This post motivated in part by the recent preprint of Fakhruddin, Khare, and Patrikis, and also by Matt’s number theory seminar at Chicago this week. (If you are interested in knowing what the calendar is for the Chicago number theory seminar this quarter, then that makes two of us. Actually, if you are giving a number theory seminar at Chicago this quarter, please leave a comment on this post with the day you are visiting, because several readers of this blog would be interested in finding out who is coming and what they are talking about.)
be a continuous representation. We now know, by the work of Emerton-Gee, that this representation admits a lift to characteristic zero representation of regular weight which is de Rham (and is even potentially diagonalizable).
On the other hand, can it be the case that there do not exist any de Rham lifts in non-regular weight?
In the most extreme case, where we demand that all the Hodge–Tate weights are zero, then there are obstructions to lifting. In this case, the image of inertia on any lift must have finite image, but the image of inertia of may already be sufficiently large to preclude this possibility. (This was exploited in the proof of Theorem 5.1 here.) So this answers the case when n = 2.
But what happens (for example) for n > 2 and HT weights = [0,…,0,1]? Or even n = 2 and replacing by a finite extension K? The first remark is that even when the residual image lands inside the Borel, there will certainly be obstructions to finding lifts inside the Borel, which means that inductive arguments will not be sufficient. On the other hand, this definitely smells like a tractable problem.
I offer an Aperol Spritz to an answer to this problem — let me do so even in the constrained version in weight [0,0,1] and