Wrong-way Frobenius reciprocity for finite groups representations

This is a typical lazy mathematician question, so do not hesitate to close it and recommend me to do my homeworks...

Let $H$ be a subgroup of a finite group $G$, and let $Res_H^G$ and $Ind_H^G$ the restriction and induction functors between the categories of linear representations of $G$ and $H$, respectively. I'm familiar with Frobenius reciprocity stating that $Ind_H^G$ is the right adjoint to $Res_H^G$. Moreover, by choosing the traditional model for the induced representation, namely,

$Ind_H^G(U)= \{f:G\to U \text{ such that } f(h g)=h\cdot f(g)\text{ for any } h\in H\}$

I also know how to write explicitly the natural isomorphism

$Hom_{Rep(H)}(Res_H^G(W),U) \stackrel{\sim}{\to} Hom_{Rep(G)}(W,Ind_H^G(U)).$

Now, I've just learnt that actually $(Res_H^G,Ind_H^G)$ is an ambidextrous adjunction, i.e. that $Ind_H^G$ is also the left adjoint to $Res_H^G$, but I'm not familiar with this fact: how is the isomorphism

$Hom_{Rep(H)}(U,Res_H^G(W)) \stackrel{\sim}{\to} Hom_{Rep(G)}(Ind_H^G(U),W)$ explicitly defined?

The point is that there are two ways to describe the restriction functor. The first way is as $\text{Hom}_{\mathbb{C}[G]}(\mathbb{C}[G], -)$, thinking of $\mathbb{C}[G]$ as a $(\mathbb{C}[G], \mathbb{C}[H])$-bimodule, which by the tensor-hom adjunction means that it has left adjoint $\mathbb{C}[G] \otimes_{\mathbb{C}[H]} (-)$. The second way is as $\mathbb{C}[G] \otimes_{\mathbb{C}[G]} (-)$, thinking of $\mathbb{C}[G]$ as a $(\mathbb{C}[H], \mathbb{C}[G]$)-bimodule, which by a second application of the tensor-hom adjunction means means that it has right adjoint $\text{Hom}_{\mathbb{C}[H]}(\mathbb{C}[G], -)$.

It remains to write down a natural isomorphism between the left and right adjoints. Very explicitly, the obvious candidate is to try something like

$$\mathbb{C}[G] \otimes_{\mathbb{C}[H]} V \ni \sum g \otimes v_g \mapsto (g \mapsto v_g) \in \text{Hom}_{\mathbb{C}[H]}(\mathbb{C}[G], V)$$

where $V$ is a left $\mathbb{C}[H]$-module and $v_g \in V$. But we need to check that this is well-defined. On the LHS we have $gh \otimes v_{gh} = g \otimes h v_{gh} = g \otimes v_g$ for any $h \in H$, hence $h^{-1} v_g = v_{gh}$, whereas on the RHS we need functions such that if $g \mapsto v_g$ then $hg \mapsto hv_g = v_{hg}$ for any $h \in H$, hence $h^{-1} v_g = v_{h^{-1} g}$. These don't quite match up, so our map is slightly wrong. It should be

$$\mathbb{C}[G] \otimes_{\mathbb{C}[H]} V \ni \sum g \otimes v_g \mapsto (g \mapsto v_{g^{-1}}) \in \text{Hom}_{\mathbb{C}[H]}(\mathbb{C}[G], V)$$

and then everything matches up; the inverse map sends a function $(g \mapsto v_g)$ to $\sum g^{-1} \otimes v_g$.

Alternatively, instead of working with representations we can work with unitary representations on Hilbert spaces (in the finite-dimensional case this recovers the same theory). Here for every pair of unitary representations $V, W$ we have a natural identification $\text{Hom}(V, W) \cong \text{Hom}(W, V)$ given by taking adjoints, so categories of unitary representations are dagger categories. Any adjunction between dagger categories is automatically ambidextrous.

In general, if $f : R \to S$ is a homomorphism of rings, we get a restriction functor $S\text{-Mod} \to R\text{-Mod}$ which can be described in either of the ways above and hence which has both a left and a right adjoint, which one might call "induction" and "coinduction" or something like that. These are left and right Kan extension along $f$ (thinking of $R, S$ as $\text{Ab}$-enriched categories with one object).

• Thanks! a really enlightening answer, putting the problem into its correct abstract setting! Yet, to complete a few computations I'm currently fighting with, what I'd need is a completely explicit version of the isomorphism $Hom_{Rep(H)}(U,Res_H^G(W)) \stackrel{\sim}{\to} Hom_{Rep(G)}(Ind_H^G(U),W)$, with the explicit model for the induced representation mentioned above. I think I can work this out from the answer, but I'll wait still a couple of days to see if there's someone knowing this on the spot (and willing to post it here :) ) – domenico fiorenza May 29 '13 at 20:39
• I added an explicit map. Something went wrong when I tried to write this map down abstractly and I'm not sure how to fix it. – Qiaochu Yuan May 29 '13 at 21:10
• Can't one just use that restriction commutes with duality and consider the dual of the induction of the dual? Or more conceptually, for general groups $G$ and $H$ there is induction and there is "compactly supported" induction, the latter contained in the former and equality when $H$ has finite index in $G$. In general these functors have opposite adjointness properties with respect to restriction (much like direct sum versus direct product), so when the two functors agree one sees this common functor having two adjointeness properties relative to restriction. – user28172 May 30 '13 at 2:43
• Qiaochu Yuan, thanks a lot for the explicit map you added to your answer. Now everything is crystal clear, thanks! To be sure I'm correctly interpreting it, you are secretely identifying $\mathbb{C}[G]\otimes_{\mathbb{C}[H]}V$ with the subspace of $\mathbb{C}[G]\otimes_{\mathbb{C}}V$ consisting of those $\sum_{g\in G} g\otimes v_g$ such that $v_{gh}=h^{-1}v_g$ for any $h\in H$, via the natural projection $\mathbb{C}[G]\otimes_{\mathbb{C}}V\to \mathbb{C}[G]\otimes_{\mathbb{C}[H]}V$, whose inverse is $g\otimes_{\mathbb{C}[H]}v\mapsto \frac{1}{|H|}\sum_{h\in H}gh^{-1}\otimes hv$, right? – domenico fiorenza May 30 '13 at 7:20
• Incidentally, I agree that there is one "most natural" isomorphism between the induction and coinduction functors, but it's of course not the only one: you may multiply by any element of $\mathbb G_m = \mathbb C^\times$. In general, the space of left adjoints to some functor has the homotopy type of a truth value (it is either empty or contractible), and as is the space of right adjoints, but the space of biadjoints is in general the intersection of two contractible sets, and so can have arbitrary homotopy type. (For functors between $1$-categories, it is never worse than a set.) ... – Theo Johnson-Freyd May 30 '13 at 17:47