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Consider a finite-dimensional $\mathbf{Q}_p$-vector space $V$ and a continuous representation $\rho : G_{\mathbf{Q}_p} \to \mathrm{GL}(V)$. Fontaine introduced various $\mathbf{Q}_p$-algebras with $G_{\mathbf{Q}_p}$-actions, notated $B_{\bullet}$ where $\bullet \in \left\{\mathrm{crys}, \mathrm{st}, \mathrm{dR}\dots \right\}$, which "classify" interesting representations $V$; we say $V$ is $\bullet$ if equality holds in the relation $\mathrm{dim}(B_{\bullet} \otimes V)^{G_{\mathbf{Q}_p}} \leq \dim{V}$.

Recently, the notion of a "trianguline" Galois representation has become increasingly important. Avoiding the precise definition, I will say rather perversely that the trianguline representations are roughly the closure of the crystalline locus in the set of all $\rho$'s as above (made precise, this is a theorem of Chenevier which is of course predicated on the actual definition of trianguline). So my question: is there a ring of periods $B_{\mathrm{tri}}$ which classifies trianguline representations in the above manner, and/or is such a ring expected to exist? Is/should it be a suitable "completion" of $B_{\mathrm{crys}}$?

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If you just close up crystalline points in a local deformation space, you will get the whole space. (See Kisin's article in the 2nd p-adic Langlands volume of Asterisque.) In order to have the closure be trianguline reps. (rather than all reps.) you need to add a refinement at every point. So whatever $B$ is (if it does exist), it will involve not just a Galois representation, but a Galois representation together with a refinement. (And note that, beyond the 2-dimensional case, it's not so clear --- at least to me --- how one should define the notion of refinement.) –  Emerton Jul 14 '11 at 3:23
Thanks for the remarks! Indeed, I swept the necessity of a refinement under the carpet with the word "roughly", though perhaps this was (ironically) too rough. –  David Hansen Jul 14 '11 at 3:36
Who came up with the word "trianguline"? Maybe not everyone should be allowed to invent words...! :) –  Mariano Suárez-Alvarez Jul 14 '11 at 7:17
Colmez invented it. Here are the man's own words "Le triangle est un instrument de musique dont le son (triangulin (?)) est presque cristallin... ". –  Laurent Berger Jul 14 '11 at 10:42
Of course, there's also the fact that a representation is trianguline if the matrices of $\phi$ and $\gamma$ on its $(\phi,\Gamma)$-module are upper triangular. –  Laurent Berger Jul 14 '11 at 10:43

1 Answer 1

up vote 13 down vote accepted

The category of trianguline representation is stable under all the usual representation-theoretic operations (subs, quotients, $\oplus$, $\otimes$), so by some general tannakian formalism, there does exist a ring $B_{tri}$. The rough idea is to look at $Q_p^{alg} \otimes B_{st} \langle \langle \log(t) \rangle \rangle$ where "$\langle \langle \log(t) \rangle \rangle$" means "power series with some non zero radius of convergence" and $t$ is the usual $t$ in this business. It's interesting to note that I first heard about this ring from Fontaine (around 2003-04 maybe - I was still at Harvard) when trianguline representations had not yet been defined. Fontaine told me at the time that repns admissible for this ring should be interesting! A few comments are in order:

  • $B_{st}$ does not have the structure of a Banach space, so you need to figure out what "radius of convergence" means
  • $\exp(\log(t))=t$, so there are relations in the definition of your ring
  • you need to decide if you want a ring for "trianguline" or "split-trianguline" repns

I thought about this again a few weeks ago and, if I remember correctly, came to the conclusion that if you take $B = Q_p(\mu_p) \otimes \hat{Q}_p^{nr} \otimes B_e \otimes Q_p \langle \langle \log(t) \rangle \rangle \otimes Q_p[\log(\tilde{p})]$ (whew!), where $B_e=B_{cris}^{\phi=1}$, then $B$-adm reps of $G_{Q_p}$ are trianguline, and conversely split trianguline reps of $G_{Q_p}$ with integer slopes are $B$-adm. This hopefully gives an idea of the kind of ring which one should be looking for.

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Dear Laurent, How does the notion of refinement fit in with this ring? (I guess I am asking how one would describe a triangulated representation --- i.e. a rep'n with a chosen triangulation.) Best wishes, Matt –  Emerton Jul 14 '11 at 6:53
Dear Matt, since a generic trianguline representation has only one possible refinement, and it's only at certain points (basically the semi-stable-up-to-twist in dim 2) that the issue arises, I'm not sure that it would be so easy to see this at the level of the ring. But maybe I'm just missing something... –  Laurent Berger Jul 14 '11 at 10:51
Dear Laurent, thanks very much for this answer. Is it really the case that none of the tensor products in your proposal should be completed tensor products? –  David Hansen Jul 14 '11 at 19:15
Dear Professor Laurent, can you please give a reference where this ring is studied and what sort of properties do we expect this ring to have? Thanks. –  Arijit Aug 15 '12 at 21:04
This ring has not been studied. At some point, my student Di Matteo was interested in it, so you could ask him. –  Laurent Berger Aug 16 '12 at 16:01

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