## The geometry of Soergel bimodules August 23, 2007

Posted by Ben Webster in Uncategorized.

OK, I’m back on to talking about my research. Hurray for self-promotion! This is a continuation of my posts on Hochschild homology and Soergel bimodules. You can read the actual paper on the arXiv.

Around a year ago, I decided I wanted to attack this stuff (by which I mean the Hochschild homology of Soergel bimodules) geometrically. This is actually a pretty obvious thing to do; there’s a very nice geometrical interpretation of Soergel bimodules, due to (who else?) Soergel. However, I asked some knowledgeable people about it and nothing seemed to be known really, though there were some enticing hints (for example, Jake Rasmussen told me the very geometrically flavored conjecture that Geordie and I solved in our paper almost a year ago in Uppsala).

The geometric interpretation of Soergel bimodules I mentioned before is as follows: Let $G$ be the complex general linear group $\mathrm{GL}_n(\mathbb{C})$ and $B$ be the subgroup of upper triangular matrices (the “B” is for “Borel,” which is what Lie theorists tend to call this subgroup). Consider the action of $B\times B$ by left and right multiplication.

Those of you who are happier thinking about linear algebra can think of this action as leftward column operations and upward row operations (these correspond to right and left multiplication, respectively). Here, I’m not including switching rows or columns as an operation.

What are the orbits of this action? Of course, you probably all remember the theorem from linear algebra that if you are also allowed to do switching rows, you can always get back to the identity matrix. Thus, it stands to reason that if you don’t allow switching, you get a matrix obtained from the identity by switching rows: a permutation matrix (it’s not particularly hard to prove this properly. I leave it as an exercise to the reader). Thus, the orbits of $B\times B$ on $G$ are all of the form $BpB$ where $p$ is a permutation matrix.

The decomposition of $G$ into these orbits is typically called “the Bruhat decomposition.” Its existence in general is a rather important fact about Lie groups (as best expressed by Jacques Tit’s theory of (B,N)-pairs). Those of you who enjoyed Noah’s post might note that this is one way of understanding the idea that the symmetric group is $\mathrm{GL}$ over the field with one element.

So, now, we have a new way of understanding the symmetric group: the geometry of these subvarieties of $G$. For example, the closure $\overline{BpB}$ is $B\times B$ invariant, and thus a union of orbits $Bq B$ for certain permuations $q$. We can put a partial order on $\mathrm{Sym}(n)$, we say $q\leq p$ if $Bq B\subset \overline{BpB}$, called the Bruhat order (the Wikipedia page gives the equivalent combinatorial definition).

Thus, these closures are rather interesting objects (more advanced readers should note that if you mod out by the [free] right action of $B$ on $\overline{BpB}$, then you will get a Schubert variety. So, these orbit closures are the same basic information as Schubert varieties, but presented a bit differently). Now, I hope, dear readers that you are well enough trained that whenever you encounter an interesting variety, you ask “Hmm, I wonder what it’s cohomology is?”

That’s a pretty good question (and one we’ll answer part of later), but fact there’s a better one (which you would ask if you were better trained, but I think that’s expecting too much) which is “Huh, that variety has a group action. I wonder what its equivariant cohomology is?” Now, from experience I know that this is the point where a lot of people start tuning out, but please, mathematicians of the world, open your minds! Equivariant cohomology is an amazing tool, and isn’t hard. It’s often actually much easier to compute than usual cohomology. The canonical example of this is probably toric varieties.

Well, it turns out that if $\overline{BpB}$ is smooth (more on the singular ones later), there’s a very good answer. One of the basic facts about equivariant cohomology is that the $B$-equivariant cohomology of a space is naturally a module over the cohomology ring of the classifying space (rather unfortunately denoted $H^*(BB)$). Thus, the equivariant cohomology of $\overline{BpB}$ is naturally a bimodule over $H^*(BB)$, which the equivariant topologists in the audience recognize as a polynomial ring ($B$ is homotopy equivalent to a product of circles, so $BB$ is a product of $\mathbb{C}P^{\infty}$‘s).

So, $R_p=H^*_{B\times B}(\overline{BpB})$ is a bimodule over a polynomial ring! Of course, it’s a Soergel bimodule; in fact an indecomposable one.

I’d known this much for quite a while (for example, Khovanov mentions it in slightly different form in his paper on Soergel bimodules and knot homology), but then got hung up on the following point: what sort of geometrical operation corresponds to taking Hochschild homology? I mean, if you’re looking at this picture from the perspective of classical homological algebra, it’s unclear what taking a projective resolution over an Ext-algebra (which is what any cohomology ring is, at heart) seems like a rather inexplicable thing to do.

The answer is pretty easy though: it’s equivariant cohomology for the conjugation action of $B$.

Theorem (W.-Williamson): If $\overline{BpB}$ is smooth, then

$H\! H^*(R_p)\cong H^*_B(\overline{BpB})\cong H^*(BB)\otimes\wedge^\bullet V$

where $V$ is a graded vector space of dimension $n$ (the grading is determined by $p$).
Now, this is not hard to prove, but it involves a few concepts I hadn’t really thought about before. The problem is that I was coming at things from a slightly wrong perspective and attempting to generalize incorrectly: really, I should have been thinking about $H^*_{B\times B}(\overline{BpB})$ as an $A_\infty$-module over $H^*(B(B\times B))$. Now, I won’t get into the details of what that is supposed to mean (I’ll leave that to The Infinite Seminar), but suffice to say, every $A_\infty$ structure in sight is formal, so it doesn’t show up in the final answer, but does in the powerful theorems one needs to state to finish the proof.

The actual geometry here comes in proving the second isomorphism, which uses some results of Akyildiz and Carell to directly compute $H^*(\overline{BpB})$ and then the theory of equivariant formality to relate that to equivariant cohomology.

Now, those of you who haven’t been beaten into submission may be wondering, what about the non-smooth cases? Does everything go to hell there? Luckily, no (unless you agree with Sartre that “hell is perverse sheaves”). But do have to change your question a little. You see, the cohomology of singular varieties isn’t so nice. Clearly not as nice as that of smooth varieties (no Poincaré duality, etc.). But it turns out that there’s another version of cohomology called intersection cohomology (denoted $I\!H^*(X)$)that matches usual cohomology on smooth varieties, but behaves differently on singular varieties (more like the cohomology of manifolds, but at the cost of things like loss of homotopy invariance and some functoriality).

So, the more general version of the theorem above is

Theorem: (various people) For any $p\in\mathrm{Sym}_n$, the equivariant intersection cohomology $I\! H_{B\times B}(\overline{BpB})$ is an indecomposable Soergel bimodule (and this is a complete and non-redundant list). The Hochschild homology of this bimodule is $I\! H_{B}(\overline{BpB})$.

I’ll just note that it is not in general true that the Hochschild homology of $I\! H_{B\times B}(X)$ is $I\! H_{B}(X)$. This depends on some formality properties of these varieties.

Unfortunately, there is no known analogue of the second isomorphism stated before for arbitrary $p$. Any statement like that would probably be a lot more complicated.

1. A.J. Tolland - August 23, 2007

The answer is pretty easy though: it’s equivariant cohomology for the conjugation action of B.

Maybe I’m crazy but this seems obvious: the conjugation action should be forced on you by Hochschild’s tendency to make things left-right symmetric.

Probably there are some subtleties I’m squinting past.

2. Ben Webster - August 23, 2007

Maybe I’m crazy but this seems obvious: the conjugation action should be forced on you by Hochschild’s tendency to make things left-right symmetric.

All mathematics is obvious if you have the right background. If you’re thinking in the framework of the equivariant derived category and dg-modules over cohomology rings of classifying spaces, then yes, it’s fairly obvious, modulo some details (as I said, it’s not in general true that the diagonal equivariant cohomology is the Hochschild homology. It requires a formality statement).

In this case, a lot of the problem is just that people think about $I\!H^*_B(\overline{BpB}/B)$ instead of $I\!H^*_{B\times B}(\overline{BpB})$, which means they don’t even notice that taking the diagonal action is a possibility. It sounds like a minor thing, I know, but little ruts can make a big difference in actually noticing stuff.