User:Jan A. Sanders/An introduction to Lie algebra cohomology/Lecture 5
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The long exact cohomology sequence
Suppose \(\mathfrak{a}\) and \(\mathfrak{b}\) are \(\mathfrak{g}\)-modules, where the representation is denoted by \(d\) in both cases.
Given \(\alpha\in Hom_{\mathfrak{g}}(\mathfrak{a},\mathfrak{b})\) (this means that \(\alpha d_1(x)=d_1(x)\alpha \) for all \(x\in\mathfrak{g}\)), one extends \(\alpha\) to a linear map from \(C^n(\mathfrak{g},\mathfrak{a})\) to \(C^n(\mathfrak{g},\mathfrak{b})\) by \[(\alpha^n a_n)(x_1,\cdots,x_n)= \alpha a_n(x_1,\cdots,x_n)\] for \(a_n\in C^n(\mathfrak{g},\mathfrak{a})\ .\)
lemma
\[ \alpha^n d_1^{n} (x)=d_1^{n}(x) \alpha^n, \quad n\geq 0\]
proof
\[\alpha^n d_1^{n}(y)a_n(x_1,\cdots,x_n)=\] \[=\alpha(d_1(y)a_n(x_1,\cdots,x_n)-\sum_{i=1}^n a_n(x_1,\cdots,[y,x_i],\cdots,x_n))\] \[=d_1(y)\alpha a_n(x_1,\cdots,x_n)-\sum_{i=1}^n \alpha a_n (x_1,\cdots,[y,x_i],\cdots,x_n)\] \[=d_1^{n}(y)\alpha^n a_n(x_1,\cdots,x_n)\quad\square\]
lemma
\( \alpha^\cdot\) maps the complex \((C^{\cdot} (\mathfrak{g},\mathfrak{a}),d^{\cdot}) \) into the complex \((C^{\cdot} (\mathfrak{g},\mathfrak{a}),d^{\cdot} )\ ,\) that is, \(\alpha^{n+1}d^n=d^n \alpha^n\ .\)
proof
The statement for \(n=0\) reduces to \[d\alpha a(x)=d_1(x)\alpha a=\alpha d_1(x) a=\alpha d a(x)=(\alpha^1 d a )(x)\] For \( n=1\) one has \[d^1 \alpha^1 a_1 (x,y)=d_1^{1}(x)\alpha^1 a_1(y)-d_1(y) \alpha^1 a_1 (y)\ :\] \[=\alpha _1 d_1^{1}(x) a_1(y)- \alpha d_-(y) a_1 (y)\ :\] \[=\alpha d^1 a_1(x,y)\ :\] \[=(\alpha^2 d^1 a_1 )(x,y)\] Assume the statement to hold for \(k< n\ .\) Then \[\iota_1^{n+1}(x)d^n \alpha^n a_n=\] \[=-d^{n-1}\iota_1^n(x) \alpha^n a_n + d_1^{n}(x) \alpha^n a_n\] \[=-d^{n-1}\alpha^{n-1}\iota_1^n(x) a_n + \alpha^n d_1^{n}(x) a_n\] \[=\alpha^n( -d^{n-1}\iota_1^n(x) + d_1^{n}(x)) a_n\] \[=\alpha^n\iota_1^{n+1}(x)d^n a_n\] \[=\iota_1^{n+1}(x)\alpha^{n+1} d^n a_n\] and the lemma follows by induction on \(n\).\(\square\)
It follows that \(\alpha^{\cdot}\) leaves cocycles and coboundaries invariant and induces a map \[[\alpha^n]:H^n(\mathfrak{g},\mathfrak{a})\rightarrow H^n(\mathfrak{g},\mathfrak{b})\ .\] One writes \([\alpha^{\cdot}]\) for this family of maps.
Let \(\mathfrak{c}\) be another \(\mathfrak{g}\)-module, and suppose \(\beta\in Hom_{\mathfrak{g}}(\mathfrak{b},\mathfrak{c})\ .\) Then \(\beta\alpha\in Hom_{\mathfrak{g}}(\mathfrak{a},\mathfrak{c})\) and \[ [(\beta\alpha)^n]=[\beta^n][\alpha^n]:H^n(\mathfrak{g},\mathfrak{a})\rightarrow H^n(\mathfrak{g},\mathfrak{c})\ .\] Suppose now that we have an exact sequence \[\tag{1} 0\rightarrow \mathfrak{a}\rightarrow \mathfrak{b} \rightarrow \mathfrak{c} \rightarrow 0,\]
where \(\alpha:\mathfrak{a}\rightarrow \mathfrak{b}\) is the injective map and \(\beta:\mathfrak{b} \rightarrow \mathfrak{c}\) the surjective.
There is an induced exact sequence \[\tag{2} 0\rightarrow H_\mathfrak{g}(\mathfrak{g},\mathfrak{a})\rightarrow H(\mathfrak{g},\mathfrak{b}) \rightarrow H(\mathfrak{g},\mathfrak{c}),\]
with (injective) \([\alpha]\) and (not necessarily surjective) \([\beta]\ .\)
Notice that the elements in \(H_\mathfrak{g}(\mathfrak{g},\cdot)\) are just the \(\mathfrak{g}\)-invariant elements in the \(\mathfrak{g}\)-module, and equivalence classes are to be identified with their representing elements, since there is nothing to divide out since \(d^{-1}=0\ .\)
Indeed, if \(0=[\alpha][a]=[\alpha a]\ ,\) then \(\alpha a=0\ ,\) which implies \(a=0\ .\)
Thus \([\alpha]\) is injective.
Suppose \([b]\in \mathrm{im} [\alpha]\ ,\) that is, there is an \(a\) such that \( [b]=[\alpha][a]\ .\)
Then \([\beta][b]=[\beta][\alpha][a]=[\beta\alpha][a]=0\ .\) Thus \(\mathrm{im} [\alpha]\subset \ker [\beta]\ .\)
On the other hand, if \([b]\in\ker[\beta]\ ,\) then \(\beta b=0\ ,\) implying \(b=\alpha a\ .\)
We check \[ 0= d^1 b= d^1 \alpha a= \alpha^1 d^1 a, \] and it follows from the injectivety of \(\alpha^1\) that \(a\in Z(\mathfrak{g},\mathfrak{a})\ ,\) or \( [a]\in H(\mathfrak{g},\mathfrak{a})\ .\)
It follows that \([b]=[\alpha][a]\ .\)
Thus the sequence is exact at \(H_\mathfrak{g}(\mathfrak{g},\mathfrak{b})\ .\)
remark
The map \([\beta]\) is not necessarily surjective. For example, suppose that \(\mathfrak{g}\) is onedimensional, with basiselement \(x\) acting on \(\mathfrak{b}=\mathbb{C}^2\) by \[ d_1(x)e_1=0\] \[d_1(x)e_2=e_1\]
Take \(\mathfrak{a}=\mathbb{C} e_1\ .\) Then \(H_\mathfrak{g}(\mathfrak{g},\mathfrak{b})=\mathbb{C} e_1\) maps to \(0\) in \(\mathfrak{c}=\mathfrak{b}/\mathfrak{a}\ ,\) but \(H_\mathfrak{g}(\mathfrak{g},\mathfrak{c})=\mathbb{C}e_2+\mathfrak{a}\) is nonzero.\(\square\)
One measures the lack of surjectivity of \([\beta]\) by constructing a connecting homomorphism that embeds the left exact sequence (2) into a long exact sequence of cohomology spaces as follows.
lemma
For \(n=0,1,\cdots\) there is a map \([\delta_n]:H^n(\mathfrak{g},\mathfrak{c})\rightarrow H^{n+1}(\mathfrak{g},\mathfrak{a})\) such that the sequence \[\tag{3} 0\rightarrow H_\mathfrak{g}(\mathfrak{g},\mathfrak{a})\rightarrow H(\mathfrak{g},\mathfrak{b}) \rightarrow H(\mathfrak{g},\mathfrak{c})\rightarrow H^1(\mathfrak{g},\mathfrak{a})\rightarrow\cdots\rightarrow H^n(\mathfrak{g},\mathfrak{a})\rightarrow H^n(\mathfrak{g},\mathfrak{b})\rightarrow H^n(\mathfrak{g},\mathfrak{c}) \rightarrow H^{n+1}(\mathfrak{g},\mathfrak{a})\rightarrow\cdots\]
is exact.
proof
First one observes that (1) gives rise to the exact sequence of \(\mathfrak{g}\)-modules \[\tag{4} 0\rightarrow C^n(\mathfrak{g},\mathfrak{a})\rightarrow C^n(\mathfrak{g},\mathfrak{b})\rightarrow C^n(\mathfrak{g},\mathfrak{c})\rightarrow 0,\]
since the maps in (4) only act on the values of the cochains.
Thus since \(B^n\) commutes with \( d^n\) we have for \(c_n\in B^n(\mathfrak{g},\mathfrak{c})\) that \[ c_n=d^{n-1} c_{n-1} =d^{n-1} \beta^{n-1} b_{n-1} = \beta^n d^{n-1} b_{n-1} \in \beta^n B^n(\mathfrak{g},\mathfrak{b}).\] or \[\tag{5} B^n(\mathfrak{g},\mathfrak{c})=\beta^n B^n(\mathfrak{g},\mathfrak{b}).\]
Given \([c_n]\in H^n(\mathfrak{g},\mathfrak{c})\ ,\) we define \([\delta^n][c_n]\) as follows: Choose \( c_n \in Z^n(\mathfrak{g},\mathfrak{c})\ .\)
By the exactness of (4) there is a cochain \(b_n\in C^n(\mathfrak{g},\mathfrak{b})\) such that \( \beta^n b_n= c_n\ .\)
Then \(d^n b_n \in B^{n+1}(\mathfrak{g},\mathfrak{b})\) satisfies \[ \beta^{n+1} d^n b_n = d^n \beta^n b_n = d^n c_n =0.\] Hence by (4) there exists an \(a_{n+1}\in C^{n+1}(\mathfrak{g},\mathfrak{a})\) such that \[ \alpha^{n+1} a_{n+1}=d^n b_n .\] Furthermore, \( d^{n+1} a_{n+1}=0,\) since \[ \alpha^{n+2} d^{n+1} a_{n+1} = d^{n+1} \alpha^{n+1} a_{n+1}= d^{n+1} d^n b_n =0.\] Define \([\delta_n][c^n]=[a_{n+1}]\in H^{n+1}(\mathfrak{g},\mathfrak{a}).\) The first thing we have to check is that this definition depends on the cohomology class \([c_n]\) only and not on the particular choice of \(b_n\) or \( c_n\ .\)
Indeed, any other choice, say \(\tilde{c}_n=\beta^n \tilde{b}_n,\) must satisfy \[ \tilde{c}_n=\beta^n \tilde{b}_n =c_n - \beta^n d^{n-1} b_{n-1}\] for some \( b_{n-1}\in C^{n-1}(\mathfrak{g},\mathfrak{b}),\) by (5).
Hence \( \beta^n(b^n-\tilde{b}_n-d^{n-1} b_{n-1})=0,\) so, by (4) there exists a unique \(a_n\in C^n(\mathfrak{g},\mathfrak{a})\) with \[ b_n-\tilde{b}_n-d^{n-1} b_{n-1}= \alpha^n a_n.\] Thus the cocycle \(\tilde{a}_{n+1}\) such that \( \alpha^{n+1} \tilde{a}_{n+1}=d^n \tilde{b}_n \) satisfies \[ \alpha^{n+1} (a_{n+1}-\tilde{a}_{n+1}) =d^n(b_n-\tilde{b}_n)=d_n \alpha^n a_n= \alpha^{n+1} d^n a_n.\] Since \(\alpha^{n+1} \) is injective by the exactness of (4), one has \[ a_{n+1}-\tilde{a}_{n+1}= d^n a_n,\] that is, \([a_{n+1}]=[\tilde{a}_{n+1}],\) as claimed.
It follows that \([\delta^n]\) is a well-defined linear map.
Next we turn to the proof of the exactness of the cohomological sequence. It follows directly from the definition of \([\delta^n]\) that \[ \mathrm{im}\ [\beta^n]\subset \ker [\delta^n],\quad \mathrm{im} [\delta^n] \subset \ker [\alpha^{n+1}].\] For indeed, take \([c_n]=[\beta^n][b_n]\) with \( d^n b_n =0\ .\)
One defines \([\delta^n]\) by constructing \(a_{n+1}\) by \(\alpha^{n+1} a_{n+1} =d^n b_n\) but this is zero.
Since \(\alpha^{n+1}\) is injective, this means that \(a_{n+1}=0,\) or, in other words, that \( [\delta^n][\beta^n][b_n]=[0].\)
Furthermore, \[ [\alpha^{n+1}][\delta^n][c]=[\alpha^{n+1}][a_{n+1}]=[\alpha^{n+1}a_{n+1}]=[d^n b_n]=[0].\]
The opposite inclusions follows from the following arguments.
Let \( [c_n]\in \ker [\delta^n]\ .\) Then \(a_{n+1}=d^n a_n\) for some \(a_n\in C^n(\mathfrak{g},\mathfrak{a})\ .\)
Hence \( d^n(b_n-\alpha^n a_n)=\alpha^{n+1}a_{n+1}-\alpha^{n+1}d^n a_n=0\ ,\) so \(\bar{b}_n=b_n-\alpha^n a_n\) is a cocycle.
But by (4) we have \( \beta^n \bar{b}_n= \beta^n b_n- \beta^n \alpha^n a_n= \beta^n b_n=c_n\ .\)
Hence \( [c_n]\in \mathrm{im} [\beta^n]\ .\)
Finally, let \([a_{n+1}]\in \ker [\alpha^{n+1}]\ .\)
This means that \( \alpha^{n+1} a_{n+1}=d^n b_n\) for some \(b_n\in C^n(\mathfrak{g},\mathfrak{b})\ .\)
Set \( c_n=\beta^n b_n\ .\)
Then \(d^n c_n = d^n \beta^n b_n= \beta^{n+1} d^n b_n = \beta^{n+1}\alpha^{n+1} a_{n+1}=0\ .\)
Thus \( c^n\) is a cocycle, and by definition, \([\delta_n][c_n]=[a_{n+1}]\ .\)
It follows that \[ \mathrm{im}\ [\beta^n]= \ker [\delta^n],\quad \mathrm{im} [\delta^n] = \ker [\alpha^{n+1}].\]
exact sequence maps
Let \(\mathfrak{a}\subset\mathfrak{b}\) and \(\tilde{\mathfrak{a}}\subset\tilde{\mathfrak{b}}\) be \(\mathfrak{g}\)-modules, and \( f\in Hom_{\mathfrak{g}}(\mathfrak{b},\tilde{\mathfrak{b}})\) such that \(f(\mathfrak{a})\subset\tilde{\mathfrak{a}}\ .\) Let \(\mathfrak{c}=\mathfrak{b}/\mathfrak{a}\) and \(\tilde{\mathfrak{c}}=\tilde{\mathfrak{b}}/\tilde{\mathfrak{a}}\ .\) Then \(f\) induces maps \(\underline{f}\in Hom_{\mathfrak{g}}(\mathfrak{a},\tilde{\mathfrak{a}})\) and \(\overline{f}\in Hom_{\mathfrak{g}}(\mathfrak{c},\tilde{\mathfrak{c}})\) by restriction and passing to the quotient, respectively.
lemma
\[ [\delta^n][\overline{f}^n]=[\underline{f}^{n+1}][\delta^n].\]
proof
Let \(\beta:\mathfrak{b}\rightarrow\mathfrak{c}\) and \(\tilde{\beta}:\tilde{\mathfrak{b}}\rightarrow\tilde{\mathfrak{c}}\) denote the quotient maps.
Given \(c_n\in Z^n(\mathfrak{g},\mathfrak{c})\ ,\) take \(b_n\in C^n(\mathfrak{g},\mathfrak{b})\) such that \( \beta^n b_n=c_n\ .\)
Then \([\delta^n][c_n]=[d^n b_n]\ ,\) and hence \([\underline{f}^{n+1}][\delta^n][c_n]=[f^{n+1}d^n b_n]\ .\)
However, \[ [\overline{f}^n][c_n]=[f^n\beta^n b_n]=\tilde{\beta}^n f^n b_n],\] so \([\delta^n][\overline{f}^n][c_n]=[d^n f^n b_n]=[f^{n+1} d^n b_n]\ .\) One can conclude that \[ [\underline{f}^{n+1}][\delta^n]=[\delta^n][\overline{f}^n]\ ,\] as claimed.\(\square\)
