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Stream: learning: questions

Topic: Virtual representations and groupoidification

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, $G$. Then comparing $G\text{-}Set$ with $G\text{-}Rep$, the linear representations of the latter are much richer than $G$ merely acting on sets.

However, if we're allowed virtual permutations representations, formal differences between $G$-sets, then things become much more comparable. The rig functor from $G\text{-}Set$ to $G\text{-}Rep$ which derives from forming the free vector space on a set, extends to a ring homomorphism from the Burnside ring (Grothendieck ring of $G\text{-}Set$ with addition and multiplication) to the representation ring. And one has to root about to find a $G$ 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 $G$ occurs in this comment by Urs Schreiber.

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

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 $E_{\infty}$-ring spectra: $\mathbb{S} = K\mathbb{F}_1 \to K \mathbb{C} \to KU$ (as explained here).

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 $G$ 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 $\mathsf{Rep}(G)$ and representation rings, you get to specify which field you're working over. People heavily focused on physics would tend to use $\mathbb{C}$, 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 $S_n$ are nice in that this effect doesn't happen: all irreducible representations are definable already over $\mathbb{Q}$. But for most groups this effect occurs, e.g. the left action of $\mathbb{Z}/3$ on itself gives a 3-dimensional representation that's irreducible over $\mathbb{Q}$, but breaks in two as soon as your field contains $\exp(2 \pi i /3)$.

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 $\mathbb{Q}$ you're done; going further to $\mathbb{C}$ or beyond doesn't help.

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!

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.

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 $\mathbb{Q}$.

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 $\mathbb{F}_1$ business.

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

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 $\mathbb{C}$, 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.]

John Baez (Feb 28 2025 at 23:02):

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

Each simple algebra over $G$ is isomorphic to a matrix algebra $M_n(D)$ where $D$ is a division algebra over $\mathbb{Q}$. Unlike over $\mathbb{R}$, there are tons of division algebras over $\mathbb{Q}$. 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 $\mathbb{Q}[G]$ corresponds to an irreducible representation of $G$.

John Baez (Feb 28 2025 at 23:19):

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

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

John Baez (Feb 28 2025 at 23:21):

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

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

John Baez (Feb 28 2025 at 23:28):

Okay, so now to Serre's example. He takes the quaternion group $Q_8$ and embeds it in the quaternions in the obvious way, but he uses the quaternions defined over the rationals, which he calls $\mathbb{H}_{\mathbb{Q}}$ - so it's a 4-dimensional rational vector space. When I say 'quaternions' I'll mean these rational quaternions.

He also embeds $\mathbb{Z}/3$ 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 $n$ 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 $\mathbb{Z}/3$ 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.

John Baez (Feb 28 2025 at 23:32):

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

John Baez (Feb 28 2025 at 23:38):

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

John Baez (Feb 28 2025 at 23:50):

He claims this representation is irreducible. I believe that already the representation of $Q_8$ on $\mathbb{H}_\mathbb{Q}$ is irreducible. If it split as the sum of a 1-dimensional rep and a 3-dimensional rep, we'd have a nonzero quaternion $q$ that's fixed by right multiplication by $i, j,$ and $k$ - impossible, because $qi, qj$ and $qk$ 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 $i, j$ and $k$ - impossible, for the same reason!

John Baez (Feb 28 2025 at 23:54):

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

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

John Baez (Feb 28 2025 at 23:56):

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

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 $K$ is related somehow to the cube root of unity $\omega$ that we picked in $\mathbb{H}_{\mathbb{Q}}$; there is probably some sense in which $K$ "is" $\mathbb{Q}[\omega]$.

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

(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 $\mathbb{H}_\mathbb{Q}$ is 4. I may be making mistakes because I'm used to representation theory over $\mathbb{C}$. While $\mathbb{H}_\mathbb{Q}$ has dimension 4 over $\mathbb{Q}$, $\mathbb{H}$ has dimension 2 over $\mathbb{C}$, and maybe that's the relevant number!)

John Baez (Mar 01 2025 at 00:27):

I thought a bit about the puzzle. Notice that

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

John Baez (Mar 01 2025 at 00:36):

• $\mathbb{Q}[\mathbb{Z}/3]$ is the free algebra (over $\mathbb{Q}$) on a cube root of unity, so it should be closely related to $\mathbb{Q}[\exp(2 \pi i /3)]$. But it's 3-dimensional over $\mathbb{Q}$ while $\mathbb{Q}[\exp(2 \pi i /3)]$ is just 2-dimensional because $\exp(2 \pi i /3)$ obeys a quadratic equation over $\mathbb{Q}$.

• $\mathbb{Q}[Q_8]$ has a basis of 8 guys obeying the quaternion relations, so it should be closely related to $\mathbb{Q}_{\mathbb{H}}$. But it's 8-dimensional over $\mathbb{Q}$ while $\mathbb{Q}_{\mathbb{H}}$ is just 4-dimensional because in there $1$ and $-1$ are linearly dependent, and so are $i$ and $-i$, and so on. It's a semisimple algebra so I'm willing to guess it's just $\mathbb{Q}_{\mathbb{H}} \oplus \mathbb{Q}_{\mathbb{H}}$. On the other hand, maybe it's $M_2(\mathbb{Q}[i])$.

John Baez (Mar 01 2025 at 00:39):

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

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.

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 $\mathbb{Z}/3$ and $Q_8$, this must also be true for $G = \mathbb{Z}/3 \times Q_8$ (using the fact that $\mathbb{Z}/3$ is a 3-group - every element has order a power of 3 while $\mathbb{Q}_8$ is a 2-group).

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 $H \subseteq \mathbb{Z}/3 \times Q_8$ is of the form $H_1 \times H_2$ where $H_1$ is a subgroup of $\mathbb{Z}/3$ and $H_2$ is a subgroup of $Q_8$. This isn't true for all pairs of groups: consider the product of $\mathbb{Z}/2$ and $\mathbb{Z}/2$. 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 $G_1 \times G_2$ is a product of a subgroup $H_1 \subseteq G_1$ and a subgroup $H_2 \subseteq G_2$ - we have

$B(G_1 \times G_2) \cong B(G_1) \otimes B(G_2)$

where $B$ means 'Burnside ring'.

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

$R(G_1 \times G_2) \cong R(G_1) \otimes R(G_2)$

Thus, if we have

$B(G_1) \cong R(G_1)$ and $B(G_2) \cong R(G_2)$

we get

$B(G_1 \times G_2) \cong R(G_1 \times G_2)$

But this would mean Serre's counterexample

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

is only possible if

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

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

So we get a contradiction.

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.

John Baez (Mar 01 2025 at 18:21):

Meanwhile I've almost worked out the representation ring and Burnside ring of $\mathbb{Z}/3$ and $Q_8$, which was a lot of fun, but there's not much point in reporting on that until I straighten out my confusion.

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 $Q_8$, and was delighted to learn that all 6 of its subgroups are normal! Three of them $H_1, H_2, H_3$ have 4 elements, so they not only give an action of $Q_8$ on a 2-element set $Q_8/H_i$ but also a 1-dimensional rational representation $Q_8 \to Q_8/H \cong \mathbb{Z}/2 \hookrightarrow \mathbb{Q}^\ast$. The fourth is its center $\{\pm 1\}$, and the fifth and sixth are the whole group $Q_8$ 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.

John Baez (Mar 01 2025 at 19:11):

I feel, but have not proved, that every subgroup $H \subseteq \mathbb{Z}/3 \times Q_8$ is of the form $H_1 \times H_2$ where $H_1$ is a subgroup of $\mathbb{Z}/3$ and $H_2$ is a subgroup of $Q_8$. This isn't true for all pairs of groups: consider the product of $\mathbb{Z}/2$ and $\mathbb{Z}/2$. 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.

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.

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 $R(G) \cong B(G)$ for $G = \mathbb{Z}/3 \times Q_8$ as my perhaps flawed argument shows, that doesn't imply the canonical map from $B(G)$ to $R(G)$ is an isomorphism! It could be non-injective (hence non-surjective).

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 $\mathbb{Z}/3$ and $Q_8$, 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.

John Baez (Mar 01 2025 at 23:56):

For $\mathbb{Z}/3$:

1. We have two irreducible representations over $\mathbb{Q}$: the 1d trivial representation $R_1$ and a 2d representation $R_2$ where the generator of $\mathbb{Z}/3$ 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.

2. We have two transitive actions on finite sets: the trivial action $A_1$ and the left action of $\mathbb{Z}/3$ on itself, say $A_3$.

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

$A_1 \to R_1$

$A_3 \to R_1 \oplus R_2$.

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

John Baez (Mar 02 2025 at 00:44):

For $Q_8$:

1. We seem to have 5 irreducible representation of $Q_8$ over $\mathbb{Q}$. There's the 1d trivial representation $R_1$. There are three nontrivial 1d representations, say $R_i, R_j, R_k$. In $R_i$, for example, $\pm i \in Q_8$ acts as the identity while $\pm j$ and $\pm k$ act as $-1$. The other two are analogous. There's also a 4d representation, say $R_4$, 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.

2. We have 6 transitive actions of $Q_8$ on finite sets, corresponding to the 5 subgroups of $Q_8$, which happen to all be normal. There's the trivial action, say $A_1$. There are three actions on 2-element sets, say $A_i, A_j, A_k$, coming from three 4-element normal subgroups. For example $A_i$ is the action on $Q_8$ on $Q_8/\{1, i , -1, -i \} \cong \mathbb{Z}/2$. The other two are analogous. There's an action $A_4$ of $Q_8$ on a 4-element set: in this $Q_8$ acts on $Q_8$ modulo its center: $Q_8/\{\pm 1\}$ is the Klein 4-group. And there's the action $A_8$ of $Q_8$ on an 8-element set, namely itself.

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

$A_1 \mapsto R_1$

$A_i \mapsto R_i \oplus R_1 , \quad A_j \mapsto R_j \oplus R_1, \quad A_k \mapsto R_k \oplus R_1$

$A_4 \mapsto R_i \oplus R_j \oplus R_k \oplus R_1$ (just guessing)

$A_8 \mapsto R_4 \oplus R_4$ (just guessing)

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 $\mathbb{Z}/3$ and $Q_8$). The nLab says:

The proof of surjectivity for $p$-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.

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 $p$-group $G_1$ and a $q$-group $G_2$ where $p$ and $q$ were distinct primes, then they'd also be isomorphic for $G_1 \times G_2$. But we're already seeing they're not isomorphic for $Q_8$! :face_with_spiral_eyes: So I was completely missing the point. The point is that the maps from Burnside ring to representation rings

$B(G_i) \to R(G_i)$

are surjective for two groups $G_1$ and $G_2$, yet apparently not for their product $G_1 \times G_2$.

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 $\mathbb{Z}/3$ and $Q_8$. 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.

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 $G$ (i.e. (complex) representations which split as the multiplicity-free sum of all irreducible representations of $G$). Moreover, you have a general way to map the dual of an irreducible representation $(V, \rho)$ of a finite group $G$ to $Hom_G ( V, L^2(G) )$, where
$L^2(G)$ denotes the right regular representation of $G$, to wit :
send $\omega \in V^*$ to the map which sends $v \in V$ to
the function $g \mapsto \omega(\rho_g(v))$. This affords an isomorphism between $V^*$ and $Hom_G ( V, L^2(G) )$. Over the complex field this proves then that the multiplicity of $(V, \rho)$ 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.

John Baez (Mar 02 2025 at 01:42):

I'd like to see it!

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 $Q_8 \times \mathbb{Z}/3$. But since $Q_8$ appears to be acted on by $\mathbb{Z}/3$ by rotating the generators, I'm curious if the semi-direct product could also be a counterexample.

John Baez (Mar 02 2025 at 18:23):

Serre doesn't say why he chose the counterexample $G = \mathbb{Z}/3 \times Q_8$. 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 $\mathbb{Q}[G]$ is a matrix algebra $M_n(D)$ where $D$ is a division ring and $n \ge 2$. But I don't yet understand his logic. In this example $D$ is the field

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

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

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

• Simon Burton, Hisham Sati, Urs Schreiber, Lift of fractional D-brane charge to equivariant Cohomotopy theory

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 $\mathbb{Z}/3 \times Q_8$?" I believe

• $\mathbb{Z}/3$ is the smallest group with an irreducible representation over $\mathbb{Q}$ whose ring of endomorphisms is not merely $\mathbb{Q}$.
• $Q_8$ is the smallest nonabelian group all of whose subgroups are normal.
• $\mathbb{Z}/3 \times Q_8$ has both these properties and maybe it's the smallest group with both properties.

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

By the way, the nontrivial semidirect product $\mathbb{Z}/3 \ltimes Q_8$ is famous: it's called the binary tetrahedral group because it has a 2-1 and onto homomorphism to the group $A_4$, 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.

John Baez (Mar 03 2025 at 01:54):

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

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

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.

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

$G = \mathbb{Z}/3 \times Q_8$

But he doesn't mention the Burnside ring or representation ring! Instead, he just claims that $G$ has an irreducible representation $\rho$ whose character $\chi$ is not a linear combination of characters of representations $1^G_H$, which is his jargon for the representation of $G$ induced from the trivial representation on some subgroup $H \subseteq G$. 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!

John Baez (Mar 03 2025 at 02:08):

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

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

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

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 $M_2(K)$, and I'm happy to explain if asked, but let me skip ahead to where things get tough for me!

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 $G$ is not a linear combination of characters of representations $1^G_H$.

John Baez (Mar 03 2025 at 02:15):

Remember the previous exercise:

From Serre's Linear Representations of Finite Groups.

John Baez (Mar 03 2025 at 02:21):

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

1. $\rho$ is faithful, i.e. one-to-one. (I see how to check this in our example.)
2. Every subgroup of $G$ is normal. (I explained why this is true for $\mathbb{Z}/3$ and $Q_8$, so it's true for their product $G$.)
3. The simple algebra corresponding to $\rho$ is of the form $M_n(D)$ for $D$ a division algebra over $\mathbb{Q}$, with $n \ge 2$. (As mentioned, I think this holds in our example, with $D = \mathbb{Q}[\exp(2 \pi i /3)]$ and $n = 2$).

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!

John Baez (Mar 03 2025 at 02:31):

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

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

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

Suppose $1^G_H$ contains $\rho$ as a subrepresentation either $n$ times or not at all depending on whether $H \subseteq G$ is the trivial group or not.

Then:

• when $H = \{1\}$, the character of $1^G_H$ is a linear combination of characters of irreps where the character $\chi$ shows up with the coefficient $n$
• while for all other $H$, the character of $1^G_H$ is a linear combination of characters of irreps where $\chi$ shows up with the coefficient $0$.

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

John Baez (Mar 03 2025 at 02:38):

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

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 $\rho$ be an irreducible representation of a finite group $G$ such that

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

Then for every subgroup $H \subseteq G$, the representation $1^G_H$ contains the representation $\rho$ $n$ times if $H$ is trivial, and not at all otherwise.

John Baez (Mar 03 2025 at 02:45):

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

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

John Baez (Mar 03 2025 at 02:52):

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

$\langle \chi, \chi^G_1 \rangle = \dim \rho$

John Baez (Mar 03 2025 at 02:55):

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

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 $\rho$ of $G = \mathbb{Z}/3 \times Q_8$ on $\mathbb{H}_{\mathbb{Q}}$ has

$\dim \rho = 4$,

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

$n = 2$

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 $\mathbb{C}$, or algebraically closed fields of characteristic zero, and blindly applied it to representations over $\mathbb{Q}$. It's apparently not true that

$\langle \chi, \chi^G_1 \rangle = \dim \rho$ :warning:

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

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

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

John Baez (Mar 03 2025 at 05:50):

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

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

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

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

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

$\langle \chi, \chi \rangle = \dim \text{End}(\rho) = \dim D$

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

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

But both these fail in general.

John Baez (Mar 03 2025 at 05:53):

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

John Baez (Mar 03 2025 at 05:57):

$\chi_1^G$ 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 $\rho$ appears with multiplicity $n$ in the regular representation, so

$\langle \chi , \chi_1^G \rangle = n \langle \chi, \chi \rangle = n \text{dim}(D)$

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 $\rho$ be an irreducible representation of a finite group $G$ such that

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

Then for every subgroup $H \subseteq G$, the representation $1^G_H$ contains the representation $\rho$ $n$ times if $H$ is trivial, and not at all otherwise.

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

$\langle \chi^G_H, \chi \rangle = n \dim(D)$

when $H$ is the trivial group, and

$\langle \chi^G_H, \chi \rangle = 0$

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

John Baez (Mar 03 2025 at 06:19):

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

$\langle \chi_H^G, \chi \rangle = \langle \chi^G_1, \chi \rangle = n \dim(D)$

So I win this round! :tada:

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 $H \subseteq G$ is the trivial group. Somehow when I consider other subgroups $H$ I'll need assumption 2, which says $H$ is normal. And I guess I'll need assumption 1, which says the representation $\rho$ is faithful.

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

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 $\mathbb{Q}$, the map from the Burnside ring to the representation ring of a finite group $G$ is nonsurjective if $G$ has an irreducible representation $\rho$ such that

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

I suspect that the smallest group to which this criterion applies is our friend $\mathbb{Z}/3 \times Q_8$. 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!

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 $A$ to $B$ has something of the flavour of a difference $(B - A)$.

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 $G$ looking only at the permutation representations: the representations coming from finite $G$-sets. But any $G$-equivariant relation between $G$-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 $f: A \to B$ we get

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

John Baez (Mar 03 2025 at 15:51):

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

John Baez (Mar 03 2025 at 15:52):

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

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?

John Baez (Mar 03 2025 at 17:43):

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

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.

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

$G = \mathbb{Z}/3 \times Q_8$

But he doesn't mention the Burnside ring or representation ring! Instead, he just claims that $G$ has an irreducible representation $\rho$ whose character $\chi$ is not a linear combination of characters of representations $1^G_H$, which is his jargon for the representation of $G$ induced from the trivial representation on some subgroup $H \subseteq G$. 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 $G$ whose Gelfand Model (the multiplicity-free direct sum of all irreps of $G$) 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 $G$, 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 $G$ which are"essentially" isomorphic via a Radon transform, which is just a stochastic mapping in the category of $G$-sets. Baby example: Take the square (as a graph) and its symmetry group $G$. The natural representations of $G$ associated to the $G$-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. $G$-invariant) way to pair a vertex with an edge in the square. Does this make sense?

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

$G = \mathbb{Z}/3 \times Q_8$

But he doesn't mention the Burnside ring or representation ring! Instead, he just claims that $G$ has an irreducible representation $\rho$ whose character $\chi$ 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 $G$ 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.

Jorge Soto-Andrade (Mar 03 2025 at 21:14):

John Baez said:

Clearly we should figure this out for $G = \mathbb{Z}/3 \times Q_8$. :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 $G$, 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 $G$ 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 $G$ (a finite group of Lie type) à la Solomon-Lehrer-Tits. The corresponding baby example (which Arnold would request...) is the case of $G= SL(2, \mathbb F_q)$ and $V = Ker \Phi, \Phi : L^2(X) \rightarrow \mathbb C$, $X$ being the set of lines throught the origin in the finite plane over $\mathbb F_q$ and $\Phi$ being just the integral (summation) operator. Zero sum is just the cocycle condition.

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.

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 $G$ whose Gelfand model (the multiplicity-free direct sum of all irreps of $G$) 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 $G$, 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?

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 $G$ whose Gelfand model (the multiplicity-free direct sum of all irreps of $G$) 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 $G$, 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 $(L^2(X), \tau)$ of $G$ where $X$ is a $G$-set and $\tau_g(f) = f \circ g^{-1}$. 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 $X \mapsto L^2(X)$ 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 $G$. Then, a nice first conjecture is that a Gelfand Model for $G$ lies in the Green ring of $G$, and the same for the other symmetric functions of the irreps of $G$, the "last one" being the tensor product of all irreps. Of course, hoping that all irreps of $G$ lie in the Green ring of $G$ 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 ...

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.

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!

John Baez (Mar 04 2025 at 02:38):

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

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 $G$ is nonsurjective if $G$ has an irreducible representation $\rho$ such that

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

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

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

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

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

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

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

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

John Baez (Mar 04 2025 at 02:44):

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

John Baez (Mar 04 2025 at 02:56):

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

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

$g' : gH \to g'gH$

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

$g': Hg \to g' H g$

This lets us see that $H$ acts trivially; if $h \in H$ we have

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

John Baez (Mar 04 2025 at 02:58):

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

John Baez (Mar 04 2025 at 03:01):

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

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 $k$ of characteristic zero.

Theorem. Suppose $G$ is a finite group with a linear representation $\rho$ such that:

1. $\rho$ is irreducible and faithful

2. every subgroup of $G$ is normal

3. $\rho$ appears with multiplicity $n \ge 2$ in the regular representation of $G$.

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

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

Since every finite $G$-set is a coproduct of transitive actions of $G$, which are isomorphic to actions on $G/H$ for subgroups $H$ of $G$, every permutation representation of $G$ is a direct sum of those on spaces of the form $k[G/H]$. Thus, it suffices to show that the multiplicity of $\rho$ in the representation on $k[G/H]$ is $n$ if $H$ is the trivial group, and $0$ otherwise.

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

$h H g = H g .$

Since $H$ is nontrivial, it contains elements $h \ne 1$ that act trivially on $k[G/H]$. But no $h \ne 1$ can act trivially on $\rho$ because $\rho$ is faithful, by assumption 1. Thus $\rho$ cannot be a subrepresentation of $k[G/H]$. That is, $\rho$ appears with multiplicity $0$ in $k[G/H]$. :notebook:

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

John Baez (Mar 04 2025 at 16:54):

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

I should have known this, because I knew most groups have order a power of 2, and that $\beta$ is surjective for all 2-groups. But David Benson quickly assembled a list of groups of order $2^n \cdot p$ for which $\beta$ 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 $\mathbb{C}$. Then it's much easier to find groups where $\beta$ is not surjective, because most groups have lots of representations definable only over an abelian extension of $\mathbb{Q}$, and those can never be formal difference of permutation representation. This happens for $G = \mathbb{Z}/n$ whenever $n > 2$.

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 $Q_8 \times \mathbb{Z}/3$. But since $Q_8$ appears to be acted on by $\mathbb{Z}/3$ 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 $\mathbb{Z}/3 \ltimes Q_8$ is not a counterexample: it's the binary tetrahedral group, and for this the map $\beta$ from the Burnside ring to the representation ring is surjective.

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

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

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 $\beta$ 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.)

David Corfield (Mar 06 2025 at 10:50):

John Baez said:

First, that semidirect product $\mathbb{Z}/3 \ltimes Q_8$ is not a counterexample: it's the binary tetrahedral group, and for this the map $\beta$ 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

• @Simon Burton, Hisham Sati, Urs Schreiber, Lift of fractional D-brane charge to equivariant Cohomotopy theory

image.png

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 $\textrm{SU}(2)$ 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.

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...

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 $G$-sets $G/H$ for all possible subgroups $H$) and how it maps to the standard basis of the representation ring (given by irreducible representations of $G$).

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.

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 $SL(2, \mathbb F_q)$. So it is not reasonable to expect that the irreps are Green

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)

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.

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 $2 \cdot D_{2n}$ where $2n \le 12$, and they conjectured that this pattern continues. As I mentioned, it's always true for $2 \cdot D_{2^n}$. If it's true for all binary dihedral groups, it's true for all finite subgroups of $\text{SU}(2)$.

(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.)

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.

Jorge Soto-Andrade (Mar 07 2025 at 04:43):

I was talking of irreps over $\mathbb C$. First step if you wish, before "going down" to $\mathbb Q$ 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...)

Simon Burton (Mar 07 2025 at 10:10):

Ah right, thanks for the clarification Jorge.

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.

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.

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...

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 $G$.

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 $G$ that are induced from representations of the form $\rho: H \to \text{U}(1)$ where $H \subseteq G$ is a subgroup. $H$ can even be taken to be of a very special form, and we can replace $\text{U}(1)$ 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 $\text{U}(1)$ 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.

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.

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 $E_{\infty}$-ring spectra: $\mathbb{S} = K\mathbb{F}_1 \to K \mathbb{C} \to KU$ (as explained here),

then $\mathbb{S}$ 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 $KU$. Is $MU_G(\ast)$ just too difficult to work with?

image.png

David Corfield (Mar 08 2025 at 10:12):

That's not terribly promising:
image.png
here

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 $\text{U}(1)$. 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.

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 $G$ as virtual representations if we start from 1-dimensional representations of subgroups $H$ and induce up to the whole group. And while you can think of these 1d representations as homomorphisms

$\rho: H \to \text{U}(1)$

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

$\rho: H \to \mathbb{Z}/n$

where $n$ is big enough, e.g. the order of $H$.

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 $G$.

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 $G$, 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 $\text{Rep}(G)$ 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.

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:

• Roger Penrose, Angular momentum: an approach to combinatorial space-time.

Mathematically what this does is express a bunch of the representation theory of $\text{SU}(2)$ 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.

John Baez (Mar 08 2025 at 19:25):

Here's a goal one can have for any group $G$: describe the irreducible complex representations of $G$, and for each collection of irreps $V_1, \dots, V_m, W_1, \dots, W_m$ a basis of morphisms

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

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 $G$, where the colors are the irreducible representations.

John Baez (Mar 08 2025 at 19:26):

Penrose essentially did this for $\text{SU}(2)$, and the same ideas work for $\text{SU}(n)$. 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 $\text{SU}(2)$ first!

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.

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 $G$ as virtual representations if we start from 1-dimensional representations of subgroups $H$ and induce up to the whole group. And while you can think of these 1d representations as homomorphisms

$\rho: H \to \text{U}(1)$

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

$\rho: H \to \mathbb{Z}/n$

where $n$ is big enough, e.g. the order of $H$.

Before reading further, I would like to claim that the "right" notion of induction is not induction from a fixed subgroup $H$ of $G$ but from the groupoid $\mathcal G_X$ canonically associated to a $G$-set $X$, so we are not choosing an origin in $X$. 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 $\mathcal G_X$ is a powerful tool indeed. It allows you to construct Gelfand Models for your groups. For example, for $G = S_n$ you take as $X$ the space of involutions of $G$. For the $q$-analogue, $G = GL(n, \mathbb F_q)$, or better $G = PGL(n, \mathbb F_q)$ you take of course $X$ to be set of all symmetric matrices in $G$. This is a re interpretation of the work of Klyachko, a clever Ukrainian student of Gelfand.