Given a lie algebra $\mathfrak{g}$ over a field $k$, we can define the cohomology groups of $\mathfrak{g}$ as follows:


where $U(\mathfrak{g})$ is the universal enveloping algebra of $\mathfrak{g}$, and $k$ is the trivial $U(\mathfrak{g})$-module. There is a cup product on $H^*(\mathfrak{g})=\oplus H^n(\mathfrak{g},k)$ which gives it the structure of a graded commutative ring. By functoriality, a map of lie algebras $\mathfrak{h}\hookrightarrow\mathfrak{g}$ induces a ring map on cohomology $H^*(\mathfrak{g})\to H^*(\mathfrak{h})$, which we call the restriction map.

For a group $G$, we may replace $U(\mathfrak{g})$ with $kG$, the group algebra, to obtain group cohomology, and again, a map of groups $H\hookrightarrow G$ induces a ring map on cohomology $H^*(G)\to H^*(H)$. However, in the case that $(G:H)<\infty$, we also obtain a corestriction map $H^*(H)\to H^*(G)$. Corestriction is the composition

$$H^n(H,k)\to H^n(G,kG\otimes_{kH}k)\to H^n(G,k)$$

where the first map is from Shapiro's lemma (this is where we use the finite index condition, so that the induced and coinduced modules are isomorphic), and the second map is induced by the $kG$-module map $g\otimes a\mapsto ga=a$.

I've heard that there is no corestriction map in the lie algebra setting. Is this true? If so, what is the obstruction in trying to define such a map?


1 Answer 1


In short the difficulty is that although Eckmann-Shapiro tells you $$\operatorname{Ext}^n_{U(\mathfrak{h})}(k,k) \cong \operatorname{Ext}^n_{U(\mathfrak{g})}(k, k\Uparrow) $$ where $k\Uparrow$ is the coinduced module, unlike in the group setting there is no canonical map of $\mathfrak{g}$-modules $k\Uparrow \to k$, and indeed in some cases there isn't any nonzero map at all.

You could ask "is induction the same as coinduction for Lie algebras?", since if that had a positive answer then we could just copy the groups construction to get a corestriction mapping. But the answer is no: indeed, suppose $\mathfrak{g}$ is finite-dimensional and $M$ is a nonzero finite-dimensional $\mathfrak{h}$-module. Then $U(\mathfrak{g})$ is an countably infinitely-generated free $U(\mathfrak{h})$-module, so the coinduced module $\hom_{U(\mathfrak{h})}(U(\mathfrak{g}),M)$ is isomorphic as a vector space to $$\hom_{U(\mathfrak{h})}( \bigoplus_{n=1}^\infty U(\mathfrak{h}), M) \cong \prod _{n=1}^\infty M $$ whereas the induced module $U(\mathfrak{g})\otimes_{U(\mathfrak{h})} M$ is $\bigoplus_{n=1}^\infty M$ as a vector space. The second has countable dimension, the first uncountable.

Of course, you don't need coinduction and induction to coincide to make corestriction work. All you need is a "canonical" map of $\mathfrak{g}$-modules from the coinduced module $\hom_{U(\mathfrak{h})}(U(\mathfrak{g}), k)$ to $k$.

Even in "nice" situations such a map may not exist. Take $\mathfrak{g}=\mathfrak{sl}(2,\mathbb{C})=\langle e,f,h\rangle$ and $\mathfrak{h}=\mathfrak{b}=\langle e,h\rangle$ a Borel subalgebra. The coinduced trivial module is $$ \hom_{\mathfrak{b}}( U(\mathfrak{g}), \mathbb{C}) $$ Now as a left $\mathfrak{b}$-module, $U(\mathfrak{g}) \cong \bigoplus_{i\geq 0} U(\mathfrak{b})f^i$, so a $\mathfrak{b}$-hom from this to $\mathbb{C}$ can be thought of as an infinite sequence of complex numbers whose $i$th element records the image of $f^i$. The action of $f$ on these sequences is by left-shift because $f \cdot \alpha( f^i) = \alpha(f^i f) = \alpha(f^{i+1})$, and left shift is surjective. Thus there are no nonzero homs to the trivial module.


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