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 $E_s$.

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

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

where, as usual, a $\hat{\cdot }$ 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 $\iota \left(x\right)\left(e^1\wedge \cdots \wedge e^n\right)={\displaystyle {\sum }_{j=1}^n{\left(-1\right)}^{j+1}\langle x,e^j\rangle e^1\wedge \cdots \wedge e^{j-1}\wedge {\hat{e}}^j\wedge \cdots \wedge e^n}$, 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 $Hom\left(V,W\right)$.

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 $c\in E_i$.