# User:Jan A. Sanders/An introduction to Lie algebra cohomology/Lecture 3

## Contents

### abstract

In this lecture we define the cohomology modules $$H^n(\mathfrak{g},\mathfrak{a})$$

## Lifting the representation to the forms

### definition

$$\mathfrak{g}$$ is called an ideal in $$\mathfrak{l}$$ if $[\mathfrak{l},\mathfrak{g}]\subset \mathfrak{g}\ .$

### example

Let $$\mathfrak{g}=\ker d_1\ .$$ Then for $$x\in\mathfrak{l}$$ and $$y\in\mathfrak{g}$$ one has $d_1([x,y])=d_1(x)d_1(y)-d_1(y)d_1(x)=0$ It follows that $$\mathfrak{g}=\ker d_1$$ is an ideal in $$\mathfrak{l}\ .$$

### definition

Let $$\mathfrak{l}$$ be a Lie algebra and $$\mathfrak{g}$$ an ideal. Then $$\mathfrak{l}/\mathfrak{g}$$ is a Lie algebra with the bracket $[[x],[y]]=[[x,y]]$ where $$[x]$$ denotes the equivalence class of $$x\ .$$ This is well defined, since varying $$x$$ and $$y$$ with elements in $$\mathfrak{g}$$ does not change the answer: $[[x],[y]]=[[x+g_1,y+g_2]]=[[x,y]]+[[x,g_2]]+[[g_1,y]]+[[g_1,g_2]]=[[x,y]]$

### terminology

When $$\mathfrak{a}$$ is a module and a representation space of $$\mathfrak{l}\ ,$$ one says that $$\mathfrak{a}$$ is an $$\mathfrak{l}$$-module. If the representation is zero, $$\mathfrak{a}$$ is a trivial $$\mathfrak{l}$$-module.

### definition

Let $$\mathfrak{a}$$ be an $$\mathfrak{l}$$-module. In order to give a general definition of a coboundary operator $$d^n , n\geq 0 \ ,$$ one defines first an induced representation on $$C^n (\mathfrak{g},\mathfrak{a})$$ as follows.

Let, for $$y\in\mathfrak{l}\ ,$$ $(d_1^n(y)a^n)(x_1,\cdots,x_n)=d_1(y)a^n(x_1,\cdots,x_n)-\sum_{i=1}^n a^n(x_1,\cdots, [y,x_i],\cdots,x_n).$ This is indeed a representation. Let $$y,z\in\mathfrak{l}\ .$$ Then $d_1^n(y)d_1^n(z)a^n(x_1,\cdots,x_n)=$ $=d_1(y)d_1^n(z)a^n(x_1,\cdots,x_n)-\sum_{i=1}^n d_1^n(z) a^n(x_1,\cdots, [y,x_i],\cdots,x_n)$ $=d_1(y)d_1(z)a^n(x_1,\cdots,x_n)-\sum_{i=1}^n d_1(z) a^n(x_1,\cdots, [y,x_i],\cdots,x_n)\ :$ $-\sum_{i=1}^n d_1(y) a^n(x_1,\cdots, [z,x_i],\cdots,x_n)+\sum_{ji}a^n(x_1,\cdots,[y,x_i],\cdots, [z,x_j],\cdots,x_n)+a^n(x_1,\cdots, [z,[y,x_i]],\cdots,x_n)$ It follows that $d_1^n(y)d_1^n(z)a^n(x_1,\cdots,x_n)-d_1^n(z)d_1^n(y)a^n(x_1,\cdots,x_n)=$ $=(d_1(y)d_1(z)-d_1(z)d_1(y))a^n(x_1,\cdots,x_n)+ \sum_{i=1}^n a^n(x_1,\cdots, [z,[y,x_i]],\cdots,x_n)- \sum_{i=1}^n a^n(x_1,\cdots, [y,[z,x_i]],\cdots,x_n)$ $=d_1([y,z])a^n(x_1,\cdots,x_n)-\sum_{i=1}^n a^n(x_1,\cdots, [[y,z],x_i]],\cdots,x_n)$ $=d_1^n([y,z])a^n(x_1,\cdots,x_n)$ or, $d_2^n(y,z)=[d_1^n(y),d_1^n(z)]-d_1^n([y,z])=0\ .$

### remark

Remark that $$C^n(\mathfrak{g},\mathfrak{a})$$ is mapped into itself by $$d_1^n(x)$$ for all $$x\in\mathfrak{l}\ .$$

## Definition of the coboundary operator.

We now reformulate the definition of $$d^{i}, i=1,2,3$$ using the $$d_1^n\ .$$ First we introduce the contraction operator $$\iota_1^n(y): C^n(\mathfrak{g},\mathfrak{a})\rightarrow C^{n-1}(\mathfrak{g},\mathfrak{a})$$ by $( \iota_1^n(y)a_n)(x_1,\cdots,x_{n-1})=a_n(y,x_1,\cdots,x_{n-1})\ .$

Recall the following definitions of the coboundary operators.

• $$d a_0(x)=d_1(x)a_0\ .$$
• $$d^1 a_1(x,y)=d_1(x)a_1(y)-d_1(y)a_1(x)-a_1([x,y])=(d_1^1(x)a_1)(y)-d_1(y)\iota_1^1(x)a_1=(d_1^1(x)a_1)(y)-d \iota_1^1(x)a_1 (y)\ .$$
• $$d^2 a_2(x,y,z)=d_1(x)a_2(y,z)-d_1(y)a_2(x,z)+d_1(z)a_2(x,y)-a_2([x,y],z)-a_2(y,[x,z])+a_2(x,[y,z])\ :$$
$=(d_1^2(x)a_2)(y,z)-d_1^1(y)\iota_1^2(x)a_2(z)+d_1(z)\iota_1^1(y)\iota_1^2(x)a_2\ :$
$=(d_1^2(x)a_2)(y,z)-d^1 \iota_1^2(x)a_2 (y,z)\ .$

### definition

This strongly suggests the following recursive definition:

• $$\iota_1^1(x)d=d_1(x)$$
• $$\iota_1^{n+1}(x)d^n+d^{n-1}\iota_1^n(x)=d_1^n(x),\quad n>0$$

### lemma

Let $$y\in\mathfrak{l}$$ and $$z\in\mathfrak{g}\ .$$ Then

• $$\iota_1^n(z)d_1^n(y)-d_1^{n-1}(y)\iota_1^n(z)=-\iota_1^{n}([y,z]).$$

### proof

Consider $(\iota_1^n(z)d_1^{n}(y)-d_1^{n-1}(y)\iota_1^n(z))a_n(x_1,\cdots,x_{n-1})$ $= d_1^{n}(y)a_n(z,x_1,\cdots,x_{n-1})-d_1^{n-1}(y)\iota_1^n(z)a_n(x_1,\cdots,x_{n-1})$ $=d_1(y)a_n(z,x_1,\cdots,x_{n-1})-a_n([y,z],x_1,\cdots,x_{n-1})-\sum_{i=1}^{n-1}a_n(z,x_1,\cdots,[y,x_i],\cdots,x_n)\ :$ $-d_1(y)a_n(z,x_1,\cdots,x_n)+\sum_{i=1}^{n-1}a_n(z,x_1,\cdots,[y,x_i],\cdots,x_{n-1})$ $=-a_n([y,z],x_1,\cdots,x_{n-1})$ $=-\iota_1^{n}([y,z])a_n(x_1,\cdots,x_{n-1})\quad\square\ .$

### lemma

Let $$y\in\mathfrak{l}\ .$$ Then $d_1^{n+1}(y)d^{n}=d^{n}d_1^{n}(y),\quad n\geq 0\ .$

### proof

For $$n=0$$ one has $d_1^{1}(x)d a(y)-dd_1(x)a(y)=d_1(x)d_1(y)a-d_1([x,y])a-d_1(y)d_1(x)a=0\ .$ For $$n>0$$ one has, with $$z\in\mathfrak{g}$$ and $$n>0\ ,$$ that $\iota_1^{n+1}(z)(d_1^{n+1}(y)d^{n}-d^{n}d_1^{n}(y))=$ $=-\iota_1^{n+1}([y,z])d^{n}+d_1^{n}(y)\iota_1^{n+1}(z)d^n-d_1^{n}(z)d_1^{n}(y)+d^{n-1}\iota_1^{n}(z)d_1^{n}(y)$ $=-\iota_1^{n+1}([y,z])d^{n}+d_1^{n}(y)(d_1^{n}(z)-d^{n-1}\iota_1^n(z)-d_1^{n}(z)d_1^{n}(y)+d^{n-1}(d_1^{n-1}(y)\iota_1^n(z)-\iota_1^n([y,z]))$ $=-d_1^{n}([y,z])+d_1^{n}(y)d_1^{n}(z)-d_1^{n}(z)d_1^{n}(y)+(d^{n-1}d_1^{n-1}(y)-d_1^{n}(y)d^{n-1})\iota_1^n(z)$ $=d_2^n(y,z)+(d^{n-1}d_1^{n-1}(y)-d_1^{n}(y)d^{n-1})\iota_1^n(z)$ $=(d^{n-1}d_1^{n-1}(y)-d_1^{n}(y)d^{n-1})\iota_1^n(z)$ This implies the statement of the lemma by induction.$$\square$$

### theorem - coboundary operator

$$d^{\cdot}$$ is a coboundary operator.

### proof

One computes $\iota_1^{n+2}(y)d^{n+1}d^{n}=d_1^{n+1}(y)d^{n}-d^{n}\iota_1^{n+1}(y)d^{n}$ $=d^{n}d_1^{n}(y)-d^{n}(d_1^{n}(y)-d^{n-1}\iota_1^{n}(y))$ $=d^{n}d^{n-1}\iota_1^{n}(y).$ Again, since $$d^1d^0=0\ ,$$ it follows by induction that

$d^{n+1}d^{n}=0.$

This shows that $$d^i, i\in\mathbb{N}$$ is a coboundary operator.

### proposition

$d^n \omega_n(x_1,\cdots,x_{n+1})=\sum_{i=1}^{n+1} (-1)^{i-1} d_1(x_i) \omega_n(x_1,\cdots,\hat{x}_i,\cdots,x_{n+1}) +\sum_{i<j} (-1)^i\omega_n(x_1,\cdots,\hat{x}_i,\cdots,[x_i,x_j],\cdots,x_{n+1})$

### corollary

$$d^n$$ maps $$C_{\wedge}^n(\mathfrak{g},\mathfrak{a})$$ to $$C_{\wedge}^{n+1}(\mathfrak{g},\mathfrak{a})\ .$$

## Cohomology

Define $$Z^n(\mathfrak{g},\mathfrak{a})=\ker d^n\ ,$$ the space of cocycles, and $$B^n(\mathfrak{g},\mathfrak{a})=\mathrm{im\ }d^{n-1}\ ,$$ the space of coboundaries.

Since $$\mathrm{im\ }d^{n-1}\subset\ker d^{n}\ ,$$ one can define $H^n(\mathfrak{g},\mathfrak{a})=Z^n(\mathfrak{g},\mathfrak{a})/B^n(\mathfrak{g},\mathfrak{a})\ ,$ the $$n$$-cohomology module of $$\mathfrak{g}$$ with values in $$\mathfrak{a}\ .$$

If $$a_n\in C^n(\mathfrak{g},\mathfrak{a})\ ,$$ the equivalence class in $$H^n(\mathfrak{g},\mathfrak{a})$$ is denoted by $$[a_n]\ .$$ Elements in the zero equivalence class, the image of $$d^{n-1}\ ,$$ are called trivial.

### remark

For $$n=0\ ,$$ $$H^0(\mathfrak{g},\mathfrak{a})=Z^0(\mathfrak{g},\mathfrak{a})=\ker d^0\ ,$$ that is, is consists of all elements in $$\mathfrak{a}$$ which are $$\mathfrak{g}$$-invariant.

This indicates that computing the cohomology can be a formidable problem, since it contains for instance classical invariant theory.

Cohomology theory itself does not provide the answers, it just asks the right questions and removes the trivial answers.

### theorem

$$H^n(\mathfrak{g},\mathfrak{a}), n>0 ,$$ is invariant under the action (by $$d_1^{n}$$) of $$\mathfrak{g}\ .$$ So one could say that $$H^n(\mathfrak{g},\mathfrak{a})$$ is a trivial $$\mathfrak{g}$$-module and an $$\mathfrak{l}/\mathfrak{g}$$-module.

### proof

Indeed, since $$d^n a_n=0\ ,$$ $d_1^{n}(y)[a_n]=[d_1^{n}(y)a_n]=[\iota_1^{n+1}(y)d^{n}a_n+d^{n-1}\iota_1^{n}(y)a_n]=.\quad\square$

### lemma

If $$d^n a_n=0$$ and $$d_1^n(y)a_n=0\ ,$$ then, with $$b_{n-1}^y=\iota_1^n(y)a_n\ ,$$ one has $$d^{n-1}b_{n-1}^y=0\ .$$

### corollary

If under these conditions $$H^{n-1}(\mathfrak{g},\mathfrak{a})=0\ ,$$ there exists $$c_{n-2}^y$$ such that $$\iota_1^n(y)a_n=d^{n-2} c_{n-2}^y\ .$$

In the case $$n=2$$ this form is known as the Hamiltonian, and $$a_n$$ is the symplectic form.

Since $$d_1(x)c^y=\iota^{1}(x)d c^y=\iota^{1}(x)b_1^y=b_1^y(x)=a_2(y,x)\ ,$$ one sees that $$c_0^y$$ is invariant under $$y$$ if $$a_2$$ is antisymmetric,

which is the usual assumption on symplectic forms.

### moment map

Assume $$[a_2]\in H_\wedge(\mathfrak{g},\mathfrak{a})$$ and the existence of an index set $$I$$ such that $$y_\iota, \iota\in I$$ is the maximal set of linearly independent elements $$y_\iota\in\mathfrak{g}$$ with $$\iota_1^2(y_\iota)a_2=0$$ and $$a_2(y_{\iota_1},y_{\iota_2})=0, \iota_1,\iota_2\in I\ .$$

Let $$\mathfrak{h}=\langle y^\iota\rangle_{\iota\in I}\ .$$

Then the map $$\mathfrak{h}\rightarrow\mathfrak{a}^I$$ is called the moment(um) map.

Notice that $$d_1(y^{\iota_1})c^{y^{\iota_2}}=0,\iota_1,\iota_2\in I,$$ by construction.

### definition (conformally) symplectic

If $$a_2$$ is a symplectic form, an element $$y\in\mathfrak{g}$$ is called conformally symplectic if $$d_1^2(y)a_2=c_1(y)a_2\ ,$$ where $$c_1$$ is a one form with values in a commutative ring.

If $$c_1(y)$$ is invertible, $$y$$ is called a scaling conformally symplectic form

If $$c_1(y)=0\ ,$$ $$y$$ is called symplectic.

The commutator of two conformally symplectic elements is symplectic.

### scaling lemma

Suppose there exists an element $$s\in\mathfrak{g}$$ such that $d_1^{n}(s)a_n=\lambda(a_n)a_n\ ,$ with $$\lambda\in C^1(C^n(\mathfrak{g},\mathfrak{a}),R)$$ and $$R$$ the ring of the module $$\mathfrak{a}\ .$$

Then for $$a_n\in Z^n(\mathfrak{g},\mathfrak{a})$$ one has $\lambda(a_n)a^n=d_1^{n}(s)a_n=d^{n-1}\iota_1^n(s)a_n\ ,$ that is, if $$\lambda(a_n)$$ is invertible, then $$a_n=d^{n-1}\lambda(a_n)^{-1}\iota_1^n(s)a_n\in B^n(\mathfrak{g},\mathfrak{a})\ .$$

In practice, this is very useful in computing cohomology, since it allows one to restrict the attention to those $$a_n\in Z^n(\mathfrak{g},\mathfrak{a})$$ which have a noninvertible $$\lambda(a_n)\ .$$

Notice that the argument does not work for $$s\in\mathfrak{l}\ .$$

### the homotopy formula

If $$R$$ equals $$\R$$ or $$\mathbb{C}$$ there is an explicit formula, the homotopy formula, to compute the preimage, at least on the span of the eigenforms.

Let $$d_1^{n}(s) a_n^\iota=\lambda_\iota a_n^\iota$$ and let $$S$$ be the span of all such $$a_n^\iota\in Z^n(\mathfrak{g},\mathfrak{a})$$ with $$\lambda_\iota\neq 0\ .$$ Then if $$a_n\in S\ ,$$ one defines $$\tau^{s}a_n^\iota=\tau^{\lambda_\iota}a_n^\iota\ .$$

This defines $$\tau^{s}a_n$$ by linearity. Let $P a_n = \left.\int \tau^{s} a_n \frac{d\tau}{\tau}\right|_{\tau=1}\ .$

Then for $$a_n\in S$$ one has $a_n=d^{n-1} \iota_1^n(s)P a_n\ .$

Here the meaning of the integral is $\int \frac{d\tau}{\tau}=\log(\tau)$ and with $$\lambda\neq 0\ ,$$ $\int \tau^\lambda\frac{d\tau}{\tau}=\frac{1}{\lambda}\tau^\lambda.$

### proof

Let $$a_n=\sum_\iota \alpha_\iota a_n^\iota\ .$$ Then $d^{n-1} \iota_1^n(s)P a_n=d^{n-1} \iota_1^n(s)\sum_\iota \alpha_\iota P a_n^\iota\ :$ $=d^{n-1} \iota_1^n(s)\sum_{\iota} \alpha_\iota \frac{1}{\lambda_\iota} a_n^\iota\ :$ $=\sum_{\iota} \alpha_\iota a_\iota^n\ :$ $= a_n$

### corollary

A scaling conformally symplectic form can be used to integrate the symplectic form.

### pseudodifferential symbols - example of a closed 2-form

Let $$a_2(f\delta^n,g\delta^k)=\mathrm{tr}([\log(\delta),f\delta^n]g\delta^k)\ .$$