# Galois Deformations 3: Tangent and Obstruction

Recall last time that if we start with some finite field ${k}$ and some absolutely irreducible residual Galois representation ${\overline{\rho}: G_S\rightarrow GL_n(k)}$, the deformation functor is (pro)representable by some universal deformation ring which we will denote ${\mathcal{R}(\overline{\rho})}$ (again, this was known even in the original papers to be true in a much more general situation than this one). Today we’ll try to figure out some properties of this ring.

First we note something that better be true. We really only consider representations up to equivalence, so if ${\overline{\rho}}$ and ${\overline{\rho}'}$ are equivalent then there is a matrix that conjugates one to the other. By functoriality of everything involved we can consider the natural transformation of deformation functors induced by the group scheme homomorphism given by this conjugation action. By representability the natural transformation gives a homomorphism of the universal rings ${\mathcal{R}(\overline{\rho})\rightarrow \mathcal{R}(\overline{\rho}')}$. It can be checked that this is an isomorphism. Geometrically this means that the “local space” of deformations is the same for equivalent representations.

By similar arguments to the last time I talked about deformation theory, we can compute that the tangent space to the functor given by ${Def_{\overline{\rho}}(k[\epsilon])}$ is isomorphic to ${H^1(G_S, Ad(\overline{\rho}))}$ where ${Ad(\overline{\rho})}$ just means the adjoint representation. So as in the geometric situation, if ${\mathcal{R}}$ is smooth, we can compute the dimension of the space of deformations by computing the dimension of some first cohomology. Thus if the deformations are completely unobstructed, then we know that ${\mathcal{R}\simeq \Lambda[[x_1, \ldots, x_d]]}$ where ${d=\dim H^1(G_S, Ad(\overline{\rho}))}$.

The standard next question is to ask what is the obstruction space. Very standard arguments that can be made with almost any deformation functor show us the following. I’ll leave out the details, but tell you what to fill in. By an obstruction, we mean take a map in our deformation ring category ${R_1\rightarrow R_0}$ that has kernel ${I}$ with the property that ${I\cdot m_{R_1}=0}$ so that it is naturally a ${k}$-vector space. Take some representation ${\rho: G_S\rightarrow GL_n(R_0)}$. We want to know if there is some ${\rho': G_S\rightarrow GL_n(R_1)}$ so that upon composing ${GL_n(R_1)\rightarrow GL_n(R_0)}$ we get ${\rho}$ back. If you’ve seen this geometrically, then your intuition should be to show that “locally” you can do it, and then the ability to glue to get a global object is some class in ${H^2}$ causing the obstruction. This is again the idea here. You formally make the lift set-theoretically. The obstruction to being a homomorphism is a class ${Ob(\rho)\in H^2(G_S, Ad(\overline{\rho})\otimes_k I)}$. If the class is 0, then we can lift the representation, otherwise it is obstructed. We call ${H^2(G_S, Ad(\overline{\rho}))}$ the “obstruction space” to the deformation functor.

This gives us an expected dimension for the deformation space. In some sense, we’d expect that each dimension of the obstruction space would cut the dimension of the space by one. If ${d_1=\dim H^1(G_S, Ad(\overline{\rho}))}$ and ${d_2=\dim H^2(G_S, Ad(\overline{\rho}))}$, then we’d expect ${Kdim(\mathcal{R}/m_{\Lambda}\mathcal{R})=d_1-d_2}$, but this is still an open conjecture and is probably hard since it generalizes Leopoldt’s conjecture. What we do get is that ${Kdim(\mathcal{R}/m_{\Lambda}\mathcal{R})\geq d_1-d_2}$. The rough idea behind having an inequality is that each dimension of the obstruction space is introducing a relation, and so the relation at most cuts the dimension down by one. But maybe we get to the fifth relation and it is redundant, then the dimension won’t go down.

Using some hard algebraic number theory we can actually convert ${d_1-d_2}$ to something only involving ${H^0}$, but it is beyond the scope of these posts. We’ll end here for today. Next time we’ll specialize to Galois representations that are modular, since that was our motivation and see what more information we can get in that case (this means if you are following along with the Gouvea article, we will be skipping lectures 5 and 6).