Induction of tensor product vs. tensor product of inductions - MathOverflow

Induction of tensor product vs. tensor product of inductions

2010-03-15T19:35:46Z

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.

Answer by Johannes Hahn for Induction of tensor product vs. tensor product of inductions

2010-03-15T20:19:16Z

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.

Answer by Marty for Induction of tensor product vs. tensor product of inductions

2010-03-15T21:01:55Z

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.

Answer by Robin Chapman for Induction of tensor product vs. tensor product of inductions

2010-03-15T21:06:20Z

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.

Answer by darij grinberg for Induction of tensor product vs. tensor product of inductions

2010-03-24T13:58:10Z

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.