## Robert Coleman

I was very sad to learn that, after a long illness with multiple sclerosis, Robert Coleman has just died.

Robert’s influence on mathematics is certainly obvious to all of us in the field. Most of my personal interaction with him was during my last two years as a graduate student at Berkeley. We would chat in his office, and sometimes have lunch at Nefeli caffe. Kevin and I had recently made some modest progress on Kevin’s crazy slope conjectures, and much of that time with Robert was spent with me presenting crazy ideas and predictions on the white board in Evans Hall while Robert looked on with his classic look of amused skepticism. There would also be the occasional wine and cheese in his office, especially if an old visitor was in town.

I certainly didn’t know him as well as many others did, but I felt very honored that he asked me to accompany him (as a grad student assistant) to China for his ICM address. As it happened, the relevant hotels in China would not allow him to bring Bishop (his guide dog) along with him, so he didn’t end up going.

Mathematically, Robert was very original. I have no plans to attempt to summarize his research, but I just want to discuss one problem which he had thought about in recent years, namely, what the eigencurve looked like at the boundary of weight space — especially in light of the description given by Kevin and Lloyd Kilford when $N = 1$ and $p = 2$. Suppose one is given a Fredholm determinant

$\mathrm{det}(1 - U T) = P(T) = 1 + \sum_{n=1}^{\infty} a_n T^n$

where $a_n \in \Lambda = \mathbf{Z}_p[[X]]$, and one wants to understand the spectrum of $U$ at the “boundary” of weight space, that is, when the valuation of $X$ goes to zero. For example, an interesting collection of points near the boundary are the classical points with highly ramified nebentypus character. If $a_n$ is not divisible by $p$, then the valuation of $a_n$ at a specialization of $X$ close to one will co-incide with the valuation of the reduction mod-$p$ of $a_n$ as an element of the discrete valuation ring $\mathbf{F}_p[[T]]$, that is, it will be determined by the smallest non-zero coefficient of $a_n$ modulo $p$. Robert’s idea was to study the “halo” of the eigencurve, which intuitively speaking, should be an object cut out by a compact operator $U_{\chi}$ in characteristic $p$ with characteristic power series $P(X) \mod p$. If the valuations of the elements $a_n(X) \mod p$ define a Newton Polygon $N$, then the Newton Polygon at some point on the eigencurve which is sufficiently close to the boundary should be a simple multiple of $N$. This is one of my favourite problems! I know Robert has some ideas on how to approach this problem, but unfortunately I don’t know exactly what they were or how much progress he had made. One natural question is whether this structure will ultimately be purely explainable in terms of $p$-adic local Langlands. One even more basic question is what happens numerically on components of the eigencurve corresponding to a representation $\overline{\rho}$ which is absolutely irreducible after restriction to a decomposition group at $p$; I presume one sees the same behavior, but has anyone checked this? Perhaps the easiest example to check would be to compute the slopes of forms on $S_2(\Gamma_1(11 \cdot 2^n),\chi)$, where $\chi$ has conductor $2^n$.

Matt Baker has some further recollections of Robert here, and he also invites his readers to share there memories there.

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### 2 Responses to Robert Coleman

1. TG says:

I never met Robert Coleman, but hearing you guys talking about him made me wish I had.

It won’t surprise you to hear that I wanted to attack this problem with p-adic local Langlands and R=T, but we never got anywhere. I think KB and I did think about this a little in 2006 – of course these representations are still trianguline, so you can look in Colmez and see a concrete description of p-adic LL, and then try to compute reductions mod p. Given that the expected answer is so simple, you might hope that there was some nice structure that you’d see that would explain it, but we didn’t spot anything. Then again, I think we were sufficiently disillusioned with the whole approach to these kinds of questions that we didn’t even explicitly bash out a single example, which is presumably possible.