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I'm writing a paper and want to cite some references to efficiently prove that over any field $k$ of characteristic zero, every irreducible representation of a product of symmetric groups, say

$$ S_{n_1} \times \cdots \times S_{n_p} $$

is isomorphic to a tensor product $\rho_1 \otimes \cdots \otimes \rho_p$ where $\rho_i$ is an irreducible representation of $S_{n_i}$.

I have an open mind about this, but I'm imagining doing it by finding references for these two claims:

  1. If $k$ is an algebraically closed field of characteristic zero, every irreducible representation of a product $G_1 \times G_2$ of finite groups is of the form $\rho_1 \otimes \rho_2$ where $\rho_i$ is an irreducible representation of $G_i$.

  2. If $k$ has characteristic zero and $\overline{k}$ is its algebraic closure, every finite-dimensional representation of $S_n$ over $\overline{k}$ is isomorphic to one of the form $\overline{k} \otimes_k \rho$ where $\rho$ is a representation of $S_n$ over $k$.

Serre's book Linear Representations of Finite Groups states the first fact for $k = \mathbb{C}$ but apparently not for a general algebraically closed field of characteristic zero. (It's Theorem 10.) It could be true already for any field of characteristic zero, which would simplify my life.

The second fact should be equivalent to saying that $\mathbb{Q}$ is a splitting field for any symmetric group, which seems to be something everyone knows - yet I haven't found a good reference.

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    $\begingroup$ I wonder if (1) and (2) together are enough to imply the desired claim -- conceivably the product could have a non-factorizable representation that becomes isomorphic to a factorizable one over the algebraic closure. (making up the word "factorizable"to mean "satisfying the conclusion of (1)"). $\endgroup$ – hunter Apr 10 at 23:46
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    $\begingroup$ Couldn't (1) and (2) be subsumed by showing that the representations of the form $\overline k \otimes_k (\rho_1 \otimes \dotsb \otimes \rho_p)$ give a decomposition of the regular representation, e.g., character-theoretically? (I don't know if this is true, but am optimistic.) $\endgroup$ – LSpice Apr 11 at 17:24
  • $\begingroup$ Also, for (2), every $S_n$-representation already carries a $\mathbb Z$-structure, so that should be even stronger. $\endgroup$ – LSpice Apr 11 at 17:25
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    $\begingroup$ Various people have essentially said this, but just to state it explicitly: Perhaps the reason you "haven't found a good reference" for "$\mathbb{Q}$ is a splitting field for any symmetric group" is that what the books tend to prove is a much stronger statement: "Here is an explicit combinatorial construction of the all the finite-dimensional irreducible representations of $S_n$ that uses only integers." There are plenty of good references for this latter fact, of which the fact you're interested in is an immediate corollary. $\endgroup$ – Timothy Chow Apr 12 at 16:30
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    $\begingroup$ Right, @TimothyChow. I'm trying to write a paper for category theorists who aren't experts in representation theory, so ideally I could just say what a splitting field is and point to Corollary 41.6 in The Great Tome of Representation Theory, which says "$\mathbb{Q}$ is a splitting field for $S_n$", instead of making them wade through a construction and figure out for themselves that $\mathbb{Q}$ is a splitting field for $S_n$. Luckily Corollary 4.16 of Lorenz's A Tour of Representation Theory says exactly what I want, right after he does the construction. $\endgroup$ – John Baez Apr 12 at 20:56
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Question 1 is a special case of the following statement for finite dimensional algebras: Let $A$ and $B$ be a finite dimensional algebras over a field $K$ such that $A/rad(A)$ and $B/rad(B)$ are isomorphic to a direct product of matrix algebras over $K$ (which is always true when $K$ is algebraically closed).

When $e_i$ and $e_i'$ are pariwise orthogonal primitive idempotents which sum to 1 for $A$ and $B$ respectively, then $e_i \otimes_K e_i'$ are pairwise orthogonal primitive idempotents which sum to 1 for $A \otimes_K B$. Thus when $S_i$ and $S_i'$ are the simple $A$ and $B$-modules respectively, then $S_i \otimes S_i'$ are the simple $A \otimes_K B$ modules. This is proved for $A=B$ in the book "Frobenius algebras I" by Skowronski and Yamagata in chapter IV. as proposition 11.3. , but the proof works in exactly the same way when $A \neq B$.

For question 2, one can find in section 4.5. corollary 4.16 in the book "A tour of repreesntation theory" by Martin Lorenz the fact that $\mathbb{Q}$ is a splitting field for the symmetric group. The whole section 4 in this book is dedicated to the representation theory of the symmetric group in characteristic 0 and might be one of the nicest modern approaches to this problem.

Thus since $\mathbb{Q}$ is a splitting field for the symmetric group, it is true for any field $K$ of characteristic 0 (not just algebraically closed fields) that the irreducible representations of a direct product of symmetric groups is given as a tensor product of the irreducible representations of the single symmetric groups.

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A positive answer to the first question for $S_{n}$ follows from the work of Alfred Young. He proved that every irreducible representation of $S_{n}$ over $\mathbb{C}$ is afforded by an explicit $\mathbb{Q}G$-module. Another way of saying this is that any irreducible $\mathbb{Q}S_{n}$-module $V$ is absolutely irreducible, and we have ${\rm End}_{\mathbb{Q}S_{n}}(V) \cong \mathbb{Q}.$ This last property is already enough to answer the question (positively) for symmetric groups for the field $k = \mathbb{Q}$, and then for every characteristic zero field $k$, since the prime subfield of $k$ is isomorphic to $\mathbb{Q}.$ This answers 2 positively.

The tensor decomposition for irreducible modules of general direct products is a consequence of Schur's Lemma: more precisely, it holds whenever the endomorphism algebra of each irreducible $kG_{i}$-module is isomorphic to $k$. This certainly holds for any algebraically closed field $k$ of characteristic zero.

I think that all this should be covered in Curtis and Reiner's book (1962). If not, an explanation of Young's constructions can be found in almost any book on the representation theory of $S_{n}$ (eg that of G. de B. Robinson, or more recent books of James, James and Kerber, or Fulton and Harris).

Later edit: Perhaps a more succinct way to say all this is that $\mathbb{Q}$ is a splitting field for $S_{n}$, and consequently, any field $k$ of characteristic zero is a splitting field for $S_{n}$. On the other hand, if field $k$ of characteristic zero is a splitting field for each finite group $G_{i}: 1 \leq i \leq n$, then $k$ is a splitting field for $G_{1} \times G_{2} \times \ldots \times G_{n}$, and any irreducible $kG$-module is the tensor product of $n$ irreducible modules, one for each $G_{i}$. This is a consequence of Schur's Lemma and Clifford's Theorem.

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To summarize the situation given in the other answers (no real new content here) it is classical theory going back to Young that the complex irreducible representations of $S_n$ can be defined over $\mathbb Q$ (i.e., written with $\mathbb Q$-coefficients, or written as $\mathbb C\otimes_{\mathbb Q}V$ with $V$ a $\mathbb QS_n$-irreducible module) and references were given; i.e., the $\mathbb Q$-irreducibles are absolutely irreducible. This can be done via Young symmetrizers and anti-symmetrizers, polytabloids, or a number of other approaches and I have nothing to add to the discussion.

What this means concretely is that $\mathbb QS_n\cong \prod_{i=1}^{p_n}M_{d_i}(\mathbb Q)$ where $p_n$ is the number of partitions of $n$ and $d_i$ is the dimension of the $i^{th}$-irreducible representations (and of course all these $d_i$ are well known through tableaux combinatorics and involve counting semi-standard Young tableaux). One way to see this is to use that if $V$ is a finite dimensional $\mathbb QS_n$-module, then $\mathrm{End}_{\mathbb CS_n}(\mathbb C\otimes_{\mathbb Q} V)\cong \mathbb C\otimes_{\mathbb Q}\mathrm{End}_{\mathbb QS_n}(V)$ by standard arguments and so by Schur's lemma, if $\mathbb C\otimes_{\mathbb Q}V$ is irreducible, then $\mathrm{End}_{\mathbb CS_n}(\mathbb C\otimes_{\mathbb Q} V)$ one-dimensional over $\mathbb C$ and hence $\mathrm{End}_{\mathbb QS_n}(V)$ is one-dimensional over $\mathbb Q$ and so apply Wedderburn-Artin to $\mathbb QS_n$ to get the statement.

Now, let's just handle $\mathbb Q[S_n\times S_m]\cong \mathbb QS_n\otimes_{\mathbb Q}\mathbb QS_m$. Then by the above, we have that this tensor product is isomorphic to $$\prod_{i=1}^{p_n}\prod_{j=1}^{p_m}M_{d_i}(\mathbb Q)\otimes_{\mathbb Q} M_{c_j}(\mathbb Q)$$ where I introduced $c_j$ for the dimensions of the $S_m$-irreducibles over $\mathbb Q$. Obviously $$M_{d_i}(\mathbb Q)\otimes_{\mathbb Q}M_{c_j}(\mathbb Q)\cong M_{d_i}(M_{c_j}(\mathbb Q))\cong M_{d_ic_j}(\mathbb Q)$$ and hence $M_{d_i}(\mathbb Q)\otimes_{\mathbb Q}M_{c_j}(\mathbb Q)$ is simple with a unique simple module (up to isomorphism) which has dimension $d_ic_j$ (and this dimension characterizes the simple module). Consequently the tensor product of the unique simple modules of the two tensor factors is the unique simple module for this tensor product, e.g., by dimension consideration. Putting it all together, we get that the simple $\mathbb Q[S_n\times S_m]$-modules are the tensor products of the simple $\mathbb QS_n$-modules and $\mathbb QS_m$-modules. Now since $K\otimes_{\mathbb Q} M_r(\mathbb Q)\cong M_r(K)$ and $K\otimes_{\mathbb Q}\mathbb Q^r\cong K^r$, the situation doesn't change when we extend the scalars. We get the same number of irreducibles and they are obtained by extending the scalars from those of $\mathbb Q[S_n\times S_m]$. Of course the argument is the same for any finite number of factors.

Tiny update. Although John didn't ask for this, it is also well known that the $p$-element field $\mathbb F_p$ is a splitting field for the symmetric group in characteristic $p$. From this one can again deduce that all irreducible representations of products of symmetric groups are tensor products of irreducible representations of the factors over any field. Likely there is a %100 direct proof using tableaux and the like. But here is one possible proof. First note that if $G$ is any finite group, $K$ is an algebraically closed field of characteristic $p$ and $|G|=p^nm$ with $\gcd(p,m)=1$, then one can show that each character $\chi$ of $G$ over $K$ takes values that are sums of $m^{th}$-roots of unity since $1$ is the only $p^{th}$-root of unity in $K$. Hence the character field of $\chi$ is a finite field. The theory of Schur indices together with Wedderburn's theorem that there are no finite division rings, then tells you that $\chi$ is realizable over the character field $\mathbb F_p(\chi)$. Moreover, a character of $G$ is well known to be determined by its values on the $p$-regular elements of $G$ (the elements of order prime to $p$). Now in $S_n$, every element $g$ of order prime to $p$ is conjugate to $g^p$. Hence $\chi(g)=\chi(g^p) = \Phi(\chi(g))$ where $\Phi$ is the Frobenius automorphism $x\mapsto x^p$. Therefore, $\chi$ takes values in $\mathbb F_p$ and so is realizable over $\mathbb F_p$.

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I wrote a note giving a simple character-theory free proof that over an algebraically closed field of char 0, irreducible representations of products are tensor products of irreducible representations. See here. As the example after Theorem A of my note shows, this is false if you don't assume the field is algebraically closed.

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    $\begingroup$ @MaximeRamzi: I could be missing something very obvious, but why can we assume that the $v_i$ are linearly independent over the endomorphism ring? And as a related question (I suspect that I will understand this once I understand the answer to my previous one), why can you then turn around that do the same to the $w_i$ after you've reduced to the case where there is only one $v_i$? $\endgroup$ – Andy Putman Apr 11 at 21:14
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    $\begingroup$ Right no I think I went too quickly and just made a mistake (that's why I should write things down !) $\endgroup$ – Maxime Ramzi Apr 11 at 21:23
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    $\begingroup$ @Andy Is the example in your note not the tensor product of the nontrivial irreducible representations of $\mathbb{Z}/3$ and $\mathbb{Z}/2$ over $\mathbb{R}$? $\endgroup$ – Jeremy Rickard Apr 11 at 23:12
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    $\begingroup$ @JeremyRickard: Whoops, you’re right! I should have used the rotation action of $\mathbb{Z}/15$ on $\mathbb{R}^2$ instead. I will fix this next time I am in my office. Thanks! $\endgroup$ – Andy Putman Apr 12 at 1:29
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    $\begingroup$ Andy your proof just needs the representations are absolutely irreducible which means that it works for direct products of symmetric groups over Q. It’s also essentially the semi simple version of @Mare’s argument without talking about idempotents because you are going one simple component at a time $\endgroup$ – Benjamin Steinberg Apr 12 at 2:56
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This is not quite what you're looking for, but here's a theorem (followed by a reference) which justifies the heuristic that the representation theory of a finite group over an algebraically closed field of characteristic $0$ "doesn't depend on the field" :

Suppose $K$ is a field in which every irreducible representation of $G$ is absolutely irreducible. Then for any field extension $K'/K$, the induced morphism on representation rings $R_K(G)\to R_{K'}(G)$ is an isomorphism.

This is in the first paragraph of section 14.6 in Serre's book.

For instance, here's how it can help for statement 1. : if $K$ is an algebraically closed field of characteristic $0$, then

a) all irreducible representations are absolutely irreducible, by Schur's lemma and the existence of eigenvalues

b) $K$ has a common field extension with $\mathbb C$.

From there it's not hard to see that if statement 1. holds over $\mathbb C$, it does so over any algebraically closed field of characteristic $0$ (indeed note that the morphism $R_K(G)\to R_{K'}(G)$ maps the "positive part" to the positive part $R_K^+(G)\to R_{K'}^+(G)$, so if it's an isomorphism so must the latter be; and free commutative monoids have at most one basis - so this induces a bijection between the irreducible representations - apply this to $G_1,G_2$ and $G_1\times G_2$)

You can also deduce 2. from the similar fact over $\overline{\mathbb Q}$ or $\mathbb C$ if you use the other statement from the same paragraph of Serre's book, namely that $R_K(G)\to R_{K'}(G)$ is always injective, no matter what the field extension $K'/K$ is.

Indeed, your statement 2. is then saying that $R_k(S_n)\to R_\overline k(S_n)$ is an isomorphism, but this follows from the following chain of morphisms, where $K$ is a common extension of $\overline k$ and $\mathbb C$: $R_\mathbb Q(S_n)\to R_k(S_n)\to R_\overline k(S_n)\to R_K(S_n)$. By the earlier statement, the last morphism is an isomorphism, but also the composite (if you already know the surjectivity statement for $\mathbb C$), therefore by injectivity so is the middle one.

In other words, the slogan works in these situations, and you can find the appropriate precise statements in this first paragraph of section 14.6 of Serre's book.

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Every complex representation of $S_n$ is defined over $\mathbb Z$, but, as suggested in the comments, it might be more natural to remember only that they are defined over $\mathbb Q$.

Let $k$ be any characteristic-$0$ field, and consider the representations $k \otimes_{\mathbb Q} (\rho_1 \otimes \dotsb \otimes \rho_p)$. Their characters $\chi_{k \otimes_{\mathbb Q} (\rho_1 \otimes \dotsb \otimes \rho_p)} = \chi_1 \otimes \dotsb \chi_p$ satisfy $$ \sum_{\rho_1, \dotsc, \rho_p} d_1\dotsm d_p\chi_1(g_1)\dotsm\chi_p(g_p) = \prod_{j = 1}^p \sum_{\rho_j} d_j\chi_j(g_j) = \prod_{j = 1}^p n_j![g_j = 1] = n_1!\dotsm n_p![(g_1, \dotsc, g_p) = (1, \dotsc, 1)], $$ using the Iverson bracket, since they do so over $\mathbb C$. In particular, if $\chi$ is any character of an irreducible representation of $S_{n_1} \times \dotsb \times S_{n_p}$ over $k$, then the inner product $\chi(1)$ with the right-hand side is non-$0$, so the inner product of $\chi$ with some $\chi_1 \otimes \dotsb \otimes \chi_p$ is non-$0$, so $\chi$ equals $\chi_1 \otimes \dotsb \otimes \chi_p$.

(Since we're working over a random characteristic-$0$ field $k$, it may not be obvious what I mean by "inner product". I'll define $\langle f_1, f_2\rangle = \frac1{\lvert G\rvert}\sum_{g \in G} f_1(g^{-1})f_2(g)$. Note, though, that it doesn't really matter that much, since all the functions with which we're dealing are valued in the rational numbers—even the integers.)

I'm inclined to think character theoretically, but here's a re-phrasing in entirely representation-theoretic terms. I'll continue to work over $\mathbb Q$; in particular, I'll regard all representation spaces as at first being over $\mathbb Q$. Then the map \begin{equation} \label{Plancherel} \tag{*} \bigoplus_{\rho_1, \dotsc, \rho_p} (V_1 \otimes \dotsb \otimes V_p)^* \otimes (V_1 \otimes \dotsb \otimes V_p) \to \mathbb Q[S_{n_1} \times \dotsb \times S_{n_p}] \end{equation} sending $v^* \otimes v$ to the matrix coefficient $g \mapsto \langle v^*, g\cdot v\rangle$ becomes an isomorphism when tensored with $\mathbb C$, so is already an isomorphism over $\mathbb Q$, so remains an isomorphism when tensored with $k$. Of course, it is $S_{n_1} \times \dotsb \times S_{n_p}$-equivariant. Now any representation $\rho$ of $G = S_{n_1} \times \dotsb \times S_{n_p}$ over $k$ injects into $k[G]$, by taking matrix coefficients as above, and so, by \eqref{Plancherel}${}\otimes_{\mathbb Q} k$, admits a non-$0$ quotient map to some $k \otimes_{\mathbb Q} (\rho_1 \otimes \dotsb \otimes \rho_p)$.

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  • $\begingroup$ What does summation over $\rho_1, \dots, \rho_p$ mean? $\endgroup$ – John Baez Apr 11 at 17:42
  • $\begingroup$ It is a sum over the equivalence classes of irreducible representations $\rho_j$ of $S_{n_j}$ (of which there are finitely many). I don't know off the top of my head if equivalence over $\mathbb Z$ is stronger than equivalence over $\mathbb C$ in this special case, so let's say that I mean equivalence over $\mathbb C$ (which is enough to give that the characters are the same, so you could view it as a sum over characters if that's more palatable). $\endgroup$ – LSpice Apr 11 at 17:43
  • $\begingroup$ Okay, for some reason I imagined the $\rho_j$ as held fixed at the start of your argument. Btw, classifying irreps of symmetric groups over finite fields is supposed to be hard, they're not just classified by Young diagrams as they are over $\mathbb{Q}$ or $\mathbb{C}$. Doesn't that imply the representation theory of $S_n$ over $\mathbb{Z}$ must be quite different than over $\mathbb{C}$? $\endgroup$ – John Baez Apr 11 at 18:07
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    $\begingroup$ I would find $\mathbb{Q}$ less scary. Btw, I'm a bit nervous about the step where you say something works over an algebraically closed field of characteristic zero because it works that way over $\mathbb{C}$. There's probably some general principles that let one make arguments like that, but what's the easiest explanation in this case? $\endgroup$ – John Baez Apr 11 at 19:02
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    $\begingroup$ I have written a re-phrased version that works only with representations, not with characters, and so hopefully makes it more explicit when we're changing base, and why it's OK to do so (namely, flatness of field extensions—I think is the jargon?). $\endgroup$ – LSpice Apr 11 at 19:58
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For number 1, when a statement in the representation theory of finite groups is true over $\mathbb{C}$, I think it should be true for any algebraically closed field of characteristic zero. That's because the development of the representation theory of finite groups proceeds the same way over any algebraically closed field of characteristic zero. I don't know a precise statement of this, but in algebraic geometry, a similar idea is called the "Lefschetz principle".

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  • $\begingroup$ I'm sure your statement is true, but I need a reference if that's the road I want to go down. None of this stuff should be new: I need it for some proofs in a paper I'm writing. $\endgroup$ – John Baez Apr 11 at 4:46
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    $\begingroup$ To be a bit more blunt, I don't really want to say in a paper "it's true for $\mathbb{C}$, so it's true for any algebraically closed field of characteristic zero" when someone has probably just proved it for any algebraically closed field of characteristic zero (or maybe even more generally). $\endgroup$ – John Baez Apr 11 at 17:41

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