User:Jan A. Sanders/An Introduction to Leibniz 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 \(d^{(0)}\) in both cases. Given \(\alpha^0\in Hom_{\mathfrak{g}}(\mathfrak{a},\mathfrak{b})\) (this means that \(\alpha^0 d_\pm^{(0)}(x)=d_\pm^{(0)}(x)\alpha^0 \) for all \(x\in\mathfrak{g}\)), one extends \(\alpha^0\) to a linear map from \(C^n(\mathfrak{g},\mathfrak{a})\) to \(C^n(\mathfrak{g},\mathfrak{b})\) by \[(\alpha^n a^n)(x_1,\cdots,x_n)=\alpha^0 a^n(x_1,\cdots,x_n)\] for \(a^n\in C^n(\mathfrak{g},\mathfrak{a})\ .\)
lemma
\[ \alpha^n d^{(n)} (x)=d^{(n)}(x) \alpha^n, \quad n\geq 0\]
proof
\[\alpha^n d^{(n)}(y)a^n(x_1,\cdots,x_n)=\] \[=\alpha^0(d_+^{(0)}(y)a^n(x_1,\cdots,x_n)-\sum_{i=1}^n a^n(x_1,\cdots,[y,x_i],\cdots,x_n))\] \[=d_+^{(0)}(y)\alpha^0 a^n(x_1,\cdots,x_n)-\sum_{i=1}^n \alpha^0 a^n (x_1,\cdots,[y,x_i],\cdots,x_n)\] \[=d^{(n)}(y)\alpha^n a^n(x\_1,\cdots,x_n)\quad\square\]
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 \[d^0\alpha^0 a(x)=d_-^{(0)}(x)\alpha^0 a=\alpha^0 d_-^{(0)}(x) a=\alpha^0 d^0 a(x)=(\alpha^1 d^0 a )(x)\] For \( n=1\) one has \[d^1 \alpha^1 a^1 (x,y)=d^{(1)}(x)\alpha^1 a^1(y)-d_-^{(0)}(y) \alpha^1 a^1 (y)\ :\] \[=\alpha^1 d^{(1)}(x) a^1(y)-\alpha^0 d_-^{(0)}(y) a^1 (y)\ :\] \[=\alpha^0 d^1 a^1(x,y)\ :\] \[=(\alpha^2 d^1 a^1 )(x,y)\] Assume the statement to hold for \(k< n\ .\) Then \[\iota^{n+1}(x)d^n \alpha^n a^n=\] \[=-d^{n-1}\iota^n(x) \alpha^n a^n + d^{(n)}(x) \alpha^n a^n\] \[=-d^{n-1}\alpha^{n-1}\iota^n(x) a^n + \alpha^n d^{(n)}(x) a^n\] \[=\alpha^n( -d^{n-1}\iota^n(x) + d^{(n)}(x)) a^n\] \[=\alpha^n\iota^{n+1}(x)d^n a^n\] \[=\iota^{n+1}(x)\alpha^{n+1} d^n a^n\] 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 \(\beta^0\in Hom_{\mathfrak{g}}(\mathfrak{b},\mathfrak{c})\ .\) Then \(\beta^0\alpha^0\in Hom_{\mathfrak{g}}(\mathfrak{a},\mathfrak{c})\) and \[ [(\beta^0\alpha^0)^n]=[\beta^n][\alpha^n]:H^n(\mathfrak{g},\mathfrak{a})\rightarrow H^n(\mathfrak{g},\mathfrak{c})\ .\] Suppose now that we have an exact sequence \[\tag{1} 0\rightarrow \mathfrak{a}\rightarrow \mathfrak{b} \rightarrow \mathfrak{c} \rightarrow 0,\]
where \(\alpha^0\) is the injective map and \(\beta^0\) the surjective. There is an induced exact sequence \[\tag{2} 0\rightarrow H^0(\mathfrak{g},\mathfrak{a})\rightarrow H^0(\mathfrak{g},\mathfrak{b}) \rightarrow H^0(\mathfrak{g},\mathfrak{c}),\]
with (injective) \([\alpha^0]\) and (not necessarily surjective) \([\beta^0]\ .\) Notice that the elements in \(H^0(\mathfrak{g},\cdot)\) 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 \(0=[\alpha^0][a^0]=[\alpha^0 a^0]\ ,\) then \(\alpha^0 a^0=0\ ,\) which implies \(a^0=0\ .\) Thus \([\alpha^0]\) is injective. Suppose \([b^0]\in \mathrm{im} [\alpha^0]\ ,\) that is, there is an \(a^0\) such that \( [b^0]=[\alpha^0][a^0]\ .\) Then \([\beta^0][b^0]=[\beta^0][\alpha^0][a^0]=[\beta^0\alpha^0][a^0]=0\ .\) Thus \(\mathrm{im} [\alpha^0]\subset \ker [\beta^0]\ .\) On the other hand, if \([b^0]\in\ker[\beta^0]\ ,\) then \(\beta^0 b^0=0\ ,\) implying \(b^0=\alpha^0 a^0\ .\) We check \[ 0= d^1 b^0= d^1 \alpha^0 a^0= \alpha^1 d^1 a^0, \] and it follows from the injectivety of \(\alpha^1\) that \(a^0\in Z^0(\mathfrak{g},\mathfrak{a})\ ,\) or \( [a^0]\in H^0(\mathfrak{g},\mathfrak{a})\ .\) It follows that \([b]=[\alpha^0][a^0]\ .\) Thus the sequence is exact at \(H^0(\mathfrak{g},\mathfrak{b})\ .\)
remark
The map \([\beta^0]\) is not necessarily surjective. For example, suppose that \(\mathfrak{g}\) is onedimensional, with basiselement \(x\) acting on \(\mathfrak{b}=\mathbb{C}^2\) by \[ d_\pm^{(0)}(x)e_1=0\] \[d_\pm^{(0)}(x)e_2=e_1\]
Remark that necessarily \([x,x]=0\) (this is automatically true in the Lie algebra case), since \([x,x]=\lambda x\) implies \(0=[ad(x),ad(x)]=ad([x,x])=\lambda ad(x)\ .\) Take \(\mathfrak{a}=\mathbb{C} e_1\ .\) Then \(H^0(\mathfrak{g},\mathfrak{b})=\mathbb{C} e_1\) maps to \(0\) in \(\mathfrak{c}=\mathfrak{b}/\mathfrak{a}\ ,\) but \(H^0(\mathfrak{g},\mathfrak{c})=\mathbb{C}e_2+\mathfrak{a}\) is nonzero.\(\square\)
One measures the lack of surjectivity of \([B^0]\) 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 \[\tag{3} 0\rightarrow H^0(\mathfrak{g},\mathfrak{a})\rightarrow H^0(\mathfrak{g},\mathfrak{b}) \rightarrow H^0(\mathfrak{g},\mathfrak{c})\rightarrow H^1(\mathfrak{g},\mathfrak{a})\rightarrow\cdots\rightarrow H^n(\mathfrak{g},\mathfrak{a})\rightarrow H^n(\mathfrak{g},\mathfrak{b})\rightarrow H^n(\mathfrak{g},\mathfrak{c}) \rightarrow H^{n+1}(\mathfrak{g},\mathfrak{a})\rightarrow\cdots\]
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 \( B^n b^n= c^n\ .\) 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 \[ \alpha^{n+2} d^{n+1} a^{n+1} = d^{n+1} A^{n+1} a^{n+1}= d^{n+1} d^n b^n =0.\] 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 \[ \alpha^{n+1} (a^{n+1}-\tilde{a}^{n+1}) =d^n(b^n-\tilde{b}^n)=a^n A^n a^n= A^{n+1} d^n a^n.\] 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 \(\bar{b}^n=b^n-A^n a^n\) 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 \(\beta^0:\mathfrak{b}\rightarrow\mathfrak{c}\) and \(\tilde{\beta}^0:\tilde{\mathfrak{b}}\rightarrow\tilde{\mathfrak{c}}\) 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.\(\square\)
