Lilong Kang, Yu Wang, Caiyu Du
College of Mathematics and Statistics, Sichuan University of Science and Engineering, Zigong, China.
DOI: 10.4236/jamp.2024.127141   PDF    HTML   XML   90 Downloads   358 Views   Citations

Abstract

Because homology on compact homogeneous nilpotent manifolds is closely related to homology on Lie algebras, studying homology on Lie algebras is helpful for further studying homology on compact homogeneous nilpotent manifolds. So we start with the differential sequence of Lie algebras. The Lie algebra g has the differential sequence $E_0,E_1,\cdots ,E_s\cdots$ , which leads to the chain complex $E_s^0\stackrel{{\Delta }_s^0}{\to }{E}_s^s\stackrel{{\Delta }_s^1}{\to }\cdots \stackrel{{\Delta }_s^i}{\to }{E}_s^{\left(i+1\right)s}\stackrel{{\Delta }_s^{i+1}}{\to }\cdots$ of $E_s$ by discussing the chain complex $E_1^0\stackrel{{\Delta }_1^0}{\to }{E}_1^1\stackrel{{\Delta }_1^1}{\to }\cdots \stackrel{{\Delta }_1^{r-1}}{\to }{E}_1^r\stackrel{{\Delta }_1^r}{\to }\cdots$ of $E_1$ and proves that $E_{s+1}^i\cong H^i\left(E_s\right)=Ker{\text{Δ}}_s^{i+1}/Im{\text{Δ}}_s^i$ and therefore $E_{s+1}\cong H\left(E_s\right)$ by the chain complex of $E_s$ (see Theorem 2).

Keywords

Lie Algebra, Differential Sequence, Differential Fractional Algebra, Cohomology

Share and Cite:

1. Introduction

In 1951, Matsushima [1] showed that if the homogeneous space M of a connected nilpotent Lie group is compact, then M is homeomorphic to G/D (where G is a simply connected nilpotent Lie group and D is a discrete subgroup of G). In 1954, Nomizu [2] defined $S_1={\displaystyle {\sum }_r{S}_1^r}$ (see Definition 4) by the differential graded algebra $C={\displaystyle {\sum }_p{C}^p}$ of an abelian Lie algebra, and thus $E_1={\displaystyle {\sum }_r{E}_1^r}$ (see Definition 5). Since the edge operator ${\Delta }_1$ exists on the chain complex of $E_1$ , the chain complex $E_1^0\stackrel{{\Delta }_1^0}{\to }{E}_1^1\stackrel{{\Delta }_1^1}{\to }\cdots \stackrel{{\Delta }_1^{r-1}}{\to }{E}_1^r\stackrel{{\Delta }_1^r}{\to }\cdots$ can be obtained, and then it is proved that the De Rham cohomology of the compact homogeneous nilpotent connected manifold $M=G/D$ is isomorphic to the homology of the Lie algebra $g=LieG$ , i.e., $H^{*}\left(M\right)=H^{*}\left(G\right)$ . In 2000, although Cordero [3] and others studied Dolbeault homology on compact nilpotent manifolds with nilpotent complex structure by differential bi-graded algebra, Dolbeault homology on compact nilpotent manifolds in general is still an unsolved problem, so it is meaningful to study differential sequences. Therefore, the chain complex $E_1^0\stackrel{{\Delta }_1^0}{\to }{E}_1^1\stackrel{{\Delta }_1^1}{\to }\cdots \stackrel{{\Delta }_1^{r-1}}{\to }{E}_1^r\stackrel{{\Delta }_1^r}{\to }\cdots$ of $E_1$ in reference [2] is discussed in this paper, and then the chain complex $E_s^0\stackrel{{\Delta }_s^0}{\to }{E}_s^s\stackrel{{\Delta }_s^1}{\to }\cdots \stackrel{{\Delta }_s^i}{\to }{E}_s^{\left(i+1\right)s}\stackrel{{\Delta }_s^{i+1}}{\to }\cdots$ of $E_s$ is obtained, and the conclusion $E_{s+1}^i\cong H^i\left(E_s\right)$ is obtained, then there is $E_{s+1}\cong H\left(E_s\right)$ .

This paper will be divided into two parts, the first part is the preparatory knowledge: introduces the basic knowledge and properties of $C^p$ , C, $B^p$ , B, $S_r^p$ , $S_r$ , $D_s^r$ , $Z_s^r$ , $E_s^r$ and $E_s$ ; the second part gives a proof of $H\left(E_1\right)\cong Hom\left(A\left(g\right),H\left(B\right)\right)$ (see Theorem 1) and proves $E_{s+1}\cong H\left(E_s\right)$ (see Theorem 2).

Previous studies only focused on the chain complex of $E_1$ and did not draw conclusions on the chain complex of $E_s$ . The difficulty in the research lies in constructing the chain complex of $E_s$ .

2. Preparatory Knowledge

Let g be a Lie algebra, where $g^{*}$ is the dual space of g, $\left\{e^i|1\le i\le n\right\}$ is the basis of $g^{*}$ , and $C={\displaystyle {\sum }_p{C}^p}$ is the differential graded algebra of g, where $C^p={\wedge }^p\left(g^{*}\right)$ , $\left\{e^{i_1}\wedge \cdots \wedge e^{i_p}|1\le i_1<\cdots <i_p\le n\right\}$ is the basis of ${\wedge }^p\left(g^{*}\right)$ (cf. [4] [5]).

Definition 1 [4]. Let g be an n-dimensional Lie algebra, and let $\left\{e_i|1\le i\le n\right\}$ be a basis for g, and define a derivation $\theta :g\to Aut{\wedge }^p\left(g^{*}\right)$ on g, i.e.,

$\theta \left(x\right)\alpha \left(e_1,\cdots ,e_p\right)={\displaystyle {\sum }_{i=1}^p\alpha \left(e_1,\cdots ,e_{i-1},\left[x,e_i\right],e_{i+1},\cdots ,e_p\right)}$ , for $x\in g,\alpha \in C^p$ . (1)

g acts on C by its derivative θ.

Definition 2 [6]. Let g be an n-dimensional Lie algebra, $g^{*}$ be the dual space of g, and $\left\{e^i|1\le i\le n\right\}$ be the basis of $g^{*}$ . Define the skew derivation $\iota \left(x\right):{\wedge }^p\left(g^{*}\right)\to {\wedge }^{p-1}\left(g^{*}\right)$ on g, as follows: for any $e^1\wedge \cdots \wedge e^p\in {\wedge }^p\left(g^{*}\right)$ , and $x\in g$ , we have

$\iota \left(x\right)\left(e^1\wedge \cdots \wedge e^p\right)={\displaystyle {\sum }_{j=1}^p{\left(-1\right)}^{j+1}x},e^j{e}^1\wedge \cdots \wedge e^{j-1}\wedge {\widehat{e}}^j\wedge \cdots \wedge e^p$, (2)

where, as usual, a $\widehat{\cdot }$ over a symbol means deletion.

Proposition 1 [4]. Let g be a Lie algebra that satisfies Definition 1 and Definition 2, then there is

$\theta \left(\left[x,y\right]\right)=\left[\theta \left(x\right),\theta \left(y\right)\right]$ ; $\left[\theta \left(x\right),\iota \left(y\right)\right]=\iota \left(\left[x,y\right]\right)$ ; (3)

$\theta \left(x\right)=\iota \left(x\right)d+d\iota \left(x\right)$ , ${\left(\iota \left(x\right)\right)}^2=0$ for $x,y\in g$ . (4)

Definition 3 [7]. Let g be a Lie algebra and $g^{*}$ be the dual space of g. Define the connection mapping $f:{\wedge }^p\left(g^{*}\right)\to C$ , i.e.,

$f\left(\iota \left(x\right)a\right)=\iota \left(x\right)f\left(a\right)$ , $f\left(\theta \left(x\right)a\right)=\theta \left(x\right)f\left(a\right)$ , for $a\in {\wedge }^p\left(g^{*}\right)$ , $x\in g$ . (5)

If g is an abelian Lie algebra, it is easy to prove $\theta \left(x\right)c=0$ , for $x\in g$ , $c\in C$ by Definition 1.

Definition 4 [8]. Let g be a Lie algebra and $C={\displaystyle {\sum }_p{C}^p}$ be a differential graded algebra of g. Define the basic element subcomplex $B={\displaystyle {\sum }_p{B}^p}$ and $S_r={\displaystyle {\sum }_p{S}_r^p}$ of C, where

$B^p=\left\{c\in {\text{C}}^p|\iota \left(x\right)c=0,\forall x\in g\right\}$ , (6)

$S_r^p=\left\{c\in {\text{C}}^p|\iota \left(x_1\right)\cdots \iota \left(x_k\right)c=0,\forall x_1,\cdots ,x_k\in g,k>p-r\right\}$ . (7)

Proposition 2. Let g be a Lie algebra. If $S_r={\displaystyle {\sum }_p{S}_r^p}$ satisfies Definition 3, then$S_{r+1}^p\subset S_r^p$ .

Proof. Because of $S_{r+1}^p=\left\{c\in {\text{C}}^p|\iota \left(x_1\right)\cdots \iota \left(x_k\right)c=0,\forall x_1,\cdots ,x_k\in g,k>p-r-1\right\}$ , so any $c\in S_{r+1}^p$ , there is $\iota \left(x_1\right)\cdots \iota \left(x_k\right)c=0$ , for $x_1,\cdots ,x_k\in g$ , $k>p-r-1$ , then $\iota \left(x\right)\iota \left(x_1\right)\cdots \iota \left(x_k\right)c=0$ , for $x\in g$ .

Because of $S_r^p=\left\{c\in {\text{C}}^p|\iota \left(x_1\right)\cdots \iota \left(x_k\right)c=0,\forall x_1,\cdots ,x_k\in g,k>p-r\right\}$ , so $c\in S_r^p$ .

Proposition 3 [9]. Let g be an n-dimensional Lie algebra. If $B={\displaystyle {\sum }_p{B}^p}$ and $S_r={\displaystyle {\sum }_p{S}_r^p}$ satisfy Definition 3, then

$S_0=C\supset S_1\supset \cdots \supset S_r\supset \cdots$ ; ${\displaystyle {\cap }_r{S}_r}=\left\{0\right\}$ ;(8)

$\iota \left(x\right)S_r^p\subset S_r^{p-1}\left(x\in g\right)$ ; $S_r^r=B^r$ . (9)

If g is an abelian Lie algebra, then $d{S}_r^p\subset S_{r+1}^{p+1}\subset S_r^{p+1}$ .

Proof. 1) Let us first prove that $S_0=C\supset S_1\supset \cdots \supset S_r\supset \cdots$ is true.

Property 2 tells us that $S_{r+1}^p\subset S_r^p$ , so $S_{r+1}\subset S_r$ . Since $S_0={\displaystyle {\sum }_p{S}_0^p}$ , and $S_0^p=\left\{c\in {\text{C}}^p|\iota \left(x_1\right)\cdots \iota \left(x_p\right)\iota \left(x_{p+1}\right)c=0\right\}=C^p$ , $S_0={\displaystyle {\sum }_p{C}^p}=C$ .

That is, $S_0=C\supset S_1\supset \cdots \supset S_r\supset \cdots$ .

2) The following is $\iota \left(x\right)S_r^p\subset S_r^{p-1}$ for $x\in g$ .

For any $c\in S_r^p$ , you get $\iota \left(x_1\right)\cdots \iota \left(x_k\right)c=0$ , for $x_1,\cdots ,x_k\in g$ , $k>p-r$ , $c\in C^p$ , and because $\iota \left(x\right)c\in C^{p-1}$ , $\iota \left(x_1\right)\cdots \iota \left(x_{k-1}\right)\left(\iota \left(x\right)c\right)=0$ , and therefore $\iota \left(x\right)S_r^p\subset S_r^{p-1}$ .

3) If g is an abelian Lie algebra, then $d{S}_r^p\subset S_{r+1}^{p+1}\subset S_r^{p+1}$ .

For any $c\in S_r^p$ , you get $c\in C^p$ , because d is the outer differential, so $d:C^p\to C^{p+1}$ , so $dc\in C^{p+1}$ . And since g is an abelian Lie algebra, so $\theta \left(x\right)c=0$ , for $x\in g$ , and by Property 1 we know $\theta \left(x\right)=\iota \left(x\right)d+d\iota \left(x\right)$ , so $\iota \left(x\right)d=-d\iota \left(x\right)$ , so $\iota \left(x_1\right)\cdots \iota \left(x_k\right)dc={\left(-1\right)}^{k}d\iota \left(x_1\right)\cdots \iota \left(x_k\right)c=0$ , for $x_1,\cdots ,x_k\in g$ , $k>p-r$ . And because of $S_{r+1}^{p+1}=\left\{c\in {\text{C}}^{p+1}|\iota \left(x_1\right)\cdots \iota \left(x_k\right)c=0,\forall x_1,\cdots ,x_k\in g,k>p-r\right\}$ , therefore $dc\in S_{r+1}^{p+1}$ , therefore $d{S}_r^p\subset S_{r+1}^{p+1}$ . $S_{r+1}^{p+1}\subset S_r^{p+1}$ is known by Property 2, which completes the proof.

4) Let us see that $S_r^r=B^r$ .

$\begin{array}{c}{S}_r^r=\left\{c\in {\text{C}}^r|\iota \left(x_1\right)\cdots \iota \left(x_k\right)c=0,\forall x_1,\cdots ,x_k\in g,k>0\right\}\\ =\left\{c\in {\text{C}}^r|\iota \left(x\right)c=0,\forall x\in g\right\}\end{array}$ , is the proof.

5) Let us prove ${\displaystyle {\cap }_r{S}_r}=\left\{0\right\}$ .

Let $\left\{e^i|1\le i\le n\right\}$ be the basis of $g^{*}$ .

From $S_n={\displaystyle {\sum }_p{S}_n^p}=S_n^n$ , $\left\{c\in {\text{C}}^n|\iota \left(x_1\right)\cdots \iota \left(x_k\right)c=0,\forall x_1,\cdots ,x_k\in g,k>0\right\}=B^n$ , we know that any $c\in B^n$ , $\iota \left(x\right)c=0$ , for any $x\in g$ . And by Definition 2 we know that $\iota \left(x\right)\left(e^1\wedge \cdots \wedge e^n\right)={\displaystyle {\sum }_{j=1}^n{\left(-1\right)}^{j+1}\langle x,e^j\rangle e^1\wedge \cdots \wedge e^{j-1}\wedge {\widehat{e}}^j\wedge \cdots \wedge e^n}$ , and $e_j\in g$ is the dual element of $e^j$ , so $e_j,e^j\ne 0$ , $e_j,e^i\ne 0$ , for $i\ne j$ , then $c=0$ if and only if $\iota \left(e_j\right)c=0$ , so $B^n=\left\{0\right\}$ , then ${\displaystyle {\cap }_r{S}_r}=\left\{0\right\}$ .

Definition 5 [10]. Let g be a Lie algebra and $C={\displaystyle {\sum }_p{C}^p}$ the differential fractional algebra of g. Let’s define

$Z_s^r=\left\{c\in S_r|dc\in S_{r+s}\right\}$ ;(10)

$D_s^r=\left\{dc|c\in Z_s^{r-s}\right\}$ ; (11)

$E_s^r=Z_s^r/\left(Z_{s-1}^{r+1}+D_{s-1}^r\right)$ ; (12)

$E_s={\displaystyle {\sum }_r{E}_s^r}$ ; (13)

where $E_0,E_1,\cdots ,E_s,\cdots$ is the differential sequence.

Lemma 1 [11]. If $E_0,E_1,\cdots ,E_s,\cdots$ is a differential sequence of an abelian Lie algebra g, it is true that $E_0=E_1={\displaystyle {\sum }_r{S}_r/S_{r+1}}$ is true.

Proof. Let’s first prove that $E_0={\displaystyle {\sum }_r{S}_r/S_{r+1}}$ .

We know by Definition 5 that $Z_0^r=\left\{c\in S_r|dc\in S_r\right\}$ , $Z_{-1}^{r+1}=\left\{c\in S_{r+1}|dc\in S_r\right\}$ , and $D_{-1}^r=d{Z}_{-1}^{r+1}=\left\{dc|c\in S_{r+1},dc\in S_r\right\}$ .

Because of $d{S}_r={\displaystyle {\sum }_{p}d{S}_r^p}\subset {\displaystyle {\sum }_p{S}_{r+1}^{p+1}}=S_{r+1}\subset S_r$ , for $p>1$ , so $Z_0^r=S_r$ , and because of$d{S}_{r+1}\subset S_{r+1}\subset S_r$ , so $D_{-1}^r\subset Z_{-1}^{r+1}$ . Because of $E_0={\displaystyle {\sum }_r{E}_0^r}={\displaystyle {\sum }_r{Z}_0^r/\left(Z_{-1}^{r+1}+D_{-1}^r\right)}$ , so $E_0={\displaystyle {\sum }_r{S}_r/S_{r+1}}$ .

Let me prove that $E_1={\displaystyle {\sum }_r{S}_r/S_{r+1}}$ .

We know $Z_1^r=\left\{c\in S_r|dc\in S_{s+1}\right\}$ and $Z_0^{r+1}=\left\{c\in S_{r+1}|dc\in S_{s+1}\right\}$ from Equation (10), so $Z_1^r=S_r$ and $Z_0^{r+1}=S_{r+1}$ . And because $D_0^r=\left\{dc|c\in Z_0^r\right\}=\left\{dc|c\in S_r\right\}=d{S}_r\subset S_{r+1}$ , and because $E_1={\displaystyle {\sum }_r{E}_1^r}={\displaystyle {\sum }_r{Z}_1^r/\left(Z_0^{r+1}+D_0^r\right) }$ , $E_1={\displaystyle {\sum }_r{S}_r/S_{r+1}}$ , that is proof.

Lemma 2. If the n-dimensional Lie algebra g satisfies Definition 5, then$\cdots \supset Z_s^p\supset Z_{s+1}^p\supset \cdots \supset D_{m+1}^p\supset D_m^p\supset \cdots$ is true.

Proof. Let us prove that $Z_s^p\supset Z_{s+1}^p$ .

If we take $c\in Z_{s+1}^p$ , we know $Z_s^p=\left\{c\in S_p|dc\in S_{p+s}\right\}$ , $Z_{s+1}^p=\left\{c\in S_p|dc\in S_{p+s+1}\right\}$ according to Equation (10), so $dc\in S_{p+s+1}\subset S_{p+s}$ , then $c\in Z_s^p$ , therefore $Z_s^p\supset Z_{s+1}^p$ .

Let’s prove that $D_{m+1}^p\supset D_m^p$ .

If we take $c\in Z_m^{p-m}$ , we know $D_m^p=\left\{dc|c\in Z_m^{p-m}\right\}$ from Equation (11), so $dc\in D_m^p$ . Since $D_{m+1}^p=\left\{dc|c\in Z_{m+1}^{p-m-1}\right\}$ , we only need to prove $c\in Z_{m+1}^{p-m-1}$ , that is, $dc\in D_{m+1}^p$ .

Because $Z_m^{p-m}=\left\{c\in S_{p-m}|dc\in S_p\right\}$ , so $dc\in S_p\subset S_{p-1}$ , then $c\in S_{p-m-1}$ , so $c\in Z_{m+1}^{p-m-1}$ , that is, $D_{m+1}^p\supset D_m^p$ .

It is shown that $Z_s^p$ is minimum when $p+s=n$ .

Because $S_n={\displaystyle {\sum }_p{S}_n^p}=S_n^n=\left\{c\in {\text{C}}^n|\iota \left(x_1\right)\cdots \iota \left(x_k\right)c=0,\forall x_1,\cdots ,x_k\in g,k>0\right\}=B^n$ and $Z_s^p=\left\{c\in S_p|dc\in S_{p+s}=S_n\right\}$ , $Z_s^p$ is minimum when $p+s=n$ .

It is shown below that $D_m^p$ is the largest when $p=m$ .

Because $S_{-1}\subset C={\displaystyle {\sum }_p{S}_0^p}$ , and $S_{-1}={\displaystyle {\sum }_p{S}_{-1}^p}$ , $S_{-1}^p\subset S_0^p$ . According to Property 2, $S_0^p\subset S_{-1}^p$ , then $S_{-1}^p=S_0^p$ , and therefore $S_{-1}=S_0$ . According to the equation (11) know $D_{m+1}^p=\left\{dc|c\in Z_m^{-1}\right\}$ and $Z_m^{-1}=\left\{c\in S_{-1}|dc\in S_m\right\}=\left\{c\in S_0|dc\in S_m\right\}=Z_m^0$ , so $D_m^p=D_{m+1}^p$ , namely when $p=m$ , $D_m^p$ is the largest and $D_m^p=D_r^p\left(r\ge m\right)$ .

Let’s prove $D_m^p\subset Z_m^p$ .

If we take $c\in D_m^p$ , we know from Equation (11) that $D_m^p=\left\{dc|c\in Z_m^{p-m}\right\}$ , so there exists $x\in Z_m^{p-m}$ , such that $dx=c$ . And according to Equation (10), $Z_m^p=\left\{c\in S_p|dc\in S_{p+m}\right\}$ , $Z_m^{p-m}=\left\{c\in S_{p-m}|dc\in S_p\right\}$ , so $dx=c\in S_p$ , then $dc=d\left(dx\right)=0\in S_{p+m}$ , so $c\in Z_m^p$ , that is, $D_m^p\subset Z_m^p$ .

That is $\cdots \supset Z_s^p\supset Z_{s+1}^p\supset \cdots \supset D_{m+1}^p\supset D_m^p\supset \cdots$ .

Definition 5 [12]. Let g be a Lie algebra over a field F, and V a vector space over F. Suppose there is a map from $g\times V$ to $V:\left(x,v\right)\to x\cdot v$ , $x\in g$ , $v\in V$ satisfying

1) $x\cdot \left(k_1{v}_1+k_2{v}_2\right)=k_{1}x\cdot v_1+k_{2}x\cdot v_2$ ,

2) $\left(k_1{x}_1+k_2{x}_2\right)\cdot v=k_1{x}_1\cdot v+k_2{x}_2\cdot v$ ,

3) $\left[x_1,x_2\right]\cdot v=x_1\cdot \left(x_2\cdot v\right)-x_2\cdot \left(x_1\cdot v\right)$ , $\forall k_1,k_2\in F,x,x_1,x_2\in g,v,v_1,v_2\in V$ .

Then V is called a (left) g-module, also referred to as g acting on V.

Definition 6. Let g be a Lie algebra, and V and W be g-modules. If a linear mapping $\Phi :V\to W$ satisfies $\Phi \left(x\cdot v\right)=x\cdot \Phi \left(v\right)$ ,$\forall x\in g$ , $v\in V$ , then Φ is called a homomorphism or intertwining operator from the g-module V to the g-module W. The set of all linear maps from V to W, denoted by $Hom\left(V,W\right)$ , is a linear space. Meanwhile, the set of all modular isomorphisms (intertwining operators) from V to W, denoted by $Ho{m}_g\left(V,W\right)$ , is a subspace of $Hom\left(V,W\right)$ .

3. Homology Theory on Differential Sequences

In this section we first introduce $H\left(E_1\right)\cong Hom\left(A\left(g\right),H\left(B\right)\right)$ and then give the proof for $E_{s+1}\cong H\left(E_s\right)$ . Next, we first give the edge operators ${\Delta }_r$ and δ.

Let g be an n-dimensional Lie algebra and $C={\displaystyle {\sum }_p{C}^p}$ be a differential graded algebra of g. Define the homomorphic map ${\eta }_r^p:Z_r^p\to E_r^p$ , satisfying

$\left({\eta }_r^{p}a\right)\left({\eta }_r^{p}b\right)={\eta }_r^{p+q}\left(a\wedge b\right)$ , for $a\in Z_r^p$ , $b\in Z_r^q$ .(14)

By mapping ${\eta }_r^p$ , we can define the edge operator

${\Delta }_r{\eta }_r^p={\eta }_r^{p+r}d$ (15)

of $E_r$ , meaning that any $da\in S_{p+r}$ , for $a\in Z_r^p\subset S_r$ , with

${\Delta }_r{\eta }_r^{p}a={\eta }_r^{p+r}da$ .(16)

In particular, ${\Delta }_r{\eta }_r^{p}a={\eta }_r^{p+r}da$ for $a\in Z_r^p$ , $da\in S_{p+r}$ , there is ${\eta }_r^{p}a\in E_r^p$ , ${\eta }_r^{p+r}da\in E_r^{p+r}$ , i.e., ${\Delta }_r:E_r^p\to E_r^{p+r}$ . And because ${\Delta }_r\left({\Delta }_r{\eta }_r^{p}a\right)={\Delta }_r\left({\eta }_r^{p+r}da\right)={\eta }_r^{p+2r}d\left(da\right)=0$ , and ${\left({\Delta }_r\right)}^2=0$ , we conclude that ${\Delta }_r$ is the edge operator on $E_r$ .

Let ι be the skew derivative on C. Let $f:{\wedge }^{p-r}\left(g^{*}\right)\to C^{p-r}$ be the concatenation map, define the map

$\phi :S_r^p\to Hom\left({\wedge }^{p-r}\left(g^{*}\right),B^r\right)$ ,(17)

that is, $\phi \left(c\right)\left(x_1\wedge \cdots \wedge x_{p-r}\right)={\left(-1\right)}^{p\left(p-r\right)}\iota \left(x_1\right)\cdots \iota \left(x_{p-r}\right)c$ , where $x_1,\cdots ,x_{p-r}\in g$ , $c\in S_r^p$ . Additionally, when $k\ne p-r$ , $\phi \left(c\right)\left({\wedge }^{k}g\right)=0$ .

Remark 1. Since $c\in S_r^p$ , then $\iota \left(x\right)\left(\iota \left(x_1\right)\cdots \iota \left(x_{p-r}\right)c\right)=0$ for $x\in g$ , which implies $\iota \left(x_1\right)\cdots \iota \left(x_{p-r}\right)c\in C^r$ , hence $\iota \left(x_1\right)\cdots \iota \left(x_{p-r}\right)c\in B^r$ . This indicates that the mapping $\phi$ is meaningful.

Define the mapping

$\psi :Hom\left({\wedge }^{p-r}\left(g\right),B^r\right)=B^r\otimes {\wedge }^{p-r}\left(g^{*}\right)\to S_r^p$ ,(18)

that is, $\psi \left(a\cdot b\right)=bf\left(a\right)$ and $\iota \left(x\right)\left(bf\left(a\right)\right)=\left(\iota \left(x\right)b\right)f\left(a\right)+{\left(-1\right)}^{r}b\left(\iota \left(x\right)f\left(a\right)\right)$ , for $b\in B^r$ , $a\in {\wedge }^{p-r}\left(g^{*}\right)$ , $x\in g$ .

Remark 2. Since $\iota \left(x\right)\left(bf\left(a\right)\right)=\left(\iota \left(x\right)b\right)f\left(a\right)+{\left(-1\right)}^{r}b\left(\iota \left(x\right)f\left(a\right)\right)={\left(-1\right)}^{r}bf\left(\iota \left(x\right)\right)a$ and $\iota \left(x\right)b=0$ , therefore $\iota \left(x\right)\left(bf\left(a\right)\right)={\left(-1\right)}^{r}b\left(\iota \left(x\right)f\left(a\right)\right)$ , then for $k>p-r$ , there is $\iota \left(x_1\right)\cdots \iota \left(x_k\right)\left(bf\left(a\right)\right)={\left(-1\right)}^{kr}bf\left(\iota \left(x_1\right)\cdots \iota \left(x_k\right)a\right)=0$ , where $x_1,\cdots ,x_k\in g$ , therefore $bf\left(a\right)\in S_r^p$ . Thus, it makes sense to define the mapping $\psi$ .

When g is an abelian Lie algebra, let $c\in S_r^p$ , such that $\phi \left(dc\right)\left(x_1\wedge \cdots \wedge x_{p-r}\right)={\left(-1\right)}^{p\left(p-r\right)}\iota \left(x_1\right)\cdots \iota \left(x_{p-r}\right)dc$ , where $dc\in S_{r+1}^{p+1}$ , and because $\iota \left(x\right)d=-d\iota \left(x\right)$ for $x\in g$ ,

$\begin{array}{c}\phi \left(dc\right)\left(x_1\wedge \cdots \wedge x_{p-r}\right)={\left(-1\right)}^{\left(p+1\right)\left(p-r\right)}{\left(-1\right)}^{p-r}d\left(\iota \left(x_1\right)\cdots \iota \left(x_{p-r}\right)c\right)\\ ={\left(-1\right)}^{\left(p+2\right)\left(p-r\right)}{\left(-1\right)}^{p\left(p-r\right)}d\left(\phi \left(c\right)\left(x_1\wedge \cdots \wedge x_{p-r}\right)\right)\\ =d\phi \left(c\right)\left(x_1\wedge \cdots \wedge x_{p-r}\right),\end{array}$ (19)

which implies $\phi \left(dc\right)\left(u\right)=\left(d\phi \left(c\right)\right)u$ for $u\in A\left(g\right)$ . Thus, upper edge operator can be defined in $Hom\left(A\left(g\right),B\right)$ .

Let $A\left(g\right)={\displaystyle {\sum }_p{\wedge }^p\left(g\right)}$ and $B={\displaystyle {\sum }_p{B}^p}$ , then, the upper edge operator

$\left(\delta F\right)\left(u\right)=d\left(Fu\right)$ , for $u\in A\left(g\right)$ , $F\in Hom\left(A\left(g\right),B\right)$ (20)

is defined in $Hom\left(A\left(g\right),B\right)$ .

Taking $\phi \left(c\right)\in Hom\left(A\left(g\right),B^r\right)$ and $u=x_1\wedge \cdots \wedge x_{p-r}\in {\wedge }^{p-r}\left(g\right)$ , we have $\begin{array}{c}\delta \phi \left(dc\right)u=\left(d\phi \left(c\right)\right)u=d\left({\left(-1\right)}^{p\left(p-r\right)}\iota \left(x_1\right)\cdots \iota \left(x_{p-r}\right)c\right)\\ ={\left(-1\right)}^{p\left(p-r\right)}{\left(-1\right)}^{p-r}\iota \left(x_1\right)\cdots \iota \left(x_{p-r}\right)dc\end{array}$ , for $c\in S_r^p$ .

Since $dc\in S_{r+1}^{p+1}\subset S_r^{p+1}\subset C^{p+1}$ , and $\iota \left(x\right)\left(\delta \phi \left(c\right)\left(u\right)\right)=0$ , $\delta \phi \left(c\right)\in Hom\left(A\left(g\right),B^{r+1}\right)$ , the upper edge operator $\delta$ becomes meaningful (ref. [2] pp. 533-534).

Nomizu [2] briefly introduces $H\left(E_1\right)\cong Hom\left(A\left(g\right),H\left(B\right)\right)$ , and then presents the proof of $H\left(E_1\right)\cong Hom\left(A\left(g\right),H\left(B\right)\right)$ based on Nomizu’s introduction.

Theorem 1 [2]. If g is a commutative Lie algebra and $H\left(B\right)$ is the cohomology of a subcomplex B, then $H\left(E_1\right)\cong Hom\left(A\left(g\right),H\left(B\right)\right)$ .

Proof. Since $Hom\left(A\left(g\right),B^r\right)=B^r\otimes A\left(g^{*}\right)$ it is straightforward to show that $H\left(E_1\right)\cong Hom\left(A\left(g\right),H\left(B\right)\right)$ is equivalent to proving $\left(E_1,{\Delta }_1\right)\cong \left(Hom\left(A\left(g\right),B\right),\delta \right)$ , where ${\Delta }_1$ and $\delta$ are the upper edge operators of $E_1$ and $Hom\left(A\left(g\right),B\right)$ , respectively.

1) First, let’s prove that $E_1={\displaystyle {\sum }_r{S}_r/S_{r+1}}\cong Hom\left(A\left(g\right),B\right)$ .

From Lemma 1 we know that $E_1={\displaystyle {\sum }_r{S}_r/S_{r+1}}$ .

Mapping $\phi :S_r^p\to Hom\left({\wedge }^{p-r}\left(g\right),B^r\right)$ , and extending $\phi$ to$S_r$ , i.e., $\tilde{\phi }:S_r/S_{r+1}\to Hom\left(A\left(g\right),B^r\right)$ , we have $S_{r+1}\subset Ker\tilde{\phi }$ . Take $c\in S_r^p$ such that $\phi \left(c\right)=0$ , i.e., $\phi \left(c\right)\left(x_1\wedge \cdots \wedge x_{p-r}\right)={\left(-1\right)}^{p\left(p-r\right)}\iota \left(x_1\right)\cdots \iota \left(x_{p-r}\right)c=0$ , for $x_1,\cdots ,x_{p-r}\in g$ .

When $k=p-r$ , take $c\in S_r^p$ , so c can only be in $S_{r+1}^p$ , so $Ker\tilde{\phi }\subset S_{r+1}$ . From the above proof we get $Ker\tilde{\phi }=S_{r+1}$ , then $S_r/S_{r+1}=S_r/Ker\tilde{\phi }\cong \tilde{\phi }\left(S_r\right)\subset Hom\left(A\left(g\right),B^r\right)$ .

Now, let’s prove $S_r/S_{r+1}\cong Hom\left(A\left(g\right),B^r\right)$ .

The mapping $\psi :Hom\left({\wedge }^{p-r}\left(g\right),B^r\right)=B^r\otimes {\wedge }^{p-r}\left(g^{*}\right)\to S_r^p$ is extended to $\tilde{\psi }:Hom\left(A\left(g\right),B^r\right)\to S_r$ and satisfies $\tilde{\psi }\tilde{\phi }=I$ , that is, $\tilde{\phi }$ is an isomorphic mapping of $S_r/S_{r+1}\to Hom\left(A\left(g\right),B^r\right)$ .

According to $Hom\left(A\left(g\right),B\right)=Hom\left(A\left(g\right),{\displaystyle {\sum }_r{B}^r}\right)={\displaystyle {\sum }_{r}Hom\left(A\left(g\right),B^r\right)}$ , we have $E_1={\displaystyle {\sum }_r{S}_r/S_{r+1}}\cong Hom\left(A\left(g\right),B\right)$ .

2) It follows that ${\phi }_{i+1}{\Delta }_1^{i}c={\delta }_i{\phi }_{i}c$ for $c\in E_i$ is true.

From the previous conclusions and ${\Delta }_1^i$ and ${\delta }_i$ definitions, ${\Delta }_1^i:E_1^i\to E_1^{i+1}$ ; ${\delta }_i:Hom\left(A\left(g\right),B^i\right)\to Hom\left(A\left(g\right),B^{i+1}\right)$ ; ${\phi }_i:E_1^{i-1}=S_{i-1}/S_i\to Hom\left(A\left(g\right),B^{i-1}\right)$ is known to be isomorphic.

Take any $c\in E_1^i$ , and if $c\in S_i^p/S_{i+1}^p$ , then $dc\in S_{i+1}^{p+1}/S_{i+2}^{p+1}$ , according to $Z_1^i=S_i$ , then ${\eta }_1^i:S_i\to E_1^i=S_i/S_{i+1}$ is also subjective, so there exists $a\in S_i^p$ such that ${\eta }_1^{i}a=c$ and $da\in d{Z}_1^i=D_1^{i+1}$ . Because ${\Delta }_1^i:E_1^i\to E_1^{i+1}$ , so ${\Delta }_1^{i}c\in S_{i+1}^{p+1}/S_{i+2}^{p+1}\subset S_i^{p+1}/S_{i+2}^{p+1}$ . We know from Equation (17) that

$\begin{array}{c}{\phi }_{i+1}{\Delta }_1^{i}c\left(x_1\wedge \cdots \wedge x_{p-i}\right)={\left(-1\right)}^{\left(p+1\right)\left(p-i\right)}\left(\iota \left(x_1\right)\cdots \iota \left(x_{p-i}\right){\Delta }_1^{i}c\right)\\ ={\left(-1\right)}^{\left(p+1\right)\left(p-i\right)}\left(\iota \left(x_1\right)\cdots \iota \left(x_{p-i}\right){\Delta }_1^i{\eta }_1^{i}a\right)\\ ={\left(-1\right)}^{\left(p+1\right)\left(p-i\right)}\left(\iota \left(x_1\right)\cdots \iota \left(x_{p-i}\right){\eta }_1^{i+1}da\right)\end{array}$ , for $x_1,\cdots ,x_{p-i}\in g$ .

We know from Equation (19) that

$\begin{array}{c}{\sigma }_i{\phi }_{i}c\left(x_1\wedge \cdots \wedge x_{p-i}\right)=d\left({\phi }_{i}c\left(x_1\wedge \cdots \wedge x_{p-i}\right)\right)\\ ={\phi }_i\left(dc\right)\left(x_1\wedge \cdots \wedge x_{p-i}\right)\\ ={\left(-1\right)}^{\left(p+1\right)\left(p-i\right)}\left(\iota \left(x_1\right)\cdots \iota \left(x_{p-i}\right)dc\right)\\ ={\left(-1\right)}^{\left(p+1\right)\left(p-i\right)}\left(\iota \left(x_1\right)\cdots \iota \left(x_{p-i}\right)d{\eta }_1^{i}c\right).\end{array}$

Thus, only $d{\eta }_1^{i}a={\eta }_1^{i+1}da$ is needed to prove ${\phi }_{i+1}{\Delta }_1^{i}c={\delta }_i{\phi }_{i}c$ , for $c\in E_i$ .

Let’s prove that $d{\eta }_1^{i}a={\eta }_1^{i+1}da$ .

Since ${\eta }_1^i:S_i\to E_1^i=S_i/S_{i+1}$ and ${\eta }_1^{i+1}:S_{i+1}\to E_1^{i+1}=S_{i+1}/S_{i+2}$ know ${\eta }_1^i\left(S_{i+1}\right)=\left\{0\right\}$ and ${\eta }_1^{i+1}\left(S_{i+2}\right)=\left\{0\right\}$ . If $a\in S_{i+1}^p$ , then $da\in S_{i+2}$ , then $d{\eta }_1^{i}a=0$ and ${\eta }_1^{i+1}da=0$ , so $d{\eta }_1^{i}a={\eta }_1^{i+1}da$ .

If $a\in {S^{\prime }}_i^p=\left\{a\in S_i^p,a\notin S_{i+1}^p\right\}$ , then ${\eta }_1^i:{S^{\prime }}_i^p\to S_i^p/S_{i+1}^p$ is an isomorphic mapping.

From the proof of (1) we know that there is a mapping ${{\phi }^{\prime }}_i:{S^{\prime }}_i^p\to Hom\left(A\left(g\right),B^i\right)$ satisfying ${{\phi }^{\prime }}_{i+1}\left(da\right)=d{{\phi }^{\prime }}_i\left(a\right)$ and ${{\phi }^{\prime }}_i={\phi }_i{\eta }_1^i$ , so ${{\phi }^{\prime }}_i$ , ${{\phi }^{\prime }}_{i+1}$ is an isomorphic mapping, so ${\phi }_{i+1}{\eta }_1^{i+1}\left(da\right)=d{\phi }_i{\eta }_1^i\left(a\right)={\phi }_{i+1}d{\eta }_1^i\left(a\right)$ , there is $d{\eta }_1^{i}a={\eta }_1^{i+1}da$ .

3) Let’s prove that $\left(E_1,{\Delta }_1\right)\cong \left(Hom\left(A\left(g\right),B\right),\delta \right)$ .

From the previous conclusion, knowing that

$E_1^0\stackrel{{\Delta }_1^0}{\to }{E}_1^1\stackrel{{\Delta }_1^1}{\to }{E}_1^2\stackrel{{\Delta }_1^2}{\to }\cdots \stackrel{{\Delta }_1^i}{\to }{E}_1^{i+1}\stackrel{{\Delta }_1^{i+1}}{\to }\cdots$ ,

$Hom\left(A\left(g\right),B^0\right)\stackrel{{\delta }_0}{\to }Hom\left(A\left(g\right),B^1\right)\stackrel{{\delta }_1}{\to }\cdots \stackrel{{\delta }_i}{\to }Hom\left(A\left(g\right),B^{i+1}\right)\stackrel{{\delta }_{i+1}}{\to }\cdots$ ,

and by (2) we obtain a commutative diagram(see Figure 1).

Figure 1. Commutative diagram.

So ${\phi }_{i+1}{\Delta }_1^i={\delta }_i{\phi }_i$ , and since ${\phi }_i$ and ${\phi }_{i+1}$ are isomorphic mappings, $\left(E_1^r,{\Delta }_1\right)$ and $\left(Hom\left(A\left(g\right),B^r\right),\delta \right)$ are isomorphic.

Nomizu gives the definition of the differential sequence $E_0,E_1,\cdots ,E_s,\cdots$ of a Lie algebra and proves that $E_2\cong H\left(E_1\right)$ [2]. Based on Nomizu’s proof for $E_2\cong H\left(E_1\right)$ , the proof for $E_{s+1}\cong H\left(E_s\right)$ is given.

Theorem 2. Let g be a Lie algebra, $E_0,E_1,\cdots ,E_s,\cdots$ be the differential sequence of g, and $H\left(E_s\right)$ be the cohomology of $E_s$ , then $E_{s+1}\cong H\left(E_s\right)$ .

Proof. Let’s first prove that $Ker{\Delta }_s=Im{\eta }_s^p$ .

1) Let’s prove that $Im{\eta }_s^p\subset Ker{\Delta }_s$ .

According to Lemma 2 and the definitions of ${\eta }_s^p$ and ${\Delta }_s$ ,

$Z_{s+1}^p\subset Z_s^p$ ; ${\eta }_s^p:Z_s^p\to E_s^p$ ; ${\Delta }_s:E_s^p\to E_s^{p+s}$ .

So if you need proof ${\eta }_s^{p+r}a\in Ker{\Delta }_s$ , for $a\in Z_{s+1}^p$ , you need proof ${\Delta }_s{\eta }_s^{p}a={\eta }_s^{p+s}da=0$ . Since ${\eta }_s^{p+s}:Z_s^{p+s}\to E_s^{p+s}=Z_s^{p+s}/\left(Z_{s-1}^{p+s+1}+D_{s-1}^{p+s}\right)$ , and $da\in d{Z}_{s+1}^p=D_{s+1}^{p+s+1}\subset Z_{s-1}^{p+s+1}$ , ${\eta }_s^{p+s}da=0$ , hence $Im{\eta }_s^p\subset Ker{\Delta }_s$ .

2) Let’s prove that $Im{\eta }_s^p\supset Ker{\Delta }_s$ .

If you take $\alpha \in Ker{\Delta }_s\subset E_s^p$ , since ${\eta }_s^p:Z_s^p\to E_s^p$ is subjective, there exists $a\in Z_s^p$ such that $\alpha ={\eta }_s^{p}a$ .

Because ${\Delta }_s\alpha ={\Delta }_s{\eta }_s^{p}a={\eta }_s^{p+s}da=0$ and ${\eta }_s^{p+s}da\in Z_s^{p+s}/\left(Z_{s-1}^{p+s+1}+D_{s-1}^{p+s}\right)$ , $da\in Z_{s-1}^{p+s+1}+D_{s-1}^{p+s}$ .

It is necessary to prove $a\in Z_{s+1}^p$ , that is ${\eta }_s^{p}a\in {\eta }_s^p{Z}_{s+1}^p$ .

If $da\in Z_{s-1}^{p+s+1}$ , and according to Equation (10) know $Z_{s-1}^{p+s+1}\subset S_{p+s+1}$ , so $da\in S_{p+s+1}$ , and because $a\in Z_s^p\subset S_p$ , so $a\in Z_{s+1}^p=\left\{c\in S_p|dc\in S_{p+s+1}\right\}$ .

So we just need to prove that $da\in Z_{s-1}^{p+s+1}$ .

Take $db\in D_{s-1}^{p+s}=d{Z}_{s-1}^{p+1}$ , then $b\in Z_{s-1}^{p+1}\subset Z_s^p$ , such that$d\left(a+b\right)\in Z_{s-1}^{p+s+1}\subset S_{p+s+1}$ , therefore $a+b\in Z_{s+1}^p$ . Because ${\eta }_s^{p}b\in {\eta }_s^p{Z}_{s-1}^{p+1}=Z_{s-1}^{p+1}/\left(Z_{s-1}^{p+1}+D_{s-1}^p\right)=\left\{0\right\}$ , so ${\eta }_s^p\left(a+b\right)={\eta }_s^{p}a\in {\eta }_s^p{Z}_{s+1}^p$ . Therefore $Ker{\Delta }_s=Im{\eta }_s^p$ .

Let’s prove that ${\eta }_s^p\left(Z_s^{p+1}+D_s^p\right)={\Delta }_s{E}_s^{p-s}$ .

Because of ${\eta }_s^p{Z}_s^{p+1}\subset {\eta }_s^p{Z}_{s-1}^{p+1}=Z_{s-1}^{p+1}/\left(Z_{s-1}^{p+1}+D_{s-1}^p\right)=\left\{0\right\}$ , take $a\in Z_{s+1}^p+D_s^p$ , make ${\eta }_s^{p}a\in {\eta }_s^p\left(Z_s^{p+1}+D_s^p\right)={\eta }_s^p{D}_s^p={\eta }_s^{p}d{Z}_s^{p-s}={\Delta }_s{\eta }_s^{p-s}{Z}_s^{p-s}={\Delta }_s{E}_s^{p-s}$ , then ${\eta }_s^p\left(Z_s^{p+1}+D_s^p\right)\subset {\Delta }_s{E}_s^{p-s}$ .

Conversely, we have $e\in {\Delta }_s{E}_s^{p-s}={\Delta }_s{\eta }_s^{p-s}{Z}_s^{p-s}={\eta }_s^{p}d{Z}_s^{p-s}={\eta }_s^p{D}_s^p$ , so we have $e\in {\eta }_s^p\left(Z_{s-1}^{p+1}+D_s^p\right)$ , which is ${\eta }_s^p\left(Z_s^{p+1}+D_s^p\right)={\Delta }_s{E}_s^{p-s}$ .

Let’s prove $E_{s+1}^i\cong H^i\left(E_s\right)$ .

Since ${\Delta }_s$ is the upper edge operator of $E_s$ and ${\Delta }_s^i:E_s^{is}\to E_s^{\left(i+1\right)s}$ , there is a chain complex

$E_s^0\stackrel{{\Delta }_s^0}{\to }{E}_s^s\stackrel{{\Delta }_s^1}{\to }\cdots \stackrel{{\Delta }_s^i}{\to }{E}_s^{\left(i+1\right)s}\stackrel{{\Delta }_s^{i+1}}{\to }\cdots$ ,

then $Ker{\Delta }_s^{i+1}={\eta }_s^{is}\left(Z_{s+1}^{is}\right)$ and $Im{\Delta }_s^i={\eta }_s^{is}\left(Z_s^{is+1}+D_s^{is}\right)$ , and therefore $H^i\left(E_s\right)=Ker{\Delta }_s^{i+1}/Im{\Delta }_s^i={\eta }_s^{is}\left(Z_{s+1}^{is}\right)/{\eta }_s^{is}\left(Z_s^{is+1}+D_s^{is}\right)$ .

Let $E_{s+1}^i=E_{s+1}^{is}$ .

Since $E_{s+1}^{is}=Z_{s+1}^{is}/\left(Z_s^{is+1}+D_s^{is}\right)$ and ${\eta }_s^{is}\left(Z_{s+1}^{is}\right)=Z_{s+1}^{is}/\left(Z_s^{is+1}+D_s^{is}\right)$ , and therefore ${\eta }_s^{is}\left(Z_s^{is+1}+D_s^{is}\right)=\left(Z_s^{is+1}+D_s^{is}\right)/\left(Z_{s-1}^{is+1}+D_{s-1}^{is}\right)$ , there is $E_{s+1}^i\cong H^i\left(E_s\right)$ , and therefore $E_{s+1}\cong H\left(E_s\right)$ .

4. Conclusion

Let g be an abelian Lie algebra and $C={\displaystyle {\sum }_p{C}^p}$ a differential graded algebra of g, define $S_1={\displaystyle {\sum }_r{S}_1^r}$ , and then define the differential sequence $E_0,E_1,\cdots ,E_s,\cdots$ , since the edge operator ${\Delta }_s$ exists on the chain complex of $E_s$ , the chain complex $E_s^0\stackrel{{\Delta }_s^0}{\to }{E}_s^s\stackrel{{\Delta }_s^1}{\to }\cdots \stackrel{{\Delta }_s^i}{\to }{E}_s^{\left(i+1\right)s}\stackrel{{\Delta }_s^{i+1}}{\to }\cdots$ can be obtained, and through this chain complex $E_{s+1}^i\cong H^i\left(E_s\right)$ can be obtained, so $E_{s+1}\cong H\left(E_s\right)$ .