Taking a page from Gil Kalai, here is a question to test your intuition about 2-dimensional crystalline deformation rings.
Fix a representation:
\(\rho: G_{\mathbf{Q}_p} \rightarrow \mathrm{GL}_2(\overline{\mathbf{F}}_p)\)
after twisting, let me assume that this representation has a crystalline lift of weight \([0,k]\) for some \(1 \le k \le p\). Let \(R\) denote the universal framed local deformation ring with fixed determinant. Now consider positive integers \(n \equiv k \bmod p-1\), and let \(R_n\) denote the Kisin crystalline deformation ring also with fixed determinant. Global considerations suggest that for \(n \equiv m \equiv k \bmod p-1\) and \(n \ge m\), there should be a surjection \(R_n/p \rightarrow R_m/p\), and quite possibly one even knows this to be true. Global considerations also suggest that any representation can be seen in high enough weight, which leads to the following problem:
Question: How large does \(n\) have to be to see the entire tangent space of the unrestricted local deformation ring \(R\)? That is, how large does \(n\) have to be for the map
\(R/(p,\mathfrak{m}^2) \rightarrow R_n/(p,\mathfrak{m}^2)\)
to be an isomorphism? Naturally, one can also ask the same question with \(\mathfrak{m}^2\) replaced by \(\mathfrak{m}^k\) for any \(k \ge 2\).
The first question came up in a discussion with my student Chengyang. I made a guess, and then we proceeded (during our meeting) to do a test computation on magma, where my prediction utterly failed, but in retrospect my computation itself may have been dodgy so now I’m doubly confused.
Matt remarked that this question is not entirely unrelated in spirit to the Breuil-Mezard conjecture. Instead of counting multiplicities of geometric cycles, one is measuring the Hilbert-Samuel function and its “convergence” to that of the free module. Also, if you know everything about \(\mathrm{GL}_2(\mathbf{Q}_p)\) and \(2\)-dimensional Galois representations then you should be able to answer this question too.
Of course I could have re-done the initial computation for this blog post, but I think at least some readers are happier when I ask questions for which I don’t know the answer…
But if you let \(R^{\mathrm{ord}}_n\) denote the ordinary locus in \(R_n\), (in particular I assume that rhobar is reducible) then \(R^{\mathrm{ord}}_n\) will satisfy the same heuristics as \(R_n\), but the embedding dimension is \(1\) for all \(n\). So it is not obvious that such \(n\) will exist.
The global heuristic I was referring to is the fact is that all modular Galois representations (over finite rings) of level \(Np^m\) also come from modular forms in level \(N\) prime to \(p\) possibly in some larger weight. In particular, you have big \(R = \mathbf{T}\) theorems where \(R\) is the unrestricted deformation ring and \(\mathbf{T}\) is the Hecke algebra acting on the space of all \(p\)-adic modular forms (which are limits of forms of level prime to \(p\)). I think (by globalizing any \(\rho\) to a totally real field in which \(p\) splits completely) this actually ends up giving a proof of the assertion that such an \(n\) exists, although purists may reasonably object to such a global argument (though I don’t see any local argument).
But if we add the adjective “ordinary” everywhere, don’t you also have big ordinary R= ordinary T a similar statement about Galois reps over finite rings, but the assertion about the embedding dimensions does not hold? The somewhat easier, but analogous setting is deformation ring of $1$-dimensional trivial rep of $Gal_{Q_p}$: there the crystalline locus if we take all the weights is dense, but the assertion about embedding dimensions does not hold.
To make my inquiry more precise: Let \(1\) be the trivial 1-dimensional representation of \(\mathrm{Gal}_{\mathbf{Q}_p}\), let \(R\) be its universal deformation ring. We globalize \(1\) to be the trivial representation of \(\mathrm{Gal}(\mathbf{Q}(\zeta_p^{\infty})/\mathbf{Q})\) and let \(R^{\mathrm{glob}}\) be its universal deformation ring.
Then \(R^{\mathrm{glob}}\) is the completed group algebra of \(G:=\mathrm{Gal}(\mathbf{Q}(\zeta_p^{\infty})/\mathbf{Q}(\zeta_p))\), which is isomorphic to \(1+p \mathbf{Z}_p\). Now the characters \(\chi_{\mathrm{cyclotomic}}^{(p-1)n}\), \(n \in \mathbf{Z}\) will be \(p\)-adically dense in the Banach space \(C(G, \mathbf{Q}_p)\), this will imply the analog of “all modular Galois representations (over finite rings) of level \(Npm\) also come from modular forms in level \(N\) prime to \(p\) possibly in some larger weight” in your comment above.
Since \(\mathrm{Gal}(\mathbf{Q}_p(\zeta_p^{\infty})/\mathbf{Q}_p(\zeta_p)) = \mathrm{Gal}(\mathbf{Q}(\zeta_p^{\infty})/\mathbf{Q}(\zeta_p))\), we have \(R^{\mathrm{loc}} = R^{\mathrm{glob}}[[x]]\), where \(x\) is the unramified direction.
Now the assertion about the embedding dimensions in this \(1\)-dimensional case is false: the image of \(x\) always spans the tangent space of \(R_n\).
So here is my question: Where does the argument that you have in mind for \(\mathrm{GL}_2\) setting goes wrong in \(\mathrm{GL}_1\) setting?
The facetious answer is that (in the ordinary and in the one-dimensional setting) the assertions do hold for the fixed determinant deformation rings. (In the ordinary case this has epsilon content — one has to take \(n > 2\)).
More seriously, I think the point is more or less as follows. Globally, you have the following objects:
1. The big global deformation ring \(R\) with no condition on the determinant.
2. The ring of divided congruences \(D\) of modular forms.
Now more or less (more if you make requisite Taylor-Wiles assumptions) the ring \(R\) acts faithfully on \(D\). Now you certainly can’t see all of \(R\) in any fixed weight, because the determinants do indeed vary non-trivially mod-\(p\) even in \(D/p\). This is more or less the same as what you noted in dimension one.
OTOH, \(D\) has an action of \(\mathbf{Z}^{\times}_p\) which cuts out the \(p\)-adic weight in the sense of Serre. The part where this group acts by any given character cuts out the part of \(R\) with the corresponding determinant. Similarly, one can consider the action of \(\mathbf{Z}^{\times}_p\) on \(D/p\). Once again you can consider the part of this fixed (say) by the action of \(1 + p \mathbf{Z}_p\). Here \(R\) will more or less factor through the fixed determinant unrestricted deformation ring modulo \(p\). So now the claim is that this invariant part is actually “seen” by the union of the mod \(p\) reductions of modular forms of fixed weight as you vary over all weights \(k\). But this is a theorem of Katz.
Interesting! Are you saying that one can convert this theorem of Katz to a commutative algebra assertion about the quotients of global deformation ring, which then implies the statement about the embedding dimension?
If yes, what is this assertion precisely?
I don’t know what “embedding dimension” is, but I think you might be happy with a reference, e.g. this paper
http://people.brandeis.edu/~jbellaic/preprint/Heckealgebra6.pdf
explicitly contains all the statements invoked here (see also Theorem II).
By “embedding dimension” I just mean
$dim_k m/(p, m^2)$, this is probably not quite correct usage of the term, sorry.
Let $\mathfrak{a}_n$ be the kernel of $R\rightarrow R_n$, in the notation of your post. I think I know how to prove the following (even for N-dimensional representations).
Prop. Let $\alpha:R\twoheadrightarrow A$ be a quotient with $A$ a finite ring. Then there are integers $M$ and $N$ such that the ideal
$( p ^M, \bigcap_{n=2}^N \mathfrak a_n )$
is contained in the kernel of $\alpha$.
If I take $A= R/(p, m^2)$ then this will give some statement of the flavour that you are after, but
in this Proposition I really have to take the intersection in the range from 2 to N, and it seems that a similar thing is happening in the paper that you have linked, where the authors consider a sum of $S_k(\mathbb F)$. (The proposition stated works both globally and locally and follows from density of algebraic vectors in completed cohomology and the patched module respectively, also for GL1).
What you seem to be saying that instead of taking the intersection in the above proposition we can take a single weight N large enough. It would be great if you could explain how you do that.
For modular forms, there is an inclusion \(S_k(\mathbf{F}_p) \rightarrow S_{k+p-1}(\mathbf{F}_p)\) given by multiplying by the Hasse invariant.
I agree with the statement, but I don’t see how it implies for example that $R_{n+(p-1)}/ p$ surjects onto $R_n/p$.
If you are complaining about the possible lack of faithfullness of \(S_k/p\) as an \(R_k/p\)-module, then why not replace \(R_k/p\) by the image inside the endomorphisms of \(S_k/p\); these obviously surject onto each other and their inverse limit is still the entire mod-p fixed determinant global deformation ring.
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