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

From Scholarpedia
Jump to: navigation, search

    Back to the second lecture

    On to the fourth lecture



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

    Lifting the representation to the forms


    \(\mathfrak{g}\) is called an ideal in \(\mathfrak{l}\) if \[[\mathfrak{l},\mathfrak{g}]\subset \mathfrak{g}\ .\]


    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 \(\mathfrak{l}\ .\)


    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: \[ [[x],[y]]=[[x+g_1,y+g_2]]=[[x,y]]+[[x,g_2]]+[[g_1,y]]+[[g_1,g_2]]=[[x,y]]\]


    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.


    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 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.

    • \(d a_0(x)=d_1(x)a_0\ .\)
    • \(d^1 a_1(x,y)=d_1(x)a_1(y)-d_1(y)a_1(x)-a_1([x,y])=(d_1^1(x)a_1)(y)-d_1(y)\iota_1^1(x)a_1=(d_1^1(x)a_1)(y)-d \iota_1^1(x)a_1 (y)\ .\)
    • \(d^2 a_2(x,y,z)=d_1(x)a_2(y,z)-d_1(y)a_2(x,z)+d_1(z)a_2(x,y)-a_2([x,y],z)-a_2(y,[x,z])+a_2(x,[y,z])\ :\)
    \[=(d_1^2(x)a_2)(y,z)-d_1^1(y)\iota_1^2(x)a_2(z)+d_1(z)\iota_1^1(y)\iota_1^2(x)a_2\ :\]
    \[=(d_1^2(x)a_2)(y,z)-d^1 \iota_1^2(x)a_2 (y,z)\ .\]


    This strongly suggests the following recursive definition:

    • \(\iota_1^1(x)d=d_1(x)\)
    • \(\iota_1^{n+1}(x)d^n+d^{n-1}\iota_1^n(x)=d_1^n(x),\quad n>0\)


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

    • \(\iota_1^n(z)d_1^n(y)-d_1^{n-1}(y)\iota_1^n(z)=-\iota_1^{n}([y,z]).\)


    Consider \[ (\iota_1^n(z)d_1^{n}(y)-d_1^{n-1}(y)\iota_1^n(z))a_n(x_1,\cdots,x_{n-1})\] \[= d_1^{n}(y)a_n(z,x_1,\cdots,x_{n-1})-d_1^{n-1}(y)\iota_1^n(z)a_n(x_1,\cdots,x_{n-1})\] \[=d_1(y)a_n(z,x_1,\cdots,x_{n-1})-a_n([y,z],x_1,\cdots,x_{n-1})-\sum_{i=1}^{n-1}a_n(z,x_1,\cdots,[y,x_i],\cdots,x_n)\ :\] \[-d_1(y)a_n(z,x_1,\cdots,x_n)+\sum_{i=1}^{n-1}a_n(z,x_1,\cdots,[y,x_i],\cdots,x_{n-1})\] \[=-a_n([y,z],x_1,\cdots,x_{n-1})\] \[=-\iota_1^{n}([y,z])a_n(x_1,\cdots,x_{n-1})\quad\square\ .\]


    Let \( y\in\mathfrak{l}\ .\) Then \[d_1^{n+1}(y)d^{n}=d^{n}d_1^{n}(y),\quad n\geq 0\ .\]


    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.


    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


    This shows that \(d^i, i\in\mathbb{N}\) is a coboundary operator.


    \[ d^n \omega_n(x_1,\cdots,x_{n+1})=\sum_{i=1}^{n+1} (-1)^{i-1} d_1(x_i) \omega_n(x_1,\cdots,\hat{x}_i,\cdots,x_{n+1}) +\sum_{i<j} (-1)^i\omega_n(x_1,\cdots,\hat{x}_i,\cdots,[x_i,x_j],\cdots,x_{n+1})\]


    \(d^n\) maps \(C_{\wedge}^n(\mathfrak{g},\mathfrak{a})\) to \(C_{\wedge}^{n+1}(\mathfrak{g},\mathfrak{a})\ .\)


    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.


    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.


    \( H^n(\mathfrak{g},\mathfrak{a}), n>0 ,\) 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.


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


    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 \( d^{n-1}b_{n-1}^y=0\ .\)


    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 \(\lambda\neq 0\ ,\) \[ \int \tau^\lambda\frac{d\tau}{\tau}=\frac{1}{\lambda}\tau^\lambda.\]


    Let \(a_n=\sum_\iota \alpha_\iota a_n^\iota\ .\) Then \[ d^{n-1} \iota_1^n(s)P a_n=d^{n-1} \iota_1^n(s)\sum_\iota \alpha_\iota P a_n^\iota\ :\] \[ =d^{n-1} \iota_1^n(s)\sum_{\iota} \alpha_\iota \frac{1}{\lambda_\iota} a_n^\iota\ :\] \[ =\sum_{\iota} \alpha_\iota a_\iota^n\ :\] \[ = a_n\]


    A scaling conformally symplectic form can be used to integrate the symplectic form.

    pseudodifferential symbols - example of a closed 2-form

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

    On to the fourth lecture

    Back to the second lecture

    Personal tools