## June 11, 2011

### Christopher Walker on Hall Algebras

#### Posted by John Baez

On Friday May 20th my student Christopher Walker successfully defended his thesis, which is now available here:

Now he’s looking for a job. If you know anyone who needs a good algebraist who is also a topnotch teacher (he has extensive teaching experience), please let them know about Christopher. Or let him know about them!

But let me tell you what he’s done.

A Hall algebra is a magically efficient way to read off a quantum group… or at least ‘half’ a quantum group… directly from a Dynkin diagram. This trick works for the simply-laced Dynkin diagrams: $A_n, D_n, E_6, E_7$, and the magnificent $E_8$:

We spoke earlier here on the n-Café about a paper in which Christopher Walker elegantly described Hall algebras as Hopf algebras in braided monoidal categories.

In his thesis, Walker categorifies this idea, describing Hall algebras as living inside certain braided monoidal bicategories. This is a very natural thing to do, since Hall algebras are obtained in the first place by a process of decategorification, or more precisely, ‘degroupoidification’. This process turns groupoids into vector spaces.

This means that while a Hall algebra is normally thought of as a vector space equipped with extra structure—multiplication, comultiplication and so on—at a more fundamental level it’s really a groupoid. The trappings of linear algebra removed, we see the bare bones here.

What is this groupoid that secretly underlies the Hall algebra? To get it, first we take our Dynkin diagram and draw arrows on its edges. We call this a ‘quiver’… because now it’s a picture of a category, and you must be quivering with anticipation to see what’ll happen next.

The next thing we do is look at representations of our quiver on finite-dimensional vector spaces over $\mathbb{F}_q$, the finite field with $q$ elements. This is how the beloved ‘$q$’ of quantum group theory gets into the game!

A ‘representation’ of a quiver is a simple thing: it’s just a vector space for each dot of our quiver, and a linear map for each arrow. Or, if you think of a quiver as a category, a representation is a functor from the quiver to the category of vector spaces.

There’s an obvious notion of a map between quiver representations, which you can surely guess if you understood the previous paragraph. So we get a category of quiver representations. If our quiver is called $Q$, let’s call this $Rep(Q)$.

At this point you should really spend a month going through examples to see what $Rep(Q)$ is like. We had a little seminar at UC Riverside where we did just that. You should work out the indecomposable representations, and the irreducible ones, and study short exact sequences of representations. You’ll see this is a fascinating subject… and you’ll see why the simply-laced Dynkin diagrams are special: these are the ones that give quivers with finitely many indecomposable representations! If you want some help, try these nice notes:

All the ideas I just mentioned make heavy use of noninvertible maps between quiver representations. But then we can throw out all the maps except the invertible ones. We’re left with a groupoid of representations: let’s call this $Rep(Q)_0$. This is the groupoid that secretly—or perhaps not-so-secretly—underlies the Hall algebra.

Degroupoidification is a process that turns groupoids into vector spaces. The basic idea is pathetically simple: we just take our groupoid, look at its set of isomorphism classes of objects, and use that as the basis for a vector space! But the interesting part is not how any groupoid gives a vector space. It’s how any span of groupoids gives a linear operator between vector spaces. A ‘span’ of groupoids looks like this: $\begin{matrix} &&S&\\ &\swarrow& &\searrow&\\ Y&& &&X\\ \end{matrix}$ where $X,Y$ and $S$ are groupoids and the arrows are functors. This sort of gadget gives us a linear operator from the vector space associated to $X$ to the one associated to $Y$. (Or vice versa: the charm of spans is that they’re symmetric.)

I explained how this works back in This Week’s Finds, and we described it more thoroughly here:

We won’t need the details now, if you can take my word for it: all sort of basic linear algebra stuff can also be done with spans of groupoids, but whenever you like you can snap your fingers, ‘degroupoidify’, and turn them into linear operators.

In particular, the Hall algebra gets a multiplication from this span of groupoids: $\begin{matrix} &&SES(Rep(Q))_0&\\ &p\swarrow& & \searrow q&\\ Rep(Q)_0&&&&Rep(Q)_0 \times Rep(Q)_0 \\ \end{matrix}$ Here $SES(Rep(Q))$ is the category of short exact sequences of quiver representations, and again the little subscript $0$ means we take the underlying groupoid, where we throw out all the morphisms except isomorphisms. Given a short sequence of quiver representations $0 \to N \to E \to M \to 0$ the functor $p : SES(Rep(Q))_0 \to Rep(Q)_0$ picks out the representation $E$ in the middle, while the functor $q: SES(Rep(Q))_0 \to Rep(Q)_0 \times Rep(Q)_0$ picks out the representations $M$ and $N$ at the two ends.

Since we get the Hall algebra by degroupoidifying $Rep(Q)_0$, any quiver representation gives a basis vector in the Hall algebra. And what I’m saying above basically amounts to this: to multiply the basis vector coming from $M$ by the basis vector coming from $N$, we take a sum over all representations $E$ that fit into a short exact sequence $\begin{matrix} 0 &\to& N &\to& E &\to& M &\to& 0 \\ \end{matrix}$ This sum has coefficients that need to be chosen just right, but degroupoidification takes care of that automatically! When we degroupoidify $Rep(Q)_0$ we get the Hall algebra, and degroupoidifying the above span gives its multiplication.

Similarly, we get the comultiplication in the Hall algebra by degroupoidifying the turned-around span $\begin{matrix} &&SES(Rep(Q))_0&\\ &q\swarrow& & \searrow p&\\ Rep(Q)_0 \times Rep(Q)_0 &&&&Rep(Q)_0 \\ \end{matrix}$

Now, at first glance, the multiplication and comultiplication in the Hall algebra don’t obey the equation we need to get a Hopf algebra:

where the green blob is the multiplication and the red blob is the comultiplication. However, this condition involves a braiding: in the diagram at left, one wire needs to cross over the other! It turns out that the Hall algebra becomes a Hopf algebra very neatly if we choose the right braiding. Otherwise we need to do a bunch of ad hoc mucking around, which is what most people do. Christopher explained this here:

The trick is not to work in the category of vector spaces with its usual braiding, but the category of $K$-graded vector spaces with a carefully chosen braiding. What’s $K$? It’s the ‘Grothendieck group’ of $Rep(Q)$: that is, the abelian group with a generator for each quiver representation, and a relation $E = M + N$ for each short short exact sequence $\begin{matrix} 0 &\to& N &\to& E &\to& M &\to& 0 \\ \end{matrix}$ The Hall algebra is naturally a $K$-graded vector space, since each quiver representation gives a basis vector for the Hall algebra, but it also gives an element of $K$, which we take to be that basis vector’s grade.

The category of $K$-graded vector spaces is a braided monoidal category in a very nice way, which Christopher describes in the paper above. But in fact, like so much about Hall algebras, this idea really wants to be explained at the level of groupoids rather than vector spaces! In his thesis, instead of working with $K$-graded vector spaces, Christopher works with groupoids over $Rep(Q)_0$: that is, groupoids equipped with a functor to the groupoid of quiver representations, like this: $\begin{matrix} X\\ \downarrow \\ Rep(Q)_0 \\ \end{matrix}$ Any such groupoid gives, not just a vector space, but a $K$-graded vector space. Why? Because any object of $X$ gives a basis vector of the degroupoidification of $X$, and the grade of that basis vector comes from the corresponding object down below in $Rep(Q)_0$.

Christopher’s main result is that groupoids over $Rep(Q)_0$ form a braided monoidal bicategory in a nontrivial and interesting way. Snapping our fingers and degroupoidifying, this explains why $K$-graded vector spaces form a braided monoidal category! For the details you should read his thesis, notably Chapter 4.

But to whet your appetite, I’ll say that a key idea is to consider $EXT(M,N)$, the groupoid of extensions of $M$ by $N$, where $M$ and $N$ are quiver representations. This has short exact sequences $\begin{matrix} 0 &\to& N &\to& E &\to& M &\to& 0 \\ \end{matrix}$ as objects, and diagrams like this: $\begin{matrix} 0 &\to& N &\to& E &\to& M &\to& 0 \\ & & \downarrow && \downarrow & & \downarrow & & \\ 0 &\to& N &\to& E' &\to& M &\to& 0 \\ \end{matrix}$ as morphisms. There’s a concept of cardinality for groupoids, and the cardinality of $EXT(M,N)$ is related to the so-called ‘Euler form’ $\langle M, N \rangle = dim(Hom(M,N)) - dim(Ext^1(M,N))$ More precisely, $|EXT(M,N)| = \displaystyle {\frac{q^{-\langle M, N \rangle}}{|Aut(M)| \, |Aut(N)| } }$ The bilinearity of the Euler form is crucial to making $K$-graded vector spaces into a braided monoidal category… but this bilinearity emerges from properties of $EXT(M,N)$ that are precisely what Christopher needs to make groupoids over $Rep(Q)_0$ braided monoidal bicategory!

And $Rep(Q)_0$ lives in this bicategory, because it’s a groupoid over itself in an obvious way: $\begin{matrix} Rep(Q)_0\\ \downarrow \\ Rep(Q)_0 \\ \end{matrix}$ where the vertical arrow is the identity morphism.

At this point, the obvious conjecture is that $Rep(Q)_0$ is a ‘Hopf 2-algebra object’ living in the braided monoidal bicategory of groupoids over $Rep(Q)_0$. Due to the pressure of time, Christopher hasn’t proved this yet. But at this point it’s almost a foregone conclusion. He has the multiplication and comultiplication on $Rep(Q)_0$, and also the unit and counit. He’s shown these degroupoidify to give the Hall algebra as a Hopf object in the braided monoidal category of $K$-graded vector spaces. So, he just needs to check the Hopf 2-algebra axioms.

But he also needs a job!

Posted at June 11, 2011 2:18 AM UTC

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### Re: Christopher Walker on Hall Algebras

Happy birthday, John!

Posted by: Tom Leinster on June 12, 2011 6:07 PM | Permalink | Reply to this

### Re: Christopher Walker on Hall Algebras

Thanks, Tom! I celebrated by taking a $13 \frac{1}{2}$ hour plane trip, and now I’m in Zurich feeling severely jet-lagged and half a century old. But it’s nice here, and I expect to have fun.

Posted by: John Baez on June 13, 2011 10:02 AM | Permalink | Reply to this

### Re: Christopher Walker on Hall Algebras

I can’t tell if the absence of substantive comments on this blog entry is due to me explaining the ideas poorly, or everybody else feeling bad and not knowing what to say. It’s a pity, because the ideas here are quite beautiful.

Anyway, the second excuse is now gone. Christopher Walker got a job!

It’s a tenure-track position at Odessa College. Together with his wife and kids, he’ll move there in August.

Posted by: John Baez on June 26, 2011 12:14 PM | Permalink | Reply to this

### Re: Christopher Walker on Hall Algebras

I can’t tell if the absence of substantive comments on this blog entry is due to me explaining the ideas poorly, or everybody else feeling bad and not knowing what to say.

For my part, I think it’s a question of losing sight of what this area of mathematics is for, not that there needs to be a clear purpose for everything. One dream, I take it, is to find a categorification of the whole of a quantum group, not just its positive part? And this might have some bearing on quantum topology? Any idea what stands in the way of approaching the negative part?

By the way, I see that someone has found a set of 26 relations to impose on an algebraic category to make it equivalent to the category of connected surfaces with one boundary component, and their cobordisms. Apparently, this is reported by one of you Riverside colleagues, Marta Asaeda.

Posted by: David Corfield on June 27, 2011 2:45 PM | Permalink | Reply to this

### Re: Christopher Walker on Hall Algebras

People trying to understand Christopher Walker’s work on categorified Hall algebras may like to compare Ben Webster’s ideas here:

To get a Hall algebra we do three things.

1) We choose a category $Q$.

2) We form the category

$Rep(Q) = hom(Q, FinVect_q)$

where $FinVect_q$ is the category of finite-dimensional vector spaces over the finite field with $q$ elements.

3) We form the vector space whose basis consists of isomorphism classes of objects of $Rep(Q)$. This is the Hall algebra of $Q$.

Then the magic starts: under certain conditions, the Hall algebra of $Q$ is not just a vector space, it’s an algebra. And with a bit more work (described above), it becomes a kind of Hopf algebra.

But the real fun comes from the examples. Say we take a ‘simply-laced Dynkin diagram’, meaning a picture like this:

Think of the circles as objects. Draw arrows on the edges any way you like and think of them as morphisms. Let these objects and morphisms freely generate a category $Q$. Then the Hall algebra of $Q$ is a famous thing:

Theorem 1 (Ringel). If $Q$ comes from a simply-laced Dynkin diagram as above, the vector space whose basis consists of isomorphism classes in $Rep(Q)$ is the quantum group $U_q(\mathbb{n}_+)$, where $\mathbb{n}_+$ is the maximal nilpotent Lie algebra associated to our simply-laced Dynkin diagram.

For example: in the $A_n$ case, $\mathbb{n}_+$ is just the Lie algebra of strictly upper triangular $(n-1) \times (n-1)$ complex matrices. $U(\mathbb{n}_+)$ is the ‘universal enveloping algebra’ of this Lie algebra: that is, the associative algebra naturally obtained from it. But $U_q(\mathbb{n}_+)$ is a funky ‘$q$-deformed version’ of this universal enveloping algebra.

Now, since step 3) above is a form of ‘decategorification’, it’s clear that we should really skip that step and work directly with a ‘categorified’ Hall algebra of some sort. That’s what Christopher did in his thesis.

Ben Webster (and other people) do some other things. The punchline of Ben’s first blog post is:

The Hall algebra of a category is the Grothendieck group of constructible sheaves/perverse sheaves on the moduli stack of objects in the category. The Hall algebra is an algebra because the constructible derived category of the moduli stack of objects in abelian category is monoidal in a canonical way.

Note taking the Grothendieck group is a kind of decategorification process, of a subtler sort than step 3) above. Step 3) is called ‘degroupoidification’. So, Ben is saying that the degroupoidification of $Rep(Q)$ is closely related to the Grothendieck group of something else. This gives two ways to think about the project of categorifying Hall algebras.

In his second blog post, he applies the Grothendieck group approach to the examples we love so much. And he explains two theorems:

Theorem 2 (Lusztig). If $Q$ comes from a simply-laced Dynkin diagram as above, the Grothendieck group of the category of perverse sheaves on the moduli stack of representations of $Q$ is $U_q(\mathfrak{n}_+)$.

Theorem 3 (Khovanov-Lauda/Rouquier). If $Q$ comes from a simply-laced Dynkin diagram, there exists a 2-category $R$ whose monoidal category of representations has Grothendieck group $U_q(\mathfrak{n}_+)$.

So, we get three interlocking ways to think about Hall algebras as arising from decategorification… and thus three interlocking ways to categorify them!

Posted by: John Baez on June 26, 2011 1:17 PM | Permalink | Reply to this

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