lecture 7

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< An introduction to Lie algebra cohomology

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Author: Dr. Jan A. Sanders, Vrije Universiteit Amsterdam
Author: Dr. Sara Lombardo, Vrije Universiteit Amsterdam

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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 $d^n K^{p,n}\subset K^{p,n+1}$ .

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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 $r>0$ .

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remark

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

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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 $\mathfrak{a}$ .

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proof

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

$d_-^{(0)}(y)a^0=d^0a^0(y)=0$ .

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proposition

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

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proof

$a^1\in Z_1^{1,1}$ implies that for $x\in\mathfrak{g}, y\in \mathfrak{h}$ one has, since $d^1 a^1\in F^2 C^2(\mathfrak{g},\mathfrak{a})$ ,

$0=d^1a^1(x,y)=d^{(0)}(x)a^1(y)-d^{(0)}(y)a^1(x)-a^1([x,y])=-d^{(0)}(y)a^1(x)$ .

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proposition

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

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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 $a^n\in Z_p^{r,n}$ .

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proposition

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

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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 $a^n\in Z_p^{r,n}$ .

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definition

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

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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 $K^{p,n}$ .

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theorem

On $E_\cdot^{r,n}$ there is an induced coboundary operator $\delta_r^1$ such that

$H^p(E_\cdot^{r,n},\delta_r^1)=E_p^{r+1,n}$

This means that $E_\cdot^{r,n}$ is a spectral sequence.

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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

$\delta_r^1[a^n]=[d^n a^n]$

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

$Z^p(E_\cdot^{r,n},\delta_r^1)=(Z_{p}^{r+1,n}+Z_{p+1}^{r-1,n})/(d^{n-1} Z_{p-r+1}^{r-1,n-1}+Z_{p+1}^{r-1,n})$

The $p$ -coboundaries consist of the elements of $d^{n-1} Z_{p-r}^{r,n-1}$ , and one has

$B^p(E_\cdot^{r,n},\delta_r^1)=(d^{n-1} Z_{p-r}^{r,n-1}+Z_{p+1}^{r-1,n})/(d^{n-1} Z_{p-r+1}^{r-1,n-1}+Z_{p+1}^{r-1,n})$

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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

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

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proposition

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

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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

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

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proposition

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

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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 $a^n\in Z_p^{r+1,n}$ .