lecture 3

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< An introduction to Leibniz algebra cohomology

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Author: Dr. Jan A. Sanders, Vrije Universiteit Amsterdam

Back to the second lecture

On to the fourth lecture

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abstract

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

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Lifting the representation to the forms

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definition

$\mathfrak{g}$ is called a (two-sided) ideal in the $\mathfrak{l}$ $\mathfrak{l}$ if

$[\mathfrak{l},\mathfrak{g}]\subset \mathfrak{g},\quad [\mathfrak{g},\mathfrak{l}]\subset \mathfrak{g}$ .

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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 $\ker d_-^{(0)}$ .

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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:

$[[x],[y]]=[[x+g_1,y+g_2]]=[[x,y]]+[[x,g_2]]+[[g_1,y]]+[[g_1,g_2]]=[[x,y]]$

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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.

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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

$d^{(n)}(y)d^{(n)}(z)a^n(x_1,\cdots,x_n)=$

$=d_+^{(0)}(y)d^{(n)}(z)a^n(x_1,\cdots,x_n)-\sum_{i=1}^n d^{(n)}(z) a^n(x_1,\cdots, [y,x_i],\cdots,x_n)$

$=d_+^{(0)}(y)d_+^{(0)}(z)a^n(x_1,\cdots,x_n)-\sum_{i=1}^n d_+^{(0)}(z) a^n(x_1,\cdots, [y,x_i],\cdots,x_n)$

$-\sum_{i=1}^n d_+^{(0)}(y) a^n(x_1,\cdots, [z,x_i],\cdots,x_n)+\sum_{j<i}a^n(x_1,\cdots,[z,x_j],\cdots, [y,x_i],\cdots,x_n)$

$+\sum_{j>i}a^n(x_1,\cdots,[y,x_i],\cdots, [z,x_j],\cdots,x_n)+a^n(x_1,\cdots, [z,[y,x_i]],\cdots,x_n)$

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,

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

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remark

Remark that $C^n(\mathfrak{g},\mathfrak{a})$ is invariant under $d_\pm^{(n)}(x)$ for all $x\in\mathfrak{l}$ .

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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.

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

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

$=(d^{(2)}(x)a^2)(y,z)-d^1 \iota^2(x)a^2 (y,z)$ .

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definition

This strongly suggests the following recursive definition:

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

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lemma

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

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

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proof

Consider

$(\iota^n(z)d^{(n)}(y)-d^{(n-1)}(y)\iota^n(z))a^n(x_1,\cdots,x_{n-1})$

$= d^{(n)}(y)a^n(z,x_1,\cdots,x_{n-1})-d^{(n-1)}(y)\iota^n(z)a^n(x_1,\cdots,x_{n-1})$

$=d_+^{(0)}(y)a^n(z,x_1,\cdots,x_{n-1})-a^n([y,z],x_1,\cdots,x_{n-1})-\sum_{i=1}^{n-1}a^n(z,x_1,\cdots,[y,x_i],\cdots,x_n)$

$-d_+^{(0)}(y)a^n(z,x_1,\cdots,x_n)+\sum_{i=1}^{n-1}a^n(z,x_1,\cdots,[y,x_i],\cdots,x_{n-1})$

$=-a^n([y,z],x_1,\cdots,x_{n-1})$

$=-\iota^{n}([y,z])a^n(x_1,\cdots,x_{n-1})\quad\square$ .

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lemma

Let $y\in\mathfrak{l}$ . Then

$d^{(n+1)}(y)d^{n}=d^{n}d^{(n)}(y),\quad n\geq 0$ .

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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.

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proposition

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

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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.

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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.

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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.

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proof

Indeed, since $d^n a^n=0$ ,

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

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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 $s\in\mathfrak{l}$ .

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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