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