# User:Jan A. Sanders/An Introduction to Leibniz Algebra Cohomology/Lecture 5

## 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.$$\square$$