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.
