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

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

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.

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