User:Jan A. Sanders/An Introduction to Leibniz 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 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 $d^n a^n=0\ ,$ $$d^{(n)}(y)[a^n]=[d^{(n)}(y)a^n]=[\iota^{n+1}(y)d^{n}a^n+d^{n-1}\iota^{n}(y)a^n]=[0].\quad\square$$

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 $$d^{n-1} \iota(s_0)P a^n=d^{n-1} \iota(s_0)\sum_\iota \alpha_\iota P a_\iota^n\ :$$ $$=d^{n-1} \iota(s_0)\sum_{\iota} \alpha_\iota \frac{1}{\lambda_\iota} a_\iota^n\ :$$ $$=\sum_{\iota} \alpha_\iota a_\iota^n\ :$$ $$= a^n$$

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