## Harris 60

I’ve just returned from the excellent MSRI workshop which honored Michael Harris’ 60th birthday, and here is a brief summary of some of the gossip and mathematics I picked up when I was there.

First, let me take note of the nicely designed posters for the conference:

One thing that can be said about the original poster (on the left) is that it does not contain a single mistake. No word on whether a third poster is in the works.

Next up, celebrity fashion. Of course, my main source for celebrity gossip is Richard Taylor, fresh from hobnobbing on the red carpet at the recent breakthrough prize awards.

Richard Taylor and Kate Bekinsdale at the Breakthough Award ceremony

Apparently the hot news from RLT is that flip phones are now back in style. You won’t see Richard or Kate brandish a smartphone; both now proudly sport the retro flip phones as if it’s 2007. Of course, all this comes months after I finally got around to getting my first iPhone.

Finally, something not entirely silly, Akshay gave a very intriguing talk on integral structures in cohomology. It reminded me of a question that we discussed a long time ago near Washington Square Park in NYC. Recall that, for tempered automorphic forms contributing to the cohomology of $\mathrm{GL}(n)/\mathbf{Q}$, a computation with $(\mathfrak{g},K)$ cohomology shows that each such form occurs in cohomology in degrees $q_0, \ldots, q_0 + \ell_0$ and contributes (a multiple) of

$\displaystyle{\binom{\ell_0}{k}}$

dimensions in degree $q_0 + k$. Is there any analogue of this for torsion classes or in characteristic p? Assume here that the residual representation also occurs only inside the relevant ranges of cohomology, which should be the case as long as the corresponding residual representation is not Eisenstein. If $\ell_0 = 0$, there is nothing to say. If $\ell_0 = 1$, then the result follows from an Euler characteristic argument; that is, the cohomology in each of the two non-zero degrees over $\mathbf{F}_p$ will have the same dimension. If $\ell_0 = 2,$ then Poincaré duality shows that the cohomology groups in the two outer degrees will have the same dimension (again we are working with non-Eisenstein classes and trivial coefficients, so from this point of view things look compact), and then an Euler characteristic argument shows that the space appears in some multiplicities of dimensions $(1,2,1),$ as in characteristic zero. Now suppose that $\ell_0$ is arbitrary. I will assume we are in a multiplicity one situtation and that all the local deformation rings are smooth. The CG-method (under suitable hypotheses) produces a resolution $P^{\bullet}$ of $R_{\infty}$ consisting of finite free $S_{\infty}$-modules, where $R_{\infty}$ is smooth of relative dimension $q$ and $S_{\infty}$ is free of relative dimension $n:=q + \ell_0$. To recover the cohomology over a finite field, one takes the quotient of this resolution by the maximal ideal of $S_{\infty}$ and then takes cohomology. In other words, what we are really computing is

$\displaystyle{\mathrm{Ext}^*_{S_{\infty}}(R_{\infty},k)}.$

This answer depends only on the rings involved and not on the resolution. For example, suppose that $q =0$, which is the same as saying that there are no non-trivial local infinitesimal deformations. Then $R_{\infty} = \mathbf{Z}_p$, and one can compute the cohomology of this ring using the Koszul complex, and one gets the expected dimensions. Note that if $R = \mathbf{Z}_p$, then this recovers the automorphic computation, but this is already slightly interesting if $R = \mathbf{F}_p$ or even $\mathbf{Z}/p^k \mathbf{Z}.$ However, it seems a little optimistic to expect this pattern to hold in general. I spent some time trying to prove it using commutative algebra, but one problem is that it is not true in that generality. For example, suppose that $\ell_0 = 3$, and that

$\displaystyle{R_{\infty} = \mathbf{Z}_p[[y_1,y_2,y_3]], \quad S_{\infty} = \mathbf{Z}_p[[x_1,x_2,x_3,x_4,x_5,x_6]],}$

where the map from $S_{\infty}$ sends the generators to the six possibile monomials of degree two. Then the appropriate dimensions of the ext groups are $4,14,14,4$. (Thanks to Daniel Erman for this example.) Now this example actually can’t occur globally, because the same computation implies that one the Betti numbers over $\mathbf{Q}_p$ are also given by these numbers, which violates the previously referenced computation with $(\mathfrak{g},K)$-cohomology. However, one should easily be able to deform it very slightly to kill off any cohomology in characteristic zero, for example replacing $y_i$ by $y_i - \epsilon_i$ for small constants $\epsilon_i$. Of course this doesn’t disprove anything, but it does strongly suggest that the dimensions over $\mathbf{F}_p$ could be all over the place, subject to the Poincaré and Euler characteristic conditions. Akshay has also pointed out that the case $\ell_0 = 3$ is interesting from a different but related perspective: the analytic torsion will vanish in this case, which implies that, at least morally (since we have localized at a maximal ideal), that the alternating product of the the orders of the cohomology groups over $\mathbf{Z}_p$ should equal one. Is this a consequence of the TW-method? I just thought of this question right now and it may have an obvious answer which Akshay knows, I’ll ask him today and report back. A second obvious question is what happens if one looks only at $\mathfrak{m}$-torsion rather than $\mathbf{F}_p$-torsion; perhaps that is the more sensible generalization of the characteristic zero question anyway.

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