## November 29, 2007

### Geometric Representation Theory (Lecture 14)

#### Posted by John Baez

This time in the Geometric Representation Theory seminar I tried to state the Fundamental Theorem of Hecke Operators… and I screwed up. Luckily, I screwed up in an instructive way!

Pick a group $G$. We’ve seen that every $G$-invariant relation between finite $G$-sets gives an intertwining operator between the resulting permutation representations of $G$. These operators are called Hecke operators.

Any category theorist worth their salt should want to make this into a functor. Since there’s a category $FinRel^G$ with

• finite $G$-sets as objects
• $G$-invariant relations as morphisms

and a category $FinVect^G$ with

• finite-dimensional representations of $G$ on complex vector spaces as objects
• $G$-invariant operators (= intertwining operators) as morphisms

one might hope the Hecke operator trick gave a functor

$F: FinRel^G \to FinVect^G$

But, it doesn’t!

That wasn’t my mistake. I can be dumb… but I’m not that dumb.

In fact, I began my lecture explaining this problem and its origin. Fundamentally, the problem is that a relation is a matrix taking values in the rig of truth values, $\{0,1\}$, while a linear operator between vector spaces equipped with bases is a matrix taking values in the rig $\mathbb{C}$. There’s a tempting inclusion

$\{0,1\} \hookrightarrow \mathbb{C}$

and this is lets us turn relations into linear operators. By functorial abstract nonsense, it turns $G$-invariant relations into $G$-invariant linear operators… and this is the Hecke operator trick.

But, this tempting inclusion is not a rig homomorphism, because “true or true does not mean twice as true”! So the Hecke operator trick does not give a functor

$F: FinRel^G \to FinVect^G$

To get around this problem I introduced spans of finite sets, which you can think of (in a slightly wimpy way) as matrices of natural numbers. The inclusion

$\mathbb{N} \hookrightarrow \mathbb{C}$

is a rig homomorphism, so we do get a functor

$F: FinSpan^G \to FinVect^G$

where $FinSpan^G$ is the category with

• finite $G$-sets as objects
• spans of finite $G$-sets as morphisms

My problem came when I wanted to state the really cool fact about Hecke operators using this functor. The really cool fact is that Hecke operators coming from atomic $G$-invariant relations — those that can’t be broken down further using ‘or’ — form a basis of intertwining operators between the resulting permutation representations. I tried to state this as follows.

First of all, I noted that for any objects $X,Y \in FinSpan^G$, $hom(X,Y)$ is a module of the rig $\mathbb{N}$, while $hom(F X, F Y)$ is a complex vector space — that is, a module of the rig $\mathbb{C}$. That’s all true.

Then I noted that

$F: hom(X,Y) \to hom(F X , F Y)$

gives rise to a linear operator

$hom(X,Y) \otimes_{\mathbb{N}} \mathbb{C} \to hom(F X , F Y)$

That’s true too.

And then I claimed this operator was one-to-one and onto! And that’s false. As Jim pointed out right after class — damn him! — it’s onto but not one-to-one.

But, all is not lost. In later lectures, Jim used my error as a springboard to study various forms of decategorification, and to give a really deep account of the form we need in this course— namely, degroupoidification. Degroupoidification is something we wanted to talk about anyway, but now we see how it allows us to state the Fundamental Theorem of Hecke Operators in a really beautiful, conceptual way.

• Lecture 14 (Nov. 13) - John Baez on matrix mechanics and Hecke operators. Any rig $R$ gives a category $Mat(R)$ whose objects are finite sets and whose morphisms are $R$-valued matrices. Any rig homomorphism from $R$ to $R'$ gives a functor from $Mat(R)$ to $Mat(R)$. The homomorphism from $\mathbb{N}$ to $\mathbb{C}$ lets us turn spans of finite sets into linear operators between finite-dimensional vector spaces. We can thus turn $G$-invariant spans between $G$-sets into intertwining operators between finite-dimensional representations of $G$. These are Hecke operators. A flawed attempt to formally state the “Fundamental Theorem of Hecke Operators” in terms of this functor.
Posted at November 29, 2007 7:45 PM UTC

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### Re: Geometric Representation Theory (Lecture 14)

Degroupoidification is something we wanted to talk about anyway, but now we see how it allows us to state the Fundamental Theorem of Hecke Operators in a really beautiful, conceptual way.

So what’s the final answer? The suspense is killing me!

Posted by: Bruce Bartlett on December 3, 2007 2:30 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 14)

You should think of the suspense as a good thing — math classes are often short on suspense.

But, the answer is explained in the next few lectures. So, if the tension is really unbearable, you can peek ahead to see lectures that haven’t been ‘officially released’ yet. Right now it’s the last week of the Fall quarter at UCR, but I plan to dole out the lectures up to lecture 20 at a slow rate throughout the Christmas break, to keep everyone happily entertained until classes resume on January 4th.

Posted by: John Baez on December 3, 2007 9:41 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 14)

Ok I peeked ahead… but I still haven’t found an explicit statement of the groupoidy version of the theorem!

But it seems you are leading towards the same statement, except perhaps the word “span” must be interpreted as “spans of groupoids” and not just “spans of finite $G$-sets”. I guess in the latter approach, the hom’s are too restrictive…we really want to be thinking of functors between the groupoids representing the $G$-sets, not functions between the $G$-sets themselves, as making up the legs of our spans.

But if that’s true (and perhaps I am badly confused here)… then that’s just the stuff you taught us way back in the first TWF articles on the Tale of Groupoidification, viz. “Always work in the language of groupoids, and everything will be fine”.

In other words, we always knew that was the language we were supposed to use… so the lectures up to now seem very strange to me. In effect, they follow the pattern : “Look guys, here’s a way - which we can all see doesn’t follow the tao - in which we might go about our business… and look, it goes wrong for such and such reasons. So now we’re going to use the language we all knew we should have used from the start.”

Posted by: Bruce Bartlett on December 4, 2007 9:07 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 14)

Let me clarify my statement somewhat : what you appear to be telling us is that, if we work in the language of groupoids, then we must take equivalence classes of spans, and not just isomorphism classes (of the underlying maps of $G$-sets). This will give us the power to identify the spans which should be identified, and make the functor faithful.

If that understanding of it is correct, then indeed I agree it’s a very nice example about how thinking of $G$-sets as groupoids is more fundamental. But I would still say that this was clearly the approach we should have been following from the beginning anyway… according to the tao you taught us!

Posted by: Bruce Bartlett on December 4, 2007 9:17 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 14)

Bruce wrote:

In other words, we always knew that was the language we were supposed to use… so the lectures up to now seem very strange to me. In effect, they follow the pattern : “Look guys, here’s a way - which we can all see doesn’t follow the tao - in which we might go about our business… and look, it goes wrong for such and such reasons. So now we’re going to use the language we all knew we should have used from the start.”

Right. But, it’s not really so strange. Don’t forget: I have actual students — people sitting in the class — and they don’t yet understand the tao of mathematics. So, when you say “we always knew that was the language we were supposed to use”, that includes you and me, but it doesn’t include the people actually in the class. These people need to learn the tao of mathematics. And, I don’t think you can really teach the tao by saying “here’s the tao: here’s how you should do things”. You have to really see how things go wrong if you don’t follow the tao, and why following the tao makes things work more smoothly.

This may seem like needless torture, but face it: everyone thinks symmetry is about group actions. You can get very far with this viewpoint. Most 20th-century mathematicians and physicists were quite content working with group actions! But eventually you run into a wall, and then you need groupoids. So, that’s what I did in the fall quarter of this course: start by studying some famous group actions and group representations, and try to develop the theory of Hecke operators as much as possible from this perspective. It went pretty well, but eventually I hit a wall, and then I switched over to groupoids.

I knew all along that I’d need to switch eventually, but I didn’t know when. I hadn’t known exactly when I’d hit the wall. Now I do! I hit it quite hard when I tried to state the Fundamental Theorem of Hecke Operators.

Ok I peeked ahead… but I still haven’t found an explicit statement of the groupoidy version of the theorem!

Jim almost stated it today, in lecture 19. But, I’ll make sure to actually state it in the final lecture, lecture 20.

Sorry for the suspense, but there’s a bunch of stuff we needed to introduce first: groupoid cardinality, and then ‘transfer’ for 0th groupoid homology, and then degroupoidification. A bunch of this is already in This Week’s Finds, but a bunch is not, because we understand things more deeply now.

Posted by: John Baez on December 5, 2007 7:26 AM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 14)

I knew all along that I’d need to switch eventually, but I didn’t know when. I hadn’t known exactly when I’d hit the wall.

You are right. I have to admit I actually find it quite surprising that the first statement of the theorem (involving isomorphism classes of spans whose legs are $G$-maps) didn’t quite work. If you had asked me in the beginning, I would have expected it to be equivalent to taking spans of groupoids…

This gives a nice a posteriori justification for why you were calling it the Tale of Groupoidification as opposed to the Tale of Spanification which might have seemed a more appropriate name in the beginning. I guess the lesson is : spans are not enough. You need to combine them with groupoidy thinking.

Posted by: Bruce Bartlett on December 5, 2007 10:15 AM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 14)

I’ve been peeking ahead too. Is there a sense then in which this groupoidification/degroupoidification process is just part of the passage between homology and homotopy, or strict $\omega$-categories and weak $\omega$-categories?

Posted by: David Corfield on December 5, 2007 9:01 AM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 14)

Good question, David. Believe it or not, we haven’t thought too much about the $n$-groupoidal or $\infty$-groupoidal generalizations of what we’ve done so far. But, if you think of a complex algebraic variety as giving a particularly nice $\infty$-groupoid, our ideas are related to something well-known!

In algebraic geometry people like to ponder correspondences between algebraic varieties. A correspondence from a variety $X$ to a variety $Y$ is a very nice binary relation: a Zariski-closed subset of $X \times Y$. If gives a span of algebraic varieties

$X \stackrel{p}{\leftarrow} S \stackrel{q}{\rightarrow} Y$

and this span is sufficiently well-behaved that we get a map on homology groups

$H_*(X) \to H_*(Y)$

using a ‘pull–push’ strategy:

$H_*(X) \stackrel{p^!}{\to} H_*(S) \stackrel{q_*}{\to} H_*(Y)$

Here $q_*$ is the usual ‘pushforward’ on homology, while $p^!$ is the sneaky ‘pullback’ part — often called the ‘transfer map’.

The pushforward is easy to define, but we can only define the transfer map

$p^! : H_*(X) \to H_*(S)$

when the map

$p: S \to X$

is sufficiently well-behaved. The simplest case is when $p$ is a covering map… if you know about singular homology, you can easily guess how this works just by drawing a few pictures! A simplex down in $X$ lifts up to a bunch of simplices in $S$. Apparently branched covers also work.

I’d love it if an algebraic geometer would step up and say in general what kind of map between varieties (or schemes) admits a well-behaved transfer. I don’t understand this stuff very well, and some of the things I said above could be missing some technical hypotheses.

Anyway: there’s a close relation between this stuff and the concept of ‘Euler characteristic’. What Jim and I seem to have (unintentionally) done is to generalize this idea to a different context using homotopy cardinality instead of Euler characteristic!

Euler characteristic tends to be well-defined for algebraic varieties. It doesn’t tend to be well-defined for $K(G,1)$’s when $G$ is a finite group, or groupoid. But homotopy cardinality is well-defined for such spaces, and it serves as a substitute for Euler characteristic. It shows up in Jim’s definition of the ‘transfer’ map

$p^! : H_0(X) \to H_0(S)$

for a functor

$p: S \to X$

between finite groupoids. This definition will show up in lecture 19.

Posted by: John Baez on December 5, 2007 4:45 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 14)

John wrote:

Any rig $R$ gives a category $\mathrm{Mat}(R)$ whose objects are finite sets and whose morphisms are $R$-valued matrices.

Funny that you should mention this. Here’s a question that I’ve been thinking about recently: for what rigs $R$ does $\mathrm{Mat}(R)$ have all finite limits? I’ll make the situation a little bit nicer: I want $\mathrm{Mat}(R)$ to be symmetric monoidal, so $R$ in fact has to have commutative multiplication.

It’s useful that $\mathrm{Mat}(R)$ automatically has biproducts, which induces the same CMon enrichment as that arising straightforwardly from addition in $R$ (I think). So for all finite limits we just need to ask for all binary equalisers.

I would make the following conjecture: $\mathrm{Mat}(R)$ has all finite limits when the addition in $R$ is ‘cancellable’, so $a+x=b+x \Rightarrow a=b$ for all morphisms $a, b, x$ coming from the same hom-set.

There are two useful examples to keep in mind when thinking about this: $\mathrm{FinRel}$ does not have all finite limits, and $FinVect$ does have all finite limits. An obvious example in between these, which I discussed a bit with Jeffrey Morton, is $R=\mathbb{N}$.

Posted by: Jamie Vicary on December 6, 2007 4:16 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 14)

The question has a homological flavor to it. I think $Mat(R)$ need not be finitely complete even if $R$ is a ring; counterexamples occur for at least some rings of homological dimension greater than 1.

For a rig $R$, define an $R$-module in the expected way: as a commutative monoid $A$ equipped with a rig homomorphism $R \to hom(A, A)$. (In other words, a rig $R$ is nothing but a one-object $CMon$-enriched category, and a module is a enriched functor $R \to CMon$.) If $R$ is a ring, then an $R$-module in this sense agrees with the usual notion: the underlying commutative monoid is an abelian group, since $-1 \in R$ acts by additive inverse.

The free $R$-module functor takes a set $X$ to an $X$-indexed coproduct of copies of $R$. By the universal property of free modules, the category $Mat(R)$ is equivalent to the category of finitely generated free $R$-modules, $Free_R$. So the question is whether $Free_R$ admits equalizers.

But take $R = \mathbb{Z}_6$. The map $m_2: R \to R$ given by multiplication by 2 has a kernel which is not free, so there is no equalizer of the pair of maps

$m_2, 0: R \to R$

in $Free_R$. (One should convince one’s self that the inclusion $Free_R \hookrightarrow Mod_R$ preserves those equalizers which happen to exist, so that in $Free_R$ one may calculate equalizers in the “usual” way. I feel pretty sure the claim is true…)

Posted by: Todd Trimble on December 6, 2007 11:46 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 14)

Well, it’s clear in retrospect that “homological dimension” is something of a red herring here (my example $\mathbb{Z}_6$ has homological dimension 0, since all of its modules are projective). The point was rather that in simple cases, projective modules need not be free (in which case $Free_R$ is not even Cauchy complete, much less finitely complete).

Posted by: Todd Trimble on December 7, 2007 1:39 AM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 14)

Thanks Todd. I think I see why $R=\mathbb{Z}_6$ is a counterexample. It seems to me that it’s the cyclic nature of $\mathbb{Z}_6$ that’s causing the problem.

So here’s an updated conjecture: $\mathrm{Mat}(R)$ is finitely complete iff $R$ has cancellable addition and is of characteristic 0.

Posted by: Jamie Vicary on December 7, 2007 11:45 AM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 14)

Here’s another example (this time in characteristic 0): let $R = \mathbb{C}[S_n]$ be the complex group algebra on the symmetric group. Then any irreducible module is a projective module which is not free, so there exist idempotent maps in $Mat(R) \simeq Free_R$ which do not split. In particular, $Free_R$ cannot be finitely complete.

I still think that the problem has a strong homological component. Even if all projective $R$-modules are free (so that $Free_R$ is equivalent to its Cauchy or idempotent splitting completion, clearly a necessary condition), it still may not be closed under taking equalizers.

Here’s another example based on Koszul resolutions (cf. Mac Lane’s Homology, pp. 204-207). Consider for example a polynomial algebra $R = k[x, y, z]$ in three (commuting) variables, where $k$ is any commutative ring; we may consider $k$ as an $R$-module where $x$, $y$, and $z$ act by the identity on $k$. The Koszul resolution of $k$ is an $R$-free resolution

$0 \to V \wedge V \wedge V \to V \wedge V \to V \to R \to k \to 0$

where $V$ is a free $R$-module $R(u_1, u_2, u_3)$ of rank 3; this exterior algebra over $R$ becomes a differential graded algebra where the differential is determined by the map $\partial: V \to R$ which takes the basis element $u_i$ to $x_i$. The point I am trying to make here is that it is known that the homological dimension of $k$ as an $R$-module is 3 (see loc. cit. for a proof), and so the kernel of the map $\partial: V \to R$ between free $R$-modules is not a free $R$-module. (I chose this example to illustrate the fact that it is not enough just to demand that all projective modules are free, as we have by the Quillen-Suslin theorem that projective modules over a polynomial ring are free. There may be simpler examples to illustrate this point.)

It’s a curious question which may have a rather nontrivial answer; sorry this isn’t a more positive response!

Posted by: Todd Trimble on December 7, 2007 2:48 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 14)

Can I make the question even more complicated by pointing out that, to a constructive mathematician, the answer may be negative even when R is a field? This is because (for example), if you want to calculate the dimension of the kernel of x ↦ ax, then you must determine whether or not a is zero.

(Here I take the correct constructive notion of ‘field’ which is satisfied by, for example, the Dedekind reals: given any element a of the commutaive ring R, a = 0 iff a is not invertible. If you use the condition that every element is either invertible xor equal to 0 —which is a stronger condition in intuitionistic logic but is not satisfied by the ring of Dedekind real numbers in constructive analysis—, then Mat(R) does have equalisers.)

If you don’t believe in constructive mathematics, then just ask the question internal to a nonBoolean topos. But my intuition is that this is further evidence that the answer is likely to be complicated, even in the classical case. If I may be so bold, it even suggests to me that Mat(R) is simply the wrong category to use.

Posted by: Toby Bartels on February 15, 2008 10:30 PM | Permalink | Reply to this
Read the post Geometric Representation Theory (Lecture 15)
Weblog: The n-Category Café
Excerpt: James Dolan on various forms of decategorification.
Tracked: December 13, 2007 1:52 AM
Read the post Geometric Representation Theory (Lecture 16)
Weblog: The n-Category Café
Excerpt: Turning a group acting on a set into a groupoid --- the 'weak quotient' or 'action groupoid'.
Tracked: December 21, 2007 12:29 AM

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