lecture 5

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< An introduction to Leibniz algebra cohomology

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

Back to the fourth lecture

On to the sixth lecture

1 The long exact cohomology sequence
- 1.1 lemma
- 1.2 proof
- 1.3 lemma
- 1.4 proof
- 1.5 remark
- 1.6 lemma
- 1.7 proof
2 exact sequence maps
- 2.1 lemma
- 2.2 proof

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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 $a^n\in C^n(\mathfrak{g},\mathfrak{a})$ .

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lemma

$\alpha^n d^{(n)} (x)=d^{(n)}(x) \alpha^n, \quad n\geq 0$

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proof

$\alpha^n d^{(n)}(y)a^n(x_1,\cdots,x_n)=$

$=\alpha^0(d_+^{(0)}(y)a^n(x_1,\cdots,x_n)-\sum_{i=1}^n a^n(x_1,\cdots,[y,x_i],\cdots,x_n))$

$=d_+^{(0)}(y)\alpha^0 a^n(x_1,\cdots,x_n)-\sum_{i=1}^n \alpha^0 a^n (x_1,\cdots,[y,x_i],\cdots,x_n)$

$=d^{(n)}(y)\alpha^n a^n(x\_1,\cdots,x_n)\quad\square$

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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, $\alpha^{n+1}d^n=d^n \alpha^n$ .

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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

(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

(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 $H^0(\mathfrak{g},\mathfrak{b})$ .

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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.

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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

(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.

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proof

First one observes that (1) gives rise to the exact sequence of $\mathfrak{g}$ -modules

(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

$c^n=d^{n-1} c^{n-1} =d^{n-1} \beta^{n-1} b^{n-1} = \beta^n d^{n-1} b^{n-1} \in \beta^n B^n(\mathfrak{g},\mathfrak{b}).$

(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

$\mathrm{im}\ [\beta^n]= \ker [\delta^n],\quad \mathrm{im} [\delta^n] = \ker [\alpha^{n+1}].$

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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.

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lemma

$[\delta^n][\overline{f}^n]=[\underline{f}^{n+1}][\delta^n].$

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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,