## July 25, 2007

### Question About Representations of Finite Groups

#### Posted by John Baez Here’s the first of some questions that have been bugging me. Maybe you can help!

I want to know when we can define the representations of a finite group using not the full force of the complex numbers, but only some subfield, like $\mathbb{Q}[\sqrt{5}]$ or $\mathbb{R}$. If I knew the answer to this question, it might be important for the groupoidification program, where we’re trying to replace complex vector spaces by groupoids whenever possible.

Suppose $k \subseteq \mathbb{C}$ is some subfield of the complex numbers. In what follows, ‘representation’ will mean representation on a finite-dimensional complex vector space. Suppose $G$ is some group with a representation $\rho$. Let’s say $\rho$ is defined over $k$ if we can find some basis of our vector space such that the matrices corresponding to the linear transformations $\rho(g), g \in G$ all have entries lying in $k$.

Question 1. Is there a smallest subfield $k \subseteq \mathbb{C}$ such that every representation of every finite group is definable over $k$? If so, what is it?

It’s not hard to see that:

• Every representation of every finite group is definable over $k$ when $k = \overline{\mathbb{Q}}$ is the field of algebraic numbers.
• Not every representation of every finite group is definable over $k$ when $k = \mathbb{R}$. There’s an easy trick to see which ones are.
• Every representation of the symmetric group $S_n$ is definable over $k$ when $k = \mathbb{Q}$.
• Every representation of the cyclic group $\mathbb{Z}/n$ is definable over $k$ when $k = \mathbb{Q}[e^{2\pi i/n}]$ is the cyclotomic field generated by taking $\mathbb{Q}$ and throwing in a primitive $n$th root of unity.

But what I really want to know is this:

Question 2. Is every representation of every finite group definable over $k$ when $k = \mathbb{Q}^{ab}$ is the field generated by taking $\mathbb{Q}$ and throwing in all roots of unity? If not, what’s the simplest counterexample?

Here’s a pathetic shred of evidence that the answer to Question 2 is “yes”:

Theorem. Let $\rho$ be a representation of a finite group. Then for any $g \in G$, $tr(\rho(g)) \in \mathbb{Q}^{ab}.$

If this theorem were false, the answer to Question 2 would be “no”. But, that’s just a pathetic shred of evidence that the answer is “yes”. In fact, if I had to guess, I’d guess the answer is “no”!

This theorem is old. I read in here:

• Charles W. Curtis, Pioneers of Representation Theory: Frobenius, Burnside, Schur and Brauer, History of Mathematics vol. 15, AMS, Providence, Rhode Island, 1999.

that it was proved by some bigshot like Frobenius or Burnside, who then went on to ponder Question 2. So, the answers to both my questions must be known by now. But, I don’t know them!

The proof of the theorem is easy. By group averaging we can find an inner product such that $\rho(g)$ is unitary for all $g$. So, for any $g$ we can find a basis in which $\rho(g)$ is diagonal. Since our group is finite, $g^n = 1$ for some $n$. So, the diagonal entries of $\rho(g)$ are roots of unity. So, $tr(\rho(g)) \in \mathbb{Q}^{ab}$.

Note: this shows that for any one element $g$ we can find a basis for which the entries of $\rho(g)$ lie in $\mathbb{Q}^{ab}$. But, that’s far short of finding a basis that works for all $g$ at once!

Posted at July 25, 2007 1:43 PM UTC

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### Re: Question About Representations of Finite Groups

One way to prove it is as follows: The group ring R of G over Qab is a semi-simple algebra, hence by Wedderburn’s theorem a product of matrix rings over finite dimensional division algebras over Qab. But the only such division algebra is Qab itself, so R must be a product of matrix rings, which implies that all representations of G are defined over Qab.

Unfortunately, I don’t know an elementary proof (it follows from class field theory) for the statement about division algebras…

Posted by: naf on July 25, 2007 3:28 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

Sorry, the above “proof” is incomplete since the division algebras need not be central…

Posted by: naf on July 25, 2007 3:38 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

naf: It seems to me your proof does show the following:

Let L be a field extension of Q.

Every complex representation of G is definable over L if and only if each of the Wedderburn factors of R (=the group ring of G over L) is a central L algebra.

Does this seem right?

Posted by: Charles on July 25, 2007 4:56 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

It’s still ok. For example, use:

• Fong, Paul, “A note on splitting fields of representations of finite groups. “, Illinois J. Math., 7 1963 515-520

but I’m sure guys like Schur and Witt (let alone Brauer) knew this already. I’ll try to find a more appropriate reference…

Posted by: lieven on July 25, 2007 6:35 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

“Every representation of a group of exponent e is defined over Q(\sqrt[e]{1})” was conjectured by Maschke around 1900 and proved in 1945 by Brauer in:

Richard Brauer, “On the representation of a group of order g in the field of the g-th roots of unity”, Am. J. Math. 67 (1945), 461–471.

Posted by: lieven on July 26, 2007 10:27 AM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

Even though your “proof” is incomplete, you seem to be using a fact about $\mathbb{Q}^{ab}$ that I don’t know. I’d like to know it!

Maybe it’s something like this: the only division algebra over $\mathbb{Q}^{ab}$ whose center is just $\mathbb{Q}^{ab}$ is $\mathbb{Q}^{ab}$?

Is that it?

Hmm, maybe I almost do know this, with a little reminder here and there from Wikipedia. I think it goes something like this.

The division algebras over $k$ having $k$ as center form a group, the Brauer group of $k$.

But, the Brauer group of $k$ can be expressed in terms of the 2nd Galois cohomology group of $k$:

$Br(k) \cong H^2(Gal(\overline{k}/k), \overline{k}^*)$

at least when $k$ is an algebraic number field.

This I actually understand. I remember being incredibly proud of myself when I first saw how it worked. Just writing down the above equation made me feel like an expert on algebra. In fact, it still does!

And, maybe for some reason the Galois cohomology of $\mathbb{Q}^{ab}$ is trivial. This part I don’t understand — maybe I need to learn more Galois cohomology and class field theory.

Posted by: John Baez on July 25, 2007 5:10 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

For any p-adic field K i.e. a finite extension of Q_p, the Brauer group is naturally isomorphic to Q/Z. If L is a finite extension of K then the map Br(K) –> Br(L) is multiplication by the degree of L over K. (This is part of local class field theory.)

A theorem of Albert, Brauer, Hasse and Noether says that the Brauer group of a number field injects into the direct sum of the Brauer group of all its completions
(including the archimedean ones). So to kill an element of the Brauer group it suffices to kill it locally, which by the above can be done if the local extension has degree divisible by the order of the given element. The extension of Q_p obtained by adding all roots of unity of order prime to p has Galois group the profinite completion of Z, so we can get extensions of any p-adic field with degree divisible by any natural number by adjoining roots of unity.

The upshot of all this is that for K any number field, any element of Br(K) can be killed by adding roots of unity to K. As a cosequence we see that the Brauer group of any algebraic extension of Q^{ab} is trivial.

Posted by: naf on July 26, 2007 2:14 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

If I knew the answer to this question, it might be important for the groupoidification program.

Can you give us a hint?

Suppose the answer to question 2 is true for some finite group $G$. Then would that help to get something equivalent to the full linear representation theory of $G$ out of combinatorial constructions involving spans of groupoids. How?

Posted by: Urs Schreiber on July 25, 2007 3:30 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

I don’t really know the answer to your question — even though it’s exactly the question I’m interested in.

In Next Week’s Finds, I’ll start explaining how to groupoidify the entire representation theory of the symmetric groups $S_n$. The particular way this trick works implies that all representations of permutation groups are definable over $\mathbb{Q}$.

Conversely, whenever all representations of a finite group are definable over $\mathbb{Q}$, I think I can do this trick.

(I should check this converse.)

But, most finite groups aren’t so nice. Even $\mathbb{Z}/3$ isn’t that nice!

However, so far, I don’t know examples of representations of finite groups that fail to be definable over $\mathbb{Q}^{ab}$.

Since we can think of this field as the rational numbers with ‘phases’ of the form $e^{2 \pi i/ n}$ thrown in, I could hope to handle such representations using something like phased sets — or maybe something even more combinatorial, like the category of cyclic sets.

I don’t know how to do this! So, I can’t really answer your question.

The point is, I could hope to do it.

So, if there are lots of finite group representations that aren’t definable over $\mathbb{Q}^{ab}$, I’ll be a bit disappointed. But if they all are definable over $\mathbb{Q}^{ab}$, I’ll get more motivated to realize my hope!

Posted by: John Baez on July 25, 2007 4:41 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

Thanks.

Before I ask another question, it would be helpful if you could quickly confirm or else correct the following summary of my understanding of the aspects of the Tale of relevance here.

There is the category of set-representations $\rho : \Sigma G \to \mathrm{Set}$ of finite groups. This is considerably less rich than that of linear representations $\rho_{\mathrm{lin}} : \Sigma G \to \mathrm{Vect} \,.$ But we can try to refine it to a 2-category as follows:

we realize that every set-rep $\rho : \Sigma G \to \mathrm{Set}$ is entirely encoded in its action groupoid $S_\rho // G \,.$ The fact that this groupoid comes from a set-rep is witnessed by the fact that there is a faithful functor back to $\Sigma G$ $S_\rho // G \to \Sigma G$ which sends every morphism $s \stackrel{g}{\to} s \cdot g$ in the action groupoid to just the element in $G$ $\bullet \stackrel{g}{\to} \bullet$ which was acting here.

(In fact, I think another way to say this is that we even have a short exact sequence of groupoids $\mathrm{Disc}(S_\rho) \to S_\rho // G \to \Sigma G \,.$

I notice that in the special case of the “regular set-rep”, where $G$ acts on itself by right-multiplication, we have $S_\rho // G = \mathrm{INN}(G) = T \Sigma G$ and the above exact sequence becomes a special case of the exact sequence $\mathrm{Mor}(C) \to T C \to C$ for the tangent category $T C$ of any groupoid $C$. This reproduces the above by setting $C = \Sigma G$.

Not sure if this is relevant, but it occured to me. The regular rep of $G$ on itself looks like it should be important for the Tale.)

Anyway, my understanding of the Tale is that the hope is to refine $[\Sigma G, \mathrm{Set}]$ by first passing from group reps to groupoids over $\Sigma G$, following the above rationale.

This should first of all greatly enlarge the number of objects which play the role of reps: not every groupoid over $\Sigma G$ is an action groupoid of an action of $G$ on some set!

On the other hand, we also enlarge the space of morphisms: instead of just looking at functors between groupoids over $\Sigma G$, we allow (or rather: you do! :-) something like profunctors/anafunctors. More concretely: spans of groupoids over $\Sigma G$.

What was the name for the resulting 2-category again? Soemthing like $\mathrm{Span}_{\Sigma G}(\mathrm{Grpd})$ I guess.

So, I guess there is a monomorphic functor $[\Sigma G,\mathrm{Set}] \hookrightarrow \mathrm{Span}_{\Sigma G}(\mathrm{Grpd})$ and the hope is that as we pass the set-representation theory of $G$ from the left to the right along this functor, all the new objects and morphisms in $\mathrm{Span}_{\Sigma G}(\mathrm{Grpd})$ will somehow conspire to make this equivalent (or somehow “more equivalent”, maybe?) to $[\Sigma G, \mathrm{Vect}]$.

Is that roughly right?

Posted by: Urs Schreiber on July 25, 2007 5:27 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

Urs wrote:

Is that roughly right?

Yes! I’ve been trying to tell the Tale of Groupoidification slowly and gently. So, I didn’t want to hit people over the head yet with general results expressed in precise formal language. Some people have expressed frustration with that! So, I’m glad you were able to read what I’ve said so far and extract the information you just summarized.

Let me try my own hand at it. I haven’t yet said precisely how all these things work, but I’ve tried hard to make it plausible that:

1. From a groupoid $X$ we can obtain a vector space $|X|$ whose basis is the set of isomorphism classes of objects of $X$.
2. If we equip $X$ with a functor to a group $G$, we obtain a representation of $G$ on $|X|$. In this situation we call $X$ a groupoid over $G$.
3. Given a span of groupoids over $G$, meaning a weakly commuting diagram like this:
                       S
/ \
/   \
/     \
v       v
X         Y
\       /
\     /
\   /
v v
G

we get an intertwining operator $|S|: |X| \to |Y|$
4. The processes just sketched give us a functor $F: Span(G) \to Rep(G)$

Here $Span(G)$ is the category with:

• groupoids over $G$ as objects;
• equivalence classes of spans of groupoids over $G$ as morphisms.

(Note that equivalence classes are necessary to make composition of spans of groupoids over $G$ be strictly associative; we are forcing a 3-category to become a mere category here! The ultimate point is precisely not to decategorify like this, but right now I’m doing it to avoid scaring people, since I just want to compare what we have to something more traditional — the ordinary category of representations of $G$.)

Similarly, $Rep(G)$ is the category with:

• representations of $G$ on $\mathbb{Q}$-vector spaces as objects,
• intertwining operators as morphisms

(Note that I’m using vector spaces over the rational numbers here; we might prefer complex vector spaces, but that is the point behind my questions in this blog entry!)

5. And, the goal is to show that $F$ is ‘close to being an equivalence’. It’s not, but I believe if we take $\mathbb{Q}$-linear combinations of morphisms in $Span(G)$, and then split idempotents, we do get a category that’s equivalent to $Rep(G)$.
Posted by: John Baez on July 26, 2007 11:39 AM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

1) From a groupoid $X$ we can obtain a vector space $|X|$ whose basis is the set of isomorphism classes of objects of $X$.

2) If we equip $X$ with a functor to a group $G$, we obtain a representation of $G$ on $|X|$. […]

Hm, is that saying that we get only reps by diagonal matrices this way?

The set of isomorphism classes of $X$ is just the set of connected components of $X$. So $X$ is a disjoint union $X = \oplus_i X_i$ of transitive (= connected) groupoids, one for each such class.

Hence the functor $f: X \to \Sigma G$ comes from a collection of functors $f_i : X_i \to \Sigma G \,.$

So the group acts only within each connected component. But if each of these is sent as a whole to a basis element of $|X|$, it seems the rep of $G$ on $X$ is at best given by diagonal matrices in that basis.

Is that what you have in mind? Or am I missing something? (Probably I am making a dumb mistake somewhere.)

Posted by: Urs Schreiber on July 26, 2007 12:30 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

but I believe if we take $\mathbb{Q}$-linear combinations of morphisms in Span($G$), and then split idempotents, we do get a category that’s equivalent to Rep($G$).

Would it be correct to argue that, since you can get the regular representation $\rho_{reg}$, and since the regular representation is the weighted direct sum of the irreducibles,

(1)$\rho_{reg} = \oplus_i dim(\rho_i) \rho_i,$

therefore (after tensoring the hom-sets with $\mathbb{Q}$) you can get all these idempotents; thus the strategy of splitting the idempotents should indeed recover Rep($G$)? Or does one perhaps need to work over $\mathbb{C}$ for this argument to go through?

Posted by: Bruce Bartlett on July 26, 2007 2:07 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

Would it be correct to argue that, since you can get the regular representation $\rho_{\mathrm{reg}}$, and since the regular representation is the weighted direct sum of the irreducibles

Yeah, that sounds like a good way to think about it, possibly. We talked about that by email.

But I am still lacking intuition for how the black magic here is supposed to operate:

so suppose I have the regular rep of $G$, regarded as the groupoid $G // G$ over $G$ as an object in $\mathrm{Span}(G)$.

Then, what happens? How can I see, in $\mathrm{Span}(G)$, the groupoids corresponding to the linear irreps of $G$ appearing?

And by the way, just to be sure: when we say that we “split idempotents”, I guess this means that for every idempotent in our category i.e. every morphism $a \stackrel{f}{\to} a$ such that $a \stackrel{f}{\to} a \stackrel{f}{\to} a \;= \; a \stackrel{f}{\to} a$ we include the image, i.e. an object and universal morphism $a \to \mathrm{im}(f)$ in our category.

Is that right?

If so, can somebody help me see what the spans of groupoids would look like which are idempotents in $\mathrm{Span}(G)$ as morphisms $G // G \to G // G$ ?

Posted by: Urs Schreiber on July 26, 2007 2:34 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

No, I’m the one who made the dumb mistake. If we have a group $G$ acting on a set $S$ we get both a groupoid over $G$, namely

$S//G \to G,$

and a representation of $G$, namely the representation on the vector space with $S$ as basis. This vector space could be called $|S|$ using my notation above, thinking of a set as a special case of a groupoid.

My dumb mistake amounts to saying we get a representation of $G$ on $|S//G|$, when in fact we get it on $|S|$.

I made this dumb mistake because I was too quickly trying to generalize from the case of ‘a group $G$ acting on a set $S$’ to ‘a groupoid over $G$’, without checking that what I was saying was a correct translation of what I knew in the former special case. A group acting on a set is the same as a groupoid faithfully over $G$.

When I write about this stuff in This Week’s Finds, I’ll get my story straight. For now, let me try replacing my previous comment with something better:

1. Let $G$ be a group. Given a groupoid faithfully over $G$, meaning a groupoid $X$ with a faithful functor $X \to G,$ there is a way to get a representation of $G$ on some vector space we shall call $|X|$.

(Without loss of generality we can assume $X = S//G$, and then this vector space is just the space of formal linear combinations of elements of $S$.)

2. Given a span of groupoids faithfully over $G$, meaning a weakly commuting diagram like this:
                       S
/ \
/   \
/     \
v       v
X         Y
\       /
\     /
\   /
v v
G

with the bottom two arrows faithful functors, we get an intertwining operator $|S|: |X| \to |Y|$
3. The processes just sketched give us a functor $F: Span(G) \to Rep(G)$

Here $Span(G)$ is the category with:

• groupoids faithfully over $G$ as objects;
• equivalence classes of spans of groupoids faithfully over $G$ as morphisms.

(Note that equivalence classes are necessary to make composition of spans of groupoids over $G$ be strictly associative; we are forcing a 3-category to become a mere category here! The ultimate point is precisely not to decategorify like this, but right now I’m doing it to avoid scaring people, since I just want to compare what we have to something more traditional — the ordinary category of representations of $G$.)

Similarly, $Rep(G)$ is the category with:

• representations of $G$ on $\mathbb{Q}$-vector spaces as objects,
• intertwining operators as morphisms

(Note that I’m using vector spaces over the rational numbers here; we might prefer complex vector spaces, but that is the point behind my questions in this blog entry!)

4. And, the goal is to show that $F$ is ‘close to being an equivalence’. It’s not, but I believe if we take $\mathbb{Q}$-linear combinations of morphisms in $Span(G)$, and then split idempotents, we do get a category that’s equivalent to $Rep(G)$.

Of course it would be nice to eliminate some of the ‘faithfulness’ conditions above, but I’m in enough trouble already right now without trying to do that!

Posted by: John Baez on July 26, 2007 1:41 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

Okay, thanks. I should have guessed that that’s what you meant.

Of course it would be nice to eliminate some of the ‘faithfulness’ conditions above,

Just to make sure: the groupoids faithfully over $G$ are precisely the $G$-action groupoids, right?

Posted by: Urs Schreiber on July 26, 2007 2:39 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

Urs wrote:

Just to make sure: the groupoids faithfully over G are precisely the G-action groupoids, right?

Yes, that’s what I claimed back in week249 and more vehemently in week250, where I proposed the following slogan:

A GROUPOID FAITHFULLY OVER $G$
IS THE SAME AS A $G$-SET.

I didn’t prove it, but it shouldn’t be hard to prove yourself, as you long as you remember that in $n$-category theory, ‘the same as’ always means ‘equivalent to’.

In particular, given a groupoid $X$ faithfully over a group $G$:

$X \to G$ you’ll need to replace $X$ with an equivalent groupoid $Y$:

$Y \stackrel{\sim}{\to} X$

and use the obvious composite

$Y \to G$

to get a functor equal to one of the sort you get from an action of $G$ on a set $S$:

$S//G \to G$

Posted by: John Baez on July 26, 2007 3:24 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

Yes, that’s what I claimed back in week249 and more vehemently in week250

Oh, okay. Thanks again. I had forgotten this part of your serialized novel! :-)

I am feeling a certain convergence of concepts here with some of the things I was thinking about. Maybe I am just hallucinating, but this here looks suggestive:

as I have just argued #, if you hand me a trivial (for simplicity) $G$-bundle with connection $\mathrm{tra} : P_1(X) \to \Sigma G$ there is a nice canonical way to take its “differential” and obtain a functor $d \mathrm{tra} : P_1(X) \to (G // G) \,.$ This happens always to respect the faithful morphism $G // G \to \Sigma G$ in that $\array{ (G // G) && \stackrel{d\mathrm{tra}(\gamma)}{\to} && (G // G) \\ & \searrow && \swarrow \\ && \Sigma G }$ commutes for all morphisms $\gamma$.

That’s a very obvious statement, not supposed to be particularly deep or anything, but it vaguely looks suggestive in light of the Tale:

If what we talked about above turns out to work out more or less as expected, it seems that a rep of a group $G$ could be thought of as a projector $\rho : (G // G) \to (G // G) \,.$ So to get the associated bundle to my principal bundle above we’d form something like $\array{ (G // G) & \stackrel{f(\gamma))}{\to} & (G // G) \\ \downarrow^\rho && \downarrow^\rho \\ (G // G) & \stackrel{d\mathrm{tra}(\gamma)}{\to} & (G // G) }$ I suppose, and thus obtain a “vector”-transport $f$ in a purely combinatorial way which happens to have the remarkable property that it lives in one categorical dimension higher than what one might have naively expected. Which is good.

Posted by: Urs Schreiber on July 26, 2007 9:44 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

Here is a simple observation related to what we were talking about. Maybe it’s relevant.

So if we indeed allow ourselves to think of $G // G \to \Sigma G$ as something playing the role of, in particular, a vector space, then we would want to know what it’s “elements” are.

So I guess it would be interesting to think of the spans $\{\mathrm{pt}\} \to G // G \,.$ Of course these are nothing but groupoids $S$ together with a morphism $S \to G // G \,.$ Now, this seems to be pretty close to the concept of a $G$-phased set. It’s a $G$-phase groupoid, maybe, rather. But suppose for a second that $S$ is just a set (a discrete groupoid). Then functors $S \to G // G$ are in bijection with $G$-colorings of all elements in $S$.

I am just mentioning this because it seemds to nicely connect the idea that in QM we want to to be thinking of complex numbers as ($U(1)$-)phased sets. On the other hand, we know that we should be thinking of these complex numbers as elements of a 1-dimensional complex vector space, really. So maybe that’s exactly what we see coming out automatically here out of the Tale.

Posted by: Urs Schreiber on July 30, 2007 12:24 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

I really enjoy the Tale of Groupoidification… not the least reason being that many of the structures closely parallel those which occur in the study of ‘2-representations’, which is the business I am involved in. Thus I can learn tricks and techniques from John.

I am very interested in the resulting chain of inclusions that the Tale has set up. In the notation that John uses above, we have faithful (not full) functors

(1)$G-sets \hookrightarrow Span(G) \hookrightarrow Rep(G).$

I’m going to cheat a bit here and work over $\mathbb{C}$, for reasons which will become clear later. As I understand it, the first theorem of the Tale is that if we write

(2)$Span(G)_\mathbb{C}$

for the category whose morphisms are $\mathbb{C}$-linear combinations of those in $Span(G)$, then the second inclusion above, $Span(G) \hookrightarrow Rep(G)$, becomes fully faithful.

I am interested to find out more about precisely how the images of these hom-sets under the functors above live inside each other.

For instance, one fact which I often end up bashing my head against, is the fact that two permutation representations $\mathbb{C} [X // G]$ and $\mathbb{C} [Y // G]$ can be isomorphic as representations without $X$ and $Y$ being isomorphic as $G$-sets. I learnt about this from this paper.

I would love to know if the Tale can enlighten my understanding of this phenomenon.

Posted by: Bruce Bartlett on July 30, 2007 4:15 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

My recollection was that the result you’re looking for follows from Brauer’s theorem that every virtual representation is a Z-linear sum of characters of irreducible representations of p-elementary subgroups. But I can’t quite get the details to work out.

Posted by: Noah Snyder on July 25, 2007 6:21 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

Oh yeah, here’s how the details work.

By Brauer’s theorem every representation is isomorphic to a subrepresentation of a direct sum of representations induced from representations of p-elementary subgroups. Hence it is enough to prove that any irrep of a p-elementary subgroup is defined over a cyclotomic field.

The definition of a p-elementary subgroup is that it is the product of a p-group and a cyclic group. So it is enough to prove the result for p-groups and for cyclic groups. For cyclic groups it is clear, as you pointed out in your post. For p-groups, since any p-group is nilpotent (and thus supersolvable) any irrep is induced from a 1-dimensional representation of a subgroup (see Serre’s book, for example), hence it is defined over the appropriate cyclotomic field.

I think that works, I may put up a post over at the secret blogging seminar giving more details about this, becuase it’s relevant to another question I’ve been thinking about.

Posted by: Noah Snyder on July 25, 2007 6:29 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

Noah, you seem to have cheated pretty egregiously above. Just because you’re a subrepresentation of a representation defined over a field doesn’t mean that you’re defined over that field. I mean, the regular representation is defined over every field, right?

Posted by: Ben Webster on July 25, 2007 8:04 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

My understanding was that if we take e to be the exponent of the group - that is the smallest positive integer such that the e-th power of every element is the identity - then every representation is defined over the cyclotomic field containing the e-th roots of unity. This field may be larger than necessary as the example of the symmetric groups shows.

Some other (easy) observations are first that the field you want must contain the entries of the character table. Second any semisimple finite dimensional algebra over a perfect field has a splitting field. That is a finite field extension which makes the algebra a direct sum of matrix algebras (sometimes called a split algebra).

Posted by: Bruce Westbury on July 25, 2007 6:47 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

Bruce wrote:

My understanding was that if we take $e$ to be the exponent of the group - that is the smallest positive integer such that the $e$-th power of every element is the identity - then every representation is defined over the cyclotomic field containing the $e$-th roots of unity.

That would be great! It would imply the answer to Question 1 is “$\mathbb{Q}^{ab}$” and the answer to Question 2 is “yes”.

Does anyone know a reference to a proof?

Posted by: John Baez on July 25, 2007 8:40 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

i think i did a couple of hours ago…

Posted by: lieven on July 25, 2007 8:59 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

Bruce is absolutely correct; it’s an old theorem of Brauer’s.

It seems to be in a fair number of books; I found it in “Representations and Characters of Finite Groups” by M. Collins.

The proof is actually not too hard, once you know the Brauer induction theorem, and pretty close to what Noah said, though he lied about an important detail.

Let K be $\mathbb{Q}$ with a e-th primitive root of unity adjoined (where e is the exponent of G) By the Brauer induction theorem we have an identity of characters $\varphi=\psi_1-\psi_2$ where $\varphi$ is the character of the representation who we want to show is defined over K, and $\psi_1$ and $\psi_2$ are induced from elementary subgroups. As Noah said, it’s well-known that elementary subgroups of G are split over K (they’re monomial). Thus, we have representations V_1 and V_2 defined over K with characters $\psi_1$ and $\psi_2$. Better yet, since $\psi_1-\psi_2$ is the character of a representation, we can pick an injective map $V_2\to V_1$. The cokernel of this map has character $\varphi$ and is obviously defined over K. Q.E.D.

This proof makes me a little unhappy, because I feel like there should be a slicker one, but I can only spend so much of my day trying to reprove classic theorems.

Posted by: Ben Webster on July 25, 2007 10:04 PM | Permalink | Reply to this

### Re: Question About Representations of Finite Groups

I just thought I’d add the following false theorem, which I have fallen for many times:

False Theorem: Given any finite group G and any character $\chi$, there is a smallest field $k \subseteq \mathbb{C}$ such that there is a representation of G over k with character $\chi$.

Counterexample: Take the Quaternion 8-group and consider the irreducible two dimensional representation. This is defined over k if and only if the equation $a^2+b^2=-1$ has a root in k. In particular, this representation can be defined over both $\mathbb{Q}[i]$ and over $\mathbb{Q}[\sqrt{-2}]$.

Posted by: David Speyer on August 8, 2007 8:45 PM | Permalink | Reply to this
Read the post The Canonical 1-Particle, Part II
Weblog: The n-Category Café
Excerpt: More on the canonical quantization of the charged n-particle for the case of a 1-particle propagating on a lattice.
Tracked: August 15, 2007 2:45 PM

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