# How ugly is the isomorphism R[GxH] = R[G] (X) R[H] for groups G, H?

Clearly, when $G$ and $H$ are two finite groups, and $V$ and $W$ are two representations of $G$ and $H$, respectively, then $V\otimes W$ is a representation of the group $G\times H$. It is a well-known fact that over an algebraically closed field of characteristic zero, every representation of $G\times H$ is a direct sum of such $V\otimes W$'s if the groups $G$ and $H$ are finite. Here are some things I am wondering about:

1) How canonical can these $V$'s and $W$'s be chosen? EDIT: This was discussed at decomposition of representations of a product group as I see. Yet the other questions are new.

2) Do we actually need all the conditions? What if our groups are not finite, or the characteristic of the field is nonzero? The latter may mean different things - we can work in the representation groups, we can work in the Grothendieck groups and we can work in the Grothendieck groups of the projective $k\left[G\right]$-modules (i. e. in the K-theory).

Note that we cannot lift the condition that $k$ be algebraically closed.

3) Does anything improve if $G=S_a$ and $H=S_b$ for integers $a$ and $b$ ? After all, symmetric groups have the nice property that all representations over $\mathbb C$ are defined over $\mathbb Q$, and this gives us hope that the tensorands $V$ and $W$ have some meaning.

Here is why I care:

The famous Hopf algebra $R\left(S\right)=\bigoplus\limits_{n\geq 0}R\left(S_n\right)$ (where $R\left(G\right)$ denotes the (Grothendieck) group of representations of a group $G$) has its product defined by

$U\cdot V = \mathrm{Ind}_{S_a\times S_b}^{S_{a+b}} U\otimes V$

(for $U\in R\left(S_a\right)$ and $V\in R\left(S_b\right)$) and its comultiplication (I hesitate to say coproduct) defined by

$\Delta\left(U\right) = \sum\limits_{k=0}^n \mathrm{Res}_{S_k\times S_{n-k}}^{S_n} U$

(for $U\in R\left(S_n\right)$). Now, $\mathrm{Res}_{S_k\times S_{n-k}}^{S_n} U$ is not an element of $R\left[S_k\right]\otimes R\left[S_{n-k}\right]$ per se, but an element of $R\left[S_k\times S_{n-k}\right]$, and one wishes to have a canonical isomorphism $R\left[S_k\times S_{n-k}\right]\to R\left[S_k\right]\otimes R\left[S_{n-k}\right]$ here. (Of course, it is canonical on the $R\left(S\right)$ level, but it would be great if it would also work out that nicely on the level of representations - after all we're being constructive. Note that multiplication in $R\left(S\right)$ is canonical on the level of representations.)

Summary of the question: Given a Young diagram $\lambda$ with $n$ boxes, is there a "canonical" (as in, explicit formulae or nice deterministic algorithm) way to decompose the restriction of the corresponding Specht module $R_{\lambda}$ to $S_a\times S_b$ (where $a+b=n$) into tensor products of the form $\left(\text{Specht module for }S_a\right)\otimes\left(\text{Specht module for }S_b\right)$ ?

Actually this is part of a bigger question, in case anyone is willing to answer that:

4) Is there a systematic text-book like account of representation theory of $S_n$ which actually uses the modern approaches (Hopf algebras, the Okounkov-Vershik constructive theory avoiding characters, algebraic combinatorics of Young tableaux, Liulevicius' K-theoretical interpretation), doesn't shy away from difficult parts (such as plethysms) and gives modern proofs of classical results rather than just refer to them as well-known (I am not really satisfied with the Schur-Froebnius era proofs, they are rather clumsy and intricate, some of them require working over $\mathbb C$ and character theory and they are just long). I am aware of Goldschmidt (very nice but too basic), Zelevinsky (seems to become tough reading) and Liulevicius (alas, only journal articles). Is there more?

-
I don't think "how evil" is an appropriate expression to use in a title of a mathematical question. –  Victor Protsak Jun 29 '10 at 21:16
I am going to replace it by "ugly" anyway because the actual meaning of evil ( ncatlab.org/nlab/show/evil ) seems to have nothing to do with the one I want (uncanonical, unconstructive, uses restrictive assumption, requires non-elegant proofs). –  darij grinberg Jun 29 '10 at 21:45
Concerning your motivation (3), the non-canonical aspect can be turned into an advantage: it's a manifestation of non-trivial structures on various "categorifications" (cf Bernstein, Frenkel, Khovanov in the Schur-dual situation of $\mathfrak{gl}_n$-modules). –  Victor Protsak Jun 30 '10 at 2:56

1. You can somewhat lift the algebraic closedness assumption: You have to allow an auxillary ring (actually, division algebra) to act equivariantly on both representation and tensor over it.
2. Such a decomposition should hold whenever one of the groups has semi-simple representation category (the division rings in 1 are endomorphisms of simples). Then, the decomposition can be made canonical precisely up to choosing representative simple objects. If $V$ is a $G \times H$-rep, and $\rho$ are representative simples for $G$, then the natural map $$\bigoplus_{\rho} \rho \otimes_{D_\rho} Hom_G(\rho, V) \to V$$ with $D_\rho = End_G(\rho)$ will be an isomorphism of $G \times H$-modules. (Conversely, given applying such a decomposition to $k[G]$ viewed as $G \times G$-module one would have to recover a representative set of simples.)
3. For symmetric groups (in char. $0$), the endomorphism rings of simples are just the field (i.e., the simples remain irreducible over the alg. closure), so in particular you get such decompositions. Moreover, there are explicit representative simples that one can write down (the Specht modules). I don't know of the combinatorial theory to say if this gives any sort of satisfactory answer to your question 3.
-
Thanks, this answers most of question 2. –  darij grinberg Jun 30 '10 at 17:24

As a partial answer to (2): If $k$ is algebraically closed (any characteristic) and $G$ and $H$ finite, then the tensor product of any irreducible $k$--representation $V$ of $G$ with any irreducible $k$--representation $W$ of $H$ is an irreducible representation of $G\times H$, and every irreducible representation of $G\times H$ can be obtained this way. (Just count them).

If $k$ is not algebraically closed, the corresponding statement is wrong in general, the tensor products will not remain irreducible. But we can weaken the condition, requiring only

$End_{kG}(V) = k\quad$ and $\quad End_{kH}(W) = k$

If $k$ is algebraically closed, that holds automatically by Schur's Lemma.

-
What do you mean bby "just count them"? -- And I'm not asking whether the tensor products will remain irreducible. I'm asking whether any representation (I am not talking about irreduciblity) is, in the Grothendieck category (note that cancellation is possible if $k$ is not of char $0$, although already the char $0$ case is interesting enough) a $\mathbb Z$-linear combination of tensor products. –  darij grinberg Jun 29 '10 at 20:36
@darij: "just count them" refers to the fact that for finite $G$, sum of squares of dims of irred. $kG$ modules $=|G|$ provided (i) $k$ is a splitting field for $G$ and (ii) $kG$ is semissimple ($\iff$ $|G|$ is invertible in $k$). I don't see how to use counting arg if $kG$ isn't ss, though Xandi's assertion remains true. Also: taking a comp. series for $G \times H$-module $V$, it seems that question of expressing $V$ as $\mathbf{Z}$-lin comb of $\otimes$s in Groth. group can be reduced to the case of irred. $V$ and thus to $\otimes$ of irreds, provided $\otimes$s of irreds remain irred. –  George McNinch Jun 30 '10 at 14:12
@darij: in beginning of previous comment, I should have written "probably refers to". –  George McNinch Jun 30 '10 at 15:34
Thanks George no "probably" needed. The counting argument also works if the characteristic $p$ divides $|G|$: The number of irreducible representations equals the number of conjugacy classes in whose elements are of order prime to $p$. –  Xandi Tuni Jun 30 '10 at 15:55
@darij: If by "composed of" you mean "a direct sum of" then certainely not, so i am just telling that at least you get the irreducuble ones. But you get surjectivity on the level of Grothendieck rings, which do not see extensions. –  Xandi Tuni Jun 30 '10 at 19:21