An introduction to Lie algebra cohomology/klad

klad

Let \(r_p\) be the map of \(F^p C^\cdot(\mathfrak{g},\mathfrak{a})\) to \( C^\cdot(\mathfrak{h},C^p(\mathfrak{g}/\mathfrak{h},\mathfrak{a}))\) given by restricting all but the last \(p\) arguments to \(\mathfrak{h}\ .\) Then \[ [r_p]: E_p^{0,n} \rightarrow C^{n-p}(\mathfrak{h},C^p(\mathfrak{g}/\mathfrak{h},\mathfrak{a})) \] is an isomorphism. Remember that \(E_p^{0,n}=F^pC^n(\mathfrak{g},\mathfrak{a})/F^{p+1}C^n(\mathfrak{g},\mathfrak{a}) \ ,\) that is, the space of forms that are zero when \(n-p+1\) arguments are in \(\mathfrak{h}\ ,\) but dividing out those that are zero with \(n-p\) arguments in \(\mathfrak{h}\ .\)

example

Let \(\alpha^4\in F^2C^4(\mathfrak{g},\mathfrak{a})\ .\) Then \[ [r_2][\alpha^4](h_1,h_2)([g_1],[g_2])=\alpha^4(h_1,h_2,g_1,g_2)\] This is well defined. If one takes another representative of \([\alpha^4]\ ,\) say \(\alpha^4+\beta^4, \beta^4\in F^3C^4(\mathfrak{g},\mathfrak{a})\ ,\) then the contribution of \(\beta^4\) would be zero, since it is zero if two of its arguments are in \(\mathfrak{h}\ .\) It is also independent of the choice of \(g_1, g_2\) since \(\alpha^4\) vanishes with three of its arguments in \(\mathfrak{h}\ .\)

On the other hand, giving a form \(\beta^2\in C^{2}(\mathfrak{h},C^2(\mathfrak{g}/\mathfrak{h},\mathfrak{a}) \) one defines \[\rho_2\beta^2(h_1,h_2,g_1,g_2)=\beta^2(h_1,h_2)([g_1],[g_2])\] Here one has to check that \(\rho_2\beta^2\in F^2C^4(\mathfrak{g},\mathfrak{a})\ ,\) that is, it has to vanish when three of its arguments are in \(\mathfrak{h}\ .\) But this would imply that either \( g_1 \) or \(g_2\) is an element of \(\mathfrak{h}\ ,\) say \( g_1 \ ,\) in which case \([g_1]=0\) and by linearity the form vanishes.

One checks that \([r_2][\rho_2\beta^2]=\beta^2\) and \( \rho_2[r_2][\alpha^4]=\alpha^4\ .\)

proof

By definition, \(E_p^{0}=F^pC^\cdot(\mathfrak{g},\mathfrak{a})/F^{p+1}C^\cdot(\mathfrak{g},\mathfrak{a}) \) and \( E_p^{0,n}\) is the natural restriction to \(n\)-forms. The kernel of \(r_p\) is \(F^{p+1}C^\cdot(\mathfrak{g},\mathfrak{a}) \ ,\) which shows that \([r_p]\) is injective. Given a section \(\sigma:\mathfrak{g}/\mathfrak{h}\rightarrow \mathfrak{g}\ ,\) one defines \(\pi(x)=x-\sigma([x])\) with \([\pi(x)]=0\ .\) Let now for a given \(\beta^{n-p}\in C^{n-p}(\mathfrak{h},C^p(\mathfrak{g}/\mathfrak{h},\mathfrak{a})) \) \[ \rho_p\beta^{n-p}(g_1,\cdots,g_n)= \beta^{n-p}(\pi(g_1),\cdots,\pi(g_{n-p}))([g_{n-p+1}],\cdots,[g_n])\] (In the antisymmetric case, one has to shuffle the two groups of variables in order to get a completely antisymmetric form).

theorem

\[ [r_p] \delta_0^1=d_{\mathfrak{h}}^{n-p} [r_p] \]

proof

In the following one uses the fact that \(a^n\in F^p C^n(\mathfrak{g},\mathfrak{a})\) vanishes when it has \(n-p+1\) of its arguments in \(\mathfrak{h}\ .\) \[[r_p]\delta_0^1 a^n(h_1,\cdots,h_{n-p+1})([g_{1}],\cdots,[g_p])= d^n a^n(h_1,\cdots,h_{n-p+1},g_1,\cdots,g_p)\ :\] \[=\sum_{i=1}^{n-p+1} (-1)^{i+1} d^{(0)}(h_i)a^n(h_1,\cdots,\hat{h}_i,\cdots,h_{n-p+1},g_1,\cdots,g_p) +\sum_{i=1}^p (-1)^{n-p+i}d^{(0)}(g_i)a^n(h_1,\cdots,h_{n-p+1},g_1,\cdots,\hat{g}_i,\cdots,g_p)\ :\] \[-\sum_{i=1}^{n-p+1} \sum_{j=1}^{n-p+1}(-1)^{i+1} a^n(h_1,\cdots,\hat{h}_i,\cdots,[h_i,h_j],\cdots,h_{n-p+1},g_1,\cdots,g_p)\ :\] \[-\sum_{i=1}^{n-p+1}\sum_{j=1}^{p} (-1)^{i+1} a^n(h_1,\cdots,\hat{h}_i,\cdots,h_{n-p+1},g_1,\cdots,[h_i,g_j],\cdots,g_p)\ :\] \[-\sum_{i=1}^{p}\sum_{j=1}^{p} (-1)^{n-p+i} a^n(h_1,\cdots,h_{n-p+1},g_1,\cdots,\hat{g}_i,\cdots,[g_i,g_j],\cdots,g_p)\ :\] \[=\sum_{i=1}^{n-p+1} (-1)^{i+1} d^{(0)}(h_i)a^n(h_1,\cdots,\hat{h}_i,\cdots,h_{n-p+1},g_1,\cdots,g_p)\ :\] \[-\sum_{i=1}^{n-p+1} \sum_{j=1}^{n-p+1}(-1)^{i+1} a^n(h_1,\cdots,\hat{h}_i,\cdots,[h_i,h_j],\cdots,h_{n-p+1},g_1,\cdots,g_p)\ :\] \[=d_{\mathfrak{h}}^{n-p} [r_p]a^n (h_1,\cdots,h_{n-p+1})(g_1,\cdots,g_p)\]

corollary

\[ E_p^{1,n}=H^{n-p}(\mathfrak{h},C^p(\mathfrak{g}/\mathfrak{h},\mathfrak{a}))\]

corollary

\[E_n^{1,n}=H^{0}(\mathfrak{h},C^n(\mathfrak{g}/\mathfrak{h},\mathfrak{a}))= C^n(\mathfrak{g}/\mathfrak{h},\mathfrak{a})^\mathfrak{h}\]

theorem

\[ E_p^{2,n}=H^{p}(\mathfrak{g}/\mathfrak{h},H^{n-p}(\mathfrak{h},\mathfrak{a}))\]

proof

One identifies in a natural way \( C^{n-p}(\mathfrak{h},C^p(\mathfrak{g}/\mathfrak{h},\mathfrak{a}))\) and \( C^{p}(\mathfrak{g}/\mathfrak{h}, C^{n-p}(\mathfrak{h},\mathfrak{a}))\ .\) This induces an identification of \( H^{n-p}(\mathfrak{h},C^p(\mathfrak{g}/\mathfrak{h},\mathfrak{a}))\) and \( C^{p}(\mathfrak{g}/\mathfrak{h}, H^{n-p}(\mathfrak{h},\mathfrak{a}))\ .\) Since \[E_p^{2,n}=H^p(E_\cdot^{1,n},\delta_1^1)= H^{p}(\mathfrak{g}/\mathfrak{h},H^{n-p}(\mathfrak{h},\mathfrak{a}))\] the result follows.

theorem

Let \(\mathfrak{h}\) be an ideal of \(\mathfrak{g}\ .\) Then every element of \(H^1(\mathfrak{h},\mathfrak{a})^\mathfrak{g}\) has a representative cocycle which is the restriction to \(\mathfrak{h}\) of an element \( f^1\in C^1(\mathfrak{g},\mathfrak{a})\) such that \(d^1 f^1\in F^2C^2(\mathfrak{g},\mathfrak{a})\) and thus determines an element of \(H^2(\mathfrak{g}/\mathfrak{h},\mathfrak{a}^\mathfrak{h})\ .\) This element depends only on the given element of \(H^1(\mathfrak{g},\mathfrak{a})^\mathfrak{h}\ .\) If \(\tau_2\) is the resulting homomorphism of \(H^1(\mathfrak{g},\mathfrak{a})^\mathfrak{h}\) to \(H^2(\mathfrak{g}/\mathfrak{h},\mathfrak{a}^\mathfrak{h})\ ,\) the following sequence is exact: \[0\rightarrow H^1(\mathfrak{g}/\mathfrak{h},\mathfrak{a}^\mathfrak{h}) \rightarrow H^1(\mathfrak{g},\mathfrak{a}) \rightarrow H^1(\mathfrak{h},\mathfrak{a})^\mathfrak{g} \rightarrow H^2(\mathfrak{g}/\mathfrak{h},\mathfrak{a}^\mathfrak{h}) \rightarrow H^2(\mathfrak{g},\mathfrak{a})\]

proof

This can be read as \[0\rightarrow H^1(\mathfrak{g}/\mathfrak{h},H^0(\mathfrak{h},\mathfrak{a})) \rightarrow H^1(\mathfrak{g},\mathfrak{a}) \rightarrow H^0(\mathfrak{g},H^1(\mathfrak{h},\mathfrak{a})) \rightarrow H^2(\mathfrak{g}/\mathfrak{h},H^0(\mathfrak{h},\mathfrak{a})) \rightarrow H^2(\mathfrak{g},\mathfrak{a})\] or \[0\rightarrow E_1^{2,1} \rightarrow H^1(\mathfrak{g},\mathfrak{a}) \rightarrow H^0(\mathfrak{g},H^1(\mathfrak{h},\mathfrak{a})) \rightarrow E_2^{2,2} \rightarrow H^2(\mathfrak{g},\mathfrak{a})\] Let \([f^1]\in H^1(\mathfrak{h},\mathfrak{a})^\mathfrak{g}\ .\) Then \[0=d^1f^1(h_1,h_2)= d^{(1)}(h_1)f^1(h_2) -d^{(0)}(h_2)f^1(h_1)\ :\] \[= -d^{(0)}(h_2)f^1(h_1)\]

This implies \( f^1\in C^1(\mathfrak{h},\mathfrak{a}^\mathfrak{h})\ .\) Let \(\tilde{f}^1(x)=f^1(\pi(x))\ .\) Then \[d^1\tilde{f}^1(x,y)= d^{(1)}(x)f^1(\pi(y)) -d^{(0)}(y)f^1(\pi(x))\ :\] \[=-d^{(0)}(y)f^1(\pi(x))\] Then \[d^1\tilde{f}^1(x,h)=0\] and, by antisymmetry, \[d^1\tilde{f}^1\in F^2 C_\wedge^2(\mathfrak{g},\mathfrak{a}) =C_\wedge^2(\mathfrak{g}/\mathfrak{h},\mathfrak{a})\] Furthermore, \[ d^{(0)}(h)d^1\tilde{f}^1(x,y)=d^{(0)}(h)d^{(0)}(x)f^1(\pi(y)) -d^{(0)}(h)d^{(0)}(y)f^1(\pi(x))-d^{(0)}(h)f^1(\pi([x,y]))\ :\] \[=d^{(0)}(x)d^{(0)}(h)f^1(\pi(y))+d^{(0)}([h,x])f^1(\pi(y)) -d^{(0)}(y)d^{(0)}(h)f^1(\pi(x))-d^{(0)}([h,y])f^1(\pi(x))\ :\] \[=0\] It follows that \[d^1\tilde{f}^1\in C_\wedge^2(\mathfrak{g}/\mathfrak{h},\mathfrak{a}^\mathfrak{h})\] This implies, since \(d^2 d^1 \tilde{f}^1=0\) that \[d^1\tilde{f}^1\in H_\wedge^2(\mathfrak{g}/\mathfrak{h},\mathfrak{a}^\mathfrak{h})\]

So one defines \[\tau_2([f^1])=[d^1\tilde{f}^1]\]