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Let $G$ be a finite group and $\chi$ a character of $G$. The values of $\chi$ generate an abelian Galois extension $K$ of $\mathbb{Q}$, and so one can consider the conjugate $\sigma(\chi)$ of $\chi$ by any element $\sigma$ of the Galois group. What's the shortest way to prove that $\sigma(\chi)$ is also a character of $G$?

This follows from the fact that the corresponding representation $V$ is realizable over a finite extension of $K$, but this fact is somewhat annoying to prove, and on an exam I don't want to prove it from scratch if I can avoid it. Are there any shorter arguments?

Motivation: I'm looking at a representation theory exam question (for practice) which asks me to fill out a character table. If I could assume the above statement, I could conclude that the missing characters are integer-valued.

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Assuming that "representation" means "map from $G$ to $GL_n(\mathbb{C})$", it is tautologically true that $V$ is definable over an extension of $K$ -- namely, the extension $\mathbb{C}$. I suspect the fact you don't want to use is that every representation is definable over $\mathbb{Q}^{alg}$ (and hence over an algebraic extension of $K$.) –  David Speyer Jun 2 '11 at 22:15
Yes, I meant "finite extension." –  Qiaochu Yuan Jun 2 '11 at 22:24
And I assume you don't want to prove that, if $K \subseteq L$ is any extension of fields, with $L$ algebraically closed, then every Galois symmetry of $K$ extends to a symmetry of $L$. (Since this uses the axiom of choice and some tricky arguments.) :) –  David Speyer Jun 2 '11 at 22:32
Does your group have lots of automorphisms? Automorphisms can often effect a galois conjugation. –  Jack Schmidt Jun 2 '11 at 23:10
@Jack: in the context of the particular question I'm looking at, I don't know anything about the group other than a few lines of its character table and the sizes of its conjugacy classes. –  Qiaochu Yuan Jun 2 '11 at 23:14
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5 Answers

up vote 9 down vote accepted

OK, I've thought about it a bit more and I'll give an alternate proof below the first horizontal line. However, I highly deny that this proof is better, and I am not sure it is actually shorter. To my mind, the morally correct proof is to show that, if $K \subseteq L$ with $L$ algebraically closed, and a system of polynomial equations has a root in $L$, then it has a root in a finite extension of $K$.

The point, which I am sure Qiaochu understands, is that he only knows a priori that the representation is defined over $\mathbb{C}$. Once he knows that the representation is definable over an algebraic extension $K'$ of $K$, he can replace $K'$ by its normal closure, note that $\mathrm{Gal}(K', \mathbb{Q}) \to \mathrm{Gal}(K, \mathbb{Q})$ is surjective, lift any element $\sigma$ of $\mathrm{Gal}(K, \mathbb{Q})$ to some $\tilde{\sigma}$ in $\mathrm{Gal}(K', \mathbb{Q})$, and apply $\tilde{\sigma}$ to the entries of his matrices.

Part of the problem is that the representation may honestly not be defined over $K$. For example, the two dimensional representation of the quaternion eight group has character with values in $\mathbb{Q}$, but can't be defined over $\mathbb{Q}$.

Fix $G$ and a representation $V$ of $G$. For $g \in G$, let $\lambda_1(g)$, $\lambda_2(g)$, ..., $\lambda_n(g)$ be the multiset of eigenvalues of $g$ acting on $V$. These are necessarily roots of unity, since $g^N=1$ for some $N$. For any symmetric polynomial $f$, with integer coefficients, define $\chi(f,g) = f(\lambda_1(g), \ldots, \lambda_n(g))$.

Lemma: With notation as above, $g \mapsto \chi(f,g)$ is a virtual character.

Proof: If $f$ is the elementary symmetric function $e_k$, then this is the character of $\bigwedge^k V$. Any symmetric function is a polynomial (with integer coefficients) in the $e_k$'s; take the corresponding tensor product and formal difference of virtual characters.

Any Galois symmetry $\sigma$ of $\mathbb{Q}(\zeta_N)$ is of the form $\zeta_N \mapsto \zeta_N^s$, for $s$ relatively prime to $N$. Consider the power sum symmetric function $p_s := \sum x_i^s$. So $\chi(p_s, \ )$ is the Galois conjugate $\chi^{\sigma}$, and we now know that it is a virtual character.

But $\langle \chi^{\sigma}, \chi^{\sigma} \rangle = \langle \chi, \chi \rangle =1$, because the inner product is built out of polynomial operations and complex conjugation, and complex conjugation is central in the Galois group. So this virtual character must correspond to $\pm W$, for some representation $W$. Since $\chi^{\sigma}(e) = \chi(e) = \dim V$, we conclude that the positive sign is correct.

It just occurred to me that actually writing this out for some specific small values of $s$ makes some nonobvious statements about representation theory. For example, if $G$ has odd order and $V$ is a $G$-irrep, then $\bigwedge^2 V$ has a $G$-equivariant injection into $\mathrm{Sym}^2 V$. Proof: The difference of their characters is the character of $V^{\sigma}$, where $\sigma: \zeta \mapsto \zeta^2$.

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Okay, maybe you can write up the proof that $V$ is realizable over a number field in a shorter way than I can. First, by explicitly projecting from the group algebra, we can realize $(\dim V) V$ over $K$. Second, we can pick a nonzero vector and hit it with every element of $G$. But this seems sort of annoying to write up in detail. Maybe I can skimp... –  Qiaochu Yuan Jun 2 '11 at 22:41
Ah, I see what you're saying in the first paragraph. If I write down a complete set of generators and relations for $G$, I get a finite number of polynomial equations in the matrix entries. I have a solution over $\mathbb{C}$, so by the Nullstellensatz I have a solution over $\overline{\mathbb{Q}}$, and I can take the number field generated by the entries. I don't know how frowned-upon using techniques from another course is, but I think that passes muster. –  Qiaochu Yuan Jun 2 '11 at 22:52
Yes, your second comment is what I am saying. –  David Speyer Jun 2 '11 at 22:55
That said, I have not spent much time in environments where my goal was to present a prepared proof as rapidly as possible. I have taught classes and had oral exams, but in those cases clarity is more important than speed, and I have competed in the Olympiads and Putnams, but there you can't prepare an argument in advance. So my advice might not be the best. –  David Speyer Jun 2 '11 at 23:01
Well, I'm not guaranteed that I'll need to use this fact on the exam. I'd just like to be prepared if it's ever relevant. I think using the Nullstellensatz should be fine. –  Qiaochu Yuan Jun 2 '11 at 23:10
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I think the natural proof (although I think that this is the one you want to avoid) is that $\mathbb Q[G]$ is a semi-simple $\mathbb Q$-algebra, hence when extended to $\overline{\mathbb Q}$ it splits as a product of matrix rings, hence it already so splits when extended over a finite extension $K$ of $\mathbb Q$, hence all representations of $G$ are defined over this $K$. (This $K$ is not the $K$ in the statement of your question, but rather the finite extension of it that you allude to.)

The reason I am writing this out despite your request not to is just to point out that it is not that annoying to prove; in fact it is quite natural. And all the arguments that I know of the type that certains systems of eigenvalues are closed under Galois conjugation (e.g. that if $f$ is a modular form which is a Hecke eigenform with system of eigenvalues $(a_p)$, then for any Galois element $\sigma$, the system of eigenvalues $(\sigma(a_p))$ is also attached to a Hecke eigenform) are proved in the same manner. (Namely, by showing that the natural $\mathbb C$-algebra that governs the situation, whether it be the group algebra or the Hecke algebra, actually has a model over $\mathbb Q$, with the same set of generators (group elements or Hecke operators, as the case may be).)

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I'm with Matt on this one: the argument that Qiaochu seems to want to avoid seems to me (too) to be natural and even rather pretty. –  Pete L. Clark Jun 3 '11 at 0:14
I'm not avoiding it by choice! I just don't feel comfortable assuming too many things that weren't covered in the course for the purposes of this exam. (We have literally not even used the word "semisimple" to refer to a ring.) –  Qiaochu Yuan Jun 3 '11 at 9:32
@Qiaochu: of course I'm not blaming you at all, but a course on representation theory that does not make any use of Wedderburn-Artin theory strikes me as rather artificial indeed. The course on finite groups and representations that I took as an undergraduate definitely covered these things (the course was taught by J.L. Alperin and used the book Groups and Representations by Alperin and Bell: I recommend it). Admittedly you're in Cambridge, not Chicago, so certain concessions are going to have to be made. :) –  Pete L. Clark Jun 3 '11 at 10:50
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It's a bit late, but you could argue like this. By Schur's Lemma, when $\chi$ is an irreducible complex character of the finite group $G$, and $\sigma$ is a representation affording $\chi$, then for each $z \in Z(\mathbb{C}G$, we have $z\sigma = \lambda(z)I$ for some scalar $\lambda(z).$ Taking traces tells us that $\lambda(z) = \frac{\chi(z)}{\chi(1)}.$ Also, $\lambda$ defines an algebra homomorphism from $Z(\mathbb{C}G)$ to $\mathbb{C}.$ Since the dimension of $Z(\mathbb{C}G)$ is $k= k(G)$, the number of conjugacy classes of $G$, and since there are $k$ different irreducible characters of $G$ (I am assuming the orthogonality relations as given), which give (using the orthogonality relations) $k$ distinct algebra homomorphisms from $Z(\mathbb{C}G)$ to $\mathbb{C}$, we see that all such algebra homomorphisms come from irreducible characters.

We use the natural "integral" basis of $Z(\mathbb{C}G)$ (that is, the basis of conjugacy class sums) to show that for $\lambda$ as above, $\lambda^{\sigma}$ is also an algebra homomorphism for each automorphism $\sigma \in {\rm Gal}(\mathbb{Q}[\omega]/\mathbb{Q}$, where $\omega$ is a primitive complex $|G|$-th root of unity. Notice that $\lambda(C) \in \mathbb{Q}[\omega]$ for each class sum $C$. Furthermore, for class sums $C_{r}$ and $C_{s}$, there are integers $a_{rst}$ such that $\lambda(C_r C_s) = \sum_{t=1}^{k}a_{rst}C_t$. That $\lambda$ is an algebra homomorphism is encapsulated precisely by the fact that we have $\lambda(C_r) \lambda(C_s) = \sum_{t=1}^{k} a_{rst}\lambda(C_t)$. Since the $a_{rst}$ are rational integers, we can apply $\sigma$ to this equation to conclude that $C_r \to \lambda^{\sigma}(C_r)$ for each class sum, (and extending by $\mathbb{C}$-linearity to $\mathbb{C}$-combinations of class sums) is an algebra homomorphism from $Z(\mathbb{C}G)$ to $\mathbb{C}$. Associated to this is, as discussed above, is a complex irreducible character $\mu$ of $G$ such that $\frac{\mu(C)}{\mu(1)} = \frac{\chi^{\sigma}(C)}{\chi(1)}$ for each class sum $C$. Hence $\mu(g) = \mu(1)\frac{\chi^{\sigma}(g)}{\chi(1)}$ for each $g \in G.$ Since $\{ \chi(g): g \in G\}$ is $\sigma$-stable, we have $\sum_{g \in G}|\chi^{\sigma}(g)|^{2} = |G|$, so that $\chi(1) = \mu(1)$ and $\chi^{\sigma} = \mu.$

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This is a very nice completely elementary argument! I have to admit that when I taught a representation theory course last semester, I stayed away from the result Qiaochu wanted to prove, precisely because I didn't want to introduce number theoretic machinery (I didn't even assume any Galois theory). But this argument would have fit in perfectly into the general spirit of my course! Noted for next time. –  Alex B. Aug 12 '11 at 3:12
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Well, wait a minute. I guess we can agree that every complex representation $\rho$ of $G$ is defined over $\mathbb{C}$! Let $\sigma$ be an automorphism of the character field $K$. Then $\sigma$ extends to an automorphism of $\mathbb{C}$: for a proof, see for instance $\S 10$ of my field theory notes (the numbering of the results and the sections is not stable as yet, so search for automorphism extension theorem to find it quickly.) The map $g \mapsto \sigma( \rho(g))$ is clearly a representation of $G$ with character $\sigma(\chi)$, where $\chi$ is the character of $\rho$. Aren't we done?

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Is it really easier to prove that every automorphism of a number field extends to one of $\mathbb{C}$ than it is to finish up one of the other suggested arguments? I seemed to remember that this was quite a pain. –  David Speyer Jun 3 '11 at 1:18
Another technique is to notice G has the same number of irreducibles over C as over Q-bar. So all of these are realized over Q-bar, and then I assume the galois extension is more obvious. –  Jack Schmidt Jun 3 '11 at 1:30
@David: This fact is proven in my (work in progress) field theory notes: I included a reference. Whether it's easier than the proof which actually meets the rationality questions head-on depends on the beholder, I think. I guess to me this is a sort of general nonsense piece of background knowledge which manifestly has nothing to do with representation theory, unlike appealing, say, to Schur indices. I can honestly say that if I were examining a student and s/he used this, I wouldn't ask for a proof of it! –  Pete L. Clark Jun 3 '11 at 1:32
I, for one, am such a beholder who finds this approach easier, for the same reason as Pete: I've encountered extensions of field automorphisms about two years before I knew anything about representation theory, so this seems very natural. –  Alon Amit Jun 3 '11 at 1:58
Not to sound like a broken record, but this also seems to rely on material that's not in the Tripos (I don't think we've ever actually talked about transcendence degree, nor have we actually extended automorphisms to algebraic closures). I am working under somewhat artificial constraints here... –  Qiaochu Yuan Jun 3 '11 at 9:39
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This is just a slightly more quantitative version of Matt E's answer. I agree also that this is the obvious and implicit argument.

CG is a direct product of matrix rings of size χ(1). The primitive central idempotents lie in KG, so KG is a direct product of the same number of matrix rings, but over division K-algebras and of size only dχ dividing χ(1). Each division algebra has a splitting field of degree χ(1)/dχ, and so G has a splitting field of degree at most the product of those numbers (called indices of the division algebras and Schur indices of the characters).

Zhen Lin's answer is then the correct answer.

You may also keep in mind that if G is not p-solvable, then Galois conjugates of Brauer characters may or may not also be Brauer characters for a fixed maximal ideal of the algebraic integers. The pth power obviously works, since the Frobenius automorphism works on the underlying representation. The −1st power works by taking dual modules. However, the powers not in this subgroup may fail to be automorphisms of the Brauer table.

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Again, I am pretty sure I can't assume Artin-Wedderburn for the purposes of this exam. –  Qiaochu Yuan Jun 2 '11 at 23:09
The fact that CG is a direct product of matrix rings is called Maschke's theorem and Schur's lemma. The splitting fields are called Schur's lemma. –  Jack Schmidt Jun 2 '11 at 23:16
I am sure this argument makes perfect sense to you, but this was a course on representation theory, not ring theory, and I don't know what theorems you're implicitly citing in any sentence after the first one in the second paragraph. I know that $\mathbb{C}[G]$ is a direct product of matrix rings, and I know that the primitive central idempotents lie in $K[G]$. Why does the next assertion follow? –  Qiaochu Yuan Jun 2 '11 at 23:21
(I call this module theory, not ring theory.) KG is a direct product of matrix rings because it is the endomorphism ring of its regular module which is semisimple (by Maschke) and the endoring of a semisimple module is a direct product of matrix rings over End(M) by Schur. The number stays the same because the PCIs all exist (otherwise you get clumps of non-isomorphic but Galois conjugate simple modules). The splitting fields are just the centralizers of the group inside the matrix ring. Tensor with the endomorphism ring, and then by Schur everything is inner, and so central. –  Jack Schmidt Jun 2 '11 at 23:34
I believe Feit's Character Theory should cover something along these lines in the first 10 or 15 pages. This is just an exercise in the "Hi, we're characters!" chapter (2) of Isaacs's Character Theory; chapter 1 is Wedderburn (but no Artin) theory. Generally one should be familiar with first 6 chapters of Isaacs's book to avoid being too surprised by character theory. The whole book is awesome though, so read on if you can find the time. James-Liebeck has lots of exercises on partial tables too, so makes a good "workbook" to go with it. –  Jack Schmidt Jun 3 '11 at 0:05
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