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Stream: learning: questions

Topic: Virtual representations and groupoidification


view this post on Zulip David Corfield (Feb 28 2025 at 11:03):

I was reminded on the discussion of @Matteo Capucci (he/him) on the positive bias of that construction that relates permutation representations to linear representations. Consider a finite group, GG. Then comparing G-SetG\text{-}Set with G-RepG\text{-}Rep, the linear representations of the latter are much richer than GG merely acting on sets.

However, if we're allowed virtual permutations representations, formal differences between GG-sets, then things become much more comparable. The rig functor from G-SetG\text{-}Set to G-RepG\text{-}Rep which derives from forming the free vector space on a set, extends to a ring homomorphism from the Burnside ring (Grothendieck ring of G-SetG\text{-}Set with addition and multiplication) to the representation ring. And one has to root about to find a GG for which this map isn't surjective. One reasonably low order case is provided by the direct product of the cyclic group of order 3 with the quaternion group of order 8.

Now anyone whose mathematical education was shaped by @John Baez around the first decade of this millennium will recall a different approach to this problem, namely, groupoidification (which includes an amusing imagined dialogue between William and Georg, standing for Burnside and Frobenius). A pithy summary of what's at stake in the use of spans of groupoids over GG occurs in this comment by Urs Schreiber.

So now I'm wondering what might be said about any relationship between these two approaches.

view this post on Zulip David Corfield (Feb 28 2025 at 11:28):

As evidence that the virtual approach is not just a trick and comes with a good mathematical pedigree:

The Burnside to representation ring comparison functor is secretly a comparison of the equivariant stable cohomotopy of the point with the equivariant K-theory of the point, ultimately deriving from maps of EE_{\infty}-ring spectra: S=KF1KCKU\mathbb{S} = K\mathbb{F}_1 \to K \mathbb{C} \to KU (as explained here).

view this post on Zulip John Baez (Feb 28 2025 at 18:23):

I'll have to think about how this is connected to groupoidification. Where can I read about the following example?

And one has to root about to find a GG for which this map isn't surjective. One reasonably low order case is provided by the direct product of the cyclic group of order 3 with the quaternion group of order 8.

One thing I feel like mentioning is that when you talk about Rep(G)\mathsf{Rep}(G) and representation rings, you get to specify which field you're working over. People heavily focused on physics would tend to use C\mathbb{C}, but even if you're committed to characteristic zero it's a lot of fun to look at how you can chop a permutation representation into more irreducible representations when you make the field larger. The symmetric groups SnS_n are nice in that this effect doesn't happen: all irreducible representations are definable already over Q\mathbb{Q}. But for most groups this effect occurs, e.g. the left action of Z/3\mathbb{Z}/3 on itself gives a 3-dimensional representation that's irreducible over Q\mathbb{Q}, but breaks in two as soon as your field contains exp(2πi/3)\exp(2 \pi i /3).

view this post on Zulip John Baez (Feb 28 2025 at 18:26):

This is all somewhat orthogonal to your point, but it means I think of the passage from actions of groups on sets to actions of groups on vector spaces as a bit more of a 'continuum' than a yes-no thing (where I'm using 'continuum' in a very non-mathematical sense... I'd say 'spectrum' but that too is ambiguous). The bigger your field, the more power you have to grind up group actions into smaller 'atoms'. Once you hit the algebraic closure of Q\mathbb{Q} you're done; going further to C\mathbb{C} or beyond doesn't help.

view this post on Zulip John Baez (Feb 28 2025 at 18:33):

This stuff was studied intensively by Brauer, and he showed you're done already when your field contains all roots of unity!

view this post on Zulip David Corfield (Feb 28 2025 at 19:14):

John Baez said:

Where can I read about the following example?

It's mentioned in Prop. 3.3 here. It seems the example is discussed by Serre on p. 77 of Linear Representations of Finite Groups.

view this post on Zulip David Corfield (Feb 28 2025 at 19:16):

Prop 3.3, which gives examples of groups where the comparison map is surjective, is working over Q\mathbb{Q}.

view this post on Zulip John Baez (Feb 28 2025 at 20:31):

Thanks! I seem to recall a lot of earlier ponderings about whether that comparison map from the Burnside ring to the representation ring was surjective, so it's nice to finally see a counterexample.

I feel in my bones that it's surjective for all Coxeter groups, but I don't know if that's true. That would be a nice addition to the list, given how large Coxeter groups loop in the whole F1\mathbb{F}_1 business.

view this post on Zulip John Baez (Feb 28 2025 at 22:52):

Okay, I checked out Serre's counterexample. I would like to understand it in detail. It's the second exercise here, but it refers back to the previous one:

Serre: Linear representations of finite groups

view this post on Zulip John Baez (Feb 28 2025 at 22:55):

I'm actually at the stage of wanting to learn everything in this book - earlier it would have been impossible for me to be so interested in representation theory over fields other than C\mathbb{C}, but after years of dabbling in algebra it now seems interesting... but still intimidating.

[EDIT: actually, looking at the book some more, it looks quite enticing.]

view this post on Zulip John Baez (Feb 28 2025 at 23:02):

So, we're working over Q\mathbb{Q}. Any finite group GG has a group algebra Q[G]\mathbb{Q}[G] that is semisimple, and thus a product of finitely many finite-dimensional simple algebras over Q\mathbb{Q}. (I'm going to stop saying "finite-dimensional" now - everything is finite-dimensional.)

Each simple algebra over GG is isomorphic to a matrix algebra Mn(D)M_n(D) where DD is a division algebra over Q\mathbb{Q}. Unlike over R\mathbb{R}, there are tons of division algebras over Q\mathbb{Q}. These were classified by our friends Brauer, Noether, Hasse and Albert.

Serre assumes we know how each of the simple algebras that are factors of Q[G]\mathbb{Q}[G] corresponds to an irreducible representation of GG.

view this post on Zulip John Baez (Feb 28 2025 at 23:19):

I think it goes like this: GG has a representation on Q[G]\mathbb{Q}[G] by left multiplication: this is called the regular representation. Every irreducible representation ρ\rho of GG appears as a summand of the regular representation, but it may appear repeatedly, say nn times. In fact it appears with a multiplicity equal to its dimension, so nn is the dimension of ρ\rho.

There's a canonical way to pick out a subrepresentation ρ~\tilde{\rho} of Q[G]\mathbb{Q}[G] that contains all these copies of ρ\rho. This subrepresentation is actually a subalgebra of Q[G]\mathbb{Q}[G]. And it's a simple algebra! And every simple algebra that appears as a factor of the semisimple algebra Q[G]\mathbb{Q}[G] is of this sort!

view this post on Zulip John Baez (Feb 28 2025 at 23:21):

Since ρ~\tilde{\rho} is a simple algebra it's isomorphic to a matrix algebra over a division algebra. But we can say more: it's isomorphic to Mn(D)M_n(D) where nn is the dimension of ρ\rho and DD is the algebra of endomorphisms of the representation ρ\rho.

(The algebra of endomorphisms of an irreducible representation is a division algebra DD, by Schur's lemma. Since ρ~\tilde{\rho} is the sum of nn copies of ρ\rho, its algebra of endomorpisms is Mn(D)M_n(D).)

view this post on Zulip John Baez (Feb 28 2025 at 23:28):

Okay, so now to Serre's example. He takes the quaternion group Q8Q_8 and embeds it in the quaternions in the obvious way, but he uses the quaternions defined over the rationals, which he calls HQ\mathbb{H}_{\mathbb{Q}} - so it's a 4-dimensional rational vector space. When I say 'quaternions' I'll mean these rational quaternions.

He also embeds Z/3\mathbb{Z}/3 in the quaternions in an unspecified way. If we were working over the reals it would be easy for me to see that all such embeddings are related by automorphisms of the quaternions: every nn th root of unity in the quaternions sits inside some copy of the complex numbers in the quaternions, and all such copies are related by automorphisms of the quaternions.

But we're working over the rationals! To embed Z/3\mathbb{Z}/3 in the rational quaternions we need to find a rational quaternion that's a cube root of unity! Are all these related by automorphisms of the rational quaternions? Since Serre blithely tells us to pick one, I bet they must be. But let me think about how we do this. Maybe someone else out there would like to do this too.

view this post on Zulip John Baez (Feb 28 2025 at 23:32):

Puzzle. Find a nontrivial cube root of 1 in the rational quaternions HQ\mathbb{H}_{\mathbb{Q}}.

Answer.

view this post on Zulip John Baez (Feb 28 2025 at 23:38):

Having picked a cube root of 1 in the rational quaternions - call it ω\omega - we get Z/3\mathbb{Z}/3 as a multiplicative subgroup of the quaternions. Serre tells us to let this Z/3\mathbb{Z}/3 act by left multiplication and Q8Q_8 by right multiplication on the rational quaternions. This gives a representation of Z/3×Q8\mathbb{Z}/3 \times Q_8 on HQ\mathbb{H}_\mathbb{Q}.

view this post on Zulip John Baez (Feb 28 2025 at 23:50):

He claims this representation is irreducible. I believe that already the representation of Q8Q_8 on HQ\mathbb{H}_\mathbb{Q} is irreducible. If it split as the sum of a 1-dimensional rep and a 3-dimensional rep, we'd have a nonzero quaternion qq that's fixed by right multiplication by i,j,i, j, and kk - impossible, because qi,qjqi, qj and qkqk are always linearly independent! If it split as the sum of a 2-dimensional rep and a 2-dimensional rep, we'd have a 2d subspace of rational quaternions that's preserved by right multiplication by i,ji, j and kk - impossible, for the same reason!

view this post on Zulip John Baez (Feb 28 2025 at 23:54):

So okay: HQ\mathbb{H}_{\mathbb{Q}} is an irreducible representation of G=Z/3×Q8G = \mathbb{Z}/3 \times Q_8.

By stuff I said earlier, it must correspond to some simple subalgebra of Q[G]\mathbb{Q}[G]. Serre claims this is M2(K)M_2(K) where KK is the extension of the rationals by a cube root of unity, e.g. K=Q[exp(2πi/3)]K = \mathbb{Q}[\exp(2\pi i /3)].

view this post on Zulip John Baez (Feb 28 2025 at 23:56):

For this to be true, the irreducible representation HQ\mathbb{H}_\mathbb{Q} must show up twice in the regular representation of GG on Q[G]\mathbb{Q}[G], and the endomorphisms of each copy must be KK.

view this post on Zulip John Baez (Feb 28 2025 at 23:59):

It would take me some time and/or a burst of cleverness to show this. Clearly the field KK is related somehow to the cube root of unity ω\omega that we picked in HQ\mathbb{H}_{\mathbb{Q}}; there is probably some sense in which KK "is" Q[ω]\mathbb{Q}[\omega].

view this post on Zulip John Baez (Mar 01 2025 at 00:02):

Let me just leave it as a puzzle to myself (and anyone who wants to help out):

Puzzle. If G=Z/3×Q8G = \mathbb{Z}/3 \times Q_8, show that the above irreducible representation of GG on HQ\mathbb{H}_Q shows up with multiplicity two in the regular representation of GG.

(Now I'm confused, because I said each irrep appears in the regular representation with a multiplicity equal to its dimension, but the dimension of HQ\mathbb{H}_\mathbb{Q} is 4. I may be making mistakes because I'm used to representation theory over C\mathbb{C}. While HQ\mathbb{H}_\mathbb{Q} has dimension 4 over Q\mathbb{Q}, H\mathbb{H} has dimension 2 over C\mathbb{C}, and maybe that's the relevant number!)

view this post on Zulip John Baez (Mar 01 2025 at 00:27):

I thought a bit about the puzzle. Notice that

Q[G]Q[Z/3]Q[Q8]\mathbb{Q}[G] \cong \mathbb{Q}[\mathbb{Z/3}] \otimes \mathbb{Q}[Q_8]

view this post on Zulip John Baez (Mar 01 2025 at 00:36):

view this post on Zulip John Baez (Mar 01 2025 at 00:39):

All this should eventually let me see why QH\mathbb{Q}_{\mathbb{H}} shows up with mutiplicity two as a subrepresentation of Q[G]\mathbb{Q}[G].

view this post on Zulip David Corfield (Mar 01 2025 at 08:03):

Excellent work being done overnight.

I wonder if eventually there's a story to emerge of the general conditions which prevent subjectivity.

view this post on Zulip John Baez (Mar 01 2025 at 17:53):

I've been flailing around and seem to have the outlines of proof that since the map from the Burnside ring to the representation ring is surjective for both Z/3\mathbb{Z}/3 and Q8Q_8, this must also be true for G=Z/3×Q8G = \mathbb{Z}/3 \times Q_8 (using the fact that Z/3\mathbb{Z}/3 is a 3-group - every element has order a power of 3 while Q8\mathbb{Q}_8 is a 2-group).

view this post on Zulip John Baez (Mar 01 2025 at 18:07):

This contradicts what Serre says, so there must be a flaw, and perhaps the flaw is here:

I feel, but have not proved, that every subgroup HZ/3×Q8H \subseteq \mathbb{Z}/3 \times Q_8 is of the form H1×H2H_1 \times H_2 where H1H_1 is a subgroup of Z/3\mathbb{Z}/3 and H2H_2 is a subgroup of Q8Q_8. This isn't true for all pairs of groups: consider the product of Z/2\mathbb{Z}/2 and Z/2\mathbb{Z}/2. But I feel it's true when you take the product of a p-group and a q-group for distinct primes p,q.

Then, I feel that when this condition holds - every subgroup of a product G1×G2G_1 \times G_2 is a product of a subgroup H1G1H_1 \subseteq G_1 and a subgroup H2G2H_2 \subseteq G_2 - we have

B(G1×G2)B(G1)B(G2) B(G_1 \times G_2) \cong B(G_1) \otimes B(G_2)

where BB means 'Burnside ring'.

I have an argument that representation rings of finite groups always obey this law:

R(G1×G2)R(G1)R(G2) R(G_1 \times G_2) \cong R(G_1) \otimes R(G_2)

Thus, if we have

B(G1)R(G1) B(G_1) \cong R(G_1) and B(G2)R(G2)B(G_2) \cong R(G_2)

we get

B(G1×G2)R(G1×G2) B(G_1 \times G_2) \cong R(G_1 \times G_2)

But this would mean Serre's counterexample

B(Z/3×Q8)R(Z/3×Q8) B(\mathbb{Z}/3 \times Q_8) \ncong R(\mathbb{Z}/3 \times Q_8)

is only possible if

B(Z/3)R(Z/3) B(\mathbb{Z}/3) \ncong R(\mathbb{Z}/3) or B(Q8)R(Q8) B(Q_8) \ncong R(Q_8)

But if that were true, Serre would have used one of these simpler counterexamples! (Proof by psychology.)

So we get a contradiction.

view this post on Zulip John Baez (Mar 01 2025 at 18:19):

Conclusion: I should examine the various things I "feel" are true, and look for one to be false. I'll probably cheat and read Serre's book to see if he talks about representation rings or Burnside rings of products of groups.

view this post on Zulip John Baez (Mar 01 2025 at 18:21):

Meanwhile I've almost worked out the representation ring and Burnside ring of Z/3\mathbb{Z}/3 and Q8Q_8, which was a lot of fun, but there's not much point in reporting on that until I straighten out my confusion.

view this post on Zulip John Baez (Mar 01 2025 at 18:31):

Well, I can't resist saying a little. I 'cheated' and used Groupprops to read about subgroups of the quaternion group Q8Q_8, and was delighted to learn that all 6 of its subgroups are normal! Three of them H1,H2,H3H_1, H_2, H_3 have 4 elements, so they not only give an action of Q8Q_8 on a 2-element set Q8/HiQ_8/H_i but also a 1-dimensional rational representation Q8Q8/HZ/2QQ_8 \to Q_8/H \cong \mathbb{Z}/2 \hookrightarrow \mathbb{Q}^\ast . The fourth is its center {±1}\{\pm 1\}, and the fifth and sixth are the whole group Q8Q_8 and the trivial group.

I haven't studied the quaternion group much, but given how much I've studied the quaternions all these facts are instant delights, like eating salt peanuts.

view this post on Zulip John Baez (Mar 01 2025 at 19:11):

I feel, but have not proved, that every subgroup HZ/3×Q8H \subseteq \mathbb{Z}/3 \times Q_8 is of the form H1×H2H_1 \times H_2 where H1H_1 is a subgroup of Z/3\mathbb{Z}/3 and H2H_2 is a subgroup of Q8Q_8. This isn't true for all pairs of groups: consider the product of Z/2\mathbb{Z}/2 and Z/2\mathbb{Z}/2. But I feel it's true when you take the product of a p-group and a q-group for distinct primes p,q.

Okay, this feeling seems to be correct: someone on Math Stackexchange says all subgroups of a product of groups are products of subgroups iff the two groups have no non-trivial common subquotient. This will be true when one is a p-group and the other is a q-group for distinct primes p and q, for simple numerical reasons.

view this post on Zulip John Baez (Mar 01 2025 at 23:21):

David Corfield said:

I wonder if eventually there's a story to emerge of the general conditions which prevent surjectivity.

There must be a beautiful story here, but it could already exist without me knowing. I'm reading around a bit now.

view this post on Zulip John Baez (Mar 01 2025 at 23:24):

I haven't gone ahead and used Serre's further hints on how to solve his exercise, and soon I should. But I see one thing: even if we have R(G)B(G)R(G) \cong B(G) for G=Z/3×Q8G = \mathbb{Z}/3 \times Q_8 as my perhaps flawed argument shows, that doesn't imply the canonical map from B(G)B(G) to R(G)R(G) is an isomorphism! It could be non-injective (hence non-surjective).

view this post on Zulip John Baez (Mar 01 2025 at 23:25):

And I feel like boring everyone and talking a bit about the representation ring and the Burnside ring of Z/3\mathbb{Z}/3 and Q8Q_8, because it's good to have some concrete knowledge of a situation before theorizing, and I want to write this stuff down before I forget it.

view this post on Zulip John Baez (Mar 01 2025 at 23:56):

For Z/3\mathbb{Z}/3:

  1. We have two irreducible representations over Q\mathbb{Q}: the 1d trivial representation R1R_1 and a 2d representation R2R_2 where the generator of Z/3\mathbb{Z}/3 turns the plane around a 1/3 turn. We have to choose a basis intelligently to define the latter rep over the rationals! But we can do it.

Here we see the irreps are not occuring in the regular rep with a multiplicity equal to their dimension - apparently that's only true over an algebraically closed field. Each of these reps occurs with multiplicity one in the regular rep.

  1. We have two transitive actions on finite sets: the trivial action A1A_1 and the left action of Z/3\mathbb{Z}/3 on itself, say A3A_3.

So, the map B(Z/3)R(Z/3)B(\mathbb{Z}/3) \to R(\mathbb{Z}/3) is an isomorphism, doing this:

A1R1A_1 \to R_1

A3R1R2A_3 \to R_1 \oplus R_2.

(The map from the Burnside ring to the representation ring is apparently always an isomorphism for cyclic groups.)

view this post on Zulip John Baez (Mar 02 2025 at 00:44):

For Q8Q_8:

  1. We seem to have 5 irreducible representation of Q8Q_8 over Q\mathbb{Q}. There's the 1d trivial representation R1R_1. There are three nontrivial 1d representations, say Ri,Rj,RkR_i, R_j, R_k. In RiR_i, for example, ±iQ8\pm i \in Q_8 acts as the identity while ±j\pm j and ±k\pm k act as 1-1. The other two are analogous. There's also a 4d representation, say R4R_4, where we think of the quaternion group as sitting inside the quaternions in the obvious way, and let them act by left multiplication.

Note that the dimensions of these irreps add up to 1+1+1+1+4 = 8, making it plausible that the regular rep breaks up as a sum of these irreps, each appearing with multiplicity one.

  1. We have 6 transitive actions of Q8Q_8 on finite sets, corresponding to the 5 subgroups of Q8Q_8, which happen to all be normal. There's the trivial action, say A1A_1. There are three actions on 2-element sets, say Ai,Aj,AkA_i, A_j, A_k, coming from three 4-element normal subgroups. For example AiA_i is the action on Q8Q_8 on Q8/{1,i,1,i}Z/2Q_8/\{1, i , -1, -i \} \cong \mathbb{Z}/2. The other two are analogous. There's an action A4A_4 of Q8Q_8 on a 4-element set: in this Q8Q_8 acts on Q8Q_8 modulo its center: Q8/{±1}Q_8/\{\pm 1\} is the Klein 4-group. And there's the action A8A_8 of Q8Q_8 on an 8-element set, namely itself.

It looks like the map from the Burnside ring to the representation ring does this:

A1R1A_1 \mapsto R_1

AiRiR1,AjRjR1,AkRkR1A_i \mapsto R_i \oplus R_1 , \quad A_j \mapsto R_j \oplus R_1, \quad A_k \mapsto R_k \oplus R_1

A4RiRjRkR1A_4 \mapsto R_i \oplus R_j \oplus R_k \oplus R_1 (just guessing)

A8R4R4A_8 \mapsto R_4 \oplus R_4 (just guessing)

view this post on Zulip John Baez (Mar 02 2025 at 00:55):

If so, this seems to be a surjection but not an isomorphism, since it's mapping a 6-dimensional ring to a 5-dimensional one. Apparently the map from the Burnside ring to the representation ring is already surjective for p-groups (such as Z/3\mathbb{Z}/3 and Q8Q_8). The nLab says:

The proof of surjectivity for pp-primary groups is due to Segal 72. (As Segal remarks on his first page, it may also be deduced from Feit 67 (14.3). See also Ritter 72.) The proof is recalled as tom Dieck 79, Theorem 4.4.1.

view this post on Zulip John Baez (Mar 02 2025 at 00:59):

So now I can clearly see a flaw in my sketchy argument: I was arguing that if the Burnside ring and representation ring were isomorphic for a pp-group G1G_1 and a qq-group G2G_2 where pp and qq were distinct primes, then they'd also be isomorphic for G1×G2G_1 \times G_2. But we're already seeing they're not isomorphic for Q8Q_8! :face_with_spiral_eyes: So I was completely missing the point. The point is that the maps from Burnside ring to representation rings

B(Gi)R(Gi) B(G_i) \to R(G_i)

are surjective for two groups G1G_1 and G2G_2, yet apparently not for their product G1×G2G_1 \times G_2.

view this post on Zulip John Baez (Mar 02 2025 at 01:39):

I should now follow Serre's hints, but I found it very fun and enlightening to work out the map from Burnside to representation ring for Z/3\mathbb{Z}/3 and Q8Q_8. This is one of those things that's more fun to do than to read, I imagine, but the point is: in both these cases the representation ring feels similar to the Burnside ring in a bunch of tantalizing ways that are hard to formalize. I imagine that for a 'typical' group the relationship is less strong.

view this post on Zulip Jorge Soto-Andrade (Mar 02 2025 at 01:39):

I would like to mention that this old arXiv preprint of us https://drive.google.com/file/d/19bXhejuA75lzVxGqocwvmp8kb2d7HcZm/view?usp=sharing
addresses related questions and provides - it seems - some counterexamples relevant for the above discussion. We are interested there in the geometric construction of Gelfand Models for finite groups GG (i.e. (complex) representations which split as the multiplicity-free sum of all irreducible representations of GG). Moreover, you have a general way to map the dual of an irreducible representation (V,ρ) (V, \rho) of a finite group GG to HomG(V,L2(G)) Hom_G ( V, L^2(G) ) , where
L2(G) L^2(G) denotes the right regular representation of GG, to wit :
send ωV\omega \in V^* to the map which sends vVv \in V to
the function gω(ρg(v)) g \mapsto \omega(\rho_g(v)) . This affords an isomorphism between V V^* and HomG(V,L2(G)) Hom_G ( V, L^2(G) ) . Over the complex field this proves then that the multiplicity of (V,ρ)(V, \rho) in the regular representation equals its dimension, and it should throw some light on the obstruction to this equality in the case of a non algebraically closed field.

view this post on Zulip John Baez (Mar 02 2025 at 01:42):

I'd like to see it!

view this post on Zulip Morgan Rogers (he/him) (Mar 02 2025 at 11:07):

I don't have a clue from this discussion how someone came up with the particular counterexample of Q8×Z/3Q_8 \times \mathbb{Z}/3. But since Q8Q_8 appears to be acted on by Z/3\mathbb{Z}/3 by rotating the generators, I'm curious if the semi-direct product could also be a counterexample.

view this post on Zulip John Baez (Mar 02 2025 at 18:23):

Serre doesn't say why he chose the counterexample G=Z/3×Q8G = \mathbb{Z}/3 \times Q_8. Since he's smart and presumably not malicious (or in Einstein's words, speaking of someone else, raffiniert aber nicht boshaft), there must not be much simpler examples. But I'll have to finish doing his homework exercise to get a sense for why this example even works at all:

From Serre's "Linear representations of finite groups"

Two properties he uses are that every subgroup of this group is normal, and it has a faithful irreducible representation whose corresponding simple factor in the group algebra Q[G]\mathbb{Q}[G] is a matrix algebra Mn(D)M_n(D) where DD is a division ring and n2n \ge 2. But I don't yet understand his logic. In this example DD is the field

D=Q[exp(2πi/3)] D = \mathbb{Q}[\exp(2 \pi i / 3)]

and that's connected to the Z/3\mathbb{Z}/3 factor in GG, but I still don't get what this factor accomplishes.

view this post on Zulip David Corfield (Mar 02 2025 at 20:49):

I should have recalled the reason this material appeared on the nLab was a physics issue that required some result about surjectivity. Lots of useful information in the bulk of

view this post on Zulip John Baez (Mar 02 2025 at 22:12):

Thanks!

I have a slightly more worked-out guess about @Morgan Rogers (he/him)'s question "why Z/3×Q8\mathbb{Z}/3 \times Q_8?" I believe

The nontrivial semidirect product Z/3Q8\mathbb{Z}/3 \ltimes Q_8 might also have both properties, but arguably Z/3×Q8\mathbb{Z}/3 \times Q_8 is easier to work with.

By the way, the nontrivial semidirect product Z/3Q8\mathbb{Z}/3 \ltimes Q_8 is famous: it's called the binary tetrahedral group because it has a 2-1 and onto homomorphism to the group A4A_4, which acts as rotational symmetries of the tetrahedron. You can think of it as a subgroup of the unit-norm quaternions, and it has lots of great properties.

view this post on Zulip John Baez (Mar 03 2025 at 01:54):

Both Z/3×Q8\mathbb{Z}/3 \times Q_8 and the binary tetrahedral group Z/3Q8\mathbb{Z}/3 \ltimes Q_8 have a 4-dimensional irreducible representation on the rational quaternions

HQQ4\mathbb{H}_{\mathbb{Q}} \cong \mathbb{Q}^4

These groups are strikingly similar, but not isomorphic! There are 15 different groups of order 24, so they say, but these two have a lot in common.

view this post on Zulip John Baez (Mar 03 2025 at 02:00):

Okay, let me get serious and try to solve Serre's exercise. I won't quite succeed, but I hope this motivates someone to fill in some of the gaps.

He's secretly showing that the god-given map from the [[Burnside ring]] to the [[representation ring]] is not surjective in the case of

G=Z/3×Q8G = \mathbb{Z}/3 \times Q_8

But he doesn't mention the Burnside ring or representation ring! Instead, he just claims that GG has an irreducible representation ρ\rho whose character χ\chi is not a linear combination of characters of representations 1HG1^G_H, which is his jargon for the representation of GG induced from the trivial representation on some subgroup HGH \subseteq G. If you want, I can explain why this is equivalent to the map from the Burnside ring to the representation not being surjective. But we won't need that to solve this homework problem!

view this post on Zulip John Baez (Mar 03 2025 at 02:08):

As I already explained, Serre directs us to embed both Z/3\mathbb{Z}/3 and the quaternion group in the multiplicative group of the rational quaternions HQ\mathbb{H}_{\mathbb{Q}}, which is a 4-dimensional [[quaternion algebra]] over Q\mathbb{Q}. I showed how to do this. He notes that by letting Z/3\mathbb{Z}/3 act via left multiplication and Q8Q_8 via right multiplication, we can get a representation of G=Z/3×Q8G = \mathbb{Z}/3 \times Q_8 on HQ\mathbb{H}_\mathbb{Q}. He claims that this representation is irreducible and I showed that.

Then he says "the corresponding simple algebra is M2(K)M_2(K)" where KK is the field

K=Q[exp(2πi/3)]K = \mathbb{Q}[\exp(2 \pi i / 3)].

I explained how irreducible representations of a group give simple algebras, which is the hard part. I think I can show that this particular irrep gives M2(K)M_2(K), and I'm happy to explain if asked, but let me skip ahead to where things get tough for me!

view this post on Zulip John Baez (Mar 03 2025 at 02:14):

He then says to use the previous exercise to conclude that the character of this particular irreducible representation of GG is not a linear combination of characters of representations 1HG1^G_H.

view this post on Zulip John Baez (Mar 03 2025 at 02:15):

Remember the previous exercise:

From Serre's Linear Representations of Finite Groups.

view this post on Zulip John Baez (Mar 03 2025 at 02:21):

Here he gives 3 conditions which supposedly guarantee that the character χ\chi of an irreducible representation ρ\rho of a finite group GG is not a linear combination of characters of representations 1HG1^G_H. Namely:

  1. ρ\rho is faithful, i.e. one-to-one. (I see how to check this in our example.)
  2. Every subgroup of GG is normal. (I explained why this is true for Z/3\mathbb{Z}/3 and Q8Q_8, so it's true for their product GG.)
  3. The simple algebra corresponding to ρ\rho is of the form Mn(D)M_n(D) for DD a division algebra over Q\mathbb{Q}, with n2n \ge 2. (As mentioned, I think this holds in our example, with D=Q[exp(2πi/3)]D = \mathbb{Q}[\exp(2 \pi i /3)] and n=2n = 2).

view this post on Zulip John Baez (Mar 03 2025 at 02:22):

So, the only place where I'm running into trouble is his argument for why these 3 conditions are sufficient to guarantee what Serre claims they do!

view this post on Zulip John Baez (Mar 03 2025 at 02:31):

He gives a cryptic hint: "Show the permutation representation on G/HG/H contains the representation ρ\rho nn times if H={1}H = \{1\} and 0 times otherwise."

This is another way of saying the representation 1HG1^G_H contains ρ\rho as a subrepresentation either nn times or not at all depending on whether HGH \subseteq G is the trivial group or not.

AHA - I SEE NOW WHY PROVING THIS IS THE KEY TO SOLVING THE EXERCISE! :light_bulb:

Suppose 1HG1^G_H contains ρ\rho as a subrepresentation either nn times or not at all depending on whether HGH \subseteq G is the trivial group or not.

Then:

Thus, if n2n \ge 2, we can't get χ\chi as a linear combination of characters of the reps 1HG1^G_H. The best we could hope to do is get nχn \chi!

view this post on Zulip John Baez (Mar 03 2025 at 02:38):

And in our example we know n=2n = 2. So we will be done.

view this post on Zulip John Baez (Mar 03 2025 at 02:43):

Okay, so I now see that to solve Serre's exercise, it's enough to prove this:

Claim. Let ρ\rho be an irreducible representation of a finite group GG such that

  1. ρ\rho is faithful, i.e. one-to-one.
  2. Every subgroup of GG is normal.
  3. The simple algebra corresponding to ρ\rho is of the form Mn(D)M_n(D) for DD a division algebra over Q\mathbb{Q} with n2n \ge 2.

Then for every subgroup HGH \subseteq G, the representation 1HG1^G_H contains the representation ρ\rho nn times if HH is trivial, and not at all otherwise.

view this post on Zulip John Baez (Mar 03 2025 at 02:45):

To prove this claim we should use characters, I bet. Let χHG\chi^G_H be the character of the representation 1HG1^G_H, and let χ\chi be the character of ρ\rho. It suffices to show the inner product χ,χHG \langle \chi, \chi^G_H \rangle is nn if HH is trivial, and zero otherwise.

[EDIT: This is based on a mistake that I catch and fix later.]

view this post on Zulip John Baez (Mar 03 2025 at 02:52):

Let me do the easy case, where H=1H = 1 is the trivial group. Then 1HG=11G1^G_H = 1^G_1 should be the regular representation of GG, i.e. its representation on Q[G]\mathbb{Q}[G]. We then have

χ,χ1G=dimρ \langle \chi, \chi^G_1 \rangle = \dim \rho

view this post on Zulip John Baez (Mar 03 2025 at 02:55):

So, in this case I just need to check that dimρ=n\dim \rho = n, where nn was defined earlier.

view this post on Zulip John Baez (Mar 03 2025 at 05:27):

Whoops - this seems to be false in the example we're actually interested in! :sweating:

The representation ρ\rho of G=Z/3×Q8G = \mathbb{Z}/3 \times Q_8 on HQ\mathbb{H}_{\mathbb{Q}} has

dimρ=4\dim \rho = 4,

but the corresponding simple algebra is M2(Q[exp(2πi/3)M_2(\mathbb{Q}[\exp(2 \pi i /3) so

n=2 n = 2

And by the way, Serre says both these things, not just me!

So I think the problem yet again is that I've taken a fact that holds for representations over C\mathbb{C}, or algebraically closed fields of characteristic zero, and blindly applied it to representations over Q\mathbb{Q}. It's apparently not true that

χ,χ1G=dimρ \langle \chi, \chi^G_1 \rangle = \dim \rho :warning:

when we work over a non-algebraically-closed field of characteristic zero.

view this post on Zulip John Baez (Mar 03 2025 at 05:36):

I think it's really the same mistake I made before, namely this:

John Baez said:

I think it goes like this: GG has a representation on Q[G]\mathbb{Q}[G] by left multiplication: this is called the regular representation. Every irreducible representation ρ\rho of GG appears as a summand of the regular representation, but it may appear repeatedly, say nn times. In fact each irreducible representation appears with a multiplicity equal to its dimension, so nn is the dimension of ρ\rho.

[EDIT: no, this last sentence is true over the complex numbers but not over the rationals! We will see counterexamples later.]

view this post on Zulip John Baez (Mar 03 2025 at 05:50):

Over any field kk, the endomorphism algebra of any irreducible group representation ρ\rho is a division ring DD over kk by [[Schur's Lemma]] Thus if ρn\rho^n is the direct sum of nn copies of ρ\rho we have

End(ρn)=Mn(D)\text{End}(\rho^n) = M_n(D)

Now let's assume kk has characteristic zero so we can do the usual things with the group algebra k[G]k[G] of a finite group. if ρ\rho appears with multiplicity nn in the regular representation of GG (i.e. its representation on the group algebra), the group algebra k[G]k[G] will contain a simple subalgebra isomorphic to

End(ρn)Mn(D)\text{End}(\rho^n) \cong M_n(D).

Serre also tells us that in Proposition 32 of Chapter 12 that the character χ\chi of ρ\rho obeys

χ,χ=dimEnd(ρ)=dimD \langle \chi, \chi \rangle = \dim \text{End}(\rho) = \dim D

I am used to the case k=Ck = \mathbb{C}. Since this is algebraically closed the division ring is the field itself, so D=CD = \mathbb{C} and things simplify:

  1. k[G]k[G] contains a copy of Mn(k)M_n(k) and
  2. χ,χ=1\langle \chi, \chi \rangle = 1.

But both these fail in general.

view this post on Zulip John Baez (Mar 03 2025 at 05:53):

So let me recompute χ,χ1G\langle \chi, \chi_1^G \rangle in the case of a general field of characteristic zero! I'll assume ρ\rho appears with multiplicity nn in the regular representation and End(ρ)=D\text{End}(\rho) = D.

view this post on Zulip John Baez (Mar 03 2025 at 05:57):

χ1G\chi_1^G is the character of the regular representation; it's the sum of the characters of all the irreducible representations appearing in the regular representation, counted with multiplicity. The characters of nonisomorphic irreducible representations are orthogonal, and we're assuming ρ\rho appears with multiplicity nn in the regular representation, so

χ,χ1G=nχ,χ=ndim(D) \langle \chi , \chi_1^G \rangle = n \langle \chi, \chi \rangle = n \text{dim}(D)

view this post on Zulip John Baez (Mar 03 2025 at 06:04):

This is the correct generalization of the equation that held only in the algebraically closed case.

Let's see if this makes my job any easier. What was I trying to do? I was trying to show this:

John Baez said:

Claim. Let ρ\rho be an irreducible representation of a finite group GG such that

  1. ρ\rho is faithful, i.e. one-to-one.
  2. Every subgroup of GG is normal.
  3. The simple algebra corresponding to ρ\rho is of the form Mn(D)M_n(D) for DD a division algebra over Q\mathbb{Q} with n2n \ge 2.

Then for every subgroup HGH \subseteq G, the representation 1HG1^G_H contains the representation ρ\rho nn times if HH is trivial, and not at all otherwise.

I now see that to prove the "Then...", it's sufficient to show

χHG,χ=ndim(D) \langle \chi^G_H, \chi \rangle = n \dim(D)

when HH is the trivial group, and

χHG,χ=0 \langle \chi^G_H, \chi \rangle = 0

otherwise. (The first formula here is the one I'd gotten wrong in my earlier attempt.)

view this post on Zulip John Baez (Mar 03 2025 at 06:19):

And yes, when H=1H = 1 we do indeed have

χHG,χ=χ1G,χ=ndim(D) \langle \chi_H^G, \chi \rangle = \langle \chi^G_1, \chi \rangle = n \dim(D)

So I win this round! :tada:

view this post on Zulip John Baez (Mar 03 2025 at 06:30):

Notice that so far I haven't even used assumptions 1 and 2. I'm reminded of some students I see who do a lot of work trying to prove a theorem in their homework, and then get stuck and ask me for me help. I ask them if they've tried using all the hypotheses of the theorem and sometimes they say no. :laughter_tears:

But I'm not surprised I didn't need assumptions 1 and 2 yet, since so far I've only tackled the easy case where HGH \subseteq G is the trivial group. Somehow when I consider other subgroups HH I'll need assumption 2, which says HH is normal. And I guess I'll need assumption 1, which says the representation ρ\rho is faithful.

(I've used assumption 3 elsewhere: it says that n2n \ge 2, and that assumption was crucial for my lightbulb of realization here.)

view this post on Zulip John Baez (Mar 03 2025 at 06:42):

By the way, @David Corfield, I might write up this argument in a much clearer way on the nLab when I'm done. Right now I'm just fighting my way through the woods, making some mistakes, and I can't expect anyone else to read what I'm writing.

In the end, my answer to Serre's exercise will give a preliminary answer to your question:

David Corfield said:

I wonder if eventually there's a story to emerge of the general conditions which prevent surjectivity.

He's giving a criterion for non-surjectivity of the map from the Burnside ring to the representation ring:

Claim. Working over Q\mathbb{Q}, the map from the Burnside ring to the representation ring of a finite group GG is nonsurjective if GG has an irreducible representation ρ\rho such that

  1. ρ\rho is faithful, i.e. one-to-one.
  2. Every subgroup of GG is normal.
  3. At least two distinct copies of the representation ρ\rho appear as summands in the regular representation of GG.

I suspect that the smallest group to which this criterion applies is our friend Z/3×Q8\mathbb{Z}/3 \times Q_8. I don't fully understand why, but by the time I finish this homework exercise I will!

This is a very specialized criterion, mainly due to condition 2. It's probably part of some bigger theory. I'm willing to bet that the map from the Burnside ring to the representation ring is non-surjective for 'almost all' finite groups. I wonder if someone has shown that!

view this post on Zulip David Corfield (Mar 03 2025 at 07:42):

John Baez said:

By the way, @David Corfield, I might write up this argument in a much clearer way on the nLab when I'm done

That would be great. To circle back to the original post, I wonder then if it might be that the span-ish groupoidal approach would struggle in the same place for similar reasons. I guess a span as a passage from two objects AA to BB has something of the flavour of a difference (BA)(B - A).

view this post on Zulip John Baez (Mar 03 2025 at 15:48):

I was going to say "yes, exactly" - but now I realize I'm fuzzy about some points.

For those who don't know what David and I are talking about:

James Dolan has a whole philosophy of how to grind permutation representations against each other, breaking them down into smaller pieces: the irreducible representations.

The idea is that we don't directly get the irreducible representations of a finite group GG looking only at the permutation representations: the representations coming from finite GG-sets. But any GG-equivariant relation between GG-sets gives an equivariant linear operatorsbetween the corresponding permutation representations, and the kernels or cokernels of such operators can be the smaller irreducible representations hiding in the permutation representations.

This works wonders for symmetric groups: we can account for the whole representation theory of symmetric groups this way, at least over fields of characteristic zero. Right now James is figuring out how it works in the Langlands program (see Daniel Bump's book Lie Groups, the sections "Hecke algebras" and "The philosophy of cusp forms"). But I feel I'm about to see - finally - why it doesn't work for all finite groups. That's why I'm doing all these calculations in the thread above.

Here's the point I'm fuzzy about. It's pretty basic! If a representation is a kernel or cokernel of an equivariant linear operator between permutation representations, does that mean that in the representation ring it's a formal difference of permutation representations? I don't see why it should be. An exact sequence of representations says that some alternating sum in the representation ring is zero. But that doesn't seem to do that job: it just means that if we have an equivariant linear operator f:ABf: A \to B we get

[kerf][A]+[B][cokerf]=0 [\text{ker} f] - [A] + [B] - [\text{coker} f] = 0

view this post on Zulip John Baez (Mar 03 2025 at 15:51):

So if AA and BB are permutation representations, [cokerf][kerf][\text{coker} f] - [\text{ker} f] is a formal difference of permutation representations - a "virtual permutation representation". But that doesn't imply that either [kerf][\text{ker} f] or [coker f][\text{coker f}] is! So if we allow ourselves access to kernels and cokernels, perhaps we can get more.

view this post on Zulip John Baez (Mar 03 2025 at 15:52):

Clearly we should figure this out for G=Z/3×Q8G = \mathbb{Z}/3 \times Q_8. :upside_down:

view this post on Zulip David Corfield (Mar 03 2025 at 16:58):

John Baez said:

James Dolan has a whole philosophy of how to grind permutation representations against each other, breaking them down into smaller pieces: the irreducible representations.

As I recall, this approach took as fundamental the incidence relations encoded in the double cosets of the transformation group. So in Euclidean geometry, equivalence classes of the double cosets corresponding to a subgroup fixing a point and one fixing a line would amount to possible perpendicular distances from a point to a line.

Will any of that story show up here, I wonder?

view this post on Zulip John Baez (Mar 03 2025 at 17:43):

When I was talking about GG-equivariant relations between GG-sets just now... you can get those from double cosets, which are a way of talking about GG-equivariant spans. So I'm talking about things like your example.

view this post on Zulip David Corfield (Mar 03 2025 at 18:42):

Right, but presumably failure of surjectivity will show up as some kind of inadequacy in the collection of invariant relations.

view this post on Zulip Jorge Soto-Andrade (Mar 03 2025 at 20:35):

John Baez said:

Okay, let me get serious and try to solve Serre's exercise. I won't quite succeed, but I hope this motivates someone to fill in some of the gaps.

He's secretly showing that the god-given map from the [[Burnside ring]] to the [[representation ring]] is not surjective in the case of

G=Z/3×Q8G = \mathbb{Z}/3 \times Q_8

But he doesn't mention the Burnside ring or representation ring! Instead, he just claims that GG has an irreducible representation ρ\rho whose character χ\chi is not a linear combination of characters of representations 1HG1^G_H, which is his jargon for the representation of GG induced from the trivial representation on some subgroup HGH \subseteq G. If you want, I can explain why this is equivalent to the map from the Burnside ring to the representation not being surjective. But we won't need that to solve this homework problem!

Sorry for jumping in late. When you say "linear combination" do you always mean "linear combination with integer coefficients"? In our old arXiv preprint mentioned above ( https://drive.google.com/file/d/19bXhejuA75lzVxGqocwvmp8kb2d7HcZm/view?usp=sharing), see 3.5,
we give some very simple examples of groups GG whose Gelfand Model (the multiplicity-free direct sum of all irreps of GG) does not lie in the Green ring of G, which entails the non surjectivity of the canonical map from the Burnside ring to the Representation ring. The latter map factors naturally through the Green ring of GG, of course. Interestingly the canonical map from Burnside to Green is not injective up to isomorphism, because you may have two linear natural (permutation) representations of GG which are"essentially" isomorphic via a Radon transform, which is just a stochastic mapping in the category of GG-sets. Baby example: Take the square (as a graph) and its symmetry group GG. The natural representations of GG associated to the GG-sets of all four vertices and all four edges are almost isomorphic via Radon (caveat: Radon does not "see" the sign representation in each case, i.e. it "kills" it), but there is no intrinsic ( i. e. GG-invariant) way to pair a vertex with an edge in the square. Does this make sense?

view this post on Zulip John Baez (Mar 03 2025 at 20:50):

Jorge Soto-Andrade said:

John Baez said approximately:

Serre is secretly showing that the god-given map from the [[Burnside ring]] to the [[representation ring]] is not surjective in the case of

G=Z/3×Q8G = \mathbb{Z}/3 \times Q_8

But he doesn't mention the Burnside ring or representation ring! Instead, he just claims that GG has an irreducible representation ρ\rho whose character χ\chi is not a linear combination of characters of permutation representations.

When you say "linear combination" do you always mean "linear combination with integer coefficients"?

Yes. Indeed Serre's argument should let us find an irreducible representation of GG that shows up with muliplicity 0 or 2 in every permutation representation of that group. Thus, its character can't be an integer linear combination of characters of permutation representations, but it might be a rational linear combination.

Your preprint sounds interesting! More later.

view this post on Zulip Jorge Soto-Andrade (Mar 03 2025 at 21:14):

John Baez said:

Clearly we should figure this out for G=Z/3×Q8G = \mathbb{Z}/3 \times Q_8. :upside_down:

I like Dolan's approach, but I would say that the name of the game in geometric group representation theory is to spot multiplicity - free linear natural (permutation) representations, work in their intertwining operator algebra, which would be a commutative operator algebra (per Gelfand and Dirac) and figure out its characters (which amount to the corresponding spherical functions, and provide the decomposition of your multiplicity free representation). All this for a finite GG, but in the spirit of Gelfand. Moreover, your ker , coker musings "stink" cohomology (this is not pejorative at all...). We have interesting examples where you get a Gelfand Model for GG as an alternating sum (à la Lefschetz) of natural representations. This alternating sum is just a top cohomology space of a G-space whose lower cohomology vanishes. A related example of this is the cohomological construction of the Steinberg representation of GG (a finite group of Lie type) à la Solomon-Lehrer-Tits. The corresponding baby example (which Arnold would request...) is the case of G=SL(2,Fq)G= SL(2, \mathbb F_q) and V=KerΦ,Φ:L2(X)CV = Ker \Phi, \Phi : L^2(X) \rightarrow \mathbb C , XX being the set of lines throught the origin in the finite plane over Fq \mathbb F_q and Φ\Phi being just the integral (summation) operator. Zero sum is just the cocycle condition.

view this post on Zulip John Baez (Mar 03 2025 at 23:46):

I like Dolan's approach, but I would say that the name of the game in geometric group representation theory is to spot multiplicity - free linear natural (permutation) representations, work in their intertwining operator algebra, which would be a commutative operator algebra (per Gelfand and Dirac) and figure out its characters (which amount to the corresponding spherical functions, and provide the decomposition of your multiplicity free representation).

That's interesting! Of course, whenever someone says something is "the name of the game", it means I should do something else, because there are already enough people doing that. But still it would be good for me to learn about this.

view this post on Zulip John Baez (Mar 03 2025 at 23:49):

Jorge Soto-Andrade said:

In our old arXiv preprint mentioned above (https://drive.google.com/file/d/19bXhejuA75lzVxGqocwvmp8kb2d7HcZm/view?usp=sharing), see 3.5,
we give some very simple examples of groups GG whose Gelfand model (the multiplicity-free direct sum of all irreps of GG) does not lie in the Green ring of G, which entails the non surjectivity of the canonical map from the Burnside ring to the representation ring. The latter map factors naturally through the Green ring of GG, of course.

What do you mean by 'Green ring'? Wikipedia and a paper by Benson use it as synonymous with 'representation ring'. Maybe you're using it to mean the image of the Burnside ring in the representation ring?

view this post on Zulip Jorge Soto-Andrade (Mar 04 2025 at 01:03):

John Baez said:

Jorge Soto-Andrade said:

In our old arXiv preprint mentioned above (https://drive.google.com/file/d/19bXhejuA75lzVxGqocwvmp8kb2d7HcZm/view?usp=sharing), see 3.5,
we give some very simple examples of groups GG whose Gelfand model (the multiplicity-free direct sum of all irreps of GG) does not lie in the Green ring of G, which entails the non surjectivity of the canonical map from the Burnside ring to the representation ring. The latter map factors naturally through the Green ring of GG, of course.

What do you mean by 'Green ring'? Wikipedia and a paper by Benson use it as synonymous with 'representation ring'. Maybe you're using it to mean the image of the Burnside ring in the representation ring?

I mean the ring generated by the natural representations (L2(X),τ) (L^2(X), \tau) of GG where XX is a GG-set and τg(f)=fg1 \tau_g(f) = f \circ g^{-1} . So, if you wish you could say that it is the image of the canonical embedding of the Burnside ring in the representation ring, via the mapping XL2(X) X \mapsto L^2(X) etc This map could be called the Koopman map, by the way... I would prefer to focus on the Green ring as a bona fide ring associated to GG. Then, a nice first conjecture is that a Gelfand Model for GG lies in the Green ring of GG, and the same for the other symmetric functions of the irreps of GG, the "last one" being the tensor product of all irreps. Of course, hoping that all irreps of GG lie in the Green ring of GG is too optimistic, although it holds for the symmetric groups.
Re the name of the game: in fact not so many people work in this way in finite group representation theory. Most people think in terms of characters, and compute characters and their inner products, instead of thinking in terms of function spaces, linear operators and intertwining operators. So my claim was rather that this should be the name of the game. I agree with your point ...

view this post on Zulip John Baez (Mar 04 2025 at 02:25):

Okay, so you're using 'Green ring' to mean the image of the canonical map from the Burnside ring to the representation ring. Okay!

Re the name of the game: in fact not so many people work in this way in finite group representation theory. [...] my claim was rather that this should be the name of the game.

Okay! Where I come from, people say "name of the game" when they're trying to pressure me to do what lots of other people are doing. So I'm glad that you're using it a different way: expressing your own view about what should be the name of the game.

view this post on Zulip John Baez (Mar 04 2025 at 02:26):

I have a lot more to say about what you've told me, but first... I want to finish solving Serre's homework exercise!

view this post on Zulip John Baez (Mar 04 2025 at 02:38):

In what follows all representations are over Q\mathbb{Q}.

I'm trying to prove this claim, since it lets us get a concrete example of a finite group where the canonical map from its [[Burnside ring]] to its [[representation ring]] is not surjective:

Claim 1. The map from the Burnside ring to the representation ring of a finite group GG is nonsurjective if GG has an irreducible representation ρ\rho such that

  1. ρ\rho is faithful, i.e. one-to-one.
  2. Every subgroup of GG is normal.
  3. The representation ρ\rho appears at least twice in the regular representation of GG.

I've showed that to prove Claim 1 it's enough to prove this:

Claim 2. If GG is a finite group with an irreducible representation such that

  1. ρ\rho is faithful, i.e. one-to-one.
  2. Every subgroup of GG is normal.
  3. At least two distinct copies of the representation ρ\rho appear as summands in the regular representation of GG.

then the permutation representation of GG on Q[G/H]\mathbb{Q}[G/H] contains at least two copies of ρ\rho as a subrepresentation if H={1}H = \{1\}, but it does not contain ρ\rho at all as a all for other subgroups HH.

(Here Q[G/H]\mathbb{Q}[G/H] is the free vector space on the set G/HG/H, and it's a representation of GG since G/HG/H is a GG-set.)

The way I'm stating things now, the "at least twice if H={1}H = \{1\}" clause follows instantly from assumption 2, because then the representation of GG on Q[G/H]\mathbb{Q}[G/H] is the regular representation.

So the only hard part is to prove the clause "but not at all for other subgroups HH". And that's what I'll do now! :tada:

view this post on Zulip John Baez (Mar 04 2025 at 02:44):

Suppose HH is a nontrivial subgroup of GG. We need to show that the permutation representation of GG on the vector space Q[G/H]\mathbb{Q}[G/H] does not contain ρ\rho as a subrepresentation.

view this post on Zulip John Baez (Mar 04 2025 at 02:56):

Since HH is normal, HH acts trivially on Q[G/H]\mathbb{Q}[G/H]. Let's see how that goes, exactly!

By definition, the representation Q[G/H]\mathbb{Q}[G/H] has a basis given by right cosets gHgH, and gGg' \in G acts on any of these cosets by left multiplication:

g:gHggH g' : gH \to g'gH

By condition 2, HH is normal, so gH=HggH = Hg . Thus we can also describe the representation Q[G/H]\mathbb{Q}[G/H] by

g:HggHg g': Hg \to g' H g

This lets us see that HH acts trivially; if hHh \in H we have

h:HghHg=Hg h : Hg \to h H g = H g.

view this post on Zulip John Baez (Mar 04 2025 at 02:58):

If ρ\rho were a subrepresentation of Q[G/H]\mathbb{Q}[G/H] then HH would act trivially on ρ\rho, too. But since HH contains elements other than the identity, this contradicts condition 1, that ρ\rho is faithful, which implies all non-identity group elements act nontrivially on ρ\rho. QED.

view this post on Zulip John Baez (Mar 04 2025 at 03:01):

I think I'm now ready to polish up this whole business for the nLab.

view this post on Zulip John Baez (Mar 04 2025 at 06:45):

Okay, here is the cleaned-up proof which I put on the nLab. It's simple now that I understand it.

In what follows all representations are taken over a field kk of characteristic zero.

Theorem. Suppose GG is a finite group with a linear representation ρ\rho such that:

  1. ρ\rho is irreducible and faithful

  2. every subgroup of GG is normal

  3. ρ\rho appears with multiplicity n2n \ge 2 in the regular representation of GG.

Then the map from the [[Burnside ring]] of GG to the [[representation ring]] R(G)R(G) of GG is not surjective.

Proof. It suffices to prove that the multiplicity of ρ\rho in any permutation representation of GG is a multiple of nn, so that the class [ρ]R(G)[\rho] \in R(G) cannot be in the image of R(G)R(G).

Since every finite GG-set is a coproduct of transitive actions of GG, which are isomorphic to actions on G/HG/H for subgroups HH of GG, every permutation representation of GG is a direct sum of those on spaces of the form k[G/H]k[G/H]. Thus, it suffices to show that the multiplicity of ρ\rho in the representation on k[G/H]k[G/H] is nn if HH is the trivial group, and 00 otherwise.

The former holds by assumption 3. For the latter, suppose HH is a nontrivial subgroup of GG. Because HH is normal by assumption 2, every element hHh \in H acts trivially on k[G/H]k[G/H]: we can see this by letting hh act on an arbitrary basis element gH=HgG/Hg H = H g \in G/H:

hHg=Hg. h H g = H g .

Since HH is nontrivial, it contains elements h1h \ne 1 that act trivially on k[G/H]k[G/H]. But no h1h \ne 1 can act trivially on ρ\rho because ρ\rho is faithful, by assumption 1. Thus ρ\rho cannot be a subrepresentation of k[G/H]k[G/H]. That is, ρ\rho appears with multiplicity 00 in k[G/H]k[G/H]. :notebook:

It is an exercise in Serre's Linear Representations of Finite Groups to show that the group G=Z/3×Q8G = \mathbb{Z}/3 \times Q_8 obeys the conditions of this proposition, so that β:A(G)R(G)\beta : A(G) \to R(G) is nonsurjective. As a hint, Serre suggests to embed Z/3\mathbb{Z}/3 and Q8Q_8 in the multiplicative group of the algebra HQ\mathbb{H}_{\mathbb{Q}} (the quaternions defined over Q\mathbb{Q}). By letting Z/3\mathbb{Z}/3 act by left multiplication and Q8Q_8 act by right multiplication, one obtains a 4-dimensional irreducible representation ρ\rho of GG which appears with multiplicity n2n \ge 2 in the regular representation. Furthermore ρ\rho is faithful and irreducible, and every subgroup of GG is normal.

view this post on Zulip John Baez (Mar 04 2025 at 16:54):

I asked on MathOverflow and found out that for most groups the map β\beta from the Burnside ring to the representation ring is surjective (taking representations over Q\mathbb{Q}).

I should have known this, because I knew most groups have order a power of 2, and that β\beta is surjective for all 2-groups. But David Benson quickly assembled a list of groups of order 2np2^n \cdot p for which β\beta is not surjective, and he says these are rare among groups of that kind!

Yves de Cornulier pointed out something I also could have known, but had neglected: the situation is dramatically different if we use representations over C\mathbb{C}. Then it's much easier to find groups where β\beta is not surjective, because most groups have lots of representations definable only over an abelian extension of Q\mathbb{Q}, and those can never be formal difference of permutation representation. This happens for G=Z/nG = \mathbb{Z}/n whenever n>2n > 2.

view this post on Zulip John Baez (Mar 05 2025 at 21:05):

Morgan Rogers (he/him) said:

I don't have a clue from this discussion how someone came up with the particular counterexample of Q8×Z/3Q_8 \times \mathbb{Z}/3. But since Q8Q_8 appears to be acted on by Z/3\mathbb{Z}/3 by rotating the generators, I'm curious if the semi-direct product could also be a counterexample.

I now know more about this.

First, that semidirect product Z/3Q8\mathbb{Z}/3 \ltimes Q_8 is not a counterexample: it's the binary tetrahedral group, and for this the map β\beta from the Burnside ring to the representation ring is surjective.

Second, the smallest groups for which β\beta is nonsurjective have order 24. So Serre was not being extravagant in using Z/3×Q8\mathbb{Z}/3 \times Q_8.

Third, there are exactly two groups of order 24 for which β\beta is nonsurjective. The other one is apparently the nontrivial semidirect product Z/3Z/8\mathbb{Z}/3 \ltimes \mathbb{Z}/8.

As you can see I'm at the stage of learning more about this issue than I ever wanted to know. I will disperse my knowledge and then forget most of it. But at least I'm not as far gone as David Benson, who computed that there are 532 groups of order 320 for which β\beta is nonsurjective.

(He wrote some great books on the cohomology of groups, representation theory and other matters, so this is friendly teasing: I would love to have his knowledge of algebra.)

view this post on Zulip David Corfield (Mar 06 2025 at 10:50):

John Baez said:

First, that semidirect product Z/3Q8\mathbb{Z}/3 \ltimes Q_8 is not a counterexample: it's the binary tetrahedral group, and for this the map β\beta from the Burnside ring to the representation ring is surjective.

This and a handful of others is just the reason for that paper I mentioned above

image.png

view this post on Zulip John Baez (Mar 06 2025 at 17:19):

Yes, this is quite cool! I haven't studied this paper or orbifolds in string theory, and I probably won't since string theory has produced too much math to follow unless one is committed to it, and I have other ideas about what's interesting, but the finite subgroups of SU(2)\textrm{SU}(2) show up all over the place, most notably the [[McKay correspondence]], which shows up in string theory.

By the way, I bet those authors could have gotten a finite group theorist to determine whether the map from the Burnside ring to the representation ring is surjective for all the dihedral groups. It looks like they handled the first 12 or so.

Maybe I'll ask about this on MathOverflow.

view this post on Zulip Simon Burton (Mar 06 2025 at 19:21):

I did make some progress on this question.. I think it's possible to write down an explicit description of the Burnside ring of an arbitrary dihedral group...

view this post on Zulip John Baez (Mar 06 2025 at 20:24):

It should be fairly easy as groups go, especially since the patterns followed by small dihedral groups should persist for larger ones, and dihedral groups are just cyclic groups with a twist thrown in

Also, to resolve the question of whether the map from the Burnside ring to the representation ring is onto, we don't need to understand the ring structure. As you probably know, we just need to think about the standard basis of the Burnside ring (given by the GG-sets G/HG/H for all possible subgroups HH) and how it maps to the standard basis of the representation ring (given by irreducible representations of GG).

So, it should be pretty fun for someone who has time to think about it.

Amusingly, we already know the map from the Burnside ring to the representation ring is onto for dihedral groups whose order is a power of 2, by a result of Graeme Segal.

view this post on Zulip Jorge Soto-Andrade (Mar 06 2025 at 21:04):

Simon Burton said:

I did make some progress on this question.. I think it's possible to write down an explicit description of the Burnside ring of an arbitrary dihedral group...

I am afraid that the irred characteres of the dihedral groups involve the real part of the n-th roots of unity, which is rarely an integer But the character values should be integers for the characters to be Green. ( lie inthe image of Burnside) By the way the same happens for juicy finite groups like SL(2,Fq) SL(2, \mathbb F_q). So it is not reasonable to expect that the irreps are Green

view this post on Zulip Jorge Soto-Andrade (Mar 06 2025 at 21:12):

However it makes sense to expect the gelfand model to be Green. After all its character (the sum of all irreducible characters of the group) looks very much as permutation character (integer values, almost always non negative , mean value 1). But in general it cannot be a permutation character ! An interesting question would be to characterize the groups for which the gelfand model is Green despite the irreps not being Green (not lying in the article mage of the burnside ring)

view this post on Zulip Simon Burton (Mar 06 2025 at 21:38):

John is just talking about the rational representations of the dihedral groups, and i'm pretty sure these are all built by adding and "subtracting" of permutation representations. That is the conjecture, anyway.

view this post on Zulip John Baez (Mar 06 2025 at 21:52):

John was talking about rational representations because we're talking about how Simon and friends showed that in this case the map from the Burnside ring to representation ring is surjective for the binary dihedral groups 2D2n2 \cdot D_{2n} where 2n122n \le 12, and they conjectured that this pattern continues. As I mentioned, it's always true for 2D2n2 \cdot D_{2^n}. If it's true for all binary dihedral groups, it's true for all finite subgroups of SU(2)\text{SU}(2).

(Earlier I was talking about dihedral groups, but I should have been talking about their double covers, the 'binary dihedral groups'. I don't know much about these! For all I know, they are isomorphic to dihedral groups.)

view this post on Zulip Jorge Soto-Andrade (Mar 07 2025 at 04:36):

Simon Burton said:

John is just talking about the rational representations of the dihedral groups, and i'm pretty sure these are all built by adding and "subtracting" of permutation representations. That is the conjecture, anyway.

view this post on Zulip Jorge Soto-Andrade (Mar 07 2025 at 04:43):

I was talking of irreps over C\mathbb C. First step if you wish, before "going down" to Q\mathbb Q if you are interested. Anyway the real part of the fifth roots of unity is not rational! Anyway I would claim that a natural first question is whether a Gelfand model is Green (lies in the image of Burnside). This happens quite often. It maybe that all symmetric functions of the irreps are Green. In that case, you have a nice universal polynomial with coefficients in the Green ring (the image of the Burnside ring), whose roots are the irreps, which usually do not like to lie in the Green ring. Notice the analogy with Galois theory. A sort of non commutative Galois theory (the case of conmutative groups should more or less boil down to classical Galois theory...)

view this post on Zulip Simon Burton (Mar 07 2025 at 10:10):

Ah right, thanks for the clarification Jorge.

view this post on Zulip Simon Burton (Mar 07 2025 at 10:19):

A lot of my (very meagre!) knowledge of permutation representations comes from this paper "The Subgroups of M24, or How to Compute the Table of Marks of a Finite Group" from 1997. Looking at this again recently it strikes me how much of it is decategorified, or rather, begging to be categorified. In much the same vein as how combinatorics/mobius inversion can be categorified.

view this post on Zulip Simon Burton (Mar 07 2025 at 10:29):

And I guess that is also the goal of the "groupoidification" program that began this topic. Much of this literature very quickly ascends the categorical levels.

view this post on Zulip Simon Burton (Mar 07 2025 at 11:19):

Just the last few days I've been rediscovering Hecke algebras in this context and it is newly mind-boggling & i have so many questions about it...

view this post on Zulip David Corfield (Mar 07 2025 at 12:03):

Simon Burton said:

Just the last few days I've been rediscovering Hecke algebras in this context

All very Baez-ian and friends, New Paper on the Hecke Bicategory

This paper represents my take on the fundamental theorem of Hecke operators, including a very simple categorification of the category of permutation representations of a finite group GG.

view this post on Zulip John Baez (Mar 07 2025 at 23:20):

Right now one thing I'm interested in is how to go beyond the limited world where the map from the Burnside ring to the representation ring is surjective, while still keeping some of the Set-based flavor of the Burnside ring. For this I'm trying to understand Artin's theorem on induced characters and Brauer's induction theorem. These point at the importance of representations of a finite group GG that are induced from representations of the form ρ:HU(1)\rho: H \to \text{U}(1) where HGH \subseteq G is a subgroup. HH can even be taken to be of a very special form, and we can replace U(1)\text{U}(1) by the group of nth roots of unity for large enough n, but right now the evocative thing to me is that we are bringing phases (elements of U(1)\text{U}(1) into the picture. This reeks of quantum mechanics, and reminds me of how Jeffrey Morton categorified the harmonic oscillator using a generalization of species involving finite sets where elements were equipped with phases.

view this post on Zulip John Baez (Mar 07 2025 at 23:22):

I also love how Artin got into his theorem on induced characters by trying to understand L-functions (which is something I'm also trying to understand).

But I want to start out humbly by improving the proof of Artin's theorem on induced characters on the nLab. The proof should be quite simple, but right now I feel it's worded quite opaquely.

view this post on Zulip David Corfield (Mar 08 2025 at 08:30):

I remember with that work on phases that Jeffrey and you carried out having the vague thought that you'd groupoidified to avoid the vector spaces of representation theory only to put something like them, phases in the complex plane, back in by hand.

If, as I mentioned,

David Corfield said:

The Burnside to representation ring comparison functor is secretly a comparison of the equivariant stable cohomotopy of the point with the equivariant K-theory of the point, ultimately deriving from maps of EE_{\infty}-ring spectra: S=KF1KCKU\mathbb{S} = K\mathbb{F}_1 \to K \mathbb{C} \to KU (as explained here),

then S\mathbb{S} being initial, we could map it wherever.

Why are we always drawn back to topological complex K-theory? Presumably there are informative comparison maps to be had to other cohomology theories.

I guess it's all to do with [[chromatic homotopy theory]]
image.png.

At least we get to level 1 with KUKU. Is MUG()MU_G(\ast) just too difficult to work with?

image.png

view this post on Zulip David Corfield (Mar 08 2025 at 10:12):

That's not terribly promising:
image.png
here

view this post on Zulip John Baez (Mar 08 2025 at 18:01):

What I was trying to do is escape from the tyranny of vector spaces: to do quantum mechanics without using vector spaces... to see where those vector spaces come from.

Groupoidification gets us a certain way, but we can get a lot further if we allow ourselves the use of the circle group. You said

you carried out having the vague thought that you'd groupoidified to avoid the vector spaces of representation theory only to put something like them, phases in the complex plane, back in by hand.

but I would contest the phrase in the complex plane. The circle doesn't need to know it's in the complex plane. Perhaps I gave up too much philosophical ground when earlier in this conversation I called it U(1)\text{U}(1). Let me just call it the circle group.

This may seem like nitpicking, but I have to make this subtle distinction to show that a tiny bit of progress has been made: we can do a bunch of perturbative quantum field theory, using Feynman diagrams, in a categorified way, without using complex vector spaces or Hilbert spaces. We just need to use "phased sets" - finite sets where each point is equipped with an element of the circle group.

I'm not saying this is fully satisfactory, but it's something.

view this post on Zulip John Baez (Mar 08 2025 at 18:27):

One interesting thing about Brauer's induction theorem is that we can get all finite-dimensional complex representations of a finite group GG as virtual representations if we start from 1-dimensional representations of subgroups HH and induce up to the whole group. And while you can think of these 1d representations as homomorphisms

ρ:HU(1) \rho: H \to \text{U}(1)

they never hit all of U(1)\text{U}(1), obviously, because HH is finite. You can think of them as representations

ρ:HZ/n \rho: H \to \mathbb{Z}/n

where nn is big enough, e.g. the order of HH.

view this post on Zulip John Baez (Mar 08 2025 at 18:34):

So it's possible we can understand the complex representations of finite groups in a categorified way using just finite groups and some categorified concept of 'virtual representation'.

Virtual representations are formal differences of isomorphism classes of objects, but that's not really good for categorification. A better idea might be to use the derived category of the category of complex representations of GG.

In short, there may be some purely combinatorial bicategory (or something like that) which is a good substitute for the derived category of representations of GG, built using Brauer's induction theorem. By 'purely combinatorial', I mean built using only finite structures, no mention of the continuum. If we do that there's no way we can describe all morphisms in Rep(G)\text{Rep}(G) or its derived category, since there's a continuum of those. But we may be able to get enough of them to 'do group representation theory', whatever that means.

view this post on Zulip John Baez (Mar 08 2025 at 18:39):

All this may sound a bit quirky, because the original motivating goal is a bit quirky: it's to do quantum physics in a purely combinatorial way. I got interested in this from thinking about spin networks and papers like this:

Mathematically what this does is express a bunch of the representation theory of SU(2)\text{SU}(2) in combinatorial terms. But there's also a bunch of charming talk about how we might try to understand quantum gravity in a purely combinatorial way.

Penrose didn't know enough category theory to say what he'd done in a really elegant mathematical way, and as far as I know nobody has said that.

view this post on Zulip John Baez (Mar 08 2025 at 19:25):

Here's a goal one can have for any group GG: describe the irreducible complex representations of GG, and for each collection of irreps V1,,Vm,W1,,WmV_1, \dots, V_m, W_1, \dots, W_m a basis of morphisms

V1VmW1Wn V_1 \otimes \cdots \otimes V_m \to W_1 \otimes \cdots \otimes W_n

and how these basis morphisms compose and change under the action of the symmetric groups, all in purely combinatorial terms, without reference to the continuum.

In other words: give a purely combinatorial description of the colored prop of irreducible complex representations of GG, where the colors are the irreducible representations.

view this post on Zulip John Baez (Mar 08 2025 at 19:26):

Penrose essentially did this for SU(2)\text{SU}(2), and the same ideas work for SU(n)\text{SU}(n). But I've never seen anyone clearly state what he did in the way I'm vaguely outlining here.

I'm fantasizing about doing the same sort of thing for any finite group using Brauer's induction theorem. But maybe I should go back and do it for SU(2)\text{SU}(2) first!

view this post on Zulip Jorge Soto-Andrade (Mar 08 2025 at 23:06):

John Baez said:

I also love how Artin got into his theorem on induced characters by trying to understand L-functions (which is something I'm also trying to understand).

But I want to start out humbly by improving the proof of Artin's theorem on induced characters on the nLab. The proof should be quite simple, but right now I feel it's worded quite opaquely.

view this post on Zulip Jorge Soto-Andrade (Mar 08 2025 at 23:16):

John Baez said:

One interesting thing about Brauer's induction theorem is that we can get all finite-dimensional complex representations of a finite group GG as virtual representations if we start from 1-dimensional representations of subgroups HH and induce up to the whole group. And while you can think of these 1d representations as homomorphisms

ρ:HU(1) \rho: H \to \text{U}(1)

they never hit all of U(1)\text{U}(1), obviously, because HH is finite. You can think of them as representations

ρ:HZ/n \rho: H \to \mathbb{Z}/n

where nn is big enough, e.g. the order of HH.

Before reading further, I would like to claim that the "right" notion of induction is not induction from a fixed subgroup HH of GG but from the groupoid GX\mathcal G_X canonically associated to a GG-set XX, so we are not choosing an origin in XX. This is like looking at the fundamental groupoid instead of the fundamental group at a point of a given space. Inducing from linear characters of GX\mathcal G_X is a powerful tool indeed. It allows you to construct Gelfand Models for your groups. For example, for G=SnG = S_n you take as XX the space of involutions of GG. For the qq-analogue, G=GL(n,Fq)G = GL(n, \mathbb F_q) , or better G=PGL(n,Fq)G = PGL(n, \mathbb F_q) you take of course XX to be set of all symmetric matrices in GG. This is a re interpretation of the work of Klyachko, a clever Ukrainian student of Gelfand.

view this post on Zulip Jorge Soto-Andrade (Mar 08 2025 at 23:24):

People interested in geometric representation theory of finite groups of Lie type, are not so fond of Brauer's theorem. To get a good grasp of the situation, I think that you should look at G=GL(2,Fq)G = GL(2, \mathbb F_q) or G=SL(2,Fq) G = SL(2, \mathbb F_q) say, and construct all its complex irreps via Brauer and compare with our favourite construction via Weil representations (associated to the cuadratic spaces given by the two algebra Galois extensions of Fq\mathbb F_q, to wit: Fq×\mathbb F_q \times \mathbb F_q$$ and Fq2\mathbb F{_q^2}, each endowed with its Galois norm. The latter construction comes right away from quantum mechanics (Segal-Shale-Weil oscillator representation of the symplectic group). To me Brauer is orthogonal to quantum mechanics...

view this post on Zulip Jorge Soto-Andrade (Mar 08 2025 at 23:26):

Remark: notice that groupoid induction work as well for non transitive GG-sets.

view this post on Zulip Jorge Soto-Andrade (Mar 09 2025 at 00:50):

John Baez said:

Here's a goal one can have for any group GG: describe the irreducible complex representations of GG, and for each collection of irreps V1,,Vm,W1,,WmV_1, \dots, V_m, W_1, \dots, W_m a basis of morphisms

V1VmW1Wn V_1 \otimes \cdots \otimes V_m \to W_1 \otimes \cdots \otimes W_n

and how these basis morphisms compose and change under the action of the symmetric groups, all in purely combinatorial terms, without reference to the continuum.

In other words: give a purely combinatorial description of the colored prop of irreducible complex representations of GG, where the colors are the irreducible representations.

Your goal above looks like a sort of extension of Schur-Weyl duality. In that case GG would be GL(n,C)GL(n, \mathbb C) . Does this make sense? I have been interested in the case of GL(n,Fq)GL(n, \mathbb F_q) instead, which seems to be trickier.
For instance (baby example a la Arnold) the flip action of S2S_2 on L2(Fq)L2(Fq)=L2(Fq×Fq)L^2(\mathbb F_q ) \otimes L^2(\mathbb F_q) = L^2(\mathbb F_q \times \mathbb F_q ) goes into the Fourier - Grassman transform in L2(Fq×Fq) L^2(\mathbb F_q \times \mathbb F_q ), which is an involution (the kernel of Fourier-Grassman is (x,y)xy (x,y) \mapsto x \wedge y).

view this post on Zulip John Baez (Mar 09 2025 at 01:41):

I'm afraid I'm so deeply locked into my conversation with @David Corfield (a conversation which has been going on for decades) that I can't easily focus on the new ideas you're contributing, @Jorge Soto-Andrade. I'd been going to respond to David's comments on spectra, and say how my thoughts connect (or don't connect) to those.

Anyway, there's a lot to think about here, and right now the first step, for me, is to learn the proof of Artin's theorem on induced characters, and continue polishing up the proof on Wikipedia.

In the longer term I can try to understand what you're talking about.

view this post on Zulip David Corfield (Mar 09 2025 at 10:05):

John Baez said:

Virtual representations are formal differences of isomorphism classes of objects, but that's not really good for categorification. A better idea might be to use the derived category of the category of complex representations of GG.

I guess everything's an approximation to something. The virtual approach is picking up on what topological K-theory sees of the cohomotopy of the sphere spectrum, here in the equivariant case:

image.png

Funny how there always seems to be yet a further peak to gain some larger perspective. Frenkel has been promoting the Langlands program as the "grand unified theory of mathematics." But then there are always hints of more, like the suggestion of a *topological* Langlands.

I've always been a sucker for the big picture.

view this post on Zulip David Corfield (Mar 09 2025 at 10:06):

Hmm, here's an interesting finding on complex orientations by Doron Grossman-Naples
image.png

view this post on Zulip Jorge Soto-Andrade (Mar 09 2025 at 16:58):

John Baez said:

I'm afraid I'm so deeply locked into my conversation with David Corfield (a conversation which has been going on for decades) that I can't easily focus on the new ideas you're contributing, Jorge Soto-Andrade. I'd been going to respond to David's comments on spectra, and say how my thoughts connect (or don't connect) to those.

Anyway, there's a lot to think about here, and right now the first step, for me, is to learn the proof of Artin's theorem on induced characters, and continue polishing up the proof on Wikipedia.

In the longer term I can try to understand what you're talking about.

Fine, @John Baez ! I share by the way your feeling regarding the classical approach to Artin's character theorem. To me, it lacks geometrical insight. I would reword property 1 of Serre as saying:
There exist a GG-set ZZ such that every gGg \in G fixes (at least) one point in ZZ.
I would call such a ZZ a spherical GG-set. An intriguing connection with Gelfand is that quite often you can construct a Gelfand Model for GG (the multiplicity free direct sum of all irreps of GG) by inducing from a suitable linear character α\alpha of the action groupoid A(Z,G)A(Z,G) associated to such a ZZ. Baby example: for G=S3G= S_3 an interesting ZZ for us is the union of the set of all three vertices of the equilateral triangle and the set of the two orientations of the triangle. Another one is the set of all involutions of S3S_3, the Identity included. For the latter, you can take α\alpha to be the sign character on each isotropy group of a flip and to be trivial on GG.
Property 2 says that for any irrep π\pi of GG you can pick a suitable virtual α\alpha such the resulting induced representation from A(Z,G)A(Z,G) to GG is a multiple of π\pi. (How does this work for S3S3 by the way?)
This rather curious fact is orthogonal to the fact that the induced representation provides a Gelfand Model. It would be nice to figure out a direct geometric proof of the equivalence of 1. and 2. Maybe it is sensible to discard the trivial case in which GG belongs to the family XX.

view this post on Zulip John Baez (Mar 09 2025 at 17:43):

Jorge Soto-Andrade said:

I share by the way your feeling regarding the classical approach to Artin's character theorem. To me, it lacks geometrical insight. I would reword property 1 of Serre as saying:
There exist a GG-set ZZ such that every gGg \in G fixes (at least) one point in ZZ.

How is Serre's property 1 equivalent to that? For those who haven't read the Wikipedia article, Serre is discussing a group GG with a collection XX of subgroups, and property 1 says that every element of GG is conjugate to an element of some subgroup HXH \in X.

view this post on Zulip Jorge Soto-Andrade (Mar 10 2025 at 22:44):

John Baez said:

Jorge Soto-Andrade said:

I share by the way your feeling regarding the classical approach to Artin's character theorem. To me, it lacks geometrical insight. I would reword property 1 of Serre as saying:
There exist a GG-set ZZ such that every gGg \in G fixes (at least) one point in ZZ.

How is Serre's property 1 equivalent to that? For those who haven't read the Wikipedia article, Serre is discussing a group GG with a collection XX of subgroups, and property 1 says that every element of GG is conjugate to an element of some subgroup HXH \in X.

view this post on Zulip John Baez (Mar 10 2025 at 23:23):

?

view this post on Zulip Jorge Soto-Andrade (Mar 11 2025 at 00:52):

Jorge Soto-Andrade said:

John Baez said:

Jorge Soto-Andrade said:

I share by the way your feeling regarding the classical approach to Artin's character theorem. To me, it lacks geometrical insight. I would reword property 1 of Serre as saying:
There exist a GG-set ZZ such that every gGg \in G fixes (at least) one point in ZZ.

How is Serre's property 1 equivalent to that? For those who haven't read the Wikipedia article, Serre is discussing a group GG with a collection XX of subgroups, and property 1 says that every element of GG is conjugate to an element of some subgroup HXH \in X.

(apologies for the slow reaction: too busy with the launching of our academic year 2025...).
Well, a subgroup HH of GG is just the isotropy group of a chosen point in a transitive GG-set, to wit the point HG/H H \in G/H. So hHh \in H means that hh fixes that point. But from a geometric perspective all points in G/H "are born equal" and the isotropy groups of other points may be described as the conjugates of HH if we are obsessed with HH. So the union of all conjugates of HH in GG consists of all gGg \in G which fix some point in G/HG/H. More intrisically, it consists of all gG g \in G which fix some point in a given transitive GG-set. Now, instead of looking at the bunch of HH 's in the collection XX, you look at the corresponding homogeneous spaces G/HG/H and you end up with a collection of transitive GG-sets, such every gGg\in G fixes (at least) some point in one of them. Those transitive GG-sets are just the GG-orbits of my GG-set ZZ. The other way around, you just choose a point in each GG-orbit in ZZ and consider its isotropy group. I would claim that whenever you are considering "all conjugates of ....'' you are missing an intrinsic viewpoint.
Also, it should be noted the characters χH\chi_H in Property 2 of Serre are allowed to be virtual. Moreover, the sum of induced representations therein is just the representation induced from a suitable (virtual) character of the Action Groupoid (Z,G)\mathcal (Z,G) to GG. It seems that we could as well work with action groupoids instead of GG-sets, by the way.

view this post on Zulip John Baez (Mar 11 2025 at 01:57):

Thanks, that's very helpful! I'll ponder it and try to use it to improve my understanding of things.

view this post on Zulip David Corfield (Mar 11 2025 at 09:36):

Sounds like Jorge should be preaching to the converted, at least the point about conjugates. I took in that idea of using groupoids to avoid making arbitrary choices from you around 30 years ago. But then there was Ronnie Brown strenuously emphasising it too.

That made it into my reasons for groupoids paper, The importance of mathematical conceptualisation, and then into my 2003 book.

view this post on Zulip John Baez (Mar 11 2025 at 21:39):

I've been focused on humbler stuff, like straightening out the proof of Artin's theorem on induced characters on Wikipedia - as a way to understand the nitty-gritty details of the proof. I'm glad someone put a version of Serre's proof on Wikipedia. I'm almost glad they put up a terribly written proof, where they do things like

view this post on Zulip John Baez (Mar 12 2025 at 01:12):

Okay, I'm done! My proof is just a spelling-out of Serre's proof, but it's here now:

view this post on Zulip David Corfield (Mar 12 2025 at 09:54):

Good stuff.

Regarding the way these induction theorems get used, I guess we're seeing Brauer's version leading to:

the values of a character of a finite group are cyclotomic integers,

as appearing in what Jack Morava is speaking about here:

In the theory of finite groups, the Adams operations are the endomorphisms of the representation ring functor. On the other hand, an old theorem of Brauer asserts that the values of a character of a finite group are cyclotomic integers, so the Galois group of the maximal abelian extension of the rationals (i.e. the cyclotomic field) acts naturally on the character ring. Serre’s argument shows that these two actions are essentially the same.

Are there similar uses for other induction theorems?

view this post on Zulip David Corfield (Mar 12 2025 at 11:36):

I see John mentioned an answer earlier:

I also love how Artin got into his theorem on induced characters by trying to understand L-functions (which is something I'm also trying to understand).

view this post on Zulip John Baez (Mar 12 2025 at 17:08):

Yes, Artin introduced a kind of L-function that arises from a finite-dimensional representation of the Galois group of a number field, now called an [[Artin L-function]]. His induction theorem implied that any such L-function is a product of rational powers of L-functions coming from 1-dimensional representations of cyclic subgroups of that Galois group. This let him show that some integer power of the Artin L-function extends to a meromorphic function on the complex plane.

Artin did this in 1927. Later, in 1947, Brauer proved his induction theorem, which implies that any Artin L-function is a product of integer powers of L-functions coming from 1-dimensional representations of elementary subgroups of that Galois group. This implies that any Artin L-function extends to a meromorphic function on the complex plane.

view this post on Zulip John Baez (Mar 12 2025 at 17:11):

So in this example it seems that Brauer's induction theorem is just better - it gives a stronger result. The rational powers above become integer powers because every element of the representation ring of the Galois group is a rational linear combination of characters of 1-dimensional representations induced from cyclic subgroups, but an integer linear combination of characters of 1-dimensional representations induced from elementary subgroups. I don't understand how Brauer (or whoever) got around the downside, which is that elementary subgroups aren't abelian.

view this post on Zulip John Baez (Mar 12 2025 at 17:20):

By the way I'm learning some of this stuff from a nice historical paper on the work of Artin:

To give a bit of the flavor, here's the start:

Emil Artin spent the first 15 years of his career in Hamburg. Andre Weil characterized
this period of Artin's career as a "love affair with the zeta function". Claude Chevalley, in his obituary of Artin, pointed out that Artin's use of zeta functions was to discover exact algebraic facts as opposed to estimates or approximate evaluations. In particular, it seems clear to me that during this period Artin was quite interested in using the Artin L-functions as a tool for finding a non-abelian class field theory, expressed as the desire to extend results from relative abelian extensions to general extensions of number fields.

view this post on Zulip John Baez (Mar 12 2025 at 17:26):

"Nonabelian class field theory" is now the topic of the "Langlands program".

Since Artin was the expert on abelian class field theory (culminating in Artin reciprocity), it's natural that when he tackled L-functions coming from representations of nonabelian Galois groups, he tried to break them down into L-functions coming from representation of abelian subgroups... leading him to prove the Artin induction theorem, a result in representation theory that doesn't even mention L-functions. Brauer's result is a lot subtler, and it brings in nonabelian subgroups, namely "elementary" subgroups.

view this post on Zulip John Baez (Mar 12 2025 at 17:34):

There may be some ways in which Artin's induction theorem is "better" - leads to more powerful consequences - because it only involves cyclic subgroups. I don't think I've seen them! So far, its main advantage is that the proof is much easier, so by now I completely understand it.

By the way, I really love how number theory repeatedly forced advances in group theory and group representation theory, leading to theorems that are a lot easier to explain if you skip all the number theory!

view this post on Zulip Jorge Soto-Andrade (Mar 14 2025 at 01:48):

David Corfield said:

Sounds like Jorge should be preaching to the converted, at least the point about conjugates. I took in that idea of using groupoids to avoid making arbitrary choices from you around 30 years ago. But then there was Ronnie Brown strenuously emphasising it too.

That made it into my reasons for groupoids paper, The importance of mathematical conceptualisation, and then into my 2003 book.

Thanks David, for the inspiring references, I was not aware of (shame on me)! Do you mean your book Toward a philo or real math, btw? Yes, the fundamental groupoid and van Kampen theorem (I learned that from Michel Zisman, in the 70's...)! You might be interested in this interdisc paper (with whose authors I am arguing about groupoids being better adapted to what they try to do than groups)
https://www.dropbox.com/scl/fi/qjqfu2l92r8r8inweoz46/Kallen.Macpherson.Miles.Richardson.2021.Symmetries-of-social-performance-environment-systems.pdf?rlkey=aogho49wsbbxlvbkgl2403a9s&dl=0

view this post on Zulip Jorge Soto-Andrade (Mar 14 2025 at 01:53):

Btw, I am still trying to understand Serre's proof of Artin in terms of groupoids (the action groupoid associated to a GG-set where each gGg \in G fixes (at least) a point and induced representations from this groupoid to GG). It seems that induction from the action groupoid associated to any GG-set is not so popular as it should be... I have not yet read John's proof, anyway.

view this post on Zulip John Baez (Mar 14 2025 at 02:17):

My proof is just an expository version of Serre's proof for people who haven't read the previous material in his book.

view this post on Zulip David Corfield (Mar 14 2025 at 07:26):

Jorge Soto-Andrade said:

Do you mean your book Toward a philo or real math, btw?

Yes, Chap. 9 of that. I was intrigued to have an example of a concept that seemed to be the subject of great contention as to its worth.

David Corfield - Towards a Philosophy of Real Mathematics Chap 9.pdf

Thanks for the link!

view this post on Zulip Jorge Soto-Andrade (Mar 14 2025 at 23:09):

David Corfield said:

Jorge Soto-Andrade said:

Do you mean your book Toward a philo or real math, btw?

Yes, Chap. 9 of that. I was intrigued to have an example of a concept that seemed to be the subject of great contention as to its worth.

David Corfield - Towards a Philosophy of Real Mathematics Chap 9.pdf

Thanks for the link!

Sad news. Basil Hiley just passed away. Did you know him? He was a British quantum physicist fond of groupoids, Clifford Algebras and category theory:
https://www.cs.ox.ac.uk/conferences/categorieslogicphysics/clap2/clap2-basilhiley.pdf
We discussed a bit a couple of years ago about those instances where the boundaries between operator and argument (and value) become blurred...