# User:Jan A. Sanders/An introduction to Lie algebra cohomology/Lecture 7

## Spectral sequences

In the examples, one uses $$K^{p,n}=F^pC^n(\mathfrak{g},\mathfrak{a})\ ,$$ but in the construction of the spectral sequence one only assumes $$K^{p,n}\supset K^{p+1,n}$$ and $$d^n K^{p,n}\subset K^{p,n+1}\ .$$

### definition

Let $Z_p^{r,n}=K^{p,n}$ for $$r\leq 0\ ,$$ $Z_p^{r,n}=\{a_n\in K^{p,n}| d^n a_n\in K^{p+r,n+1}\}$ for $$r>0\ .$$

### remark

If $$p>n$$ then $$Z_p^{r,n}=0\ .$$

### proposition

$Z_0^{1,0}=\mathfrak{a}^\mathfrak{h}=H^0(\mathfrak{g},\mathfrak{a})\ ,$ where $$\mathfrak{a}^\mathfrak{h}$$ denote the $$\mathfrak{h}$$-invariant elements (under $$d_-^{(0)}$$) in $$\mathfrak{a}\ .$$

### proof

$$a\in Z_0^{1,0}$$ implies $$da \in F^1 C^1(\mathfrak{g},\mathfrak{a})\ ,$$ that is, for $$y\in\mathfrak{h}$$ one has $d_1(y)a=da(y)=0\ .$

### proposition

$Z_1^{1,1} \subset F^1 C^1(\mathfrak{g},\mathfrak{a}^\mathfrak{h})$

### proof

$$a_1\in Z_1^{1,1}$$ implies that for $$x\in\mathfrak{g}, y\in \mathfrak{h}$$ one has, since $$d^1 a_1\in F^2 C^2(\mathfrak{g},\mathfrak{a})\ ,$$ $0=d^1a_1(x,y)=d_1(x)a_1(y)-d_1(y)a_1(x)-a_1([x,y])=-d_1(y)a_1(x)\ .$

### proposition

$Z_{p+1}^{r-1,n}\subset Z_p^{r,n}$

### proof

Let $$a_n\in Z_{p+1}^{r-1,n}\ .$$ Then $$a_n\in K^{p+1,n}\subset K^{p,n}$$ and $$d^n a_n \in K^{p+r,n+1}\ .$$ But this immediately implies that $$a_n\in Z_p^{r,n}\ .$$

### proposition

$d^{n-1} Z_{p-r+1}^{r-1,n-1}\subset Z_p^{r,n}$

### proof

Let $$a_{n}\in d^{n-1} Z_{p-r+1}^{r-1,n-1}\ .$$

Then one can write $$a_n$$ as $$d^{n-1}a_{n-1}\ ,$$ with $$a_{n-1}\in Z_{p-r+1}^{r-1,n-1}\ ,$$ that is to say, $$a_{n-1}\in K^{p-r+1,n-1}$$ and $$a_n=d^{n-1}a_{n-1}\in K^{p,n}\ .$$

Since $$d^na_n=d^n d^{n-1}a_{n-1}=0\ ,$$ it follows that $$a_n\in Z_p^{r,n}\ .$$

### definition

$E_p^{r,n}=Z_p^{r,n}/(d^{n-1} Z_{p-r+1}^{r-1,n-1}+Z_{p+1}^{r-1,n})$

### remark

In particular, $$E_p^{0,n}=Z_p^{0,n}/Z_{p+1}^{0,n}\ ,$$ that is, the graded version of the filtered sequence $$K^{p,n}\ .$$

### theorem

On $$E_\cdot^{r,n}$$ there is an induced coboundary operator $$\mathbf{d}^n$$ such that $H^p(E_\cdot^{r,n},\mathbf{d}^n)=E_p^{r+1,n}$ This means that $$E_\cdot^{r,n}$$ is a spectral sequence.

### proof

$$d^n$$ maps $$Z_p^{r,n}$$ to $$Z_{p+r}^{r,n+1}$$ and $$d^{n-1} Z_{p-r+1}^{r-1,n-1}+Z_{p+1}^{r-1,n}$$ to $$d^{n} Z_{p+1}^{r-1,n}\ .$$

Let $$[a_n]\in E_p^{r,n}\ .$$ Define $\mathbf{d}^n[a_n]=[d^n a_n]$ One has $$[d^n a_n]\in E_{p+r}^{r,n+1}\ .$$ Suppose $$[a_n]$$ is a cocycle. This means that $$d^na_n\in d^{n} Z_{p+1}^{r-1,n}+Z_{p+r+1}^{r-1,n+1}\ .$$ That is, there exist $$\tilde{a}^n \in Z_{p+1}^{r-1,n}$$ and $$a_{n+1}\in Z_{p+r+1}^{r-1,n+1}$$ such that $d^n a_n=d^n \tilde{a}^n+a_{n+1}\ .$ Let $$\bar{a}^n=a_n-\tilde{a}^n\in Z_p^{r,n}+Z_{p+1}^{r-1,n} \subset K^{p,n}\ ,$$ with $$d^n \bar{a}^n=a_{n+1}\in Z_{p+r+1}^{r-1,n+1}\in K^{p+r+1,n+1}\ .$$

Therefore $$\bar{a}^n \in Z_{p}^{r+1,n}\ .$$

This implies that $$a_n=\bar{a}^n+\tilde{a}^n\in Z_{p}^{r+1,n}+Z_{p+1}^{r-1,n}\ .$$

It follows that $Z^p(E_\cdot^{r,n},\mathbf{d}^n)=(Z_{p}^{r+1,n}+Z_{p+1}^{r-1,n})/(d^{n-1} Z_{p-r+1}^{r-1,n-1}+Z_{p+1}^{r-1,n})$ The $$p$$-coboundaries consist of the elements of $$d^{n-1} Z_{p-r}^{r,n-1}\ ,$$ and one has $B^p(E_\cdot^{r,n},\mathbf{d}^n)=(d^{n-1} Z_{p-r}^{r,n-1}+Z_{p+1}^{r-1,n})/(d^{n-1} Z_{p-r+1}^{r-1,n-1}+Z_{p+1}^{r-1,n})$

### Noether isomorphism

If $$W\subset U$$ then $U/(W+U\cap V)\simeq (U+V)/(W+V)$ and $$(M/V)/(U/V)=M/U\ .$$

It follows that $H^p(E_\cdot^{r,n},\mathbf{d}^n)=(Z_{p}^{r+1,n}+Z_{p+1}^{r-1,n})/(d^{n-1} Z_{p-r}^{r,n-1}+Z_{p+1}^{r-1,n})$

### proposition

$d^{n-1} Z_{p-r}^{r,n-1}\subset Z_p^{r+1,n}\ .$

### proof

Let $$a_n \in d^{n-1} Z_{p-r}^{r,n-1}\ .$$

Then $$a_n=d^{n-1} a_{n-1}$$ with $$a_{n-1}\in Z_{p-r}^{r,n-1}\ .$$

Therefore $$a_n\in Z_p^{r+1,n}\ ,$$ since $$d^n a_n=0$$.$$\square$$

It follows that $H^p(E_\cdot^{r,n},\mathbf{d}^n)=Z_{p}^{r+1,n}/(d^{n-1} Z_{p-r}^{r,n-1}+Z_{p}^{r+1,n}\cap Z_{p+1}^{r-1,n})$

### proposition

$Z_p^{r+1,n}\cap Z_{p+1}^{r-1,n}=Z_{p+1}^{r,n}\ .$

### proof

Let $$a_n\in Z_p^{r+1,n}\cap Z_{p+1}^{r-1,n}\ .$$ Then $$a_n\in K^{p+1,n}$$ and $$d^n a_n \in F^{p+r+1} C^{n+1}(\mathfrak{g},\mathfrak{a})\ .$$

This implies $$a_n\in Z_{p+1}^{r,n}\ .$$

On the other hand, if $$a_n\in Z_{p+1}^{r,n}\ ,$$ we have $$a_n\in K^{p+1,n}\subset F^p C^n(\mathfrak{g},\mathfrak{a})$$ and $$d^na_n \in K^{p+r+1,n+1}\subset K^{p+r,n+1}\ .$$

Thus, $$a_n \in K^{p,n}$$ and $$d^na_n \in K^{p+r+1,n+1}\ ,$$ implying $$a_n\in Z_p^{r+1,n}\ .$$

Furthermore, $$a_n \in K^{p+1,n}$$ and $$d^na_n \in K^{p+r,n+1}\ ,$$ implying $$a_n\in Z_{p+1}^{r-1,n}\ .$$

The result follows. $$\square$$

### corollary

$H^p(E_\cdot^{r,n},\mathbf{d}^n)=Z_{p}^{r+1,n}/(d^{n-1} Z_{p-r}^{r,n-1}+Z_{p+1}^{r,n})=E_p^{r+1,n}$ This proves the theorem.