# User:Jan A. Sanders/An Introduction to Leibniz 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 a (two-sided) ideal in the $$\mathfrak{l}$$$$\mathfrak{l}$$ if $[\mathfrak{l},\mathfrak{g}]\subset \mathfrak{g},\quad [\mathfrak{g},\mathfrak{l}]\subset \mathfrak{g}\ .$

### example

Let $$\mathfrak{g}=\ker d_+^{(0)}\ .$$ Then for $$x\in\mathfrak{l}$$ and $$y\in\mathfrak{g}$$ one has $d_+^{(0)}([x,y])=d_+^{(0)}(x)d_+^{(0)}(y)-d_+^{(0)}(y)d_+^{(0)}(x)=0$ $d_+^{(0)}([y,x])=d_+^{(0)}(y)d_+^{(0)}(x)-d_+^{(0)}(x)d_+^{(0)}(y)=0$ It follows that $$\mathfrak{g}=\ker d_+^{(0)}$$ is a two-sided ideal in $$\mathfrak{l}\ .$$ Observe that this proof does not hold in the case of $$\ker d_-^{(0)}\ .$$

### definition

Let $$\mathfrak{l}$$ be a $$\mathfrak{l}$$ and $$\mathfrak{g}$$ a (two-sided) ideal. Then $$\mathfrak{l}/\mathfrak{g}$$ is a Leibniz 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 a $$\mathfrak{l}$$-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 left representation on $$C^n (\mathfrak{g},\mathfrak{a})$$ as follows. Let, for $$y\in\mathfrak{l}\ ,$$ $(d^{(n)}(y)a^n)(x_1,\cdots,x_n)=d_+^{(0)}(y)a^n(x_1,\cdots,x_n)-\sum_{i=1}^n a^n(x_1,\cdots, [y,x_i],\cdots,x_n).$ This is indeed a representation. Let $$y,z\in\mathfrak{l}\ .$$ Then $d^{(n)}(y)d^{(n)}(z)a^n(x_1,\cdots,x_n)=$ $=d_+^{(0)}(y)d^{(n)}(z)a^n(x_1,\cdots,x_n)-\sum_{i=1}^n d^{(n)}(z) a^n(x_1,\cdots, [y,x_i],\cdots,x_n)$ $=d_+^{(0)}(y)d_+^{(0)}(z)a^n(x_1,\cdots,x_n)-\sum_{i=1}^n d_+^{(0)}(z) a^n(x_1,\cdots, [y,x_i],\cdots,x_n)\ :$ $-\sum_{i=1}^n d_+^{(0)}(y) a^n(x_1,\cdots, [z,x_i],\cdots,x_n)+\sum_{ji}a^n(x_1,\cdots,[y,x_i],\cdots, [z,x_j],\cdots,x_n)+a^n(x_1,\cdots, [z,[y,x_i]],\cdots,x_n)$ It follows that $d^{(n)}(y)d^{(n)}(z)a^n(x_1,\cdots,x_n)-d^{(n)}(z)d^{(n)}(y)a^n(x_1,\cdots,x_n)=$ $=(d_+^{(0)}(y)d_+^{(0)}(z)-d_+^{(0)}(z)d_+^{(0)}(y))a^n(x_1,\cdots,x_n)+ \sum_{i=1}^n a^n(x_1,\cdots, [z,[y,x_i]],\cdots,x_n)- \sum_{i=1}^n a^n(x_1,\cdots, [y,[z,x_i]],\cdots,x_n)$ $=d_+^{(0)}([y,z])a^n(x_1,\cdots,x_n)-\sum_{i=1}^n a^n(x_1,\cdots, [[y,z],x_i]],\cdots,x_n)$ $=d^{(n)}([y,z])a^n(x_1,\cdots,x_n)$ or, $d^{(n)}([y,z])=d^{(n)}(y)d^{(n)}(z)-d^{(n)}(z)d^{(n)}(y)\ .$

### remark

Remark that $$C^n(\mathfrak{g},\mathfrak{a})$$ is invariant under $$d_\pm^{(n)}(x)$$ 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 $$d^{(n)}\ .$$ First we introduce the contraction operator $$\iota^n(y): C^n(\mathfrak{g},\mathfrak{a})\rightarrow C^{n-1}(\mathfrak{g},\mathfrak{a})$$ by $( \iota^n(y)a^n)(x_1,\cdots,x_{n-1})=a^n(y,x_1,\cdots,x_{n-1})\ .$

Recall the following definitions of the coboundary operators.

• $$d^0 a^0(x)=d_-^{(0)}(x)a^0\ .$$
• $$d^1 a^1(x,y)=d_+^{(0)}(x)a^1(y)-d_-^{(0)}(y)a^1(x)-a^1([x,y])=(d^{(1)}(x)a^1)(y)-d_-^{(0)}(y)\iota^1(x)a^1=(d^{(1)}(x)a^1)(y)-d^0 \iota^1(x)a^1 (y)\ .$$
• $$d^2 a^2(x,y,z)=d_+^{(0)}(x)a^2(y,z)-d_+^{(0)}(y)a^2(x,z)+d_-^{(0)}(z)a^2(x,y)-a^2([x,y],z)-a^2(y,[x,z])+a^2(x,[y,z])\ :$$
$=(d^{(2)}(x)a^2)(y,z)-d^{(1)}(y)\iota^2(x)a^2(z)+d_-^{(0)}(z)\iota^1(y)\iota^2(x)a^2\ :$
$=(d^{(2)}(x)a^2)(y,z)-d^1 \iota^2(x)a^2 (y,z)\ .$

### definition

This strongly suggests the following recursive definition:

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

### lemma

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

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

### proof

Consider $(\iota^n(z)d^{(n)}(y)-d^{(n-1)}(y)\iota^n(z))a^n(x_1,\cdots,x_{n-1})$ $= d^{(n)}(y)a^n(z,x_1,\cdots,x_{n-1})-d^{(n-1)}(y)\iota^n(z)a^n(x_1,\cdots,x_{n-1})$ $=d_+^{(0)}(y)a^n(z,x_1,\cdots,x_{n-1})-a^n([y,z],x_1,\cdots,x_{n-1})-\sum_{i=1}^{n-1}a^n(z,x_1,\cdots,[y,x_i],\cdots,x_n)\ :$ $-d_+^{(0)}(y)a^n(z,x_1,\cdots,x_n)+\sum_{i=1}^{n-1}a^n(z,x_1,\cdots,[y,x_i],\cdots,x_{n-1})$ $=-a^n([y,z],x_1,\cdots,x_{n-1})$ $=-\iota^{n}([y,z])a^n(x_1,\cdots,x_{n-1})\quad\square\ .$

### lemma

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

### proof

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

The final check is $\iota^{n+2}(y)d^{n+1}d^{n}=d^{(n+1)}(y)d^{n}-d^{n}\iota^{n+1}(y)d^{n}$ $=d^{n}d^{(n)}(y)-d^{n}(d^{(n)}(y)-d^{n-1}\iota^{n}(y))$ $=d^{n}d^{n-1}\iota^{n}(y).$ 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

$d^n \omega^n(x_1,\cdots,x_{n+1})=\sum_{i=1}^n (-1)^{i-1} d_+^{(0)}(x_i) \omega^n(x_1,\cdots,\hat{x}_i,\cdots,x_{n+1}) +(-1)^n d_-^{(0)}(x_{n+1})\omega^n(x_1,\cdots,x_n) +\sum_{i<j} (-1)^i\omega^n(x_1,\cdots,\hat{x}_i,\cdots,[x_i,x_j],\cdots,x_{n+1})$

## 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 $$d^{(n)}$$) 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 $$d^n a^n=0\ ,$$ $d^{(n)}(y)[a^n]=[d^{(n)}(y)a^n]=[\iota^{n+1}(y)d^{n}a^n+d^{n-1}\iota^{n}(y)a^n]=.\quad\square$

### scaling lemma

Suppose there exists an element $$s\in\mathfrak{g}$$ such that $d^{(n)}(s)a^n=\lambda(a^n)a^n\ ,$ 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 $\lambda(a^n)a^n=d^{(n)}(s)a^n=d^{n-1}\iota^n(s)a^n\ ,$ that is, if $$\lambda(a^n)$$ is invertible, then $$a^n=d^{n-1}\lambda(a^n)^{-1}\iota^n(s)a^n\in B^n(\mathfrak{g},\mathfrak{a})\ .$$

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 $$d^{(n)}(s_0) a_\iota^n=\lambda_\iota a_\iota^n$$ and let $$S$$ be the span of all such $$a_\iota^n\in Z^n(\mathfrak{g},\mathfrak{a})$$ with $$\lambda_\iota\neq 0\ .$$ Then if $$a^n\in S\ ,$$ one defines $$\tau^{s_0}a_\iota^n=\tau^{\lambda_\iota}a_\iota^n\ .$$ This defines $$\tau^{s_0}a^n$$ by linearity. Let $P a^n = \left.\int \tau^{s_0} a^n \frac{d\tau}{\tau}\right|_{\tau=1}\ .$

Then for $$a^n\in S$$ one has $a^n=d^{n-1} \iota(s_0)P a^n\ .$ 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 $$a^n=\sum_\iota \alpha_\iota a_\iota^n\ .$$ Then $d^{n-1} \iota(s_0)P a^n=d^{n-1} \iota(s_0)\sum_\iota \alpha_\iota P a_\iota^n\ :$ $=d^{n-1} \iota(s_0)\sum_{\iota} \alpha_\iota \frac{1}{\lambda_\iota} a_\iota^n\ :$ $=\sum_{\iota} \alpha_\iota a_\iota^n\ :$ $= a^n$