Galois Deformations 5: Flach’s Theorem

I’ll begin today by taking a poll. Usually I’d do this at the end, but I assume no one will be reading at that point. Since my research has a bit of an arithmetic flavor, one thing I’d like to do is post about some standard algebraic number theory where about a 10 week graduate class would leave off with an eye towards class field theory. I should probably be more comfortable with these things. My other idea is to blog about Neron models which I feel like I should understand better as well. If you strongly want to see one of these over another, then speak up soon in the comments (I’ll probably do both eventually).

The subject of today is to discuss the obstruction space a little. Recall that whenever we’ve discussed the universal deformation ring in the previous posts we had the caveat “if the deformations are unobstructed, then … “. It would be nice to have some theorems that tell us when this actually happens, or if it doesn’t happen at least get some good control on what the obstructions are.

Let’s quickly recall how this works. Fix {S} a finite set of primes and some prime {\ell} (not in {S}). Fix a an absolutely irreducible residual representation {\overline{\rho}:G_S\rightarrow GL_2(\mathbb{F}_\ell)}. In this case, we know the deformation functor is representable by some universal deformation ring denoted {\mathcal{R}(\overline{\rho})}. The structure of this ring is immensely important since it completely determines what all the deformations are. We said that the obstruction space for deforming is {H^2(G_S, Ad(\overline{\rho}))}, so if we suppose this is {0}, then we know that {\mathcal{R}(\overline{\rho})\simeq \Lambda [[x_1, \ldots, x_d]]} where {d=\dim H^1(G_S, Ad(\overline{\rho}))}. This is why unobstructedness is so important. We can completely determine {\mathcal{R}(\overline{\rho})} up to isomorphism.

A theorem of Flach in 1992 tells us one case when we get unobstructedness. Let {E/\mathbb{Q}} be an elliptic curve having good reduction at {\ell} and {S} the primes of bad reduction together with {\ell} and {\infty}. Let {\rho=\rho_E:G_S\rightarrow GL_2(\mathbb{Z}_\ell)} and {\overline{\rho}} the residual representation. Suppose we have the following three conditions satisfied as well: {\rho} is surjective, for all {p\in S\setminus \{\infty\}} we have {E[\ell]\otimes E[\ell]} has no {G_{\mathbb{Q}_p}}-invariants, and {\ell} does not divide {L(Sym^2 T_\ell E, 0)/\Omega}, then the deformation problem is unobstructed for {\overline{\rho}}.

We definitely won’t worry about what everything in that last condition is. You might be worried that this never happens, but in some sense that can be made precise these conditions are almost always satisfied. Now to finish the post off we’ll just very generally give an outline of the proof. Since we are trying to make some {H^2} vanish, it isn’t surprising that we can reduce to checking that Ш{^1(G_s, E[\ell]\otimes E[\ell])\simeq}
Ш{^1(G_s, \mu_\ell)\oplus} Ш{^1(G_S, Sym^2 E[\ell])} vanishes. We can already start to see the appearence of some of the conditions.

Using standard techniques in Galois cohomology like Hilbert 90 it can be checked directly that the first factor vanishes, so we only need to check that second one. Now define {A=Sym^2 T_\ell E\otimes \mathbb{Q}_\ell/\mathbb{Z}_\ell}. It turns out that we get an injection
Ш{^1(G_S, Sym^2 E[\ell])\hookrightarrow H_f^1(\mathbb{Q}, A)} into the Selmer group, so we can reduce to proving this Selmer group is {0}. Using a very hard, deep theorem and the fact that {E} is modular we can conclude that {\deg \phi H^1_f(\mathbb{Q}, A^*)=0} where {\phi: X_0(N)\rightarrow E} is a modular parametrization. From this we can conclude the theorem of Flach.


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