# User:Jan A. Sanders/An introduction to Lie 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 $$d$$ 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 $$[\delta^n][\beta^n][b_n]=.$$

Furthermore, $[\alpha^{n+1}][\delta^n][c]=[\alpha^{n+1}][a_{n+1}]=[\alpha^{n+1}a_{n+1}]=[d^n b_n]=.$

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.$$\square$$