Let $k$ be a field which I don't suppose to be algebraically closed. Then, the endomorphism ring of any irreducible representation of a finite group over $k$ is a division ring.

What is known about these division rings? When are they fields? When they are, are these fields Galois over $k$? At least, normal? When not, are the centers of the division rings Galois over $k$? What do we know about these centers?

If $k$ is a perfect field, do the irreducible representations themselves split into distinct irreducible representations over the algebraic closure of $k$, or can some be equal?

Sorry for this many questions that are probably well-known, but I can't find an introductory text on representation theory which takes non-algebraically closed fields seriously.

  • $\begingroup$ Touché. I was thinking about groups. $\endgroup$ – darij grinberg Jul 18 '10 at 18:50
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    $\begingroup$ See Part II of Serre's Linear Representations of Finite Groups for a nice introduction to representation theory over a nonalgebraically closed field of characteristic $0$. Some keywords: Schur indices, Schur subgroup. M. Isaacs's book on representation theory covers these topics in somewhat greater depth. $\endgroup$ – Pete L. Clark Jul 18 '10 at 18:57
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    $\begingroup$ In particular, the example of the quaternion group of order $8$ over $\mathbb{Q}$ is one to keep in mind. This is the smallest group order for which $\mathbb{Q}[G]$ has a factor which is a noncommutative division algebra. When this occurs, the corresponding irreducible reprsentation "ramifies", i.e., decomposes into multiple copies of the same geometrically irreducible representation. $\endgroup$ – Pete L. Clark Jul 18 '10 at 19:02
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    $\begingroup$ To add to Pete's comments, a reference for the determination of the Schur subgroup, i.e., the set of division algebras that appear, is Yamada: The Schur subgroup of the Brauer group, SLN 397. $\endgroup$ – Torsten Ekedahl Jul 18 '10 at 20:38
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    $\begingroup$ Every irreducible representation of a finite group in characteristic zero is realizable over a cyclotomic field (see Lang's Algebra). So the centre of any division ring that arises as the endomorphism ring of an irrep over a characteristic zero field $k$ will be an abelian extension of $k$. But over $\mathbb{Q}$ or $\mathbb{Q}_p$ the only examples of noncommutative endomorphism rings of irreps I've seen have been quaternionic. Are there examples where the degree of the division ring over its centre is greater than $4$? $\endgroup$ – Robin Chapman Jul 18 '10 at 20:38


are the centers of the division rings Galois over $k$

A finite dim'l $k$-algebra $A$ is split provided $\operatorname{End}_A(S) = k$ for every simple $A$-module $S$.

[This terminology is consistent with that used in other contexts -- $A$ is split just in case the reductive quotient of the "unit group" $A^\times$ is a split reductive algebraic group over $k$.]

Suppose that $k$ is perfect and that $A_\ell = A \otimes_k \ell$ is a split $\ell$-algebra for a finite, separable extension $\ell \supset k$. Then we have the following:

$(*)$ If $S$ is a simple $A$-module, the center $Z$ of the division $k$-algebra $\operatorname{End}_A(S)$ is a subfield of $\ell$.

To see this, I claim first that $\operatorname{End}_A(S) \otimes_k \ell$ is a split semisimple $\ell$-algebra. Indeed, since $k \subset \ell$ is separable, the $A_\ell$-module $S \otimes_k \ell = S_\ell$ is semisimple, say $S_\ell = \bigoplus_i S_i$ as $A_\ell$-module, where $S_i$ is the $T_i$-isotypic component of $S_\ell$ and where $T_i$ are distinct simple $A_\ell$-modules. Since by assumption $\operatorname{End}_{A_\ell}(T_i) = \ell$, we see that $$\operatorname{End}_A(S) \otimes_k \ell \simeq \operatorname{End}_{A_\ell}(S_\ell) = \prod_i \operatorname{End}_{A_\ell}(S_i)$$ is a product of full matrix algebras over $\ell$. Now observe that the center of a split semisimple $\ell$-algebra is a a split commutative etale $\ell$-algbra $\ell \times \cdots \times \ell$. Assertion $(*)$ now follows.

Apply this to $A = kG$ for a finite group $G$. By Torsten's comment (following Emerton's answer) we may suppose $k$ to be perfect. Then $A \otimes_k \ell$ is split for an Abelian extension $\ell$ of $k$ -- a suitable $\ell$ can be obtained by adjoining to $k$ enough roots of unity. It follows that $Z$ is always Galois over $k$ (for $A = kG$).

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  • $\begingroup$ Sounds interesting. Typo: End_k (S) = k should be End_A (S) = k. Also, I suppose A (X)_k l should be split as l-algebra, not as k-algebra? $\endgroup$ – darij grinberg Jul 18 '10 at 22:54
  • $\begingroup$ I don't see how you conclude that $\left(\mathrm{End}_AS\right)\otimes_k\ell$ is a split semisimple $\ell$-algebra, but I'll come back to this later when I have more time to seriously think about it (have to prepare talk about something quite different). $\endgroup$ – darij grinberg Jul 18 '10 at 23:00
  • $\begingroup$ Thanks for typo-spotting. Perhaps the edited version is clearer re: split semisimplicity? $\endgroup$ – George McNinch Jul 18 '10 at 23:46
  • $\begingroup$ I have not forgotten your answer, but I'm first going to read some semisimple theory before I go through it. At the moment I know the properties of semisimple algebras only over algebraically closed fields, which is definitely not enough. $\endgroup$ – darij grinberg Jul 20 '10 at 10:29
  • $\begingroup$ I think the treatment of semisimple algebras given by Curtis and Reiner [Methods of representation theory. Vol. I. With applications to finite groups and orders] takes non-alg closed fields seriously in this context. As does Lang's Algebra, I'm sure. (Well, this is all by memory -- my copies of both of those books are at my office...) $\endgroup$ – George McNinch Jul 20 '10 at 12:19

Pete's reference and other remarks in his comments are absolutely the right place to look/thing to think about. Let me make a slightly more elaborate remark, which may be helpful.

A key fact which helps one think about this sort of question is the following: if $V$ is a rep. of $G$ over $k$, and $l$ is a finite extension of $k$, then $End_{l[G]}(l\otimes_k V) = l\otimes_k End_{k[G]}(V).$ (This is easily checked.)

At least if $k$ is perfect, then when $V$ is irreducible over $k$ the base-change $l\otimes_k V$ will be semisimple over $l$ (i.e. a direct sum of irreds.). Given this, one can determine its structure (is it a sum of distinct irreds., say, or does it contain two copies of the same irrep.?) by looking at $End_{l[G]}(l\otimes_k V)$, which as I already noted we can compute as $l\otimes_k End_{k[G]}(V)$.

Let's suppose that $l = \bar{k}$, since that's probably the case of greatest interest. Then if $W$ is a direct sum of mutually non-isomorphic irreps., then $End_{\bar{k}[G]}(W)$ is a product of copies of $\bar{k}$, as many as there are summands of $W$. (If $W = \oplus_i W_i,$ then we get one copy of $\bar{k}$ for each $W_i$, since $End_{\bar{k}[G]}(W) = \bar{k}$, but there are no maps between the different $W_i$, since they are non-isomorphic by assumption.) On the other hand, if say $W$ were a direct sum $W = W_1\oplus W_2,$ then $End_{\bar{k}[G]}(W) = M_2(\bar{k})$ (since $Hom_{\bar{k}[G]}(W_i,W_j) = \bar{k}$ for any choice of $i$ and $j$).

Now using our formula $End_{\bar{k}[G]}(\bar{k}\otimes_k V) = \bar{k}\otimes_k End_{k[G]}(V),$ we see that $\bar{k}[G]\otimes_k V$ is a direct sum of distinct irreps. if and only if $End_{k[G]}(V)$ is a field (since it is precisely this case which gives a product of copies of $\bar{k}$ when we tensor up with $\bar{k}$), while $\bar{k}\otimes_k V$ will contain multiple copies of some irrep. if and only if $End_{k[G]}(V)$ is a non-commutative division ring, since then extending scalars to $\bar{k}$ will give a non-trivial matrix ring over $\bar{k}$. (This explains Pete's remark about the quaternion group in his comment above.) More precisely, we see that if $End_{k[G]}(V)$ is a division ring of dimension $n^2$ over its centre, which has say degree $d$ over $k$, then $\bar{k}\otimes_k V$ will be a product of $n$ copies each of $d$ distinct irreps. over $\bar{k}$.

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  • $\begingroup$ Hey, I was about to write that answer! $\endgroup$ – Ben Webster Jul 18 '10 at 19:57
  • $\begingroup$ The case of non-perfect k is really no problem. The group ring is defined over the prime field $\mathbb Z/p$ and it modulo its radical is a product of matrix rings (by Wedderburn's theorem) with centers that are separable extensions. This implies that the same thing is true for any field of positive characteristic $\endgroup$ – Torsten Ekedahl Jul 18 '10 at 20:23
  • $\begingroup$ Dear Torsten, Thanks for this helpful remark. $\endgroup$ – Emerton Jul 18 '10 at 20:37
  • $\begingroup$ Thanks for the detailed post, Emerton, but I already saw the argument when Pete mentioned the quaternion group. $\endgroup$ – darij grinberg Jul 18 '10 at 23:02
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    $\begingroup$ To put Emerton's concern about imperfectness in perspective, if $(V,\rho)$ is a finite-dimensional semisimple linear representation of a general group $G$ (perhaps infinite), all over a field $k$, when $G$ is infinite semisimplicity can be lost after a ground field extension when $k$ is imperfect. (Related fact: for a finite-dim. simple $k$-algebra $A$, $k' \otimes_k A$ can be non-ss if $k'/k$ is inseparable and $A$ has center inseparable over $k$.) For example, if $k' = k(a^{1/p})$ with $p = {\rm{char}}(k) > 0$ and $a \in k$ not a $p$-power, take $V = k'$ and $G = \mathbf{Z}$ with $1.v = av$. $\endgroup$ – BCnrd Jul 19 '10 at 13:19

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