A homework exercise for Oaxaca

Here’s a homework problem for those coming to Oaxaca who have a facility for working with Breuil-Kisin modules and finite flat group schemes. Let \(\mathbf{F}\) be a finite field of characteristic \(p\), and consider a Galois representation:

\(\rho: G_{\mathbf{Q}_p} \rightarrow \mathbf{GL}_2(\mathbf{F}).\)

which (one should imagine) is the local restriction of a global representation coming from a modular form. By a standard global argument, one can find a congruent form in weight \(2\), and thus a lift to a representation which is de Rham with Hodge-Tate weights \([0,1]\). For almost all such representations one can ensure that lift is potentially crystalline and hence comes from a representation which is potentially Barsotti-Tate. An immediate consequence is that the representation \(\rho\) itself is — after restriction to some finite extension \(K\) — the generic fibre of a finite flat group scheme. Without any other conditions this is obvious, since one can take \(K\) to be the splitting field of \(\rho\). However, the global argument gives a further restriction that one can take \(K/\mathbf{Q}_p\) Galois with the property that, for some \(2\)-dimensional representation \(V_K\) lifting the restriction of \(\rho\) to \(G_K\), there is a representation:

\( \varrho: \Gamma:=\mathrm{Gal}(K/\mathbf{Q}_p) \rightarrow \mathrm{GL}(D_{\mathrm{cris}}(V_K)) \rightarrow \mathrm{GL}_2(\overline{\mathbf{Q}}_p) \)

which is faithful on the inertia subgroup. In particular, this forces \(\Gamma\) and \(K\) to be “small” in some sense. One can prove this result directly without recourse to any global arguments. For example, consider the case when \(\rho\) is reducible, and, if the ratio of characters is cyclotomic, then additionally assume the extension is not très ramifée. In this case, I claim that one can take \(K\) to be the (unramified extension) of \(\mathbf{Q}_p(\zeta_p)\) which contains the fixed field of the characters on the diagonal. The restriction of \(\rho\) to \(K\) is then the extension of the generic fibre of the trivial group scheme by the multiplicative group scheme. But our assumptions imply that the Kummer extension that arises will come from the pth power of a unit and hence come from a finite flat group scheme over \(K\). The (abelian) group \(\mathrm{Gal}(K/\mathbf{Q}_p)\) has no problem admitting a representation of small dimension which is faithful on inertia.

When \(n = 3\), the automorphic picture would suggest that one can find de Rham lifts with Hodge-Tate weights \([0,1,2]\), and this is the type of thing that I guess one knows now in full generality by Emerton-Gee (but probably earlier in this case). But suppose we are still interested in whether there exist lifts of \(\rho\) which are potentially Barsotti-Tate. We can ask the weaker question: does \(\rho\) come (after restriction to \(K\)) from the generic fibre of a finite flat group scheme for a Galois extension \(K/\mathbf{Q}_p\) which admits a representation:

\( \varrho: \Gamma:=\mathrm{Gal}(K/\mathbf{Q}_p) \rightarrow \mathrm{GL}_3(\overline{\mathbf{Q}}_p) \)

which is faithful on inertia? This seems like a question which one should be able to answer. In particular, suppose that \(\rho\) is some representation with upper-triangular image. It seems possible that if \(K/\mathbf{Q}_p\) is any extension such that \(\rho\) is the generic fibre of a finite flat group scheme over \(K\) then \(K\) might be “too big” to admit such a \(\varrho\). If that were true, this would give a direct proof that \(\rho\) does not admit lift which are potentially crystalline with Hodge-Tate weights \([0,0,1]\), which would (essentially) answer the final question in this post. (I say “essentially” because one should also consider potentially semistable lifts as well. Certainly one should be able to address this by similar methods, but for now, perhaps just assume that the ratio of any two consecutive characters occurring in \(\rho\) is not cyclotomic.)

This seems to be an eminently answerable question to someone who knows what they are doing, and there are certainly some experts in this sort of computation who will be in Oaxaca in a few weeks time. So maybe one of you can work out the answer (calling the Hawk!).