## January 27, 2006

### Spans, 2-Hilbert Spaces and Module Categories

#### Posted by Urs Schreiber (This entry has been momentarily removed. I need to correct something.)

Posted at January 27, 2006 5:04 PM UTC

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/730

### Re: Spans, 2-Hilbert Spaces and Module Categories

I’m packing for tomorrow’s trip to Marseille, where I’ll talk about monads, PROPs, Lawvere’s algebraic theories and reasoning with diagrams to a bunch of people interested in the geometry of calculation. You might like the stuff about the relation between classical and quantum logic - that is, categories with finite products, versus symmetric monoidal categories!

But, for now, just one remark on spans versus bimodules. At first glance spans and bimodules seem pretty different, since a span is a map from some gadget $X$ to some gadget $A×B$, while a bimodule is some kind of action of $A×B$ on $X$.

(I want to be a bit vague about what these things $A,B$ and $X$ are, since the issue is very general. I’m also being a bit vague about when we use $B$ and when we use ${B}^{\mathrm{op}}$; for bimodules of rings we normally say we have an action of $A×{B}^{\mathrm{op}}$ on $X$. Maybe I should even be a bit vague about whether I really mean $×$ or $\otimes$! One can usually sort out all these issues without too much trouble in any specific situation.)

But, we can see spans and bimodules are actually not so different if we think about it right!

First of all, the issue of “spans” versus “bimodules” is really just a special case of the issue of “maps” versus “modules”, as we can see by calling $A×B$ above simply $P$. Then our span becomes a map $P\to X$, while our bimodule becomes an action of $X$ on $P$.

So, how can maps masquerade as actions and vice versa?

Well, here’s one way. Suppose we have a principal $G$-bundle $P\to X$. Then this is “the same” as a map $X\to \mathrm{BG}$ where $\mathrm{BG}$ is the classifying space of $G$.

And here’s another way… which is secretly the same idea. Suppose we have a functor $P\to X$. Then this is “the same” as a weak 2-functor $X\to \mathrm{Cat}$ assigning to each object of $X$ its “essential preimage” under the functor $P\to X$.

Or, decategorifying: suppose we have a function $P\to X$. Then this is “the same” as a function $X\to Set$, assigning to each point its preimage under the map $P\to X$.

Or, a special case: suppose we have a subset $P\to X$. Then this is “the same” as a function $X\to 0,1$, namely the characteristic function of this subset.

These are all examples of a general construction which I described in week223. And, they all have pretty easy “bimodule/span” versions, where you write $X$ as $A×B$.

Jeffrey uses this trick in his paper when he thinks of a stuff operator, not as a “matrix of groupoids” $A×B\to \mathrm{Gpd}$, but (equivalently) as a groupoid $P$ equipped with a functor $P\to A×B$.

Posted by: john baez on January 28, 2006 8:29 AM | Permalink | Reply to this

### Re: Spans, 2-Hilbert Spaces and Module Categories

It’s possible that nothing I said in my last comment is interesting to you or relevant to what you’re doing… I’m just gradually catching up with your train of thought, and I can’t help saying a bunch of stuff as I do so.

Your account of Ostrik’s theorem is really fascinating. I’ve mainly thought about module categories of Vect, thanks to the work of Kapranov-Voevodsky and Yetter. Indeed, under some nice assumptions these turn out to be just categories like Vectn, and Vectn is the category of modules of kn - the commutative ring of k-valued functions on the n-point set, where k is our ground field. So, Ostrik is generalizing the heck out of this…..

Of course we can also think of Vectn as the category of modules of any algebra Morita-equivalent to kn, like a direct sum of n matrix algebras. So, Morita equivalence shows up very naturally these questions.

This makes me want to digress a bit….

As you know, Morita equivalence gets along with the operation of tensoring algebras, so we get a monoid of Morita equivalence classes of algebras over k, where the monoid operation comes from tensor product. And sitting inside this monoid is the Brauer group, consisting of all the invertible elements.

The Brauer group of the real numbers, for example, is Z/2, with elements corresponding to R (real numbers) and H (the quaternions).

But, we can also play this game with superalgebras, and then it turns out the superBrauer group of R is Z/8, which is just the Bott periodicity clock! I spoke about this in week211 and week212, based on some preliminary remarks in week209.

It would be fun to fit this into your big picture somehow, especially since it’s connected to how superstrings work best in 10 dimensions.

But I’m still just rambling, not really responding to your ideas… sorry!

Posted by: john baez on January 28, 2006 7:37 PM | Permalink | Reply to this

### Re: Spans, 2-Hilbert Spaces and Module Categories

It’s possible that nothing I said in my last comment is interesting to you or relevant to what you’re doing…

Oh, it is all very interesting! But I will have to better understand this. I understand how a map $P\to X$ is equivalently encoded in a generalization of a characteristic function on $X$.

So I guess I roughly understand how we can think of a span $X\to A×B$ as a map $A×B\to \mathrm{something}$. But how is the latter like a bimodule over $A×B$?

I know one obvious action of $A$ (say) on maps $f:A\to \mathrm{something}$, namely that obtained by setting

$\left(\mathrm{af}\right)\left(\cdot \right)=f\left(a\cdot \right)$.

But can I think of every $A$-module as a function on $A$ this way?

BTW, Victor Ostrik has kindly confirmed my little conjecture, that under suitable conditions the 2-category of bimodules internal to $C$ is equivalent to the 2-category of module catgories over $C$.

For the time being, this provides the relation to spans that I was looking for. As I tried to work out in the above mentioned notes, the 2-category of spans in $C$ also embeds into the 2-category of module categories over $C$.

But certainly I would want to understand this in the more direct way which you seem to have in mind. I will have another look at the TWFs which you mention.

Posted by: urs on January 29, 2006 6:48 PM | Permalink | Reply to this

### Kapranov’s 2-vectors as a special case

I’ve mainly thought about module categories of Vect, thanks to the work of Kapranov-Voevodsky and Yetter.

Thanks for mentioning this. This is a nice example and in fact a special case in which we can easily think of bimodule as spans.

So we consider the 2-ring $R=\mathrm{Vect}$ (finite dimensional vector spaces with respect to some field). We have 2-category ${}_{R}\mathrm{Mod}$ whose

- objects are left $R$-module categories

- morphisms are compatible functors

- 2-morphisms are natural transformations .

We have a large sub-2-category sitting inside this beast (maybe even being equivalent to it) that comes from bimodules of finite hereditary algebras in $R$.

Using the quiver-language, we can think of the $R$-modules as categories of functors from quivers to $\mathrm{Vect}$.

In order to obtain the special case of Kapranov-Voevodsky 2-vector spaces we restrict attention to discrete quivers, i.e. to those which have only vertices, no edges.

A 2-vector of this sort is hence simply a tuple of $n$ vector spaces, one over each vertex of the quiver.

A dual 2-vector is a functor from the opposite quiver, which in this case is the same as the original quiver. The tensor product over the quiver algebra ${K}^{n}$ plays the role of the inner product on these 2-vectors and is simply the $\mathrm{Hom}$ operation. (Actually the $\text{Anti-Hom}$, but that’s invisible for the special case of Kapranov-Voevodsky 2-vectors).

Since there is an $n×m$-dimensional vector space of linear maps from an $n$ to an $m$-dimensional vector space, the $\mathrm{Hom}$-operator correctly reproduces the componentwise product on 2-vectors.

And so on.

Now, you have taught me that we can switch to a dual description. Since our quivers are discrete, we can regard functors

(1)$Q\to \mathrm{Vect}$

from the quiver to vector spaces equivalently as a vector bundle

(2)$E\to Q$

over the quiver. In this picture, it is the fiberwise product of vector bundles (the pullback of the respective half-spans) which reproduces the inner product on Kapranov’s 2-vectors.

Very nice.

Posted by: Urs on January 30, 2006 3:25 PM | Permalink | Reply to this
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