5
$\begingroup$

I'm having trouble understanding on which side the induced representation functor is adjoint to the restriction functor.

For simplicity I'm assuming $H$ is a subgroup of $G$ (we could also see it as a morphism $H\to G$). Let me denote $\mathbf{Rep}_H$ the category of representations of $H$ over a fixed field $K$ (similarly for $G$), let $\mathbf{Res}_H^G: \mathbf{Rep}_G \to \mathbf{Rep}_H$ be the restriction functor.

One usual description of the induced representation is the following: given $(\rho,V)$ a representation of $H$, let $\mathbf{Ind}_H^G\rho = \{f: G\to V \mid \forall h\in H, \forall x\in G, f(hx)=h\cdot f(x)\}$ with $g\cdot f(x) =f(xg)$.

With this definition it's easy to prove $Hom_G(\pi, \mathbf{Ind}_H^G\rho)\simeq Hom_H(\mathbf{Res}_H^G\pi, \rho)$, which makes $\mathbf{Ind}_H^G$ right-adjoint to $\mathbf{Res}_H^G$. If I'm not mistaken, the explicit arrows are $\lambda\mapsto (v\mapsto \lambda(v)(1))$ and in the reverse direction $f\mapsto (v\mapsto (g\mapsto f(g\cdot v)))$.

First of all this is a bit surprising as $\mathbf{Res}_H^G$ is more of a forgetful-type functor so we'd expect the "obvious functor in the other direction" to be its left adjoint, but as the proof goes through so simply we can let this surprise on the side.

However, Wiki states the following : "In the case of finite groups, they are actually both left- and right-adjoint to one another" (from the article on Frobenius Reciprocity)

Moreover in the same article they state that there is a natural isomorphism $Hom_{K[G]}(K[G]\otimes_{K[H]}V, W) \simeq Hom_{K[H]}(V,W)$ where a representation is simply seen as a $K[G]$(resp. $K[H]$-)module (and $K[G]$ is a $(K[G],K[H])$-bimodule to make sense of the tensor product and the $K[G]$-module structure on it). Once again, this isomorphism seems easy to establish : one direction is $\lambda\mapsto (v\mapsto \lambda(1\otimes v))$ and the other $f\mapsto (g\otimes v\mapsto g\cdot f(v))$.

This seems to work for arbitrary groups, not just finite ones. So I assume the sentence in the wikipedia article means that $K[G]\otimes_{K[H]}V \simeq \mathbf{Ind}_H^GV$ only if $G$ is finite (the "only" being in the sense "in general").

If that's not the case, what does this sentence mean ? If it is, is this isomorphism natural (in $V$ ? and in $(H,G)\in \mathbf{FinGrp}^\to$ ?) ? What is the isomorphism? If my interpretation is correct, can anything be said on the relationship between $K[G]\otimes_{K[H]}V$ and $\mathbf{Ind}_H^GV$ when $G$ is infinite ?

If my interpretation is not correct and the "$K[G]\otimes_{K[H]}V$" construction does not work for infinite $G$ (because of some mistake I made), then does $\mathbf{Res}_H^G$ have a left adjoint in general ?

It seems as though one can apply the general adjoint functor theorem to prove that it does, the only bit I'm not sure about being the fact that it preserves limits... But I think it preserves products and equalizers so it should suffice, right ?

If this is correct, and still working under the assumption that my interpretation isn't correct, can the aforementioned left adjoint be explicited? Does it have anything to do with $\mathbf{Ind}_H^G$ ? with $K[G]\otimes_{K[H]}-$ ?

Any correction of anything I said, besides the explicit questions, is very welcome as well as an answer to the questions !

$\endgroup$
  • 1
    $\begingroup$ As far as I know, some authors define an induced representation as the tensor product you mentioned, and so define it as the Hom space you mentioned. They are isomorphic when the group is finite, but not in general. The tensor product is left-adjoint to restriction; the Hom space is right-adjoint to restriction. When people want to distinguish between the two, they speak of "induction" and "coinduction", but I don't know if there is a standard for which is which. $\endgroup$ – darij grinberg May 14 '18 at 13:13
  • $\begingroup$ @darijgrinberg Thank you for your comment ! I've added a little bit to ask about the isomorphism in the finite case (I don't know whether the isomorphism can be explicited or at least easily constructed). $\endgroup$ – Max May 14 '18 at 13:24
  • 1
    $\begingroup$ The solution to Exercise 4.1.4 in Darij Grinberg and Victor Reiner, Hopf Algebras in Combinatorics, arXiv:1409.8356v5 (see the ancillary file for the solutions) constructs the isomorphism for finite $G$ and $H$. (Actually, it just requires $G/H$ to be finite, not $G$ and $H$.) Now, whether it is natural in $\left(H,G\right)$ is a question I've asked myself too, but I never gotten around to answer (part of the problem is that there are several reasonable choices of category). $\endgroup$ – darij grinberg May 14 '18 at 13:33
  • $\begingroup$ Oh, I've just realized you're talking about morphisms of finite groups, not just inclusion. Then, you want Exercise 4.1.14(i) instead of Exercise 4.1.4, and there is an extra complication (you need to require the size of the kernel of the morphism to be invertible in the base ring). $\endgroup$ – darij grinberg May 14 '18 at 13:38
  • $\begingroup$ @darijgrinberg for most questions I'm ok with just inclusion (first sentence of the second paragraph ;) ). Thank you for your reference ! $\endgroup$ – Max May 14 '18 at 13:43
3
$\begingroup$

Let $f : R \to S$ be a ring homomorphism, which in this special case is a homomorphism of group rings $k[H] \to k[G]$ induced by a homomorphism of groups. $f$ induces a restriction of scalars functor between categories of left modules

$$f^{\ast} : \text{Mod}(S) \to \text{Mod}(R)$$

which has both a left and a right adjoint. The left adjoint is called extension of scalars and looks like

$$f_L : \text{Mod}(R) \ni M \mapsto S \otimes_R M \in \text{Mod}(S)$$

and the right adjoint doesn't have a common name that I'm aware of, but we could call it coextension of scalars, and it looks like

$$f_R : \text{Mod}(R) \ni M \mapsto \text{Hom}_R(S, M) \in \text{Mod}(S).$$

Now, I claim that if $G$ and $H$ are finite groups, then these functors are naturally isomorphic, meaning induction is both left and right adjoint to restriction in this case. But in general they're not. Which one deserves to be called "induction" and which one deserves to be called "coinduction" is a question I haven't settled to my own satisfaction. The second one seems to generalize more cleanly to the setting where $G$ and $H$ are groups with extra structure, e.g. algebraic groups; IIRC it can be interpreted as computing the space of sections of a $G$-equivariant vector bundle on the quotient $G/H$.

$\endgroup$
  • $\begingroup$ Okay and I'm guessing the proof that these are adjoints is essentially the same as the one for group rings. Is there a condition on $R,S$ that one can express to "look like" the condition '$[G:H]$ is finite' so that the two become isomorphic ? And is there a slick proof of the natural isomorphism ? (The one in darij's paper is alright but it has an awful lot of computation and I was wondering if there's a more abstract way to see it ) $\endgroup$ – Max May 15 '18 at 8:24
  • $\begingroup$ The entirely standard (in pure algebra) convention is that tensor product with $R$ is called induction and the functor $\mathrm{Hom}_R(S,M)$ is called co-induction. They are not in general isomorphic, even for finite groups with field coefficients: if the map $f$ has a kernel and the characteristic of the field is positive then they can be different. $\endgroup$ – Stephen May 15 '18 at 11:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.