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sniffnoySo today John told me about this preprint[0] of his and Gene's wherein... well, wherein they do a whole bunch of stuff, but in particular, they give a new proof of the fact that for a finite group G and an irreducible character χ, that χ(1) divides the order of G. More importantly, the proof is entirely elementary, in that unlike the usual proof, it doesn't use the notion of algebraic integers at all. While they do some complicated things in most of the paper, the proof of this particular fact uses nothing but Schur's lemma, some counting arguments, and a divisibility argument.
Unfortunately they do not resolve the old question of Alperin about whether γ, the number of ways of writing an element as a commutator, can be found naturally as the character of some representation[3], but still, proving χ(1) divides |G| with such elementary means is still something I'd consider a big deal.
Now, their paper does a bunch of other stuff as well -- much involving more advanced tools -- but this particular result, which constitutes section 6, depends only on Lemma 3.2, Lemma 3.3, equation (27), and parts of Theorem 4.1 and Corollary 4.2 and the resulting equation (52), all of which can be proven elementarily for finite groups (the bits in section 4, though stated in terms of some topological stuff, don't have to be thought of that way). I'm not going to go over them here -- you can go read them yourself -- but I thought I would put here a few explanatory notes, so you don't have to go deciphering for yourself just what these statements mean in the finite case.
In all of what follows, G will indicate a finite group, g an element of g, w a group word on n letters (i.e., an element of the free group Fn), ρ an irreducible representation of G, χ its character, f a complex-valued function on G. For a vector g=(g1,...,gn)∈Gn, w(g) will indicate w with g1,...,gn plugged in for the variables. g will always indicate a vector with components g1,...,gn.
Lemmas 3.2 and 3.3: These are straightforward to understand -- integrating over a finite group as usual means summing over the group and then dividing by the order, and integrating over Gn means summing and dividing by |G|n. The only thing that's not obvious is vw(χρ). So let me state here what vw means in the finite context: vw(f)=1/|G|n*Σgf(w(g)). If we denote by γ(w;g) the number of ways of representing g as w(g) for g∈Gn, this is also 1/|G|nΣgf(g)γ(w;g).
Equation (27) -- OK, now it's obvious what this means, but it may not be obvious why this one is true. So let me tell you the way you want to prove this one:
1. Write it out as a sum, using the γ(w;g) form above.
2. γ(w,g) is a class function, so expand it out in terms of characters and inner products.
3. Expand out that inner product.
4. You can replace each χ with its conjugate without changing the sum; do that.
5. Notice that you've now got the right-hand side.
Theorem 4.1 and Lemma 4.2: They're thinking of this topologically, so they've written this all in terms of κ, which (if we're considering m independent commutators, with n=2m variables total) which is 2-2m. So where you see κ-1, just think 1-2m, and where you see κ-2, just think -2m. So this just goes exactly as they've done it; just ignore the topology and consider it purely as an identity as representing things as products of commutators[4]. OK. This yields equation 49 (note they seem to be defining δ1 to be |G| at the identity rather than 1 there); again, instead of -κ we can just think 2m-2. Equation 52 is then immediate; ignore the stuff inbetween.
The point is, we have now made a bunch of elementary Schur's lemma, rearranging stuff, and and counting arguments to show this one fact: That the number of writing the identity as a product of m commutators is |G|2m-1Σχχ(1)2-2m. Then section 6 is straightforward to understand; just note that in the final proof, k=m-1, so 2k+1=2m-1 and 2k=2m-2. OK, I guess that was kind of obvious.
In any case assuming you know basic representation theory I've probably explained more than enough so go read it! It's not like anything I said above makes sense outside that context.
-Harry
[0]Yes, it's from February. I only heard about it today.
[3]I'm guessing not everyone will be familiar with this question, so let me state it here. Because for every irreducible character χ, |G|/χ(1) is a whole number, we can form a character γ by including the character χ with coefficient |G|/χ(1). Why do we care? Well, the function γ(g) -- without any knowledge of the fact that it is a character -- can be shown to be equal to the number of ways of writing g as a commutator. (This certainly isn't obvious, but it's not hard. Do it yourself.) That's a pretty natural function on a group, and by above it happens to be a character! OK, well, it's a character, but what is it the character of? Well, obviously, it's the character of the direct sum of |G|/(dim ρ) copies of each irreducible representation ρ! But that's a pretty unhelpful answer; we don't know that's a valid direct sum unless we already know that each |G|/(dim ρ) is an integer. So the question is, can we -- without using the fact that each |G|/χ(1) is an integer -- prove that γ(g), the number of ways of writing g as a commutator, is a character, say by finding some natural representation that it's the character of? Because if you could, then you'd have shown that each |G|/χ(1)=<γ,χ> is an integer, and (hopefully) in a pretty elementary manner.
[4]Actually, the n=1 case they do is exactly the same as the proof of the fact stated in footnote 3. Guess I've given that one away.
Unfortunately they do not resolve the old question of Alperin about whether γ, the number of ways of writing an element as a commutator, can be found naturally as the character of some representation[3], but still, proving χ(1) divides |G| with such elementary means is still something I'd consider a big deal.
Now, their paper does a bunch of other stuff as well -- much involving more advanced tools -- but this particular result, which constitutes section 6, depends only on Lemma 3.2, Lemma 3.3, equation (27), and parts of Theorem 4.1 and Corollary 4.2 and the resulting equation (52), all of which can be proven elementarily for finite groups (the bits in section 4, though stated in terms of some topological stuff, don't have to be thought of that way). I'm not going to go over them here -- you can go read them yourself -- but I thought I would put here a few explanatory notes, so you don't have to go deciphering for yourself just what these statements mean in the finite case.
In all of what follows, G will indicate a finite group, g an element of g, w a group word on n letters (i.e., an element of the free group Fn), ρ an irreducible representation of G, χ its character, f a complex-valued function on G. For a vector g=(g1,...,gn)∈Gn, w(g) will indicate w with g1,...,gn plugged in for the variables. g will always indicate a vector with components g1,...,gn.
Lemmas 3.2 and 3.3: These are straightforward to understand -- integrating over a finite group as usual means summing over the group and then dividing by the order, and integrating over Gn means summing and dividing by |G|n. The only thing that's not obvious is vw(χρ). So let me state here what vw means in the finite context: vw(f)=1/|G|n*Σgf(w(g)). If we denote by γ(w;g) the number of ways of representing g as w(g) for g∈Gn, this is also 1/|G|nΣgf(g)γ(w;g).
Equation (27) -- OK, now it's obvious what this means, but it may not be obvious why this one is true. So let me tell you the way you want to prove this one:
1. Write it out as a sum, using the γ(w;g) form above.
2. γ(w,g) is a class function, so expand it out in terms of characters and inner products.
3. Expand out that inner product.
4. You can replace each χ with its conjugate without changing the sum; do that.
5. Notice that you've now got the right-hand side.
Theorem 4.1 and Lemma 4.2: They're thinking of this topologically, so they've written this all in terms of κ, which (if we're considering m independent commutators, with n=2m variables total) which is 2-2m. So where you see κ-1, just think 1-2m, and where you see κ-2, just think -2m. So this just goes exactly as they've done it; just ignore the topology and consider it purely as an identity as representing things as products of commutators[4]. OK. This yields equation 49 (note they seem to be defining δ1 to be |G| at the identity rather than 1 there); again, instead of -κ we can just think 2m-2. Equation 52 is then immediate; ignore the stuff inbetween.
The point is, we have now made a bunch of elementary Schur's lemma, rearranging stuff, and and counting arguments to show this one fact: That the number of writing the identity as a product of m commutators is |G|2m-1Σχχ(1)2-2m. Then section 6 is straightforward to understand; just note that in the final proof, k=m-1, so 2k+1=2m-1 and 2k=2m-2. OK, I guess that was kind of obvious.
In any case assuming you know basic representation theory I've probably explained more than enough so go read it! It's not like anything I said above makes sense outside that context.
-Harry
[0]Yes, it's from February. I only heard about it today.
[3]I'm guessing not everyone will be familiar with this question, so let me state it here. Because for every irreducible character χ, |G|/χ(1) is a whole number, we can form a character γ by including the character χ with coefficient |G|/χ(1). Why do we care? Well, the function γ(g) -- without any knowledge of the fact that it is a character -- can be shown to be equal to the number of ways of writing g as a commutator. (This certainly isn't obvious, but it's not hard. Do it yourself.) That's a pretty natural function on a group, and by above it happens to be a character! OK, well, it's a character, but what is it the character of? Well, obviously, it's the character of the direct sum of |G|/(dim ρ) copies of each irreducible representation ρ! But that's a pretty unhelpful answer; we don't know that's a valid direct sum unless we already know that each |G|/(dim ρ) is an integer. So the question is, can we -- without using the fact that each |G|/χ(1) is an integer -- prove that γ(g), the number of ways of writing g as a commutator, is a character, say by finding some natural representation that it's the character of? Because if you could, then you'd have shown that each |G|/χ(1)=<γ,χ> is an integer, and (hopefully) in a pretty elementary manner.
[4]Actually, the n=1 case they do is exactly the same as the proof of the fact stated in footnote 3. Guess I've given that one away.
