## October 4, 2010

### A Hopf Algebra Structure on Hall Algebras

#### Posted by John Baez My student Christopher Walker is groupoidifying Hall algebras. What’s a Hall algebra? We get such an algebra starting from any category with a sufficiently well-behaved concept of ‘short exact sequence’. In this algebra, the product of an object $A$ and an object $B$ is a cleverly weighted sum over all objects $X$ that fit into a short exact sequence

$0 \to A \to X \to B \to 0$

The ‘clever weighting’ is neatly explained by groupoidification, as sketched here. And if we pick our category in a nice way, the algebra we get is part of a quantum group!

But the Hall algebra is more than a mere algebra. It’s also a coalgebra! The algebra and coalgebra want to fit together to form a Hopf algebra, and they do, but only after a peculiar sort of struggle. Lately Christopher has been thinking about this, and he’s written a paper:

• Christopher Walker, A Hopf algebra structure on Hall algebras.

Abstract: One problematic feature of Hall algebras is the fact that the standard multiplication and comultiplication maps do not satisfy the bialgebra compatibility condition in the underlying symmetric monoidal category $Vect$. In the past this problem has been resolved by working with a weaker structure called a ‘twisted’ bialgebra. In this paper we solve the problem differently by first switching to a different underlying category $Vect^K$ of vector spaces graded by a group $K$ called the Grothendieck group. We equip this category with a nontrivial braiding which depends on the $K$- grading. With this braiding, we find that the Hall algebra does satisfy the bialgebra condition exactly for the standard multiplication and comultiplication, and will also become a Hopf algebra object.

The point is that a Hopf algebra is a bialgebra, so the multiplication and comultiplication need to get along nicely. As we’ve recently discussed, they need to obey a condition sort of like this: where the green blob is the multiplication and the red blob is the comultiplication. And 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.

Actually, in the picture above, you’ll note that the two wires just cross, without one visibly going over the other. That style of drawing is fine if we’re in a symmetric monoidal category, like the category of vector spaces with its usual tensor product and usual braiding. But the Hall algebra becomes a Hopf algebra in a braided monoidal category that’s not symmetric. So we need to draw the compatibility condition a bit more carefully — see Christopher’s paper.

All these ideas can be groupoidified, and that’s what Christopher is doing now.

Posted at October 4, 2010 3:58 AM UTC

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### Re: A Hopf Algebra Structure on Hall Algebras

Hi Chris. I just had a quick look at your paper over breakfast.

My first question is about your equation for the set $P^E {}_{M\,N}$ on p4 (equation numbers would be helpful here!). Won’t its cardinality be the continuum in general? Also, I have some typographical comments: I think the distinction between the two types of P is too subtle, something like $|P^E _{M\,N}|$ might be better for the cardinality; also, in the line beginning “and we call” after the displayed equation, you’re missing some punctuation.

Does all this generalize to other quivers which aren’t simply laced?

Posted by: Jamie Vicary on October 4, 2010 11:27 AM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

Hi Jamie,

Thank you for reading my paper.

The big theorem of this topic (Gabriel’s Theorem) says that the representation category of a quiver is of finite type if and only if the quiver is simply laced. I should definitely be more specific about why I can take the cardinality of this set and expect it to be finite (i.e. that I am using isomorphism classes).

As far as generalizing to non-simply laced quivers, It would take a different approach since the important sets here would not be finite. This would definitely be another direction to go. If you have every studied Lie theory, you will remember that the representation theory of lie algebras that are not simple becomes very difficult, and in many instances not even understood yet.

As for the two different forms of P, I have been going back and forth on this one, and may end up at a cardinality notion like you suggested in the end. There are some notational conflicts in the groupoidification program that I was trying to avoid.

Posted by: Christopher Walker on October 4, 2010 6:31 PM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

Jamie wrote:

My first question is about your equation for the set $P^E_{M N}$ on p. 4 (equation numbers would be helpful here!). Won’t its cardinality be the continuum in general?

It’s finite given the assumptions Christopher made on page 3. He’s looking at the category $Rep(Q)$ of representations of a simply-laced Dynkin quiver on vector spaces over a finite field. For this category the set of short exact sequences

$P^E_{M N} = \{ 0 \to N \to E \to M \to 0 \}$

is finite for all objects $M, N, E$.

If you use more general quivers, or more general fields, the set $P^E_{M N}$ could be infinite.

Christopher wrote:

I should definitely be more specific about why I can take the cardinality of this set and expect it to be finite (i.e. that I am using isomorphism classes).

That’s not the reason: you’re not taking isomorphism classes on page 4… unless I’m seriously confused.

On the other hand, thinking about these finiteness issues, I just noticed that you need to add an extra condition to make Proposition 2 and Theorem 3 true. You need to assume the groups $Ext^i(M,N)$ are finite!

By the time you get to Theorem 4 you have decided to work only with the abelian category $Rep(Q)$. You could easily prove this theorem for a larger class of abelian categories, but they’d need to satisfy three conditions:

1) they need to be hereditary,

2) the sets $P^E_{M N}$ need to be finite,

3) the groups $Ext^i(M,N)$ need to be finite.

Of course ‘hereditary’ means $Ext^i(M,N) = \{0\}$ for $i \gt 1$, so 3) is equivalent to

3’) the groups $Ext^1(M,N)$ need to be finite.

Jamie wrote:

Does all this generalize to other quivers which aren’t simply laced?

This stuff works as stated for simply laced Dynkin quivers and also simply laced affine Dynkin quivers (not mentioned in Christopher’s paper). For a general quiver $Q$, I don’t know why $Rep(Q)$ would be hereditary, and this is needed to make the Hall product associative.

But there are other tricks one can play…

Posted by: John Baez on October 5, 2010 1:54 AM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

Hi there.

One way to see that $\mathrm{Rep}(Q)$ is always hereditary is to use the standard resolution.

Given a module $M$ for the path algebra $K Q$, we can write this as a direct sum $M=\bigoplus_i M_i$ indexed by the vertices. Let $P_i$ be the projective module corresponding to the vertex $i$, so a direct summand of the regular module $K Q$. Then there is a natural epimorphism

(1)$\bigoplus_i P_i\otimes_K M_i \to M$

given by multiplication. Given an arrow $a\colon i\to j$, there is a map

(2)$P_j\otimes_K M_i \to (P_i\otimes_K M_i)\oplus(P_j\otimes_K M_j), \quad b\otimes m \mapsto (b a\otimes m,-b\otimes a m).$

Putting these together yields an exact sequence

(3)$0 \to \bigoplus_{a\colon i\to j}P_j\otimes_K M_i \to \bigoplus_i P_i\otimes_K M_i \to M \to 0.$

This is called the standard resolution, and is a projective resolution of $M$ of length 1. Hence the category is hereditary.

We can now generalise this, but the notation becomes a bit of a nightmare. My preferred way to think of it is via tensor algebras.

Let $\Lambda_0$ be a semisimple $K$-algebra and let $\Lambda_1$ be a $\Lambda_0$-bimodule on which $K$ acts centrally.

In the case of the path algebra $K Q$ we take $\Lambda_0=\prod_i K e_i$, indexed by the vertices of $Q$, and $\Lambda_1=\bigoplus_a K a$, indexed by the arrows of $Q$.

We next form the tensor algebra

(4)$\Lambda := T(\Lambda_0,\Lambda_1) = \Lambda_0\oplus\Lambda_1\oplus\Lambda_2\oplus\cdots,$

where $\Lambda_{r+1}=\Lambda_r\otimes_{\Lambda_0}\Lambda_1$.

In the case of the path algebra $K Q$, $\Lambda_r$ has basis the paths in $Q$ of length $r$.

The standard resolution now has the simple form

(5)$0 \to \Lambda_+\otimes_{\Lambda_0}M \to \Lambda\otimes_{\Lambda_0}M \to M \to 0,$

where $\Lambda_+=\Lambda_1\oplus\Lambda_2\oplus\cdots$ is the graded radical of $\Lambda$. The left-hand map comes from the identification $\Lambda_+\cong\Lambda\otimes_{\Lambda_0}\Lambda_1$, and as above we can place the $\Lambda_1$ component either on the left or on the right of the tensor product

(6)$\Lambda\otimes_{\Lambda_0}\Lambda_1\otimes_{\Lambda_0}M \to \Lambda\otimes_{\Lambda_0}M, \quad b\otimes a\otimes m \mapsto b a\otimes m-b\otimes a m.$

This is again a projective resolution of $M$. For, $\Lambda_0$ is semisimple, and so every $\Lambda_0$-module is projective. Thus the first and second terms are projective $\Lambda$ modules.

When $K$ is a finite field, this construction will get you every non-simply laced diagram (and so every symmetrisable generalised Cartan matrix). In fact, you can also get loops, and so certain symmetrisable Borcherds matrices as well.

If $K$ is a perfect field, then any finite dimensional hereditary $K$-algebra will be such a tensor algebra. On the other hand, Dlab and Ringel gave an example of a finite dimensional hereditary algebra which was not of this form.

Of course, once you’ve done algebras, you can move on to other hereditary categories such as sheaves over weighted projective lines, as Schiffmann did.

Sorry for the long post, but then again, I’ve been a follower for ages without posting.

Posted by: Andrew Hubery on October 6, 2010 12:41 AM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

Thanks! Wow, it’s great to hear from you! Christopher Walker and I have been endlessly reading and rereading your notes on Ringel–Hall algebras in the course of our work, so a post from you feels like a deus ex machina coming down from the sky to save us. Perhaps if I’d read your notes often enough I would have known what you just said…

If there’s one thing you can do to make me even happier, it’s this: put those notes of yours on the arXiv. They’re a valuable resource. Individual webpages have an uncertain future, but the arXiv will last at least as long as our current mode of civilization.

Posted by: John Baez on October 7, 2010 3:05 AM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

Hi Chris,

This is the “other Chris”.

Some typos:

p.1: “…the lie algebra…”

p.2: “We first to draw…”

p.3: “We then accomplishes…”

p.4: “Where we call aut(M) the set cardinality of the group Aut(M)” (Not a sentence)

p.4 “These are the correct factor…”

A small comment/question: You say a few times in the introduction that the incompatibility between the multiplication and comultiplication is a “problematic feature”. Why should I think of this as a problem, exactly? It seems more like an “interesting feature”, rather than a problem. (You essentially say this yourself on page 2.)

Another stupid comment: You say “As is standard, we will write multiplication…” and then write your diagrams backward in my opinion. When I’m thinking about an operad, I always write the diagrams so the inputs are at the bottom. I thought that was the standard way, no? (I have a feeling this is the kind of question that starts heated debates…my apologies if this is the case.)

Posted by: Chris Rogers on October 4, 2010 1:42 PM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

Let’s not have a heated debate! Everyone should simply agree that there’s no universally-agreed orientation.

Some people have their inputs at the bottom and work upwards; others do the opposite. Sometimes I have my inputs on the left and work rightwards. I’m sure there are situations in which I’d want to have the inputs on the right and work leftwards.

Different people do different things at different times, and that’s all there is to it.

Posted by: Tom Leinster on October 4, 2010 3:25 PM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

Can you go diagonally, or into the page?

Posted by: Tom Ellis on October 4, 2010 4:41 PM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

Yeah, you can do anything you like!

It’s very like the situation for arrows in a category, when you’re drawing commutative diagrams. The usual convention is to try to get your arrows going down and/or to the right, but sometimes it’s convenient to do otherwise.

The “arrows” we’re dealing with here are more bulky—they’re 2-dimensional—but you have a similar amount of freedom. It’s just convention.

Posted by: Tom Leinster on October 4, 2010 4:55 PM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

Trust me, I’m not looking for a debate! Thanks for the clarification.

Posted by: Chris Rogers on October 4, 2010 5:42 PM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

Hi “other” Chris,

Thank you for the catches on typos.

I think subconsciously I use “problematic” because I think anything that is not compatible is a problem. I may re-word it to sound less like an opinion.

As my wife the sociologist would say, diagram direction is just the oppressive right-handed society trying to keep us poor left-handed folk down. I mean they get to be called “right”-handed like there’s something “wrong” with the other hand. :)

In all seriousness, I will remove the reference to “standard” practice here, as there seems to be a difference of opinions.

Posted by: Christopher Walker on October 4, 2010 8:07 PM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

Chris wrote:

A small comment/question: You say a few times in the introduction that the incompatibility between the multiplication and comultiplication is a “problematic feature”. Why should I think of this as a problem, exactly?

A bialgebra is a beautiful thing: it’s a monoid in the category of comonoids in $Vect$ — or equivalently, a comonoid in the category of monoids in $Vect$ Thanks to this elegant definition, the power of category theory kicks in, and we get all sorts of wonderful results. For starters, the category of representations of any bialgebra is a monoidal category.

But with the usual approach to Hall algebra, instead of a bialgebra we get something that’s sort of close to a bialgebra, but where the compatibility condition fails, due the annoying intrusion of a mysterious ‘fudge factor’. People call this gadget a ‘twisted bialgebra’, but that’s just jargon to summarize a mystery: this gadget doesn’t fit into any clear pattern.

At least that’s what people usually say! But Christopher has shown that in fact this gadget is a bialgebra object — only not in $Vect$, but in some other braided monoidal category!

More precisely, he takes the Grothendieck group of the category of representations of a simply-laced Dynkin quiver, say $K$, and puts an interesting braiding on the category of $K$-graded vector spaces, say $Vect^K$.

Then Christopher shows the Hall algebra is a bialgebra object in $Vect^K$. In other words, it’s a monoid object in the category of comonoid objects in $Vect^K$ — or equivalently, a comonoid object in the category of monoid objects in $Vect^K$.

So, the power of category theory kicks in again! Now we can easily do all sorts of wonderful things with the Hall algebra — things we’d normally do with a bialgebra, but now with $Vect^K$ taking the place of $Vect$.

For example, now we instantly know that the Hall algebra has a monoidal category of representations in $Vect^K$. Back when it was a mere ‘twisted bialgebra’, this would be utterly nonobvious: even if you were enough of a genius to guess it, you’d have to check it by playing around with ‘fudge factors’ and discovering ‘surprising cancellations’. But now it’s an automatic consequence of general facts.

Of course, some people enjoy all sorts of ‘twisted’, ‘deformed’, or ‘warped’ mathematical gadgets. Some people enjoy complexity for its own sake, because it offers scope for cleverness. But simplicity is better, because it lets you do good things without cleverness. So it’s always best when you can take a ‘twisted’ gadget and reinterpret it as a plain old fully functional gadget in another category.

Posted by: John Baez on October 5, 2010 2:49 AM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

Maybe my question will be answered in the paper — I’m so far responding only to the abstract. I will begin by recalling some standard facts that you probably know and reference in your paper, and then ask for clarification of the final sentence in the abstract, where you write: “the Hall algebra … and will also become a Hopf algebra object.”

Let $(A,\Delta,\epsilon)$ be any coassociative counital coalgebra and $(B,m,1)$ any associative unital algebra (both objects in the same monoidal category). Then $\operatorname{Hom}(A,B)$ (hom in the underlying category) is an associative unital monoid under the “convolution product”: $f\star g = m\circ (f\otimes g) \circ \Delta$, and the identity element for $\star$ is $1\circ \epsilon$.

Now suppose that $B=A$ as objects, but I don’t impose any bialgebra condition. Then I do get a distinguished elements $\operatorname{id} \in \operatorname{Hom}(A,A)$. Inventing a word (maybe someone else has also invented it?), I would say that the data $(A,m,1,\Delta,\epsilon)$ is antipodal if $\operatorname{id}$ is left- and right-invertible in the monoid $(\operatorname{Hom}(A,A),\star)$. Notice that there is no need for any braidings, compatibility conditions, etc.

So: it seems from your sentence that the point is that the Hall algebra just is antipodal, whereas for you a Hopf algebra is the data $(A,m,1,\Delta,\epsilon)$ such that it is both antipodal and satisfies the bialgebra compatibility condition.

Is there any other content to that line that I’m missing?

Posted by: Theo on October 4, 2010 6:03 PM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

Hi Theo,

The point of the paper is the bialgebra compatibility condition. The existence of an antipode was simply a nice addition point. You are correct that the definition of an antipode does not require information about the braiding, but I have never thought about antipodes independently. I have always thought of a Hopf algebra as a bialgebra with an antipode, and so this was what I meant in the last sentence.

Posted by: Christopher Walker on October 4, 2010 7:59 PM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

Very neat! Of course you have to make a choice regarding which string to cross over and which under in drawing the bialgebra compatibility diagram in a braided monoidal category. Does that mean there are multiple distinct notions of “bialgebra” in a braided monoidal category, and the Hall algebra is only one of them?

Posted by: Mike Shulman on October 4, 2010 9:41 PM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

We can certainly define two kinds of bialgebra in a braided monoidal category, depending on which string crosses over which in this picture: Alternatively, we can pick our favorite way, use that to define our favorite kind of bialgebra, and note that any braided monoidal category has a kind of ‘opposite’, where we use this new braiding:

$B^{opp}_{X,Y} = B^{-1}_{Y,X}$

Then the other kind of bialgebra can be thought of as our favorite kind of bialgebra, but in the ‘opposite’ braided monoidal category.

Of course the terminology gets a bit confusing here. We can take the opposite of a category by reversing the arrows, we can take the ‘opposite’ of a monoidal category by reversing the tensor product:

$X \otimes_{opp} Y = Y \otimes X$

and we can take the opposite of a braided monoidal category by reversing the braiding. We can even combine these different options, for a total of $2^3$ choices. That’s because a braided monoidal category is a 3-category, so its morphisms can be drawn as 3-dimensional globes, which have a reflection symmetry group of size $2^3$.

But when you said “multiple” distinct notions of bialgebra, did you mean more than two?

I used to know whether you could take a monoid $X$ in a braided monoidal category, with multiplication

$m: X \otimes X \to X \, ,$

and define a new monoid with a “$n$-tuply twisted multiplication”:

$m_{new} = m \circ B_{X,X}^n$

The trick is checking the associative law. I forget the answer.

I don’t think I ever had the brains to check whether a braided monoidal category gives an infinite series of braided monoidal categories where we braid things around a bunch of times, e.g.:

$B^{new}_{X,Y} = B_{X,Y} B_{Y,X} B_{X,Y}$

If the answer to this question were “yes”, then the answer to the previous question would also be “yes”. Furthermore, we would get an infinite series of interesting concepts of bialgebra in a given monoidal category.

I’m too busy to draw the string diagrams needed to check this! But if I had to go by gut instinct, I’d guess the answer is “no”.

Posted by: John Baez on October 5, 2010 3:20 AM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

$m_{new} = m \circ B_{X,X}^n$

brings to mind the exotic multiplications on the 3-sphere where m is quaternionic multiplication and B is the commutator.

Some interesting non-associative phenomena arise:

MR0187242 (32 #4695) Slifker, James F. Exotic multiplications on $S^{3}$. Quart. J. MAth. Oxford Ser. (2) 16 1965 322–359.

Posted by: jim stasheff on October 5, 2010 2:14 PM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

It seems quite unlikely to me that you’d get a new braided monoidal category by twisting strings around each other multiple times. But couldn’t you at least get a different notion of “bialgebra” in a braided monoidal category that way, even if it couldn’t be identified with the original notion of bialgebra in some other braided monoidal category?

Posted by: Mike Shulman on October 5, 2010 6:24 PM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

Mike wrote:

It seems quite unlikely to me that you’d get a new braided monoidal category by twisting strings around each other multiple times.

Yeah, me too. But sometime on a long boring trip I’ll check one of the hexagon identities, just to nail this particular coffin shut.

But couldn’t you at least get a different notion of “bialgebra” in a braided monoidal category that way, even if it couldn’t be identified with the original notion of bialgebra in some other braided monoidal category?

Okay, yeah — you can write down the definition. But to me, I guess, the whole point of a bialgebra object is that its category of actions naturally becomes a monoidal category, thanks to the comultiplication. If our notion of bialgebra doesn’t give some nice result like this — or at least some nice result — I’ll probably conclude that notion is useless. Especially since I don’t know any examples.

Posted by: John Baez on October 6, 2010 7:42 AM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

Posted by: Mike Shulman on October 6, 2010 5:55 PM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

Shahn Majid has studied intensively Hopf algebras in braided categories and called them braided groups. One of the first papers was this, but look also for later papers and his book Foundations of quantum group theory.

He has lots of recipes of how to make new examples out of old. For example, making braided groups out of quasitriangular Hopf algebras by a process of “transmutation”. Majid developed also associated “braided” geometry.

Posted by: Zoran Skoda on October 5, 2010 1:35 AM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

It’s true. Christopher should cite Majid’s work when introducing the concept of a Hopf algebra in a braided monoidal category.

I also told Christopher to read and cite Joyal and Street’s paper where they get a braiding on the category of $A$-graded vector spaces from a bilinear form on the abelian group $A$. It’s one of these, or perhaps both:

• A. Joyal and R. Street, Braided monoidal categories, Macquarie Math Reports 860081 (1986). Available at http://www.maths.mq.edu.au/~street/JS1.pdf.
• A. Joyal and R. Street, Braided tensor categories, Adv. Math. 102 (1993) 20-78.

Hmm, yes, it’s Theorem 12 together with Proposition 13 on page 44 of the first, unpublished, paper. Better to cite the second one if it’s also in there — I don’t have it on me right now.

As usual, they prove this result in such great generality that it’s a bit hard to understand at first. But it’s not hard to extract the desired fact from what they prove.

An interesting fact is that Theorem 12 goes back to the work of Eilenberg and Mac Lane on the cohomology of groups. All this stuff can now be understood in terms of Postnikov towers for $n$-groupoids.

Posted by: John Baez on October 5, 2010 3:41 AM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

Another typo: on page 4, “we define set:”

Posted by: John Baez on October 5, 2010 4:30 AM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

If I understood right, the main result of this paper is long known to the experts. In

Kapranov says:

Note that because of the twist (1.4.4), the theorem does not mean that $R(A)$ is a bialgebra in the ordinary sense; it can be interpreted, however, by saying that $R(A)$ is a bialgebra in braided monoidal category of $K_0(A)$-graded vector spaces.

Posted by: Zoran Skoda on October 17, 2010 4:12 PM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

But there still seem to be some interesting differences in viewpoint. For example, in (1.4.3) Kapranov is using the symmetrized Euler form, while Christopher uses the unsymmetrized one. Kapranov could probably explain this in an instant, but it seems mysterious and interesting to me right now.

Anyway, thanks for pointing this out.

Posted by: John Baez on October 17, 2010 5:51 PM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

Don’t worry, John, I am sure the fresh and systematic point of view, which Christopher is taking with your guidance, will result in many more results on Hall algebras along the way. I personally had some interest some time ago in the sequel paper of Kapranov relating Hall algebras and Heisenberg doubles, but never really went into it. Eisenstein series title on the other hand was always repelling to me; I hope I overcome the fear from number theory once…

Posted by: Zoran Skoda on October 17, 2010 8:07 PM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

The good news is that I’ve finally found out that I like Hopf algebras and would like to go to grad school at a place that they’re covered in the US. The bad news is that it’s already too late to apply at most schools. Any ideas?

Last year I published a paper that can be rewritten in Hopf algebra language. It showed that when you resum the long time propagators for the Hopf algebra (defined by the mutually unbiased bases of the Pauli algebra), you get the usual spin-1/2 propagators (i.e. projection operators), but three copies that you can think of as generations.

The paper I’m working on now shows that when you find the propagators for the group (Hopf) algebra defined by the permutation group on three elements you discover that they map nicely on to the weak hypercharge and weak isospin quantum numbers of the elementary fermions.

Right now I understand Hopf algebra at a very low, intuitive level, but it’s clear to me that this is where I’m going to continue to work. The challenge of getting numbers out of the algebra is incredibly attractive. I’ve been working on my own, but it would be a lot easier if I were in a department; any suggestions?

Posted by: Carl on February 19, 2011 7:55 PM | Permalink | Reply to this

### Re: A Hopf Algebra Structure on Hall Algebras

Carl - I’m coming to this thread late, but your paper on path integrals via mutually unbiased bases actually looks decent to me. I’m curious, did you end up at a grad school?

Posted by: Bruce Bartlett on May 24, 2014 12:16 AM | Permalink | Reply to this
Read the post Christopher Walker on Hall Algebras
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Excerpt: Christopher Walker has successfully defended his thesis, A Categorification of Hall Algebras.
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