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

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

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

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