These are my live-TeXed notes for the course 18.786: Galois Representations taught by Sug Woo Shin at MIT, Spring 2014. References.

Any mistakes are the fault of the notetaker. Let me know if you notice any mistakes or have any comments!

[-] Contents
Deformations of Galois representations
Group-theoretic hypothesis
Liftings of mod $\ell$ representations
Some background on irreducible representations
Deformations of mod $\ell$ representations
Linear algebraic lemmas
Tangent spaces
Generic fiber of universal lifting rings
Deformation problems
Global Galois deformation problems
Presenting global deformation rings over local lifting rings
Computation of $H^i_{\mathcal{A}.T}$
Local universal lifting rings $\ell\neq p$
Local universal lifting rings $\ell=p$
The Weil-Deligne functor
Hodge-Tate weights
Potentially semistable local lifting rings

02/25/2014

TopDeformations of Galois representations

The global Langlands correspondence is roughly a correspondence between automorphic forms (representations) and $\ell$-adic Galois representations. Suppose we are given the arrow automorphic representations $\rightarrow$ $\ell$-adic Galois representations (of course this is highly nontrivial), then it is relatively easy to show this arrow is injective. The strategy to show that this is also surjective (modularity of Galois representations $\rho$) is to first show that the reduction $\rho_0=\rho \bmod{\ell}$ is the reduction of the Galois representation $\rho_f$ coming from a modular form $f$ (this is Serre's conjecture in general), then try to study infinitesimal liftings $$\{f: f\equiv f_0\bmod \ell\}\rightarrow\{\rho: \rho\equiv\rho_0 \bmod \ell\}.$$ Now both sides then have more algebro-geometric structures (like a map of schemes $\Spec \mathbb{T}\rightarrow\Spec R$). It often turns out to be a closed immersion. If one can then show (e.g., by dimension reason) this is actually an isomorphism ($R=T$ theorem), then the surjectivity will follow. Our goal in next few lectures is to study the right hand side, the deformations of Galois representations.

TopGroup-theoretic hypothesis

When deforming a representation $\rho:\Gamma\rightarrow GL_n(\mathbb{F}_\ell)$, we would like to impose a finiteness condition on the profinite $\Gamma$, in order to make sense of the the space of deformations and make ring $R$ noetherian. Fix $\ell$ a prime and $\Gamma$ a profinite group. We impose the following $\ell$-finiteness assumption.

Definition 1 Hyp($\Gamma$): for any $\Delta\subseteq\Gamma$ an open finite index subgroup,
  1. $\Delta/\langle[\Delta,\Delta],\Delta^\ell\rangle$ is finite ( = $\Hom_\mathrm{cont}(G,\mathbb{F}_\ell)$ is finite dimensional over $\mathbb{F}_\ell$). Here $\Delta^\ell=\{g^\ell :g\in\Delta\}$.
  2. The maximal pro-$\ell$ quotient of $\Delta$ is topologically finitely generated (i.e., there exists a finitely generated dense subgroup).
Remark 1 Nikolov-Segal proved that when $\Gamma$ is topologically finitely generated, any finite index subgroup $\Delta\subseteq \Gamma$ is automatically open. This may fail for groups like $\Gal(\overline{\mathbb{Q}}/\mathbb{Q})$.
Lemma 1
  1. a) $\Longleftrightarrow$ b).
  2. If $\Gamma$ is topologically finitely generated, then a) and b) are both satisfied.
Proof
  1. Use Burnside basis theorem (see [Boe] Ex 1.8.1).
  2. The same generator for $\Delta$ will work for the maximal pro-$\ell$ quotient of $\Delta$. ¡õ
Remark 2 Since every finite index subgroup of a finitely generated group is again finitely generated, it follows that if $\Gamma$ is topologically finitely generated, then any $\Delta\subseteq \Gamma$ of finite index is also topologically finitely generated.
Example 1
  1. Let $K/\mathbb{Q}_p$ be a finite extension ($p=\ell$ or $p\ne\ell$) Then $G_K=\Gal(\overline{K}/K)$ satisfies Hyp: to check a), it is the same thing to check that there exists only finitely many abelian extension of exponent $\ell$ for a given local field. This follows from Kummer theory.
  2. Let $F/\mathbb{Q}$ be a finite extension, $S$ be a finite set of finite places of $F$, $F_S$ be the maximal extension of $F$ unramified outside $S$. Then $G_{F,S}=\Gal(F_S/F)$ satisfies Hyp (by global class field theory, see [DDT] 2.41 for details). Notice that $G_{F,S}$ itself is not known to be topologically finitely generated.

TopLiftings of mod $\ell$ representations

Let $L/\mathbb{Q}_\ell$ be a finite extension, $\mathcal{O}=\mathcal{O}_L$ with uniformizer $\lambda$ and residue field $\mathbb{F}=\mathcal{O}/\lambda$.

Definition 2 Let $\mathcal{C}_\mathcal{O}$ be the following category. The objects of $\mathcal{C}_\mathcal{O}$ are complete noetherian local $\mathcal{O}$-algebras $A$ such that $A/\mathfrak{m}_A\cong \mathbb{F}$. Here $\mathfrak{m}_A$ is the unique maximal ideal of $A$ and the completeness means that $A\rightarrow\varprojlim_n A/\mathfrak{m}_A^n$ is an isomorphism. Notice the $\mathcal{O}$-algebra isomorphism $A/\mathfrak{m}_A\cong F$ is unique if it exists.

The morphisms in $\mathcal{C}_\mathcal{O}$ are morphisms of local $\mathcal{O}$-algebras $f: A\rightarrow B$ (i.e., $f(\mathfrak{m}_A)\subseteq \mathfrak{m}_B$). It follows that $f$ induces an isomorphism $A/\mathfrak{m}_A\cong B/\mathfrak{m}_B$.

Remark 3 An $\mathcal{O}$-algebra morphism $f: A\rightarrow B$ between $A,B\in\mathcal{C}_\mathcal{O}$ is automatically local: since $1\not\in f^{-1}(\mathfrak{m}_B)$, $f^{-1}(\mathfrak{m}_B)\subseteq A$ is proper ideal. Hence $A/f^{-1}(\mathfrak{m}_B)\hookrightarrow B/\mathfrak{m}_B\cong\mathbb{F}$ is a nonzero subring, so $A/f^{-1}(\mathfrak{m}_B)$ must be a field. Therefore $f^{-1}(\mathfrak{m}_B)=\mathfrak{m}_A$.
Definition 3 Let $\bar \rho:\Gamma\rightarrow GL_n(\mathbb{F})$ be a continuous representation. Define $\mathcal{R}^\Box_{\bar\rho}: \mathcal{C}_\mathcal{O}\rightarrow \mathbf{Sets}$ given by $$A\mapsto\{ \rho:\Gamma\rightarrow GL_n(A), \rho\bmod \mathfrak{m}_A=\bar \rho \text{ via } A/\mathfrak{m}_A\cong \mathbb{F}\}.$$ This is a functor and is called the liftings of $\bar \rho$.
Remark 4 No "isomorphism/equivalence class" of $\rho$ is involved in the definition! In Kisin's terminology, $\mathcal{R}^\Box_{\bar\rho}$ is called the framed deformations of $\bar\rho$, which somehow explains the $\Box$ in the notation.
Proposition 1 The functor $\mathcal{R}^\Box_{\bar\rho}$ is represented by some $R^\Box_{\bar \rho}\in \mathcal{C}_\mathcal{O}$.
Remark 5 We call $R^\Box_{\bar \rho}$ the universal lifting ring of $\bar \rho$. Explicitly, this means that $R^\Box_{\bar \rho}$ satisfies one of the following equivalent conditions:
  1. There are bijections $R^\Box_{\bar \rho}(A)\cong\Hom_{\mathcal{C}_\mathcal{O}}(R^\Box_{\bar \rho},A)$, functorial in $A$.
  2. There exists a universal lifting $\rho^\Box_{\bar\rho}: \Gamma\rightarrow GL_n(R^\Box_{\bar\rho})$ of $\bar\rho$ such that for any $A\in \mathcal{C}_\mathcal{O}$ and a lifting $\rho:\Gamma\rightarrow GL_n(A)$, there exists a unique $f_\rho: R^\Box_{\bar\rho}\rightarrow A$ such that $\rho=f_\rho\circ\rho^\Box_{\bar \rho}$
Example 2 Consider $\Gamma=\hat{\mathbb{Z}}$. Giving $\bar \rho: \Gamma\rightarrow GL_n(\mathbb{F})$ is the same as giving an element $\bar\rho(1)\in GL_n(\mathbb{F})$. Fix a lift $\widetilde{\bar\rho(1)}\in GL_n(\mathcal{O})$. Then $$\mathcal{R}^\Box_{\bar\rho}(A)=\{\rho: \hat{\mathbb{Z}}\rightarrow GL_n(A), \rho\bmod \mathfrak{m}_A=\bar\rho\}$$ is in bijection with $\widetilde{\bar\rho(1)}+ \mathfrak{m}_A M_n(A)$, given by $\rho\mapsto \rho(1)$. Hence $$\mathcal{R}^\Box_{\bar\rho}(A)\cong\mathfrak{m}_A^{n^2}\cong\Hom_{\mathcal{C}_\mathcal{O}}(\mathcal{O}[ [X_{i,j}] ]_{i,j=1}^n, A).$$ So $\mathcal{O}[ [ X_{i,j}] ]_{i,j=1}^n$ represents $\mathcal{R}^\Box_{\bar \rho}$.
Proof Step 1 By Hyp($\Gamma$), one may assume that $\Gamma$ is topologically finitely generated.

In fact, let $\Gamma_0=\ker\bar\rho\subseteq \Gamma$ (an open finite index subgroup). Let $\Delta=\ker (\Gamma_0\rightarrow\Gamma_\ell)$, here $\Gamma_\ell$ is the maximal pro-$\ell$ quotient of $\Gamma_0$. We claim that one may replace $\Gamma$ with $\Gamma/\Delta$. Then $\Gamma/\Delta$ is topologically finitely generated profinite group (though $\Gamma$ may not be): look at the exact sequence $$1\rightarrow \Gamma_0/\Delta\rightarrow \Gamma/\Delta\rightarrow\Gamma/\Gamma_0\rightarrow1.$$ Notice $\Gamma/\Gamma_0$ is finite and $\Gamma_0/\Delta$ is topologically finitely generated by Hyp($\Gamma$). It follows that $\Gamma/\Delta$ is also topologically finitely generated.

The key thing is that every lifting of $\bar\rho$ factors through $\Gamma/\Delta$. Indeed, Suppose $\rho:\Gamma\rightarrow GL_n(A)$ is such a lifting. Since $\bar\rho(\Delta)=\{1\}$, we know that $\rho(\Delta)\in\ker (GL_n(A)\rightarrow GL_n(\mathbb{F}))$, this kernel is pro-$\ell$, but by construction $\Delta$ is prime-to-$\ell$. Hence $\rho(\Delta)=\{1\}$ and we may replace $\Gamma$ with $\Gamma/\Delta$.

Step 2 Let $F_s$ be the free group on $s$ generators $g_1,\ldots,g_s$ and $\hat F_s$ be the profinite completion of $F_s$. One can choose $s$ so that there is a surjection $\phi: \hat F_s\rightarrow \Gamma$. Let $\phi_0:F_s\rightarrow\phi(F_s)$ be the restriction to the dense subgroup $F_s$. Let $H=\ker\phi$ and $H_0=\ker\phi_0$. One can check that $H_0\subseteq H$ is also dense. Now giving a lifting $\rho: \Gamma\rightarrow GL_n(A)$ is the same as giving $\tilde \rho:\hat F_s\rightarrow GL_n(A)$ (i.e. $\tilde\rho(g_i)_{i=1,\ldots,s}$) being trivial on $H_0$. Now $\mathcal{R}^\Box_{\bar\rho}$ is represented by $\mathcal{O}[ [X_{i,j}^k] ]_{i,j=1}^n{}_{k=1}^s/\{\text{relations from }H_0\}$. ¡õ

02/27/2014

TopSome background on irreducible representations

We switch notations temperately. Let $k$ be a field, $\Gamma$ be an abstract group, $V$ be an finite dimensional $k$-vector spaces. Recall that $\rho:\Gamma\rightarrow GL_k(V)$ is irreducible if there is no proper $k$-subvector space which is $\Gamma$-stable.

Lemma 2 (Schur) If $\rho$ is irreducible, then $\End_{k[\Gamma]}(\rho)=\{\phi\in\End_k(V): \phi\rho(\gamma)=\rho(\gamma)\phi,\forall \gamma\in\Gamma\}$ is a division algebra over $k$.
Proof Suppose $\alpha\in\End_{k[\Gamma]}(\rho)$ is nonzero, then $\alpha V\subseteq V$ is nonzero and $\Gamma$-stable, hence $\alpha V=V$. Therefore $\alpha$ is invertible. ¡õ
Lemma 3 (Schur) Suppose $k=\bar k$ is algebraically closed. Let $\rho_1,\rho_2$ be irreducible representations. Then $$\Hom_{k[\Gamma]}(\rho_1,\rho_2)= \begin{cases} k & \rho_1\cong\rho_2,\\ 0 & \rho_1\not\cong\rho_2. \end{cases}$$
Remark 6 The only division algebra over an algebraically closed field $k$ is $k$ itself.
Remark 7 $\rho$ is irreducible does not imply that $\rho \otimes_k \bar k$ is irreducible. A division algebra over $k$ may not be a division algebra when extending scalars to $\bar k$.
Remark 8 The converse of Schur's Lemma is not true (for non-semisimple representations): $\End_{k[\Gamma]}(\rho)=k$ does not imply that $\rho$ is irreducible. For example, consider the non-semisimple representation $\rho: \Gamma=\left(\begin{smallmatrix}a & b\\ 0& d\end{smallmatrix}\right)\hookrightarrow GL_2(k)$. Then $\End_{k[\Gamma]}(\rho)=k$ but $\rho$ is clearly reducible.
Proposition 2 (Burnside's theorem) If $\rho$ is irreducible over $k=\bar k$, then $k[\Gamma]\rightarrow \End_k(V)$ is surjective.
Definition 4 We say $\rho$ is Schur, if $\End_{k[\Gamma]}(\rho)=k$; absolutely irreducible if for any $k'/k$ a field extension, $\rho \otimes _k k'$ is irreducible.
Proposition 3 $\rho$ is absolutely irreducible if and only if $\rho \otimes_k \bar k$ is irreducible, if and only if $\rho$ is irreducible and Schur.
Proof See [CR] Section 29. ¡õ

TopDeformations of mod $\ell$ representations

We are back to the usual notation as in the section of lifting of mod $\ell$ representations.

Definition 5 Let $\bar\rho:\Gamma\rightarrow GL_n(\mathbb{F})$ be a continuous representation. We define $\mathcal{R}_{\bar\rho}:\mathcal{C}_\mathcal{O}\rightarrow\mathbf{Sets}$ the functor of deformations of $\bar\rho$ by $$A\mapsto\{\rho:\Gamma\rightarrow GL_n(A),\rho\bmod \mathfrak{m}_A=\bar\rho\}/\cong.$$ Here $\rho_1\cong\rho_2$ if there exists $a\in GL_n(A)$ such that $a\rho_1a^{-1}=\rho_2$. When $\rho_1=\rho_2\bmod \mathfrak{m}_A$, this happens if and only if $a$ can be chosen in $\ker(GL_n(A)\rightarrow GL_n(\mathbb{F}))$.
Remark 9 The isomorphism classes of liftings are not in the definition of $\mathcal{R}^\Box_{\bar \rho}$.
Proposition 4 If $\bar\rho$ is Schur, then $\mathcal{R}_{\bar \rho}$ is representable. Say by $R^\mathrm{univ}_{\bar\rho}\in\mathcal{C}_\mathcal{O}$, the universal deformation ring of $\bar\rho$.
Remark 10 The Schur condition rigidifies the moduli problem and ensures the representability.
Remark 11 For Galois representations, deformations were introduced by Mazur (1989). Later, liftings were introduced by Kisin (mid 2000) (to remove the Schur condition).
Proof Mazur's original proof uses Schlessinger's criterion of representability. One can also argue as for $\mathcal{R}^\Box_{\bar \rho}$ (see [DDT] 2.36). Kisin's approach, roughly speaking, is to show that $R^\mathrm{univ}_{\bar\rho}$ is the geometric quotient of $R^\Box_{\bar\rho}$ by $PGL_n$ (see [Boe] 2.1). Also see [Maz97] 10. ¡õ
Remark 12 As before, for any deformation $\rho$ of $\bar\rho$, there exists a unique $f_\rho: R^\mathrm{univ}_{\bar \rho}\rightarrow A$ such that $f_\rho\circ\rho_{\bar\rho}^\mathrm{univ}\cong\rho$.
Remark 13 There is a canonical map $R^\mathrm{univ}_{\bar\rho}\rightarrow R^\Box_{\bar\rho}$. It is induced by the functor $\mathcal{R}^\Box_{\bar\rho}(A)\rightarrow\mathcal{R}_{\bar\rho}(A)$ by sending $\rho$ to its isomorphism class. Another way: view $\Gamma\rightarrow GL_n(R_{\bar\rho}^\Box)$ as a deformation. We will see more precise descriptions later.

TopLinear algebraic lemmas

Lemma 4 Let $A\in\mathcal{C}_\mathcal{O}$, $\rho:\Gamma\rightarrow GL_n(A)$ be a continuous representation. Assume $\bar\rho=\rho\bmod \mathfrak{m}_A$ is absolutely irreducible.
  1. (Schur for $A$-coefficient) For $a\in GL_n(A)$, if $a\rho a^{-1}=\rho$, then $a\in A^\times$.
  2. (Carayol's Lemma) Suppose $B\subseteq A$ (in $\mathcal{C}_\mathcal{O})$ is closed and $\tr\rho(\Gamma)\subseteq B$. Then there exists $a\in\ker(GL_n(A)\rightarrow GL_n(\mathbb{F}))$ such that $a\rho a^{-1}:\Gamma\rightarrow GL_n(A)$ factors through $GL_n(B)$.
Proof One reduces to the case $A$ is Artinian local $\mathcal{O}$-algebra (since $A/\mathfrak{m}_A^n$ are of this shape). Then induct on the length of $A$. The case $A=\mathbb{F}$ is easy and one can further reduce to the case $A=\mathbb{F}[\varepsilon]/(\varepsilon^2)$.
  1. Assume $A\not\ne\mathbb{F}$. Choose a minimal nonzero ideal $I\subseteq A$: one has the filtration $$0=\mathfrak{m}_A^{e+1}\subsetneq \mathfrak{m}_A^e\subsetneq\cdots \subsetneq A,$$ where $\mathfrak{m}_A^e$ is a $\mathbb{F}$-vector space; one can choose $I\subseteq\mathfrak{m}_A^e$ to be isomorphic to $\mathbb{F}$ as an $A$-module. Let $a\in GL_n(A)$ such that $a\rho a^{-1}=\rho$. Then the induction hypothesis implies that $a\bmod I\in (A/I)^\times$. So we can write $a=\alpha \mathbf{1}_n+a_0$, where $\alpha\in A^\times$ and $a_0\in M_n(I)$. Now the equation $$(\alpha\mathbf{1}_n+a_0)\rho(\gamma)=\rho(\gamma)(\alpha\mathbf{1}_n+a_0)$$ tells us that $a_0\rho(\gamma)=\rho(\gamma)a_0$ in $M_n(A)$. Under the identification $I\cong\mathbb{F}$, we find that $a_0\bar\rho(\gamma)=\bar\rho(\gamma)a_0$ in $M_n(\mathbb{F})$, hence by Schur's lemma, $a_0$ itself must be a scalar too (we only need $\bar\rho$ to be Schur in this part).
  2. By induction hypothesis, we may assume that $\rho\bmod I$ lands in $GL_n(B/IB)$. Since $I\cap B\subseteq I\cong\mathbb{F}$ as $\mathcal{O}$-submodules, we know that either $I\cap B=0$ or $I\cap B=I$. In the latter case, it is immediate that $\rho$ lands in $GL_n(B)$. The first case is more difficult. We build the $\mathcal{O}$-algebra $B\oplus I\varepsilon$, where $\varepsilon^2=0$. Then $B$ embeds into $A$ by $(b,i\varepsilon)\mapsto b+i$. By induction hypothesis, we may assume $B\hookrightarrow A$ is indeed an isomorphism. We may replace $A,B$ by $A/\mathfrak{m}_B\cong\mathbb{F}\oplus \mathbb{F}\varepsilon$ and $B/\mathfrak{m}_B\cong\mathbb{F}$ (because quotient by any thing in $B$ is fine). This is a much more concrete problem and the rest of the proof can be found in [CHT] Lemma 2.1.10. Here is roughly how it works. Extend $\rho: \Gamma\rightarrow \mathbb{F}[\varepsilon]/\varepsilon^2$ $\mathbb{F}$-linearly and we obtain that $\rho: \mathbb{F}[\Gamma]\rightarrow M_n(\mathbb{F})\oplus M_n(\mathbb{F})\varepsilon$, $\rho(\gamma)=\bar\rho(\gamma)+\theta(\gamma)\varepsilon$. Here
    1. $\theta:\mathbb{F}[\Gamma]\rightarrow M_n(\mathbb{F})$ is $\mathbb{F}$-linear,
    2. $\theta(\gamma\delta)=\theta(\gamma)\bar\rho(\delta)+\bar\rho(\gamma)\theta(\delta)$,
    3. $\tr\theta(\gamma)=0$.

We want to get rid of $\bar\rho$ and deal with purely matrix algebra. We claim that $\theta$ factors through the surjective map $\bar\rho: \mathbb{F}[\Gamma]\rightarrow M_n(\mathbb{F})$ (here we used the absolutely irreducible assumption for the surjectivity), i.e., $\theta$ is trivial on $\ker\bar\rho$. Let $\delta\in\ker\bar\rho$, then $0=\tr\theta(\gamma\delta)=\tr(\bar\rho(\gamma)\theta(\delta))$. By the surjectivity of $\bar\rho$, we know that $\theta(\delta)=0$ because $\bar\rho(\gamma)$ can be anything in $M_n(\mathbb{F})$. This proves the claim.

Now we are looking for $a=1+a'\varepsilon\in 1+M_n(\mathbb{F})\varepsilon$ such that $(1+a'\epsilon)\bar\rho(\gamma)(1-a'\varepsilon)\in M_n(\mathbb{F})$ for any $\gamma\in\Gamma$. This is equivalent to that the coefficient of $\varepsilon$ $$\theta(\gamma)+a'\bar\rho(\gamma)-\bar\rho(\gamma)a'{}=0.$$ So we are reduced to the problem of showing that for $\theta': M_n(\mathbb{F})\rightarrow M_n(\mathbb{F})$, there exists $a'$ such that

  1. $\tr\theta'{}=0$,
  2. $\theta'(\gamma\delta)=\theta'(\gamma)\delta+\gamma\theta'(\delta)$ ,
  3. $\theta'(\gamma)=\gamma a'-a'\gamma$.

Conceptually this means that every derivation on $\mathfrak{sl}_n$ is given by the Lie bracket with some element $a'$. One can directly show that $a'=\sum_{j=1}^n\theta'(e_{j1})e_{1j}$ works. ¡õ

Lemma 5 (Brauer-Nesbitt for $A$-coefficient). Suppose $\bar\rho: \Gamma\rightarrow GL_n(\mathbb{F})$ is absolutely irreducible. If $\rho_1\cong\rho_2\cong\bar\rho$ such that $\tr\rho_1=\tr\rho_2$. Then $\rho_1\cong\rho_2$.
Proof Use similar reduction and then use Carayol's lemma when $A=\mathbb{F}[\varepsilon]/(\varepsilon^2)$. See [Boe] 2.2.1. ¡õ
Corollary 1 ([Gee] Ex 3.9) The universal lifting ring $R^\Box_{\bar\rho}$ is a power series ring in $n^2-1$ variables over the the universal deformation ring $R_{\bar\rho}^\mathrm{univ}$.

03/04/2014

TopTangent spaces

We are going to work with the universal lifting rings (the same argument works for universal deformation rings). Write $R^\Box=R^\Box_{\bar\rho}$ for short. Denote its unique maximal ideal by $\mathfrak{m}^\Box$.

Definition 6 The adjoint representation $\ad: GL_n(\mathbb{F})\rightarrow \End(M_n(\mathbb{F}))$ is given by conjugation $g\mapsto (\phi\mapsto g\phi g^{-1})$. We denote $\ad\bar\rho=\ad\circ\bar\rho$ (and the corresponding $\mathbb{F}[\Gamma]$-module).

One can analyze the tangent space of $R^\Box$ in terms of group cohomology of $\ad\bar\rho$.

Lemma 6 There exists natural bijections between
  1. $\Hom_\mathbb{F}(\mathfrak{m}^\Box/((\mathfrak{m}^\Box)^2,\lambda),\mathbb{F})$.
  2. $\Hom_{\mathcal{C}_\mathcal{O}}(R^\Box,\mathbb{F}[\varepsilon]/\varepsilon^2)$.
  3. $\mathcal{R}^\Box(\mathbb{F}[\varepsilon]/\varepsilon^2)$ (liftings of $\bar\rho$ to $\mathbb{F}[\varepsilon]/\varepsilon^2$).
  4. $Z^1(\Gamma,\ad\bar\rho)$ (continuous 1-cocycles).
Proof b) $\Longleftrightarrow$ c): by definition.

a) $\Longleftrightarrow$ b): Since $\mathcal{O}+\mathfrak{m}^\Box$ surjects on $R^\Box$. For $f\in \Hom_\mathbb{F}(\mathfrak{m}^\Box/((\mathfrak{m}^\Box)^2,\lambda),\mathbb{F})$, we define $$R^\Box\rightarrow\mathbb{F}[\varepsilon]/\varepsilon^2\quad, a+x\mapsto \bar a+f(x)\varepsilon.$$ One routinely check that this is well defined (using $\mathcal{O}\cap \mathfrak{m}^\Box=(\lambda)$ and gives the desired bijection.

c) $\Longleftrightarrow$ d). Given $\rho:\Gamma\rightarrow GL_n(\mathbb{F}[\varepsilon]/\varepsilon^2)$, we can write $\rho(\gamma)=(1+\phi(\gamma)\varepsilon)\bar\rho(\gamma)$. One can check that $\rho$ is a homomorphism if and only if $\gamma\mapsto\phi(\gamma)$ defines a continuous 1-cocycle. ¡õ

Write $d=\dim_\mathbb{F} Z^1(\Gamma,\ad\bar\rho)=\dim_\mathbb{F} \mathfrak{m}^\Box/((\mathfrak{m}^\Box)^2,\lambda)$. This is the dimension of cotangent space (ignoring the $\lambda$-direction) of $R^\Box$.

Corollary 2 $d=\dim_\mathbb{F} H^1(\Gamma,\ad\bar\rho)-\dim_\mathbb{F}H^0(\Gamma,\ad\bar\rho)+n^2$.
Proof There is an exact sequence of finite dimensional $\mathbb{F}$-vector spaces $$0\rightarrow (\ad\bar\rho)^\Gamma=H^0(\Gamma,\ad\rho)\rightarrow\ad \bar\rho\rightarrow Z^1(\Gamma,\ad\bar\rho)\rightarrow H^1(\Gamma,\ad\bar\rho)\rightarrow0.$$ Here any $\phi\in\ad\bar\rho$ gives the coboundary $\gamma\mapsto\gamma\phi-\phi$. The result immediately follows. ¡õ

So we can know how big the ring $R^\Box$ is as long as we know the dimension of $H^0$ and $H^1$ of the adjoint representation.

Corollary 3 Choose $\phi: \mathcal{O}[ [X_1,\ldots, X_d] ]\rightarrow R^\Box$ such that $\phi(x_i)\in \mathfrak{m}^\Box$ generate $\mathfrak{m}^\Box/((\mathfrak{m}^\Box)^2,\lambda)$ as $\mathbb{F}$-vector space, then $\phi$ is surjective.

This follows from a topological version of Nakayama's lemma. In nice situations, $\phi$ is an isomorphism. In general this is too optimistic but one can further control the the kernel $J=\ker\phi$. Notice that $J\subseteq (\mathfrak{m}^2,\lambda)$. Here $\mathfrak{m}=(\lambda,X_1,\ldots, X_d)$ is the maximal ideal of $\mathcal{O}[ [X_1,\ldots, X_d] ]$. We shall construct an injective map $$\Hom_\mathbb{F}(J/\mathfrak{m}J,\mathbb{F})\rightarrow H^2(\Gamma,\ad\bar\rho),\quad f\mapsto [C_f].$$ When this $H^2$ vanishes, $R^\Box$ will be an formal power series ring and hence is formally smooth of dimension $d$: there is no obstruction for liftings (controlled by $H^2$) and the space of liftings is $d$-dimensional (controlled by $H^1$ and $H^0$).

To construct $f\mapsto [C_f]$, we notice that $GL_n(\mathcal{O}[ [X_1,\ldots, X_d] ]/\mathfrak{m}J)\rightarrow GL_n(\mathcal{O}[ [X_1,\ldots, X_d] ]/J)$ is surjective. For $\gamma\in\Gamma$, we can choose a lift $\tilde\rho(\gamma)$ of $\rho(\gamma)=\rho^\Box(\gamma)$. Notice $\tilde\rho$ may not be a homomorphism and this failure is measured by $$C_f(\gamma,\delta)=f(\tilde\rho(\gamma\delta)\tilde\rho(\delta)^{-1}\tilde\rho(\gamma)^{-1}-\mathbf{1}_n).$$ Since $\rho(\gamma\delta)\rho(\delta)^{-1}\rho(\gamma)^{-1}=\mathbf{1}_n$, we know that $\tilde\rho(\gamma\delta)\tilde\rho(\delta)^{-1}\tilde\rho(\gamma)^{-1}-\mathbf{1}_n\in M_n(J/\mathfrak{m}J)$ and hence it makes sense to apply $f$ in this expression. One checks directly that $$g_1 C_f(g_2,g_3)-C_f(g_1g_2,g_3)+C_f(g_1,g_2g_3)-C_f(g_1,g_2)=0.$$ So

Lemma 7 $C_f\in Z^2(\Gamma,\ad\bar\rho)$ is a continuous 2-cocycle.
Exercise 1
  1. $C_f$ gives a well-defined class $[C_f]\in H^2(\Gamma,\ad\bar\rho)$.
  2. $[C_f]=0$ if and only if there exists a choice of $\tilde\rho$ such that $\tilde\rho$ mod $J_f$ is a homomorphism. Here $J_f:=\ker(J\rightarrow J/\mathfrak{m}J\xrightarrow{f}\mathbb{F})$ (so $\mathfrak{m} J\subseteq J_f\subseteq J$).
Lemma 8 The $\mathbb{F}$-linear map $\Hom_\mathbb{F}(J/\mathfrak{m}J,\mathbb{F})\rightarrow H^2(\Gamma,\ad\bar\rho)$ is injective.
Proof It suffices to show that if there exists $\tilde\rho$ as in b) of the previous exercise, then $f=0$. Notice $\mathcal{O}[ [X_1,\ldots, X_d] ]/J_f\twoheadrightarrow \mathcal{O} [[ X_1,\ldots,X_d] ]/J=R^\Box$. Since $\tilde \rho$ is a lifting of $\bar\rho$, we have a map $$\eta: \mathcal{O} [[ X_1,\ldots,X_d] ]/J\rightarrow \mathcal{O} [[ X_1,\ldots,X_d] ]/J_f$$ by the universal property. The composite map $$\mathcal{O} [[ X_1,\ldots,X_d] ]/J\xrightarrow{\eta} \mathcal{O} [[ X_1,\ldots,X_d] ]/J_f\twoheadrightarrow \mathcal{O} [[ X_1,\ldots,X_d] ]/J$$ is the identity map. This shows that $\eta$ is injective. Let $g(\underline{X})\in J$ (in particular $g\in (\mathfrak{m}^2,\lambda)$), we want to show that $g(\underline{X})\in J_f$. Suppose $\eta$ maps $X_i$ to $X_i+a_i$ (so $a_i\in J$). One checks directly that $g(\underline{X}+\underline{a})-g(\underline{X})\in \mathfrak{m}J\subseteq J_f$. But $g(\underline{X}+\underline{a})=\eta(g(\underline{X}))\in J_f$ by injectivity, hence $g(\underline{X})\in J_f$. Therefore so $J=J_f$ and $f=0$. ¡õ

So the number of generators of $R^\Box$ is $d$ and with the number of relations is equal to $\dim_\mathbb{F} J/\mathfrak{m}J\le\dim H^2$.

Corollary 4
  1. If $H^2=0$, then we have a non-canonical isomorphism $R^\Box\cong \mathcal{O}[ [X_1,\ldots, X_d] ]$.
  2. In general, $\dim R^\Box\ge d+1-\dim H^2$.

03/06/2014

TopGeneric fiber of universal lifting rings

The following is a basic algebraic fact.

Proposition 5 Let $R\in \mathcal{C}_\mathcal{O}$. There is a bijection between closed points of the generic fiber $R[1/\ell]$ (these are dense in the generic fiber) and pairs $(\phi',L')$, where $L'/L$ is a finite extension, $\phi': R\rightarrow \mathcal{O}_{L'}$ is continuous such that $L'{}=L(\phi'(R[1/\ell]))$.
Remark 14 A pair $(\phi',L')$ induces $\phi'[/1/\ell]: R[1/\ell]\rightarrow L'$. The kernel of this map is a maximal ideal. This gives the corresponding closed point of the generic fiber. Conversely, any closed point $x$ with maximal ideal $\mathfrak{m}_x$ gives rise to $\phi_x: R\rightarrow R[1/\ell]\rightarrow R[1/\ell]/\mathfrak{m}_x=L_x$ (which factors through $\mathcal{O}_{L_x}$).

Let $x\in \Spec R^\Box[1/\ell]$ be a closed point. Then we know from the above bijection and the universal property that $x$ corresponds to a representation $\rho_x: \Gamma\rightarrow GL_n(L_x)$. Let $\mathcal{C}_{L_x}^\mathrm{Art}$ to be the category of local Artinian $L_x$-algebra with residue field $L_x$.

Lemma 9 The ring $R^\Box[1/\ell]_x^\wedge:= \lim_j R^\Box[1/\ell]/(\ker\phi_x[1/\ell])^j$ pro-represents the functor $\mathcal{C}_{L_x}^\mathrm{Art}\rightarrow\mathbf{Sets}$ given by $$A\mapsto \{\rho:\Gamma\rightarrow GL_n(A)\text{ such that }\rho\bmod \mathfrak{m}_A=\rho_x\}.$$
Remark 15 The lemma means that $\Hom(R^\Box[1/\ell]_x^\wedge,A)=\mathcal{R}^\Box_{\rho_x}(A)$, functorial in $A$. Though $R^\Box$ is constructed as the deformations of a representation in characteristic $\ell$, the generic fiber of $R^\Box$ knows about deformations of representations in characteristic 0.
Remark 16 There is no difference to take the large category $\mathcal{C}_\mathcal{O}$ (so the functor is in fact representable in $\mathcal{C}_\mathcal{O}$). But Artinian rings are easier to deal with because the topology on finite dimensional representations are obvious (the $\ell$-adic topology).
Proof (Sketch of proof) Given $f: R^\Box[1/\ell]_x^\wedge\rightarrow A$, one simply composes to get $R^\Box\rightarrow GL_n(R^\Box)\rightarrow GL_n(R^\Box[1/\ell]_x^\wedge)\xrightarrow{f}GL_n(A)$, an object in $\mathcal{R}^\Box_{\rho_x}(A)$.

Conversely, given $(\rho,A)\in \mathcal{R}_{\rho_x}^\Box$, let $A^0$ be the $\mathcal{O}_{L_x}$-subalgebra of $A$ generated by the entries $\rho(\Gamma)$. One can show by a compactness argument that $A^0\in \mathcal{C}_{\mathcal{O}_{L_x}}$ and is finitely generated as $\mathcal{O}_{L_x}$-module. Let $\mathbb{F}_x$ be the residue field of $\mathcal{O}_{L_x}$. Reducing $\rho$ then gives $\bar\rho_x=\bar\rho \otimes _\mathbb{F} \mathbb{F}_x$. One can show (homework) that $R^\Box \otimes_\mathcal{O} \mathcal{O}_{L_x}\cong R^\Box_{\bar\rho_x}$, so the universal property gives the map $R^\Box_{\bar\rho_x}\rightarrow A^0$, hence a map $R^\Box\rightarrow R^\Box_{\bar\rho_x}\rightarrow A^0$. By the universal property of localization, it extends to a map on the generic fiber $R^\Box[1/\ell]\rightarrow A^0[1/ \ell]$. By continuity, it extends to the completion $R^\Box[1/\ell]_x^\wedge\rightarrow A$. For more details, see Brian's notes at Stanford seminar on modularity liftings. ¡õ

Remark 17 It is not surprising that the proof constructs certain integral models of the representations in characteristic 0 in order to relate the universal property of $R^\Box$.

TopDeformation problems

In practice, we are more interested in lifting Galois representations with prescribed local behaviors (e.g., requiring good reduction at certain primes for elliptic curves). So one would like to work with certain subspaces of $\Spec R^\Box$ (or in terms of rings, certain quotient rings of $R^\Box$). We would like to make a checklist for technical conditions to define nice subspaces of $\Spec R^\Box$.

Definition 7 A (framed) deformation problem $\mathcal{D}$ is a collection $\{(\rho,A)\}$, where $A\in\mathcal{C}_\mathcal{O}$, $\rho:\Gamma\rightarrow GL_n(A)$ is a lifting of $\bar\rho$ such that
  1. $(\mathbb{F},\bar\rho)\in \mathcal{D}$.
  2. If $f: A\rightarrow B$ is a map in $\mathcal{C}_\mathcal{O}$ and $(A,\rho)\in \mathcal{D}$, then $(B,f\circ \rho)\in\mathcal{D}$.
  3. (restriction) If $f: A\hookrightarrow B$ and $(B,f\circ\rho)\in\mathcal{D}$, then $(A,\rho)\in \mathcal{D}$.
  4. (gluing) Let $A_1, A_2\in \mathcal{C}_\mathcal{O}$ with ideals $I_1,I_2$ and $f: A/I_1\cong A_2/I_2$. Suppose $(A_1,\rho_1),(A_2,\rho_2)\in\mathcal{D}$ such that $f(\rho_1\bmod I_1)=\rho_2\bmod I_2$, then $(A_1\oplus_f A_2,\rho_1\oplus \rho_2)\in\mathcal{D}$. Here $$A_1\oplus A_2=\{(a_1,a_2)\in A_1 \oplus A_2: f(a_1\bmod I_1)=a_2\bmod I_2\}.$$
  5. (nested intersection) Suppose $I_1\supseteq I_2\supseteq\cdots$ is a chain of ideals of $A\in\mathcal{C}_\mathcal{O}$ such that $\cap I_i=0$. Suppose $(A/I_i, \rho \bmod I_i)\in \mathcal{D}$, then $(A,\rho)\in\mathcal{D}$.
  6. (conjugate) If $(A,\rho)\in\mathcal{D}$, $a\in\ker(GL_n(A)\rightarrow GL_n(\mathbb{F}))$, then $(A,a\rho a^{-1})\in\mathcal{D}$.
Lemma 10 We have a bijection $$\{\text{deformation problems}\}\longleftrightarrow\{\ker (GL_n(R^\Box)\rightarrow GL_n(\mathbb{F}))\text{-inv. ideals of } R^\Box\}.$$
Remark 18 Suppose $a\in \ker(GL_n(R^\Box)\rightarrow GL_n(\mathbb{F}))$, then conjugation on $\rho^\Box\rightarrow a^{-1}\rho^\Box a$ induces an action $a:R^{\Box}\rightarrow R^\Box$ by the universal property.
Remark 19 $\mathcal{D}\mapsto I(\mathcal{D})$ is characterized by $(A,\rho)\in\mathcal{D}$ if and only if $f_\rho: R^\Box\rightarrow A$ is trivial on $I(\mathcal{D})$. Conversely, when $I=\sqrt{I}$ and $I\ne \mathfrak{m}$ (these restrictions are needed to check f), for this subtlety, see [BLGHT] Lem 3.2), we have $\mathcal{D}(I)=\{(A,\rho): f_\rho(I)=0\}$.
Proof We explain why $I(\mathcal{D})$ is uniquely determined. Let $\mathcal{J}$ be the set of all ideals $I\subseteq R^\Box$ such that $(R^\Box/I,\rho^\Box\bmod I)\in \mathcal{D}$. This is nonempty by a). Using b) and c), we know $(A,\rho)\in\mathcal{D}$ if and only if $(R^\Box/\ker f_\rho,R^\Box\bmod \ker f_\rho)\in\mathcal{D}$. Moreover $\mathcal{J}$ is closed under finite intersection and nested infinite intersection by d) and e). So by Zorn's lemma, there exists a unique minimal ideal $I(\mathcal{D})\in \mathcal{J}$. It is kernel invariant by f). ¡õ

03/11/2014

Example 3 Typical deformation problems concerns local Galois representations $\bar\rho: \Gal(\bar K/K)\rightarrow GL_n(\mathbb{F})$, where $K/\mathbb{Q}_p$ is a local field. See [CHT] 2.4 for several examples: when $p=\ell$, the Fontaine-Laffaille liftings or ordinary liftings; when $p\ne \ell$, Taylor-Wiles liftings, are all deformation problems.

For applications, it is important to understand the ring theoretic properties of $R^\Box/I(\mathcal{D})$, e.g., Krull dimension, number of generators, number of relations and so on. We computed these for $R^\Box$ in terms of Galois cohomology. It is similar for $R^\Box/I(\mathcal{D})$.

Definition 8 Inside $Z^1(\Gamma,\ad\bar\rho)=\Hom_\mathbb{F}(\mathfrak{m}^\Box/((\mathfrak{m}^\Box)^2,\lambda),\mathbb{F})$, one can consider the annihilator of the image of $I(\mathcal{D}$ in $\mathfrak{m}^\Box/((\mathfrak{m}^\Box)^2,\lambda)$, $\tilde{\mathcal{L}}(\mathcal{D})=\mathrm{Ann}(I(\mathcal{D}))$ (think: the subspace cut out by $\mathcal{D}$ of the tangent space). We define $\mathcal{L}(\mathcal{D})$ be its image in $H^1(\Gamma,\ad\bar\rho)$.

Using the kernel-invariance property of $\mathcal{L}(\mathcal{D})$, one can show that

Lemma 11 $\tilde{\mathcal{L}}(\mathcal{D})$ is the full preimage of $\mathcal{L}(\mathcal{D})$.

So one can work directly at the level of cohomology instead of cocycles.

TopGlobal Galois deformation problems

We begin with a remark on fixing the determinant.

Remark 20 [Gee] Ex 3.19. Lifting the determinant of $\bar\rho$ to $\chi:\Gamma\rightarrow\mathcal{O}^\times$ such that $\chi \otimes_\mathcal{O}\mathbb{F}=\det\bar\rho$, then one obtains a subfunctor $\mathcal{R}_{\bar\rho,\chi}^\Box(A)\subseteq \mathcal{R}_{\bar\rho}^\Box(A)$ by requiring the extra condition $\det\rho=\chi$ on the liftings ($A$ is an $\mathcal{O}$-algebra, so this makes sense). This subfunctor $\mathcal{R}_{\bar\rho,\chi}^\Box$ is represented by a ring $R_{\bar\rho,\chi}^\Box\in\mathcal{C}_\mathcal{O}$ (one can construct $R^\Box_{\bar\rho,\chi}$ as the quotient of $R^\Box$ by the equation $\det\rho-\chi$). The previous discussion all carries over to this setting: the main change is that $\ad\bar\rho$ should be replaced by $\ad^0\bar\rho$ (the traceless subspace).

Now let $F$ be a number field. Let $v$ be a place of $F$ and fix $\bar F\hookrightarrow \bar F_v$. One has the local Galois group $G_v\hookrightarrow G_F$ (well-defined up to conjugation). Let $S$ be a set of finite places of $F$ and $G=G_{F,S}$.

Definition 9 A global (Galois) deformation problem is a collection $\mathcal{A}=(F,S,\mathcal{O},\bar\rho,\chi,\{\mathcal{D}_v\}_{v\in S})$, where
  1. $F$, $S$, $\mathcal{O}$ as above;
  2. $\bar\rho:G\rightarrow GL_n(\mathbb{F})$ is absolutely irreducible (the work of Skinner-Wiles and Thorne can relax this condition);
  3. $\chi:G\rightarrow\mathcal{O}^\times$ such that $\chi \otimes_\mathcal{O}\mathbb{F}=\det\bar\rho$;
  4. $\mathcal{D}_v$ is a deformation problem for $\bar\rho|_{G_v}$.

Next we will define a deformation functor and show it is representable and study its ring theoretic properties.

Definition 10 Let $T\subseteq S$ (may be empty). Define the functor $\mathcal{R}_{\mathcal{A}}^{\Box_T}:\mathcal{C}_\mathcal{O}\rightarrow\mathbf{Sets}$ given by $A\mapsto \{(\rho,\{\alpha_v\}_{v\in T}\}/\sim$. Here
  1. $\rho:G\rightarrow GL_n(A)$ is a lifting of $\bar\rho$ with $\det\rho=\chi$,
  2. $\alpha_v\in\ker(GL_n(A)\rightarrow GL_n(\mathbb{F}))$,
  3. $\rho|_{G_v}\in \mathcal{D}_v$ for $v\in S$.

The equivalence relations $\sim$ is generated by the following: for any $\beta\in \ker(GL_n(A)\rightarrow GL_n(\mathbb{F}))$, $(\rho,\{\alpha_v\})\sim(\beta\rho\beta^{-1},\{\beta\alpha_v\})$.(Think: we are imposing local conditions at $S$ and "framing at $T$"; this helps the Galois cohomology calculation and also helps achieving the final goal of comparing the deformation ring with the Hecke ring).

Lemma 12 $\mathcal{R}_\mathcal{A}^{\Box_T}$ is represented by $R_\mathcal{A}^{\Box_T}\in\mathcal{C}_\mathcal{O}$. When $T=\varnothing$, we write it as $R^{\mathrm{univ}}_\mathcal{A}$.
Proof Let $\mathcal{D}_v^0 \supseteq \mathcal{D}_v$ be all liftings of $\bar\rho|_{G_v}$ and $\mathcal{A}^0$ be the deformation problem given by $\{\mathcal{D}_v^0\}_{v\in S}$. Then $R_{\mathcal{A}^0}^{\Box_T}$ is representable (which is $R^\mathrm{univ}_{\bar\rho,\chi}$ if $T=\varnothing$). Then one can construct $R_\mathcal{A}^{\Box_T}$ as $R_{\mathcal{A}^0}^{\Box_T}/I$, where $I$ is the minimal ideal such that $\rho$ factors through $R_{\mathcal{A}^0}^{\Box_T}/I$ if and only if $\rho|_{G_v}\in\mathcal{D}_v$ for $v\in S$. For more details, see [CHT]. ¡õ

TopPresenting global deformation rings over local lifting rings

We have seen how to represent $R^\Box$ over $\mathcal{O}$ and when $R^\Box$ happens to be a power series ring over $\mathcal{O}$. We are now going to represent $R_\mathcal{A}^{\Box_T}$ over another bigger ring, the local lifting ring. This idea is due to Kisin.

Notice $R_\mathcal{A}^{\Box_T}$ has the following universal object:

  1. $\rho^{\Box_T}: G\rightarrow GL_n(R_\mathcal{A}^{\Box_T})$, and
  2. $\alpha_v\in \ker (GL_n(R_\mathcal{A}^{\Box_T})\rightarrow GL_n(\mathbb{F}))$, $v\in T$.

By the kernel-invariance property, $\alpha_v^{-1} \rho^{\Box_T}|_{G_v}\alpha_v\in\mathcal{D}_v$. Moreover, it is well-defined element independent of the choice of the representative of the equivalence class. By the universal property of $R_{\bar\rho|_{G_v},\chi}^\Box/I(\mathcal{D}_v)$, we know that $\rho^{\Box_T}$ factors through $$G\rightarrow GL_n(R_{\bar\rho|_{G_v},\chi}^\Box)\twoheadrightarrow GL_n(R_{\bar\rho|_{G_v},\chi}^\Box/I(\mathcal{D}_v))\rightarrow GL_n(R^{\Box_T}_\mathcal{A}).$$

Definition 11 Define the local lifting ring to be the completed tensor product $$R_{\mathcal{A},T}^\mathrm{loc}=\widehat{\bigotimes}_{v\in T} R_{\bar\rho_v,\chi}^\Box/I(\mathcal{D}_v).$$ We have a natural map $R_{\mathcal{A},T}^\mathrm{loc}\rightarrow R_\mathcal{A}^{\Box_T}$.

Our next goal is to find the number of the generators and relations for presenting $R_\mathcal{A}^{\Box_T}$ over $R_{\mathcal{A},T}^\mathrm{loc}$ using Galois cohomology.

As a first thought, suppose $\mathcal{D}_v$ consists of all liftings. Write $\mathfrak{m}\subseteq R_{\mathcal{A}}^{\Box_T}$ and $\mathfrak{m}^\mathrm{loc}\subseteq R_{\mathcal{A},T}^\mathrm{loc}$ be the maximal ideals. The the same argument as in Lemma 6 shows that $$\Hom_\mathbb{F}(\mathfrak{m}/(\mathfrak{m}^2,\lambda),\mathbb{F})=\mathcal{R}_\mathcal{A}^{\Box_T}(\mathbb{F}[\varepsilon]/(\varepsilon^2))\cong \frac{Z^1(G,\ad^0\bar\rho)\oplus\bigoplus_{v\in T}1+\varepsilon M_n(\mathbb{F})}{1+\varepsilon M_n(\mathbb{F})}.$$ Here $1+\varepsilon M_n(\mathbb{F})$ sits diagonally. Rewriting this as $$\coker(\ad\bar\rho\xrightarrow{(\partial,\diag)} Z^1(G,\ad^0\bar\rho)\oplus\bigoplus_{v\in T}\ad\bar\rho).$$

Two modifications are needed in general:

  1. to consider the tangent space over $R_{\mathcal{A},T}^\mathrm{loc}$: one should replace $\mathfrak{m}/(\mathfrak{m}^2,\lambda)$ by $\mathfrak{m}/(\mathfrak{m}^2,\mathfrak{m}^\mathrm{loc},\lambda)$. Concretely, this requires the liftings at $v\in T$ to be trivial, i.e., $\rho$ lies in the kernel of $$Z^1(G,\ad^0\bar\rho)\oplus\bigoplus_{v\in T} \ad \bar\rho\xrightarrow{(\mathrm{res},\partial)} \bigoplus_{v\in T}Z^1(G_v,\ad^0\bar\rho).$$ Notice the image of $(\rho,\{\alpha_v\})$ is $\{\alpha_v^{-1}\rho\alpha_v\}$, which we require to be trivial, i.e., $\alpha_v^{-1}\rho\alpha_v=\bar\rho$. Write $\alpha_v=1+\psi_v \varepsilon$ and $\rho=(1+\phi\varepsilon)\bar\rho$, then $(1-\psi_v\varepsilon)(1+\phi\varepsilon)\bar\rho(1+\psi_v\varepsilon)=\bar\rho$ if and only if $\phi-\psi_v+\bar\rho\psi_v\bar\rho^{-1}=\phi-\partial\psi_v=0$.
  2. to allow general $\mathcal{D}_v$ for $v\in S$: one requires that $$\rho\in\ker\left(Z^1(G,\ad^0\bar\rho)\xrightarrow{\mathrm{res}} \bigoplus_{v\in S\setminus T}\frac{Z^1(G_v,\ad^0\bar\rho)}{\tilde{\mathcal{L}}(\mathcal{D}_v)}\right).$$

The upshot is that $$\Hom_\mathbb{F}(\mathfrak{m}/(\mathfrak{m}^2,\mathfrak{m}^\mathrm{loc},\lambda),\mathbb{F})$$ is the $H^1$ of the complex $$\ad\bar\rho\rightarrow Z^1(G)\oplus\bigoplus_{v\in T} \ad\bar\rho\rightarrow \bigoplus_{v\in T} Z^1(G_v)\bigoplus_{v\in S\setminus T}Z^1(G_v)/\tilde{\mathcal{L}}(\mathcal{D}_v).$$

This motivates the definition of the mysterious complex in [Gee].

Definition 12 We define the complex $\mathcal{C}_{\mathcal{A},T}^\cdot$ to be $$\xymatrix@R=0cm@C=0.3cm{\mathcal{C}^0(G,\ad\bar\rho) \ar[r] \ar[rdd] & \mathcal{C}^1(G,\ad^0\bar\rho) \ar[r] \ar[rdd] & \mathcal{C}^2(G,\ad^0\bar\rho) \ar[r] & \mathcal{C}^3(G,\ad^0\bar\rho)\\ & \oplus & \oplus &\oplus \\& \mathcal{C}^0(G,\ad\bar\rho) \ar[r] \ar[rdd] & \displaystyle\bigoplus_{v\in T}\mathcal{C}^1(G_v,\ad^0\bar\rho) \ar[r] & \displaystyle\bigoplus_{v\in S}\mathcal{C}^2(G_v,\ad^0\bar\rho)\\ & & \oplus & &\\ & & \displaystyle\bigoplus_{v\in S\setminus T}\mathcal{C}^1(G_v,\ad^0\bar\rho)/\tilde{\mathcal{L}}(\mathcal{D}_v) \ar[ruu]& }$$ Here $\mathcal{C}^i(G,M)=\Hom(\underbrace{G\times \cdots\times G}_i, M)$.
Definition 13 Write $\mathcal{C}_0^\cdot(G,\ad^0\bar\rho)$ to be the complex in the first row and $\mathcal{C}^\cdot_{\mathcal{A},T,\mathrm{loc}}(G,\ad^0\bar\rho)$ to be complex in the second row. So $$\mathcal{C}_{\mathcal{A},T}^\cdot=\mathcal{C}_0^\cdot\oplus \mathcal{C}_{\mathcal{A},T,\mathrm{loc}}^{\cdot-1}.$$

We define $H_*^i(G,\ad^0\bar\rho)=H^i(\mathcal{C}_*^\cdot(G,\ad^0\bar\rho)$, for $*=0$ or $*=\mathcal{A},T,\mathrm{loc}$.

Next time we shall study the cohomology $$H_{\mathcal{A},T}^i(G,\ad^0\bar\rho)=H^i(\mathcal{C}_{\mathcal{A},T}^\cdot).$$ Similarly, the number of generators will be given by $H^1$ and the number of generators will be bounded by $H^2$. Can we compute $H^1$ and $H^2$? This needs serious input from Galois cohomology, which we will do next time.

03/13/2014

Proposition 6 There exists a surjection $R_{\mathcal{A},T}^\mathrm{loc}[ [ X_1,\ldots, X_d] ]\twoheadrightarrow R_\mathcal{A}^{\Box_T}$. Here $d=\dim_\mathbb{F}H_{\mathcal{A},T}^1$ and the number of the relation is at most $\dim H_{\mathcal{A},T}^2$.
Proof The proof goes as in Corollary 4. ¡õ

Our next goal is to compute $H^i_{\mathcal{A},T}(G,\ad^0\bar\rho)$ for $i=1,2$ in terms of

  1. the usual local and global Galois cohomology,
  2. the dimension of the local conditions $\mathcal{L}(\mathcal{O}_v)\subseteq H^1(G_v,\ad^0\bar\rho)$,
  3. the dimension of the "dual Selmer group" (as the error term).

TopComputation of $H^i_{\mathcal{A}.T}$

Assume for simplicity that

  1. $\ell>2$ ($\ell=2$ causes, e.g., problems at real places; see Kisin's modularity results on 2-adic representations),
  2. $\ell\nmid n$. This implies that there exists a splitting of Galois modules $$0\rightarrow \ad^0\bar\rho\rightarrow \ad\bar\rho\rightarrow\mathbb{F}\rightarrow0.$$
  3. all places above of $F$ above $\ell$ lies in $S$ (this is a harmless assumption).

The fact is that all cohomology groups are finite dimensional over $\mathbb{F}$ and concentrate in bounded degree. So we can define the Euler characteristic $$\chi_*(G,\ad^0\bar\rho)=\sum_{i\ge0}(-1)^{i+1}\dim_\mathbb{F}H^i_*(G,\ad^0\bar\rho).$$

There are four steps to compute $H^i_{\mathcal{A},T}$.

Step 1 We have $$\chi_{\mathcal{A},T}=\chi_0-\chi_{\mathcal{A},T,\mathrm{loc}}.$$ This is clear exact sequence of complexes $$0\rightarrow\mathcal{C}_{\mathcal{A},T,\mathrm{loc}}^{\cdot-1}\rightarrow\mathcal{C}_{\mathcal{A},T}^\cdot\rightarrow\mathcal{C}^\cdot_0\rightarrow0$$ and the fact that the Euler characteristic is additive in long exact sequences. It follows that $$\chi_{\mathcal{A},T}=\chi(G,\ad^0\bar\rho)-\chi_{\mathcal{A},T,\mathrm{loc}}+\underbrace{h^0(G,\ad^0\bar\rho)-h^0(G,\ad\bar\rho)}_{=-1}.$$ The latter is equal to $-1$ due to the existence of the splitting $\ad\bar\rho=\ad^0\bar\rho\oplus \mathbb{F}$.

Step 2 We compute $\chi_{\mathcal{A},T,\mathrm{loc}}$ in terms of usual Galois cohomology. By definition, \begin{align*} \chi_{\mathcal{A},T,\mathrm{loc}}&=\sum_{v\in S}\chi(G_v,\ad^0\bar\rho)+\sum_{v\in T}h^0(G_v,\ad^0\bar\rho)-h^0(G_v,\ad\bar\rho)\\&+\sum_{v\in S\setminus T}h^0(G_v,\ad^0\bar\rho)-\sum_{v\in S\setminus T}\dim\mathcal{L}(\mathcal{D}_v). \end{align*} Again, the second term is equal to $\sum_{v\in T}(-1)$. Therefore \begin{align*} \chi_{\mathcal{A},T}&=\chi-\chi_{\mathcal{A},T,\mathrm{loc}}-1\\&=\chi-1+\#T-\sum_{v\in S}\chi(G_v,\ad^0)-\sum_{v\in S\setminus T}h^0(G_v,\ad^0)-\dim\mathcal{L}(\mathcal{D}_v). \end{align*}

Step 3 Apply the local and global Euler-Poincare characteristic formula to $\chi(G,-)$ and $\chi(G_v,-)$ to get a formula for $\chi_{\mathcal{A},T}$.

Step 4 It turns out $H^i_{\mathcal{A},T}=0$ when $i\ge4$. $H^0_{\mathcal{A},T}$ is always easy. By the Euler-Poincare characteristic, to compute $H^1_{\mathcal{A}.T}$, it remains to compute $H^2$ and $H^3$. The Poitou-Tate duality allows one to understand $H^2$ and $H^3$ in terms of $H^1$ (this is the error term mentioned above) and $H^0$ (easy) of the dual Galois module. When the error term vanishes, $H^2_{\mathcal{A},T}$ is zero so the deformation ring is indeed a power series ring.

To execute the last two steps, We need the following facts.

Theorem 1 (Cohomological vanishing)
  1. Let $K $ be a nonarchimedean local field and $M$ be a finite $\mathbb{F}[G_K]$-module. Then $H^i(G_K,M)=0$ for $i>2$. ($K $ has cohomological dimension 2).
  2. When $K=\mathbb{R}$, $H^i(G_\mathbb{R},M)=0$ for $i>0$ (here we use the assumption that $\ell>2$).
  3. When $F$ is a number field and $M$ is a finite $\mathbb{F}[G=G_{F,S}]$-module, $H^i(G,M)=0$ for $i>2$ (here we use the assumption that $\ell>2$ as well: the $\ell$-cohomological dimension of a number field is 2 when $\ell>2$).

From the long exact sequence in cohomology, it follows that

Corollary 5 $H^i_{\mathcal{A},T}=0$ for $i\ge4$.

Another input is the determination of the Euler-Poincare characteristics.

Theorem 2 (Euler-Poincare characteristic)
  1. When $K $ is a nonarchimedean local field of characteristic 0, then $$\chi(G_K, M)=\log_{\#\mathbb{F}}(\mathcal{O}_K/\# M)=\dim_\mathbb{F}M\cdot \log_{\#\mathbb{F}}(\mathcal{O}_K/\#\mathbb{F}).$$ This is zero unless $\ell=p$, in which case is $\dim_\mathbb{F}M\cdot [K: \mathbb{Q}_p=\mathbb{Q}_\ell]$.
  2. When $F$ is a number field and $\ell>2$, then $$\chi(G_{F,S},M)=[F: \mathbb{Q}]\dim_\mathbb{F}M-\sum_{v\mid \infty}h^0(G_v,M).$$
Corollary 6 Write $M=\ad^0$, then \begin{align*} \chi(G,M)-\sum_{v\in S}\chi(G_v,M)&=[F:\mathbb{Q}]\dim_\mathbb{F}M-\sum_{v\mid \infty}h^0(G_v, M)-\sum_{v\mid \ell}\dim_\mathbb{F}M [F_v: \mathbb{Q}_\ell]\\&=-\sum_{v\mid\infty}h^0(G_v,M). \end{align*} (here we use the assumption that all primes above $\ell$ are in $S$).

The final key inputs are the local and global duality theorems.

Definition 14 Let $K $ be a local or global field. Let $M$ be a finite $\mathbb{F}[G_K]$-module. Let $M^\vee=\Hom(M,\mathbb{F})$ be the linear dual and $M^D=M^\vee \otimes_{\mathbb{Z}_\ell} \mathbb{Z}_\ell(\varepsilon_l)=M^\vee(1)$ be the Cartier dual. Here $\mathbb{Z}_\ell(\varepsilon_\ell)$ the twist of $\mathbb{Z}_\ell$ by the cyclotomic character $\varepsilon_\ell:G_K\rightarrow \mathbb{Z}_\ell^\times$.
Exercise 2 Let $M=\ad^0\bar\rho$. There is a natural perfect pairing $$M\times M\rightarrow \mathbb{F},\quad (a,b)\mapsto\tr(ab).$$ This gives identification $$M^\vee\cong M,\quad M^D\cong M(1).$$
Theorem 3 (Local duality) Suppose $K $ is a nonarchimedean local field. Then $$H^r(G,M^D)\cong H^{2-r}(G,M)^\vee,\quad r=0,1,2.$$
Remark 21 In particular, the perfect pairing $$H^1(G_v, M^D)\times H^1(G_v,M)\rightarrow\mathbb{F}$$ defines the dual $\mathcal{L}(\mathcal{D}_v)^\perp$ of $\mathcal{L}(\mathcal{D}_v)$ as the annihilator of $\mathcal{L}(\mathcal{D}_v)$.
Theorem 4 (Poitou-Tate) We have a nine term exact sequence

$$\xymatrix{0 \ar[r] & H^0(G,M) \ar[r] & \oplus_{v\in S}H^0(G_v,M) \ar[r] & H^2(G,M^D)^\vee &\\ \ar[r] & H^1(G,M) \ar[r] & \oplus_{v\in S}H^1(G_v,M) \ar[r] & H^1(G,M^D)^\vee &\\ \ar[r] & H^2(G,M) \ar[r] & \oplus_{v\in S}H^2(G_v,M) \ar[r] & H^0(G,M^D)^\vee \ar[r] & 0\\}$$

Remark 22 When $S$ is infinite, $\oplus H^i(G_v)$ needs to be replaced with the restricted product with respect to the unramified parts, which is the direct product (direct sum) when $i=0$ ($i=2$), since $H^0_\mathrm{ur}(G_v)=H^0(G_v)$ ($H^2_\mathrm{ur}(G_v)=0$); When $S$ contains infinite places, the corresponding $H^0$ should be replaced by $\hat H^0$.

Back to our situation with $M=\ad^0\bar\rho$, by definition we have the following exact sequence $$\xymatrix{ & H^1(G,M) \ar[r] & \displaystyle\bigoplus_{v\in T}H^1(G_v,M)\bigoplus_{v\in S\setminus T}H^1(G_v,M)/\mathcal{L}(\mathcal{D}_v) \ar[r] & H^2_{\mathcal{A},T} \\ \ar[r] & H^2(G,M) \ar[r] & \displaystyle\bigoplus_{v\in S}H^2(G_v,M) \ar[r] & H^3_{\mathcal{A},T}.}$$

Notice that 1, 4, 5 terms are the same as in Poitou-Tate.

Suppose we have a commutative diagram $$\xymatrix{1 \ar[r] \ar[d]^{=}& 2 \ar[r]^{\gamma} \ar[d]& 3 \ar[r] \ar[d] & 4 \ar[d]^{=}\\ 1 \ar[r] & 2/N \ar[r] & 3/\gamma(N) \ar[r] & 4}.$$ where $N\subseteq 2$ is a subspace. If the top row is exact, then the second row is still exact. Take the first row to be the Poitou-Tate exact sequence and $N=\bigoplus_{v\in S\setminus T}\mathcal{L}(\mathcal{D}_v)$. Then

\begin{align*} 3/\gamma(N)&=\coker(\bigoplus_{v\in S\setminus T}\mathcal{L}(\mathcal{D}_v)\rightarrow H^1(G,M)^\vee)\\&=\ker(H^1(G,M^D)\rightarrow\bigoplus_{v\in S\setminus T}H^1(G_v,M^D)/\mathcal{L}(\mathcal{D}_v)^\perp)^\vee\\&=: H^1_{\mathcal{A},T}(G,M^D)^\vee. \end{align*}

It follows that $$\dim H^2_{\mathcal{A},T}=\dim H^1_{\mathcal{A},T}(G,M^D),\quad \dim H^3_{\mathcal{A},T}=\dim H^0(G,M^D).$$ Since $\bar\rho$ is absolutely irreducible, we also know that $\dim H^0_{\mathcal{A},T}=\dim H^0(G,\ad\bar\rho)=1$. Combining these with the Euler-Poincare characteristics computation, we obtain desired formulas for $H^1_{\mathcal{A},T}$ and $H^2_{\mathcal{A},T}$. For explicit expressions, see [Gee] 3.24.

03/18/2014

(I was out of town for AWS 2014, this section is shameless copied from Rong Zhou's typed notes.)

The notation is as above. $K/\mathbb{Q}_p$ is a finite extension and $\bar\rho:G_K\rightarrow GL_n(\mathbb{F})$ is a continuous representations, and fix a character $\chi:G_K\rightarrow \mathcal{O}^\times$, which reduces to $\det\bar\rho$.

We constructed the ring $R^\Box_{\bar\rho,\chi}\in\mathcal{C}_\mathcal{O}$ which represents the lifting problem for $\bar\rho$ with the fixed determinant $\chi$. Its generic fiber $R^\Box_{\bar\rho,\chi}[1/\ell]$ has closed points corresponding to $\ell$-adic liftings of $\bar\rho$ with determinant $\chi$.

The goal of the next week will be to study the properties (e.g. irreducible components, dimension) of $R^\Box_{\bar\rho,\chi}[1/\ell]$ (or $R^\Box_{\bar\rho,\chi}$). We split into the two cases $\ell\neq p$ and $\ell=p$ (the second requires some background in $p$-adic hodge theory). This information (irreducible components, dimension) enters into the proof of automorphy lifting theorems. In order to control $H^i_{\mathcal{A},T}$ ($i=1,2$) or the Krull dimension of $R^\Box_{\mathcal{A},T}$, we saw last time that we need to know the $\dim_\mathbb{F}\mathcal{L}(\mathcal{D}_v)$, or the Krull dimension of $R^\Box_{\bar\rho_v,\chi}/I(\mathcal{D}_v)$.

TopLocal universal lifting rings $\ell\neq p$

Proposition 7 $\Spec R^\Box_{\bar\rho,\chi}[1/\ell]$ has finitely many irreducible components and each irreducible component is generically formally smooth (over $L$) and of dimension $n^2-1$.
Remark 23 The same is true of $R^\Box_{\bar\rho}[1/\ell]$ (here the dimension is $n^2$) and $R^\mathrm{univ}_{\bar\rho}[1/\ell]$ if $\bar\rho$ is Schur (with dimension 1).
Proof We define a closed point $x$ of $\Spec R^\Box[1/\ell]$ corresponding to the $\ell$-adic representation $\rho_x:G_K\rightarrow GL_n(L_x)$ to be smooth if $H^0(G_K,\ad^0\rho_x(1))=0$. It is shown in [BLGGT], Lemma 1.3.2 that the smooth points are Zariski dense. Thus it suffices to prove that $R^\Box[1/\ell]^\wedge_x\cong L_x[ [y_1,...,y_{n^2-1}] ]$. Notice this ring is the universal lifting ring for $\rho_x$ with coefficients in $\mathcal{C}^{\mathrm{Art}}_{L_x}$ (Lemma 9). The idea is to mimic the argument for liftings of $\bar\rho$ using tangent spaces and Galois cohomology. Define $$d:=\dim_{L_x}(\text{tangent space at }x)$$ which, if we fix the determinant, is equal to $$d=n^2-1-\dim H^0(G_K,\ad^0\rho_x)+\dim H^1(G_K,\ad^0\rho_x)$$ Hence there exists a surjection $$\phi:L_x[ [x_1,...,x_d] ]\twoheadrightarrow R^\Box[1/\ell]_x^\wedge.$$ One shows in the same way as before that $\ker\phi=0$ if $H^2(G_K,\ad^0\rho_x)=0$.

Thus it suffices to prove $\dim H^0=\dim H^1$ and $H^2=0$. This follows from the $\ell$-adic version of local duality and the Euler-Poincare formula. The first gives us $$H^2(G_K,\ad^0\rho_x)= H^0(G_K, \ad^0\rho_x(1))^\wedge=0$$ (the second equality follows from the smoothness), and the second gives $\chi(\ad^0\rho_x)=0$. These two together imply what we wanted. ¡õ

Definition 15 Let $\mathcal{C}$ be a nonempty subset of irreducible components of $\Spec R^\Box[1/\ell]$. We define $R^\Box_{\bar\rho,\chi,\mathcal{C}}=R^\Box_{\mathcal{C}}$ to be the largest quotient of $R^\Box_{\bar\rho,\chi}$ which is
  1. reduced and $\ell$-torsion free;
  2. $\Spec R^\Box_{\bar\rho,\chi,\mathcal{C}}[1/\ell]\subset \mathcal{C}$.
Remark 24 Here is a rough idea for the construction of this ring. Consider $$\overline{R}^\Box:=R^\Box/\ell\text{-torsion}\hookrightarrow R^\Box[1/\ell].$$ Then $\Spec R^\Box[1/\ell]\subset \Spec \overline{R}^\Box$ is open dense. Take the closure $\overline{\mathcal{C}}$ of $C$ in $\Spec \overline{R}^\Box$ and then take the reduced closed subscheme structure on $\overline{\mathcal{C}}$.
Lemma 13
  1. Let $I_\mathcal{C}:=\ker(R^\Box\twoheadrightarrow R^\Box_\mathcal{C})$. Then $\mathcal{D}(I_\mathcal{C})$ is a deformation problem.
  2. $R^\Box_\mathcal{C}$ is equidimensional of dimension $n^2$. (Note that $R^\Box_\mathcal{C}=R^\Box/{I_{\mathcal{C}}}=R^\Box/I(\mathcal{D}(I_\mathcal{C}))$.
Proof
  1. The non-trivial part is to show that $I_\mathcal{C}$ is $\ker(GL_n(R^\Box)\rightarrow GL_n(\mathbb{F})$ invariant (this is [BLGGT] Lemma 1.2.2).
  2. $\Spec R^\Box[1/\ell]$ is open and dense in $\Spec \overline{R}^\Box$. Let $Z$ be an irreducible component of $\Spec\overline{R}^\Box$ and define $Z':=Z\cap\Spec R^\Box[1/\ell]$. One checks that $Z'$ is an irreducible component and is non-empty with $\dim Z=\dim Z'+1.$ In fact, suppose $Z'{}=\Spec Y'$ and $Z=\Spec Y$. Take a sequence of ideals: $$0\subsetneq \mathfrak{p}_0\subsetneq\cdots\subsetneq \mathfrak{p}_{n^2-1}\subsetneq Y'.$$ Then $$0\subsetneq \mathfrak{p}_0\cap R^\Box\subsetneq\cdots\subsetneq \mathfrak{p}_{n^2-1}\cap R^\Box\subsetneq \mathfrak{m}^\Box\subsetneq Y,$$ since the quotient $\mathfrak{m}^\Box/\mathfrak{p}_{n^2-1}\cap R^\Box$ is the ring of integers in a finite extension of $L$. ¡õ

Now consider the map which takes finite dimensional Weil-Deligne representations of $W_K$ on $L$-vector space to equivalence classes of triples $(V,r_0,N)$, where $r_0:I_K\rightarrow GL(V)$ is an representation of the inertia subgroup, $N\in \End(V)$ is nilpotent, and $r_0$ and $N $ commute.

Remark 25 The map is not onto (Exercise).
Definition 16 An inertial type is any $\tau$ in the image. A Weil-Deligne representation is of type $\tau$ if it lies in the preimage of $\tau$.
Example 4 A representation is of unramified type if it is in the preimage of $\tau=[(1,0)]$ of any dimension.
Example 5 (classification for dimension 2) Say $\mathbb{C}$ (or $\overline{\mathbb{Q}_\ell}$) coefficients, there are four inertial types:
  1. Unramified up to character $\tau=[(\psi\oplus\psi,0)]$, $\psi:I_K\rightarrow \mathbb{C}^\times$ a character.
  2. Steinberg: $\tau=[(\psi\oplus\psi, \left(\begin{smallmatrix} 0& 1 \\ 0 & 0\end{smallmatrix}\right)]$.
  3. Split unramified: $\tau=[(\psi_1\oplus\psi_2,0)]$ where $\psi_1$ and $\psi_2$ are two distinct characters of $I_K$.
  4. Irreducible type: $\tau=[(r_0,0)]$ comes from an irreducible Weil-Deligne representations of dimension 2.

Notice that an irreducible Weil-Deligne representation may restrict to a reducible representation of $I_K$. To make type c) and d) disjoint, we make the additional condition that type c) comes reducible Weil-Deligne representations.

Theorem 5 ([Pil] Section 4) Suppose $L$ is sufficiently large (or we can work with geometrical irreducible components). Then
  1. Each irreducible component $\mathcal{C}$ of $\Spec R^\Box[1/\ell]$ has associated type $\tau_\mathcal{C}$ such that:
    1. Closed points corresponds to $\rho_x$ such that $\WD(\rho_x)$ has type $\tau_\mathcal{C}$ (type a), c), d)), or
    2. Closed points correspond to $\rho_x$ fitting into an exact sequence $$0\rightarrow\psi(1)\rightarrow \rho_x\rightarrow\psi\rightarrow 0$$ for some $\psi:G_K\rightarrow L_x^\times$ (type a) or b)).
  2. Each type occurs in at most one component with the following exception: if $\bar\rho=\overline{\psi}_1\oplus\overline{\psi}_2$ with $\overline{\psi}_1$ and $\overline{\psi}_2$ distinct, and $\overline{\psi}_1\overline{\psi}_2^{-1}$ is unramified, then there exists 2 components with type c).
  3. Two components intersect only when the two components are of type a) and type b), the representations in the component corresponding to type a) are of the form $\rho=\psi'\oplus\psi'(1)$ and those of type b) are of the form $$0\rightarrow \psi(1)\rightarrow \rho\rightarrow \psi\rightarrow 0$$ such that $\WD(\rho)=(\psi(1)\oplus\psi,N=\left(\begin{smallmatrix}0& a \\ 0 & 0\end{smallmatrix}\right))$. As one approaches the intersection point, $a$ approaches $0$ and $\psi'$ approaches $\psi$, the representation at the intersection point is $\psi(1)\oplus\psi$.
  4. Each component is formally smooth.
Remark 26 Here is a heuristic argument for part c). Suppose $x$ is the intersection point, the proof of the previous proposition implies that $H^0(G_K,\ad^0\rho_x(1))\neq0$. Hence $$0\neq \ad\rho_x(1)^{G_K}=\Hom_{L_x}(\rho_x,\rho_x(1))^{G_K}=\Hom_{L_x[G_K]}(\rho_x,\rho_x(1)).$$ This implies $\rho_x$ is reducible $$0\rightarrow \psi_1\rightarrow\rho_x\rightarrow\psi_2\rightarrow 0.$$ Hence $\psi_2=\psi_1(1)$ and $\rho_x$ is split, i.e.. $\rho_x\cong\psi_1\oplus\psi_1(1)$.

03/20/2014

Last time we looked at local universal lifting rings for $\ell\ne p$. We will leave the important Taylor-Wiles deformation (allowing auxiliary primes with ramification) in the homework. Another important deformation problem is the Ihara avoidance defomations due to Taylor (for $GL_2(\mathbb{Q})$, Ihara's lemma allows one to raise the level; but Ihara's lemma is not known in higher dimensional. The Ihara avoidance deformation was introduced to bypass Ihara's lemma).

TopLocal universal lifting rings $\ell=p$

As always, all Galois representations are finite dimensional on $L$-vector spaces. There are more $p$-adic representations of $G_K$ than $\ell$-adic representations $\ell\ne p$ (where the wild inertia is almost killed). The slogan of $p$-adic Hodge theory is to try to understand $p$-adic representations of $G_K$ through linear algebraic categories via equivalence (at least fully faithful embeddings) of categories. There exists a hierarchy of $p$-adic Galois representations:

crystalline $\subseteq$ semistable $\subseteq$ potentially semistable ( = de Rham) $\subseteq$ Hodge-Tate $\subseteq$ all

We will not explain these technical terms but show some analogies with $\ell$-adic ($\ell\ne p$) representations and representations comes from geometry (smooth projective varieties $X/K$).

$p$-adic crystalline semistable potentially semistable de Rham
$\ell$-adic unramified inertia acting unipotently inertia acting potentially unipotent all
smooth projective variety $X/K$ good reduction semistable reduction potentially semistable reduction all
Remark 27 $X/K$ with good reduction gives rises to unramified $\ell$-adic representations (but the converse is not true). It is still a conjecture that all smooth projective varieties are potentially semistable. Notice by Grothendieck's $\ell$-adic monodromy theorem, all $\ell$-adic representations are potentially unipotent.
Remark 28 The Langlands-Fontaine-Mazur philosophy asserts that if $\rho:G_F\rightarrow GL_n(L)$ comes from automorphic forms, then $\rho|_{G_v}$ (for $v\mid \ell$) is potentially semistable (=de Rham). Conversely, if $\rho$ is potentially semistable at $v\mid \ell$ and unramified at almost all places, then $\rho$ should be automorphic. From this philosophy it is natural to impose potentially semistable condition at $v\mid \ell$ so that $R=T$ theorem has a chance to be true.

There are two important invariants associated to potentially semistable representations: WD (a Weil-Deligne representation of $W_K$) and HT (Hodge-Tate weights, a multiset of integers).

TopThe Weil-Deligne functor

When $\ell\ne p$, taking the Weil-Deligne representation gives a functor from all $G_K$-representations to WD-representations of $W_K$. When $\ell=p$, we only define the Weil-Deligne representation functor for potentially semistable representations. Let $K'/K$ be a finite Galois extension and $K'_0\subseteq K'$ be the maximal unramified subextension over $\mathbb{Q}_p$. Take $L/\mathbb{Q}_p$ to be sufficient large (containing the Galois closure of $K'$), e.g., $L=\overline{\mathbb{Q}_p}$. Let $\phi_0$ be the absolute Frobenius on $K_0'$.

Definition 17 We define $\WD_{K'/K}$ to be the category of WD-representations $(r, N)$ of $W_K$ such that $r|_{W_{K'}}$ is unramified.
Definition 18 We define $\mathbf{Mod}_{K'/K}$ to be the category of $(\varphi, N,\Gal(K'/K))$-modules $D $. Here
  1. $D $ is a finite rank free $K_0' \otimes_{\mathbb{Q}_p} L$-module;
  2. $\varphi: D\rightarrow D$ is a bijection and $\phi_0 \otimes 1$-semilinear, i.e., $\varphi((k \otimes l)\cdot d)=\phi_0(k) \otimes l\cdot\varphi(d)$;
  3. $N\in \End_{K_0' \otimes L}(D)$ is nilpotent and $N\varphi=p\varphi N$;
  4. The $\Gal(K'/K)$-action on $D $ is semilinear and commutes with $\varphi,N$.
Theorem 6
  1. There exists a dimension preserving functor $D_{\mathrm{st},K'/K}$ from the category of potentially semistable representations $\rho$ such that $\rho|_{G_{K'}}$ is semistable to the category $\mathbf{Mod}_{K'/K}$.
  2. There exists an equivalence between $\mathbf{Mod}_{K'/K}\cong \WD_{K'/K}$.

We treat the first functor as a black box ($p$-adic Hodge-Tate theorem). The second functor is purely linear algebraic and can be described more easily. Let $\sigma_0: K_0'\hookrightarrow L$ be an embedding. Suppose $D\in\mathbf{Mod}_{K'/K}$, we define $V=D \otimes_{K_0' \otimes L} L$ via $\sigma_0 \otimes 1$ and $r(w)=w\cdot \varphi^{v_{\mathbb{Q}_p}(w)}$, here $w\in W_K$, $v_{\mathbb{Q}_p}$ is the absolute $p$-adic valuation, and $N=N \otimes 1$. Then $(V, r,N)\in \WD_{K'/K}$ and the isomorphism class does not depend on $\sigma_0$.

Allowing $K'$ to be larger and larger, we obtain a functor from potentially semistable $G_K$-representations to the category of WD-representations of $G_K$ ($L=\overline{\mathbb{Q}}_p$).

Remark 29 Write $(r,N)=WD(\rho)$. Then $\rho$ is semistable if and only if $r$ is unramified; $\rho$ is crystalline if and only if $r$ is unramified and $N=0$; $\rho$ is potentially crystalline if and only if $N=0$. Moreover, $\rho|_{G_{K'}}$ is semistable if and only if $r|_{I_{K'}}=1$ and $r(W_{K'})$ centralizes $r(W_K)$.

TopHodge-Tate weights

Given $\rho: G_K\rightarrow GL_L(V)$, we will define a collection $\{\HT_\sigma(\rho)\}_{\sigma: K\hookrightarrow L}$, where $\HT_\sigma(\rho)$ is an unordered multiset of integers. Each $i\in \mathbb{Z}$ in $\HT_\sigma(\rho)$ has the following multiplicity $$\dim_L(V(i) \otimes_{K}\hat{\bar K})^{G_K}.$$ It is known that the sum of all these multiplicities is equal to $n=\dim_L V$ (for any Hodge-Tate representation). These numbers can be read off from some natural filtration (defined after extending coefficients) attached to $D_{\mathrm{st},K'/K}(\rho)$.

Example 6 The $p$-adic cyclotomic character has Hodge-Tate weight $\HT_\sigma=\{1\}$ for any $\sigma$.
Example 7 When $A/K$ is an abelian variety and $\rho$ is given by its $p$-adic Tate module. Then $\HT_\sigma(\rho)=\{0,\ldots, 0, 1,\ldots,1\}$ with 0 and 1 each occurring $g$ times.
Example 8 If $f$ is an eigen cuspform of weight $k\ge1$. The associated Galois representation $\rho_f: G_\mathbb{Q}\rightarrow GL_2(\overline{\mathbb{Q}_p})$ has $\HT(\rho|_{G_{\mathbb{Q}_p}})=\{0,k-1\}$.
Example 9 If $\rho$ has finite image, then all Hodge-Tate weights are 0.
Remark 30 Can one recover potentially semistable $\rho$ from $\WD(\rho)$ and $\{\HT_\sigma(\rho)\}$ uniquely? This is the case when the latter data has an "admissible filtration". Breuil-Scheider conjectured a nice answer to what these should correspond to on the $GL_2(\mathbb{Q}_p)$-side, which is part of the beginning of the $p$-adic Langlands program.

TopPotentially semistable local lifting rings

Theorem 7 (Kisin) Suppose we are given $\{H_\sigma\}_{\sigma: K\hookrightarrow L}$, where each $H_\sigma$ consists of distinct integers.
  1. There exists a unique $p$-torsion-free, reduced quotient $R^{\Box}_{\bar\rho,\chi,\{H_\sigma\},K'-\mathrm{sst}/\mathrm{cris}}$ of $R_{\bar\rho,\chi}^\Box$ such that for any closed (geometric) point $x$ of $R^{\Box}_{\bar\rho,\chi}[1/p]$, $x$ factors through this quotient if and only if $\HT_\sigma(\rho_x)=H_\sigma$ and $\rho_x|_{G_{K'}}$ is semistable/crystalline.
  2. $\Spec R^{\Box}_{\bar\rho,\chi,\{H_\sigma\},K'-\mathrm{sst}/\mathrm{cris}}$ is equidimensional of dimension $n^2+[K:\mathbb{Q}_p]n(n-1)/2$. Its generic fiber is generically smooth.
Remark 31 This construction is very difficult and uses serious $p$-adic Hodge theory. The lifting ring may have several irreducible components. Compare the Fontaine-Laffaille condition using relatively easy $p$-adic Hodge theory for crystalline condition and $K'{}=K$, which has only one component.
Remark 32 When the Hodge-Tate weights are not distinct, the dimension becomes strictly smaller. The deformation space is too small for $R=T$ theorem to be true. See the recent work of Calegari-Geraghty.
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