},\mathfrak{g}]=3$

The space of cycles in $C_{w+3}^{[w]}$ :

The space of cycles in $C_{w+2}^{[w]}$ :

6.4.1 Rank of ${\mu}_{p,q,r}$

6.4.2 Rank of $\lambda^{i}_{a,b,c}$

The space of cycles in $C_{w+1}^{[w]}$ :

The space of cycles in $C_{w}^{[w]}$ :