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This is a pure curiosity question and may turn out completely devoid of substance.

Let $G$ be a finite group and $H$ a subgroup, and let $V$ and $W$ be two representations of $H$ (representations are finite-dimensional per definitionem, at least per my definitions). With $\otimes$ denoting inner tensor product, how are the two representations $\mathrm{Ind}^G_H\left(V\otimes W\right)$ and $\mathrm{Ind}^G_HV\otimes \mathrm{Ind}^G_HW$ are related to each other? There is a fairly obvious map of representations from the latter to the former, but it is neither injective nor surjective in general. I am wondering whether we can say anything about the decompositions of the two representations into irreducibles.

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Be careful about the dimensions involved: induction multiplies the dimension of a given module by the index $[G:H]$, so your second module is typically bigger than your first. –  Jim Humphreys Mar 15 '10 at 20:10
    
Oh! I knew about the different dimensions, but I confused smaller and larger. So we've got a map neither injective nor surjective (because it only maps to pure tensors), which makes the question way less attractive. –  darij grinberg Mar 15 '10 at 20:17
    
In general, it's hard to say anything helpful about the decomposition of an induced module or of a tensor product of two modules for a finite group. For groups of Lie type, much work has been done in such directions for specific types of subgroups and modules, but results usually depend strongly on special methods such as Hecke algebras. Clifford theory and Mackey theory do provide some general guidelines for arbitrary finite groups, of course. –  Jim Humphreys Mar 15 '10 at 20:58
    
Does it help to generalise? You have an inclusion $H\times H\rightarrow G\times G$ and the diagonal subgroups $H\rightarow G$. You take $V\otimes W$ as a representation of $H\times H$. Then induce to $G\times G$ followed by restriction to $G$. Alternatively restrict to $H$ and then induce to $G$. This suggests taking a commutative square of group homomorphisms and comparing induction followed by restriction with restriction followed by induction. –  Bruce Westbury Mar 15 '10 at 21:09
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4 Answers

up vote 5 down vote accepted

Surely you mean "former to latter"?

I think the natural map is injective. Let $V$ and $W$ have bases $v_1,\ldots,v_r$ and $w_1,\ldots,w_s$ respectively. Let $g_1,\ldots,g_t$ be coset representatives for $H$ in $G$. Then a basis for $\mathrm{Ind}_H^G V\otimes \mathrm{Ind}_H^G W$ consists of the $(v_i g_k)\otimes(w_j g_l)$. The image of the natural injection from $\mathrm{Ind}_H^G(V\otimes W)$ is spanned by the $(v_i g_k)\otimes(w_j g_l)$ with $k=l$. There are exactly the right number of these.

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Thanks, never expected this to be that simple. –  darij grinberg Mar 15 '10 at 22:45
    
I think my argument is essentially the same as that in Bruce's comment, but phrased in a more low-brow way. –  Robin Chapman Mar 16 '10 at 21:05
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Try using Frobenius reciprocity. Let $V$ and $W$ be two representations of $H$, and let $U$ be a representation of $G$. Consider first the space: $$Hom_G \left(U, Ind_H^G (V \otimes W) \right) \cong Hom_H \left( U, V \otimes W \right),$$ by Frobenius reciprocity.

On the other hand, one can consider the space: $$Hom_G(U, Ind_H^G V \otimes Ind_H^G W).$$ This is canonically isomorphic to $$Hom_G(U \otimes Ind_H^G V', Ind_H^G W),$$ where $V'$ denotes the dual representation of $V$. By Frobenius reciprocity again, this is isomorphic to: $$Hom_H(U \otimes Ind_H^G V', W).$$ This is canonically isomorphic to $$Hom_H(U, (Ind_H^G V) \otimes W).$$ Now, we are led to compare the two spaces: $$Hom_H(U, V \otimes W), \quad Hom_H \left( U, (Ind_H^G V) \otimes W \right).$$

There is a natural embedding of $V$ into $Res_H^G Ind_H^G V$. This gives a natural map: $$\iota: Hom_H(U, V \otimes W) \rightarrow Hom_H \left( U, (Ind_H^G V) \otimes W \right).$$

Using complete reducibility, let us (noncanonically) decompose $H$-representations: $$Res_H^G Ind_H^G V \cong V \oplus V^\perp.$$ It follows that $$Hom_H \left( U, (Ind_H^G V) \otimes W \right) \cong Hom_H \left(U, V \otimes W \right) \oplus Hom_H \left( U, V^\perp \otimes W \right).$$

It follows that $\iota$ is injective. This explains (via Yoneda, if you like) why $Ind_H^G(V \otimes W)$ is canonically a subrepresentation of $Ind_H^G V \otimes Ind_H^G W$. It also explains that computation of "the rest" of $Ind_H^G V \otimes Ind_H^G W$ -- the full decomposition into irreducibles -- requires Mackey theory: the decomposition of $Res_H^G Ind_H^G V$. There can be no neat answer, without performing this kind of Mackey theory.

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Thanks, this looks interesting. I'll reply in more detail when I have read this more carefully, but can it be that at the end you forgot that the decomposition of Res Ind V into V (+) V^perp was noncanonical? –  darij grinberg Mar 15 '10 at 22:09
    
Anyway, your post taught me that induction commutes with taking duals, so you're already got +1. –  darij grinberg Mar 15 '10 at 22:17
    
Ah, I see. You say that the decomposition is noncanonical because it isn't canonical with respect to G and H. Of course it's canonical with respect to V (it's an application of Mackey's double coset theorem). –  darij grinberg Mar 15 '10 at 22:21
    
I guess I haven't thought too carefully about the canonicity (canonicalness) of the Mackey double coset theorem here. But in any case, the injectivity of the unit (or is it counit?) map from $V$ into $Res Ind V$ is the important point. I mentioned that it is a summand, because I didn't want to rely on exactness of any functors. I was trying to be extra careful, but I might have been confusing instead. –  Marty Mar 15 '10 at 22:43
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Hi Darij.

$Ind_H^G X = R[G] \otimes_{R[H]} X$ and hence there are two canonical maps:

$R[G]\otimes V \otimes W \to R[G] \otimes V\otimes R[G] \otimes W, x\otimes v\otimes w\mapsto x\otimes v\otimes 1 \otimes w$ and

$R[G]\otimes V \otimes R[G] \otimes W \to R[G] \otimes V \otimes W, x\otimes v\otimes y \otimes w\mapsto xy \otimes v \otimes w$.

Obviously the second is a right inverse to the first. Hence the first one is injective and the second is surjective. If $R$ is a field, then there cannot be injective (surjective) maps in the other direction because the dimensions don't agree.

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Hmm. I am not convinced that the first map is a map of representations. All of these tensor signs are different kinds of tensor signs; the induction is a tensor product of "consistent" bimodules (i. e. a right R[H]-module (X) a right R[H]-module), while V (X) W is an inner tensor product of left modules. –  darij grinberg Mar 15 '10 at 20:25
    
All my tensorproducts are R[H]-tensors. What is an "inner" tensor product? –  Johannes Hahn Mar 16 '10 at 13:29
    
Johannes, not all your $\otimes$ are over $R[H]$:: for example, you write $R[G]\otimes V\otimes W$ which is $R[G]\otimes_{R[H]}(V\otimes_R W)$. –  Mariano Suárez-Alvarez Mar 16 '10 at 13:36
    
damn it... Okay, I promise to think before I post again. –  Johannes Hahn Mar 16 '10 at 13:48
    
No problem. If I would think before posting my postcount would be much smaller here... –  darij grinberg Mar 16 '10 at 14:28
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Just for the case someone is interested, here is another answer I've found. Probably it's equivalent to Robin's and Marty's answers, with the only difference that it's more abstract-nonsense than Robin's (so if you don't consider this as a virtue in itself, you don't have to read on) and shorter than Marty's (in particular, I don't switch to dual representations).

We will always abbreviate $\mathrm{Ind}^G_H$ by $\mathrm{Ind}$ and $\mathrm{Res}^G_H$ by $\mathrm{Res}$. Then, the push-pull formula states that $\mathrm{Ind}\left(U\otimes\mathrm{Res} T\right)\cong \mathrm{Ind}U\otimes T$ for any representation $T$ of $G$. Applying this to $T=\mathrm{Ind}V$, we get $\mathrm{Ind}\left(U\otimes\mathrm{Res}\mathrm{Ind}V\right)\cong \mathrm{Ind}U\otimes\mathrm{Ind}V$. Now, we can see the representation $\mathrm{Res}\mathrm{Ind}V$ as the left $k\left[H\right]$-module $k\left[G\right]\otimes _{k\left[H\right]}V$. Then, there is a canonical $H$-equivariant injection $V\to \mathrm{Res}\mathrm{Ind}V$ given by $v\mapsto 1\otimes v$, and there is a canonical $H$-equivariant projection $\mathrm{Res}\mathrm{Ind}V\to V$ given by $g\otimes v\mapsto gv$ for $g\in H$ and $g\otimes v\mapsto 0$ for $g\not\in H$. This projection splits the injection, and therefore the representation $V$ is canonically a direct summand of the representation $\mathrm{Res}\mathrm{Ind}V$. Hence, $\mathrm{Ind}\left(U\otimes V\right)$ is canonically a direct summand of $\mathrm{Ind}\left(U\otimes\mathrm{Res}\mathrm{Ind}V\right)\cong \mathrm{Ind}U\otimes\mathrm{Ind}V$.

"Canonically" means "canonically with respect to $U$ and $V$ and kind-of canonically with respect to $G$ and $H$" here. "Kind-of canonically with respect to $G$ and $H$" means that it's functorial with respect to maps which preserve both "lying in $H$" and "not lying in $H$", and I think we can't do better. As opposed to Robin's and Marty's proof, we don't need to rely on some randomly chosen system of representatives of cosets or double cosets.

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