## January 29, 2008

### Geometric Representation Theory (Lecture 23)

#### Posted by John Baez

In this session of the Geometric Representation Theory Seminar, I was stuck in the local courthouse on jury duty. (I wasn’t selected to be a juror.) So, Jim Dolan continued his story from last time: groupoidifying the Hall algebra of a quiver.

The ultimate goal is to categorify the theory of quantum groups.

• Lecture 23 (Jan. 15) - James Dolan on groupoidifying the Hall algebra of an abelian category. Spans as “nondeterministic maps”. The “twisted sum” of objects in an abelian category as a nondeterministic map. Calculating the Hall algebra coming from the category of representations of this quiver, known as $A_2$ because it has two dots in a row: $\bullet \longrightarrow \bullet$ As usual, the Hall algebra has a basis consisting of isomorphism classes of representations. Every representation of $A_2$ is a direct sum of copies of three indecomposable ones: $A = 0 \longrightarrow F_q$ $B = F_q \longrightarrow 0$ and $C= F_q \stackrel{1}{\longrightarrow} F_q$ where $F_q$ is the field with $q$ elements. Note that $A$ and $B$ are irreducible, while $C$ is not: it is a “twisted sum” of $A$ and $B$. In other words, there is a short exact sequence of quiver representations $0 \to A \to C \to B \to 0$ which does not split. Computing the product in the Hall algebra: warmup.

Posted at January 29, 2008 3:42 AM UTC

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### The jury’s out on quantum groups; Re: Geometric Representation Theory (Lecture 23)

I really do NOT want to distract from the project, which fascinates me: “categorify the theory of quantum groups.”

But I am moved to say that you may have answered questions literally in the Voir Dire (questions to potential jurors). It is more fun if you first consider the question/answer system metalinguistically, and generate word strings which increase your probability of actually being on a Jury.

There was a murder trial that looked entertaining, so I resolved th give the minimal answers that were technically true (I was under oath) but most likely to lead judge and attorneys away from seeing where I really was in mental state space.

Lawyer: “Do you have a college degree?”

JVP: “Yes.” [thinking many, many, you specialist fool]

Lawyer: “what degree?” [falsely assuming one and only one]

JVP: “English.” [not mentioning that it was a B.S. from Caltech simultaneously with a B.S. in Math (advanced Logic), followed by an M.S. in Computers (Cybernetics and AI), and then…]

Lawyer: “Why?”

So I slip through the net intended to catch and throw back into the pool those actually intelligent, and capable of critical thought.

I become Jury Foreman. The trial takes a day. The jury argues for 4 more days, and keeps sending questions to the judge. I hang the jury on Murder (the burden of proof was lacking, as was motive, and a murder weapon). I lead them to acquit on Attempted Murder (as they failed to prove that the alleged victim of that even existed, let alone was in California at the time of the manifestly fatal shooting of the alleged murder victim, let alone at the scene of the crime).

A man walked free (or, I think, back to his cell where he was serving time for something else, as the outfit and handcuffs and leg shackles suggested) – a very scary man whom I suspected of being a stone cold killer – and he gave me a huge smile and thumbs up. Because the State had failed their duty, and I followed mine.

I had an extensive discussion afterwards with both Prosecution and Defense attorneys.

They: “How did you get the Jury to ask such tricky questions about fingerprints, testimony of plea-bargained crack dealer, temperature of muzzle in firing those particular bullets, and so forth?

JVP: You were sloppy in Voir Dire. I have other degrees, and my English degree led me to be an Active Member of Mystery Writers of America.

They: “A damned Mystery Writer. We’ll never let THAT happen again. As it turns out, Prosecution lost a Motion in Limine, so the jury never got to hear that this was a crack deal gone bad, in which Defendant shot a dealer from a rival gang WHO HAD PREVIOUSLY SHOT DEFENDANT’S BROTHER.”

JVP: “I strongly urge you to try a second time to convict that guy. But this time, get your ducks in a row first. The Jury found the narrative too thin. They wanted to convict the guy, whom they agreed LOOKED like a killer. But you need to provide a REASON for his actions, and a better presentation of the circumstantial evidence which, better packages, could and should lead to conviction.”

They: “You didn’t ever go to Law School, or did you…?”

JVP: “No,” [not knowing that my son would, at age eighteen, be in USC Law School]. “Math and Physics are hard. Law is easy. So do your homework, and try again.”

Again, sorry for the digression, but Jury Duty can be a spectacular way to go beyond naive philosophical and sociological theory and grapple with the multidimensional writhing knotted paradoxical glory of The Human Condition.

Now, back to Quantum Groups – which, naively speaking, are not quite Groups as the name suggests….

The witness may now step down.

Posted by: Jonathan Vos Post on January 29, 2008 10:45 AM | Permalink | Reply to this

### Jury Duty

Jonathan wrote:

It is more fun if you first consider the question/answer system metalinguistically, and generate word strings which increase your probability of actually being on a Jury.

I didn’t want to be on a jury. I’ve got lots of stuff to do. I would never do anything to shirk jury duty, because I think it’s important. I also don’t like how trial lawyers avoid choosing academics as jurors — though in their situation, I’d do the same. But I’m teaching two grad courses this quarter, and I hate the thought of missing more days of class.

Anyway, they picked a complete jury with alternates before they got to me.

Posted by: John Baez on January 30, 2008 6:02 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 23)

I want to ask the following question somewhere in public. It is closely related to the content of the Geometric Representation Theory Seminar, if not to this particular session.

Question. For $G$ some $n$-group, $n \in \mathbb{N}$, and $\mathbf{B} G$ the corresponding 1-object $n$-groupoid; how do we characterize $n$-functors $p : C \to \mathbf{B}G$ which come from $C$ being the action $n$-groupoid of $G$ acting on something?

For $n=1$ we know we demand the functor $p$ to be faithful. What’s the corresponding condition for higher $n$?

From Lie $\infty$-algebraic considerations # I have the suspicion that a particularly elegant answer to this should be possible using $\mathbb{Z}$-groupoids: I expect that $\mathbb{Z}$-groupoids $C$ with morphisms $p : C \to \mathbf{B}G$ which are injective on positive degree Hom-spaces can be nicely related to actions of $G$ on $k$-things, the $k$-thing sitting in degrees $0, -1, \cdots, -k$ in $C$.

Notice that for the ordinary action of a group on a set (a 0-thing), this means $C$ has that thing in degree 0 and the group faithfully in degree 1, which is indeed the case.

I know from playing some Lie $\infty$-algebraic tricks that something like this is true.

I expect it is too much to hope that the answer to the above question is already known (but am prepared for pleasant surprises). But since I think it is a good and important question, somebody should think about it.

And clearly, this will become relevant to the issue Geometric Representation Theory as soon as somebody begins trying to groupoidify groupoidification, in order to describe 2-vector spaces combinatorially.

So it’s of relevance here. And in some other contexts, too, I think.

Posted by: Urs Schreiber on January 30, 2008 1:38 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 23)

Thinking more about this, I might have to take the remark on negative degrees and $\mathbb{Z}$-groupoids back.

On the other hand, the following looks like it is of relevance:

if $C$ is the action groupoid of the group $G$ acting on the set $V$, then we don’t just have the faithful morphism $C \to \mathbf{B} G$ characterizing this situation, but in fact also a functor $V \to C$ including $V = Obj(C)$, regarded as a discrete groupoid into $C$.

And their composite $V \to C \to \mathbf{B}G$ is a “sequence of groupoids” in that the image of the composite contains only identity morphisms.

For instance, if we regard the regular representation of $G$ on itself, then $C = G//G = INN(G)$ and the above sequence is the groupoid incarnation of the universal $G$-bundle $G \to G//G \to \mathbf{B} G \,.$

So this looks like a good starting point for categorification: we should characterize action $n$-groupoids $C$ of $n$-groups $\mathbf{B} G$ on $(n-1)$-groupoids $V$ by the fact that they sit in a sequence

$V \to C \to \mathbf{B}G$

with certain properties.

That’s in fact the point of view that I am adopting in the entry on $L_\infty$-associated bundles, where I am suggesting to define the action of an $L_\infty$-algebra on a chain complex as such a sequence.

More on that I’ll have in a while…

Posted by: Urs Schreiber on January 31, 2008 1:21 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 23)

Maybe some of you would enjoy this puzzle. (Others may already know the answer; if so don’t spoil it.)

Take the $A_3$ quiver:

$\bullet \longrightarrow \bullet \longrightarrow \bullet$

A representation of this quiver is a pathetically simple thing, just a gadget like this:

$V_1 \stackrel{f_1}{\longrightarrow} V_2 \stackrel{f_2}{\longrightarrow} V_3$

consisting of 3 vector spaces and 2 linear maps.

There’s an obvious notion of ‘direct sum’ for such gadgets: the direct sum of

$V_1 \stackrel{f_1}{\longrightarrow} V_2 \stackrel{f_2}{\longrightarrow} V_3$

and

$W_1 \stackrel{g_1}{\longrightarrow} W_2 \stackrel{g_2}{\longrightarrow} W_3$

is

$V_1 \oplus W_1 \stackrel{f_1 \oplus g_1}{\longrightarrow} V_2 \oplus W_2 \stackrel{f_2 \oplus g_2}{\longrightarrow} V_3 \oplus W_3$

Puzzle: How many such gadgets are there that aren’t direct sums of other such gadgets, except in a trivial way?

These are called indecomposable representations of the $A_3$ quiver.

Posted by: John Baez on January 31, 2008 12:41 AM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 23)

I reckon 6, not counting the trivial one!

For finite groups, the answer is the number of conjugacy classes of arrows. But surely there’s some generalisation of ‘conjugacy class’ that works for all finite categories, including quivers. I would guess it has something to do with sieves — families of arrows closed under precomposition — because this sort of thing is useful for similar prolems in topos theory. And it seems that $A_3$ has 6 nonempty sieves! And this also seems to give the right answer for $A_0$, $A_1$, $A_2$ and $A_4$, so maybe it works for index categories with no nonidentity invertible morphisms.

Posted by: Jamie Vicary on January 31, 2008 11:03 AM | Permalink | Reply to this

### GL(n, q), and SMBL; Re: Geometric Representation Theory (Lecture 23)

Just a hunch, but shouldn’t the number of indecomposable representations of the A_n quiver = number of conjugacy classes in GL(n,2)?

Equivalently, Unlabeled permutations of sets.

I’m thinking that the same arguments apply as in W. Feit and N. J. Fine, Pairs of commuting matrices over a finite field. Duke Math. Journal, 27 (1960) 91-94.

That is,

The number f(n) of conjugacy classes in the group GL(n, q) is the coefficient of t^n in the infinite product:
product [k=1, 2, …] (1-t^k)/(1-qt^k)

Sorry I didn’t put that in MathML. I’ve got to learn MathML now to build a huge multiscale mulidisciplinary supermodel tying togther various submidels relating the paper I’m coauthoring with the surgeon who saved my life (Dr. Thomas L. Vander Laan) at Caltech’s Beckman Institute, running in the lovely open source SBML – Systems Biology Markup Language. Anyone here into SBML?

Oh, the data:

n f(n)
0 1
1 1
2 3
3 6
4 14
5 27
6 60
7 117
8 246
9 490
10 1002
11 1998
12 4053
13 8088
14 16284
15 32559
16 65330
17 130626
18 261726
19 523374
20 1047690
21 2095314
22 4192479
23 8384808
24 16773552
25 33546736
26 67101273
27 134202258
28 268420086
29 536839446
30 1073710914
31 2147420250

Posted by: Jonathan Vos Post on February 1, 2008 4:14 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 23)

Jamie writes:

I reckon 6, not counting the trivial one!

Yes!

Jonathan wrote:

Just a hunch, but shouldn’t the number of indecomposable representations of the $A_n$ quiver = number of conjugacy classes in GL($n$,2)?

No!

Jamie wrote:

For finite groups, the answer is the number of conjugacy classes of arrows. But surely there’s some generalisation of ‘conjugacy class’ that works for all finite categories, including quivers. I would guess it has something to do with sieves — families of arrows closed under precomposition — because this sort of thing is useful for similar prolems in topos theory.

Hmm. I don’t think it’s that simple in general. How many indecomposable representations do you get for $D_4$?

             o
/
/
o-------o
\
\
o


Assume the arrows are pointing in, say — it doesn’t really matter. According to Gabriel’s theorem there should be 12. Right now I can only see 11 of them. But, I’ve done this calculation before with Jim, and remember discovering the 12th after a suspenseful search.

For $A_n$, all the indecomposable reps look like this:

$0 \to 0 \to k \stackrel{1}{\to} k \stackrel{1}{\to} k \stackrel{1}{\to} k \to 0 \to 0 \to 0 \to 0$

so there are $n$ choose 2 of them… and indeed, 3 choose 2 is 6.

Posted by: John Baez on February 1, 2008 5:38 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 23)

John said:

According to Gabriel’s theorem there should be 12. Right now I can only see 11 of them. But, I’ve done this calculation before with Jim, and remember discovering the 12th after a suspenseful search.

Wow — I didn’t realise there could be nontrivial ones. I think I’ve found the 12th — it’s big! The search was indeed suspenseful! (Certainly more suspenseful than Cloverfield, which I just watched.)

For $A_n$ … there are $n$ choose 2 [irreps].

I think it’s actually $\frac{1}{2}(n$ choose $2) + n = n(n+1)/2$, and you’ve suffered an unfortunate mathematical coincidence! (Unless you mean “choose” in the sense that objects are replaced and order does not matter.) It’s just the number of arrows in the category arising from the quiver.

A quiver is of finite type if it has finitely many non-isomorphic indecomposable representations … A (connected) quiver is of finite type if and only if its underlying graph (when the directions of the arrows are ignored) is one of the following Dynkin diagrams: An, Dn, E6, E7, E8.

So what’s the simplest quiver not of finite type? In fact, what’s the simplest category not of finite type? It seems really weird to me that a finite category could have an infinite number of irreps…

Also, I just noticed that you said

Hmm. I don’t think it’s that simple in general.

This gives me the impression that this is an open problem! Is it? I would be VERY interested in the answer — not least because I bet that “take the number of irreps” gives a functor from FinCat to FinSet! (Or at the very least, a rig homomorphism from the rig of finite categories to the rig of finite sets.)

Posted by: Jamie Vicary on February 2, 2008 2:14 AM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 23)

Jamie wrote:

I think I’ve found the 12th — it’s big!

Okay, now I think I see 12. Here’s the biggest indecomposable rep of $D_4$ that I found. Each outer dot:

             o
/
/
o-------o
\
\
o


gets a 1-dimensional vector space, say our field $k$, while the inner one gets a 2-dimensional vector space, say $k^2$. The three maps from $k$ to $k^2$ are 3 lines in general position, say:

$i_1: x \mapsto (x,0)$

$i_2: y \mapsto (0,y)$

$i_3: z \mapsto (z,z)$

This clearly can’t be written as a direct sum of two other representations in a nontrivial way.

So, my 12 are:

• the four irreducible reps with a $k$ at one dot (and $0$ elsewhere),
• the three indecomposable reps with a $k$ at one of the outer dots and and $k$ at the center dot,
• the three indecomposable reps with a $k$ at two outer dots and a $k$ at the center dot,
• the indecomposable rep with a $k$ at each dot,
• the big one mentioned above, with a $k$ at each outer dot and a $k^2$ in the middle.

Jamie wrote:

John wrote:

For $A_n$ there are $n$ choose 2 indecomposable reps.

I think it’s actually $\frac{1}{2}(n$ choose $2)+n = n(n+1)/2$

Ah, you’re right. It’s the ‘big’ triangle number not the ‘little one: namely, $(n+1)$ choose $2$, which happens to be the number of unordered pairs of elements in an $n$-element set.

It’s just the number of arrows in the category arising from the quiver.

Right!

So what’s the simplest quiver not of finite type? In fact, what’s the simplest category not of finite type?

The simplest quiver that’s not of finite type has one dot and one arrow (which starts and ends at that dot). This freely generates a category called the ‘walking endomorphism’, otherwise known as $\mathbb{Z}$.

Puzzle: which famous theorem classifies the indecomposable representations of this quiver?

A quiver with a loop of arrows can never have just finitely many indecomposable representations.

But, you may not like this example, since the category is infinite (though the quiver is finite). Maybe you’ll like $\tilde{D}_4$ better:

        o
|
|
o-----o----o
|
|
o


It doesn’t really matter which way the arrows point (that’s a curious feature of this game), but take them pointing inwards. Then you get a category with only 5 objects and 9 morphisms, with infinitely many indecomposable representations.

Why? Well, just as the reps of $D_4$ are closely related to the ‘three subspace problem’ (that is, the problem of classifying 3 subspaces of a vector space), the reps of this guy are closely related to the famously complex ‘four subspace problem’.

It seems really weird to me that a finite category could have an infinite number of irreps…

I can’t resist a nitpick: you shouldn’t really use ‘irrep’ as an abbreviation of ‘indecomposable representation’, as you are doing here. After all, one of the main issues here is the difference between indecomposable representations (nonzero representations that can’t be decomposed as a direct sum of other nonzero representations) and irreducible representations (nonzero representations that don’t have any nonzero subrepresentations).

Gabriel’s theorem classifies the quivers with finitely many indecomposable representations.

For example, $D_4$ has 12 indecomposable representations, but only 4 irreducible ones.

The ‘walking endomorphism’ has infinitely many irreducible representations, but also a bunch of indecomposable ones that aren’t irreducible.

I think $\tilde{D}_4$ with all arrows pointing inwards has just 5 irreducible representations, but infinitely many indecomposable ones.

Posted by: John Baez on February 4, 2008 6:32 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 23)

Conc. representations of quivers, I’d like to recommend D.J. Bensons “Cohomology and Representations”, whose 1st volume contains a very good to read intro to that, both volumes containing many very good intros to other interesting topics.

A look to the history of representations of quivers may give an interesting example of the unfolding of a mathematical theory:
It started with Gelfand, Ponomarev “Problems in linear Algebra and classification of quadruples of subspaces” in Coll.Math. Soc. I. Bolyai,5, 1970 , then continued with Gabriels “Unzerlegbare Darstellungen I”, then came a clearifying formalization in Bernstein, Gelfand, Ponomarev “Coxeter Functors and Gabriel’s Theorem”.

Posted by: Thomas Riepe on February 1, 2008 11:07 AM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 23)

I’ve spent a lot of time reading Benson’s books… they’re great! I recommend them to everyone who likes algebra. In his talks, Jim is groupoidifying some of the linear algebra in there.

By the way, I personally never link to material that requires the user to pay to gain access. I refer to it, but I don’t put in a hyperlink. Why? Because it’s frustrating when you click on a link (as I just did with yours) and find someone demanding payment to read it. Of course I can turn on my UC Riverside superpowers and read most of this stuff, but not everyone can.

Here is some freely accessible material on quiver representations:

There’s more in week230.

Posted by: John Baez on February 1, 2008 5:55 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 23)

Derksen has an entire course on quiver representations here.

Posted by: Aaron Bergman on February 1, 2008 8:07 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 23)

Sorry, I thought the access is free :((
Here corrections to Bensons books ( 1 , 2 , both in new ed.)

Posted by: Thomas Riepe on February 2, 2008 3:04 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 23)

Just a comment about finding the indecomposable representations - one can do this quite easily for Dynkin quivers of types ADE (and for many others too) using the Auslander-Reiten theory (refer: Elements of representation theory of associative algebras by Simson, Assem and Skowronski).

Posted by: Kavita on May 11, 2009 9:24 PM | Permalink | Reply to this

### Re: Geometric Representation Theory (Lecture 23)

Thanks! I guess I know how it works for the ADE ones.

By the way, here’s a little progress report on the theme of this lecture.

This January, Julie Bergner, Aviv Censor, Christopher Walker and myself formed a kind of working seminar on groupoidifying Hall algebras — and not just their algebra structure, but also their coalgebra structure. Now that Alex Hoffnung has returned from Columbia, he has joined in too.

As an introduction to Hall algebras, we’ve found this paper very helpful:

It’s not on the arXiv!

Posted by: John Baez on May 11, 2009 11:53 PM | Permalink | Reply to this

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