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!

02/25/2014

## Deformations of Galois representations

The global Langlands correspondence is roughly a correspondence between automorphic forms (representations) and -adic Galois representations. Suppose we are given the arrow automorphic representations -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 ) is to first show that the reduction is the reduction of the Galois representation coming from a modular form (this is Serre's conjecture in general), then try to study infinitesimal liftings Now both sides then have more algebro-geometric structures (like a map of schemes ). It often turns out to be a closed immersion. If one can then show (e.g., by dimension reason) this is actually an isomorphism ( theorem), then the surjectivity will follow. Our goal in next few lectures is to study the right hand side, the deformations of Galois representations.

## Group-theoretic hypothesis

When deforming a representation , we would like to impose a finiteness condition on the profinite , in order to make sense of the the space of deformations and make ring noetherian. Fix a prime and a profinite group. We impose the following -finiteness assumption.

Definition 1 Hyp(): for any an open finite index subgroup,
1. is finite ( = is finite dimensional over ). Here .
2. The maximal pro- quotient of is topologically finitely generated (i.e., there exists a finitely generated dense subgroup).
Remark 1 Nikolov-Segal proved that when is topologically finitely generated, any finite index subgroup is automatically open. This may fail for groups like .
Lemma 1
1. a) b).
2. If 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 will work for the maximal pro- quotient of . ¡õ
Remark 2 Since every finite index subgroup of a finitely generated group is again finitely generated, it follows that if is topologically finitely generated, then any of finite index is also topologically finitely generated.
Example 1
1. Let be a finite extension ( or ) Then satisfies Hyp: to check a), it is the same thing to check that there exists only finitely many abelian extension of exponent for a given local field. This follows from Kummer theory.
2. Let be a finite extension, be a finite set of finite places of , be the maximal extension of unramified outside . Then satisfies Hyp (by global class field theory, see [DDT] 2.41 for details). Notice that itself is not known to be topologically finitely generated.

## Liftings of mod representations

Let be a finite extension, with uniformizer and residue field .

Definition 2 Let be the following category. The objects of are complete noetherian local -algebras such that . Here is the unique maximal ideal of and the completeness means that is an isomorphism. Notice the -algebra isomorphism is unique if it exists.

The morphisms in are morphisms of local -algebras (i.e., ). It follows that induces an isomorphism .

Remark 3 An -algebra morphism between is automatically local: since , is proper ideal. Hence is a nonzero subring, so must be a field. Therefore .
Definition 3 Let be a continuous representation. Define given by This is a functor and is called the liftings of .
Remark 4 No "isomorphism/equivalence class" of is involved in the definition! In Kisin's terminology, is called the framed deformations of , which somehow explains the in the notation.
Proposition 1 The functor is represented by some .
Remark 5 We call the universal lifting ring of . Explicitly, this means that satisfies one of the following equivalent conditions:
1. There are bijections , functorial in .
2. There exists a universal lifting of such that for any and a lifting , there exists a unique such that
Example 2 Consider . Giving is the same as giving an element . Fix a lift . Then is in bijection with , given by . Hence So represents .
Proof Step 1 By Hyp(), one may assume that is topologically finitely generated.

In fact, let (an open finite index subgroup). Let , here is the maximal pro- quotient of . We claim that one may replace with . Then is topologically finitely generated profinite group (though may not be): look at the exact sequence Notice is finite and is topologically finitely generated by Hyp(). It follows that is also topologically finitely generated.

The key thing is that every lifting of factors through . Indeed, Suppose is such a lifting. Since , we know that , this kernel is pro-, but by construction is prime-to-. Hence and we may replace with .

Step 2 Let be the free group on generators and be the profinite completion of . One can choose so that there is a surjection . Let be the restriction to the dense subgroup . Let and . One can check that is also dense. Now giving a lifting is the same as giving (i.e. ) being trivial on . Now is represented by . ¡õ

02/27/2014

## Some background on irreducible representations

We switch notations temperately. Let be a field, be an abstract group, be an finite dimensional -vector spaces. Recall that is irreducible if there is no proper -subvector space which is -stable.

Lemma 2 (Schur) If is irreducible, then is a division algebra over .
Proof Suppose is nonzero, then is nonzero and -stable, hence . Therefore is invertible. ¡õ
Lemma 3 (Schur) Suppose is algebraically closed. Let be irreducible representations. Then
Remark 6 The only division algebra over an algebraically closed field is itself.
Remark 7 is irreducible does not imply that is irreducible. A division algebra over may not be a division algebra when extending scalars to .
Remark 8 The converse of Schur's Lemma is not true (for non-semisimple representations): does not imply that is irreducible. For example, consider the non-semisimple representation . Then but is clearly reducible.
Proposition 2 (Burnside's theorem) If is irreducible over , then is surjective.
Definition 4 We say is Schur, if ; absolutely irreducible if for any a field extension, is irreducible.
Proposition 3 is absolutely irreducible if and only if is irreducible, if and only if is irreducible and Schur.
Proof See [CR] Section 29. ¡õ

## Deformations of mod representations

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

Definition 5 Let be a continuous representation. We define the functor of deformations of by Here if there exists such that . When , this happens if and only if can be chosen in .
Remark 9 The isomorphism classes of liftings are not in the definition of .
Proposition 4 If is Schur, then is representable. Say by , the universal deformation ring of .
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 (see [DDT] 2.36). Kisin's approach, roughly speaking, is to show that is the geometric quotient of by (see [Boe] 2.1). Also see [Maz97] 10. ¡õ
Remark 12 As before, for any deformation of , there exists a unique such that .
Remark 13 There is a canonical map . It is induced by the functor by sending to its isomorphism class. Another way: view as a deformation. We will see more precise descriptions later.

## Linear algebraic lemmas

Lemma 4 Let , be a continuous representation. Assume is absolutely irreducible.
1. (Schur for -coefficient) For , if , then .
2. (Carayol's Lemma) Suppose (in is closed and . Then there exists such that factors through .
Proof One reduces to the case is Artinian local -algebra (since are of this shape). Then induct on the length of . The case is easy and one can further reduce to the case .
1. Assume . Choose a minimal nonzero ideal : one has the filtration where is a -vector space; one can choose to be isomorphic to as an -module. Let such that . Then the induction hypothesis implies that . So we can write , where and . Now the equation tells us that in . Under the identification , we find that in , hence by Schur's lemma, itself must be a scalar too (we only need to be Schur in this part).
2. By induction hypothesis, we may assume that lands in . Since as -submodules, we know that either or . In the latter case, it is immediate that lands in . The first case is more difficult. We build the -algebra , where . Then embeds into by . By induction hypothesis, we may assume is indeed an isomorphism. We may replace by and (because quotient by any thing in 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 -linearly and we obtain that , . Here
1. is -linear,
2. ,
3. .

We want to get rid of and deal with purely matrix algebra. We claim that factors through the surjective map (here we used the absolutely irreducible assumption for the surjectivity), i.e., is trivial on . Let , then . By the surjectivity of , we know that because can be anything in . This proves the claim.

Now we are looking for such that for any . This is equivalent to that the coefficient of So we are reduced to the problem of showing that for , there exists such that

1. ,
2. ,
3. .

Conceptually this means that every derivation on is given by the Lie bracket with some element . One can directly show that works. ¡õ

Lemma 5 (Brauer-Nesbitt for -coefficient). Suppose is absolutely irreducible. If such that . Then .
Proof Use similar reduction and then use Carayol's lemma when . See [Boe] 2.2.1. ¡õ
Corollary 1 ([Gee] Ex 3.9) The universal lifting ring is a power series ring in variables over the the universal deformation ring .

03/04/2014

## Tangent spaces

We are going to work with the universal lifting rings (the same argument works for universal deformation rings). Write for short. Denote its unique maximal ideal by .

Definition 6 The adjoint representation is given by conjugation . We denote (and the corresponding -module).

One can analyze the tangent space of in terms of group cohomology of .

Lemma 6 There exists natural bijections between
1. .
2. .
3. (liftings of to ).
4. (continuous 1-cocycles).
Proof b) c): by definition.

a) b): Since surjects on . For , we define One routinely check that this is well defined (using and gives the desired bijection.

c) d). Given , we can write . One can check that is a homomorphism if and only if defines a continuous 1-cocycle. ¡õ

Write . This is the dimension of cotangent space (ignoring the -direction) of .

Corollary 2 .
Proof There is an exact sequence of finite dimensional -vector spaces Here any gives the coboundary . The result immediately follows. ¡õ

So we can know how big the ring is as long as we know the dimension of and of the adjoint representation.

Corollary 3 Choose such that generate as -vector space, then is surjective.

This follows from a topological version of Nakayama's lemma. In nice situations, is an isomorphism. In general this is too optimistic but one can further control the the kernel . Notice that . Here is the maximal ideal of . We shall construct an injective map When this vanishes, will be an formal power series ring and hence is formally smooth of dimension : there is no obstruction for liftings (controlled by ) and the space of liftings is -dimensional (controlled by and ).

To construct , we notice that is surjective. For , we can choose a lift of . Notice may not be a homomorphism and this failure is measured by Since , we know that and hence it makes sense to apply in this expression. One checks directly that So

Lemma 7 is a continuous 2-cocycle.
Exercise 1
1. gives a well-defined class .
2. if and only if there exists a choice of such that mod is a homomorphism. Here (so ).
Lemma 8 The -linear map is injective.
Proof It suffices to show that if there exists as in b) of the previous exercise, then . Notice . Since is a lifting of , we have a map by the universal property. The composite map is the identity map. This shows that is injective. Let (in particular ), we want to show that . Suppose maps to (so ). One checks directly that . But by injectivity, hence . Therefore so and . ¡õ

So the number of generators of is and with the number of relations is equal to .

Corollary 4
1. If , then we have a non-canonical isomorphism .
2. In general, .

03/06/2014

## Generic fiber of universal lifting rings

The following is a basic algebraic fact.

Proposition 5 Let . There is a bijection between closed points of the generic fiber (these are dense in the generic fiber) and pairs , where is a finite extension, is continuous such that .
Remark 14 A pair induces . The kernel of this map is a maximal ideal. This gives the corresponding closed point of the generic fiber. Conversely, any closed point with maximal ideal gives rise to (which factors through ).

Let be a closed point. Then we know from the above bijection and the universal property that corresponds to a representation . Let to be the category of local Artinian -algebra with residue field .

Lemma 9 The ring pro-represents the functor given by
Remark 15 The lemma means that , functorial in . Though is constructed as the deformations of a representation in characteristic , the generic fiber of knows about deformations of representations in characteristic 0.
Remark 16 There is no difference to take the large category (so the functor is in fact representable in ). But Artinian rings are easier to deal with because the topology on finite dimensional representations are obvious (the -adic topology).
Proof (Sketch of proof) Given , one simply composes to get , an object in .

Conversely, given , let be the -subalgebra of generated by the entries . One can show by a compactness argument that and is finitely generated as -module. Let be the residue field of . Reducing then gives . One can show (homework) that , so the universal property gives the map , hence a map . By the universal property of localization, it extends to a map on the generic fiber . By continuity, it extends to the completion . 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 .

## Deformation 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 (or in terms of rings, certain quotient rings of ). We would like to make a checklist for technical conditions to define nice subspaces of .

Definition 7 A (framed) deformation problem is a collection , where , is a lifting of such that
1. .
2. If is a map in and , then .
3. (restriction) If and , then .
4. (gluing) Let with ideals and . Suppose such that , then . Here
5. (nested intersection) Suppose is a chain of ideals of such that . Suppose , then .
6. (conjugate) If , , then .
Lemma 10 We have a bijection
Remark 18 Suppose , then conjugation on induces an action by the universal property.
Remark 19 is characterized by if and only if is trivial on . Conversely, when and (these restrictions are needed to check f), for this subtlety, see [BLGHT] Lem 3.2), we have .
Proof We explain why is uniquely determined. Let be the set of all ideals such that . This is nonempty by a). Using b) and c), we know if and only if . Moreover is closed under finite intersection and nested infinite intersection by d) and e). So by Zorn's lemma, there exists a unique minimal ideal . It is kernel invariant by f). ¡õ

03/11/2014

Example 3 Typical deformation problems concerns local Galois representations , where is a local field. See [CHT] 2.4 for several examples: when , the Fontaine-Laffaille liftings or ordinary liftings; when , Taylor-Wiles liftings, are all deformation problems.

For applications, it is important to understand the ring theoretic properties of , e.g., Krull dimension, number of generators, number of relations and so on. We computed these for in terms of Galois cohomology. It is similar for .

Definition 8 Inside , one can consider the annihilator of the image of in , (think: the subspace cut out by of the tangent space). We define be its image in .

Using the kernel-invariance property of , one can show that

Lemma 11 is the full preimage of .

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

## Global Galois deformation problems

We begin with a remark on fixing the determinant.

Remark 20 [Gee] Ex 3.19. Lifting the determinant of to such that , then one obtains a subfunctor by requiring the extra condition on the liftings ( is an -algebra, so this makes sense). This subfunctor is represented by a ring (one can construct as the quotient of by the equation ). The previous discussion all carries over to this setting: the main change is that should be replaced by (the traceless subspace).

Now let be a number field. Let be a place of and fix . One has the local Galois group (well-defined up to conjugation). Let be a set of finite places of and .

Definition 9 A global (Galois) deformation problem is a collection , where
1. , , as above;
2. is absolutely irreducible (the work of Skinner-Wiles and Thorne can relax this condition);
3. such that ;
4. is a deformation problem for .

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

Definition 10 Let (may be empty). Define the functor given by . Here
1. is a lifting of with ,
2. ,
3. for .

The equivalence relations is generated by the following: for any , .(Think: we are imposing local conditions at and "framing at "; this helps the Galois cohomology calculation and also helps achieving the final goal of comparing the deformation ring with the Hecke ring).

Lemma 12 is represented by . When , we write it as .
Proof Let be all liftings of and be the deformation problem given by . Then is representable (which is if ). Then one can construct as , where is the minimal ideal such that factors through if and only if for . For more details, see [CHT]. ¡õ

## Presenting global deformation rings over local lifting rings

We have seen how to represent over and when happens to be a power series ring over . We are now going to represent over another bigger ring, the local lifting ring. This idea is due to Kisin.

Notice has the following universal object:

1. , and
2. , .

By the kernel-invariance property, . Moreover, it is well-defined element independent of the choice of the representative of the equivalence class. By the universal property of , we know that factors through

Definition 11 Define the local lifting ring to be the completed tensor product We have a natural map .

Our next goal is to find the number of the generators and relations for presenting over using Galois cohomology.

As a first thought, suppose consists of all liftings. Write and be the maximal ideals. The the same argument as in Lemma 6 shows that Here sits diagonally. Rewriting this as

Two modifications are needed in general:

1. to consider the tangent space over : one should replace by . Concretely, this requires the liftings at to be trivial, i.e., lies in the kernel of Notice the image of is , which we require to be trivial, i.e., . Write and , then if and only if .
2. to allow general for : one requires that

The upshot is that is the of the complex

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

Definition 12 We define the complex to be Here .
Definition 13 Write to be the complex in the first row and to be complex in the second row. So

We define , for or .

Next time we shall study the cohomology Similarly, the number of generators will be given by and the number of generators will be bounded by . Can we compute and ? This needs serious input from Galois cohomology, which we will do next time.

03/13/2014

Proposition 6 There exists a surjection . Here and the number of the relation is at most .
Proof The proof goes as in Corollary 4. ¡õ

Our next goal is to compute for in terms of

1. the usual local and global Galois cohomology,
2. the dimension of the local conditions ,
3. the dimension of the "dual Selmer group" (as the error term).

## Computation of

Assume for simplicity that

1. ( causes, e.g., problems at real places; see Kisin's modularity results on 2-adic representations),
2. . This implies that there exists a splitting of Galois modules
3. all places above of above lies in (this is a harmless assumption).

The fact is that all cohomology groups are finite dimensional over and concentrate in bounded degree. So we can define the Euler characteristic

There are four steps to compute .

Step 1 We have This is clear exact sequence of complexes and the fact that the Euler characteristic is additive in long exact sequences. It follows that The latter is equal to due to the existence of the splitting .

Step 2 We compute in terms of usual Galois cohomology. By definition, Again, the second term is equal to . Therefore

Step 3 Apply the local and global Euler-Poincare characteristic formula to and to get a formula for .

Step 4 It turns out when . is always easy. By the Euler-Poincare characteristic, to compute , it remains to compute and . The Poitou-Tate duality allows one to understand and in terms of (this is the error term mentioned above) and (easy) of the dual Galois module. When the error term vanishes, 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 be a nonarchimedean local field and be a finite -module. Then for . ( has cohomological dimension 2).
2. When , for (here we use the assumption that ).
3. When is a number field and is a finite -module, for (here we use the assumption that as well: the -cohomological dimension of a number field is 2 when ).

From the long exact sequence in cohomology, it follows that

Corollary 5 for .

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

Theorem 2 (Euler-Poincare characteristic)
1. When is a nonarchimedean local field of characteristic 0, then This is zero unless , in which case is .
2. When is a number field and , then
Corollary 6 Write , then (here we use the assumption that all primes above are in ).

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

Definition 14 Let be a local or global field. Let be a finite -module. Let be the linear dual and be the Cartier dual. Here the twist of by the cyclotomic character .
Exercise 2 Let . There is a natural perfect pairing This gives identification
Theorem 3 (Local duality) Suppose is a nonarchimedean local field. Then
Remark 21 In particular, the perfect pairing defines the dual of as the annihilator of .
Theorem 4 (Poitou-Tate) We have a nine term exact sequence

Remark 22 When is infinite, needs to be replaced with the restricted product with respect to the unramified parts, which is the direct product (direct sum) when (), since (); When contains infinite places, the corresponding should be replaced by .

Back to our situation with , by definition we have the following exact sequence

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

Suppose we have a commutative diagram where 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 . Then

It follows that Since is absolutely irreducible, we also know that . Combining these with the Euler-Poincare characteristics computation, we obtain desired formulas for and . 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. is a finite extension and is a continuous representations, and fix a character , which reduces to .

We constructed the ring which represents the lifting problem for with the fixed determinant . Its generic fiber has closed points corresponding to -adic liftings of with determinant .

The goal of the next week will be to study the properties (e.g. irreducible components, dimension) of (or ). We split into the two cases and (the second requires some background in -adic hodge theory). This information (irreducible components, dimension) enters into the proof of automorphy lifting theorems. In order to control () or the Krull dimension of , we saw last time that we need to know the , or the Krull dimension of .

## Local universal lifting rings

Proposition 7 has finitely many irreducible components and each irreducible component is generically formally smooth (over ) and of dimension .
Remark 23 The same is true of (here the dimension is ) and if is Schur (with dimension 1).
Proof We define a closed point of corresponding to the -adic representation to be smooth if . It is shown in [BLGGT], Lemma 1.3.2 that the smooth points are Zariski dense. Thus it suffices to prove that . Notice this ring is the universal lifting ring for with coefficients in (Lemma 9). The idea is to mimic the argument for liftings of using tangent spaces and Galois cohomology. Define which, if we fix the determinant, is equal to Hence there exists a surjection One shows in the same way as before that if .

Thus it suffices to prove and . This follows from the -adic version of local duality and the Euler-Poincare formula. The first gives us (the second equality follows from the smoothness), and the second gives . These two together imply what we wanted. ¡õ

Definition 15 Let be a nonempty subset of irreducible components of . We define to be the largest quotient of which is
1. reduced and -torsion free;
2. .
Remark 24 Here is a rough idea for the construction of this ring. Consider Then is open dense. Take the closure of in and then take the reduced closed subscheme structure on .
Lemma 13
1. Let . Then is a deformation problem.
2. is equidimensional of dimension . (Note that .
Proof
1. The non-trivial part is to show that is invariant (this is [BLGGT] Lemma 1.2.2).
2. is open and dense in . Let be an irreducible component of and define . One checks that is an irreducible component and is non-empty with In fact, suppose and . Take a sequence of ideals: Then since the quotient is the ring of integers in a finite extension of . ¡õ

Now consider the map which takes finite dimensional Weil-Deligne representations of on -vector space to equivalence classes of triples , where is an representation of the inertia subgroup, is nilpotent, and and commute.

Remark 25 The map is not onto (Exercise).
Definition 16 An inertial type is any in the image. A Weil-Deligne representation is of type if it lies in the preimage of .
Example 4 A representation is of unramified type if it is in the preimage of of any dimension.
Example 5 (classification for dimension 2) Say (or ) coefficients, there are four inertial types:
1. Unramified up to character , a character.
2. Steinberg: .
3. Split unramified: where and are two distinct characters of .
4. Irreducible type: comes from an irreducible Weil-Deligne representations of dimension 2.

Notice that an irreducible Weil-Deligne representation may restrict to a reducible representation of . 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 is sufficiently large (or we can work with geometrical irreducible components). Then
1. Each irreducible component of has associated type such that:
1. Closed points corresponds to such that has type (type a), c), d)), or
2. Closed points correspond to fitting into an exact sequence for some (type a) or b)).
2. Each type occurs in at most one component with the following exception: if with and distinct, and 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 and those of type b) are of the form such that . As one approaches the intersection point, approaches and approaches , the representation at the intersection point is .
4. Each component is formally smooth.
Remark 26 Here is a heuristic argument for part c). Suppose is the intersection point, the proof of the previous proposition implies that . Hence This implies is reducible Hence and is split, i.e.. .

03/20/2014

Last time we looked at local universal lifting rings for . 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 , 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).

## Local universal lifting rings

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

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

We will not explain these technical terms but show some analogies with -adic () representations and representations comes from geometry (smooth projective varieties ).

 -adic crystalline semistable potentially semistable de Rham -adic unramified inertia acting unipotently inertia acting potentially unipotent all smooth projective variety good reduction semistable reduction potentially semistable reduction all
Remark 27 with good reduction gives rises to unramified -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 -adic monodromy theorem, all -adic representations are potentially unipotent.
Remark 28 The Langlands-Fontaine-Mazur philosophy asserts that if comes from automorphic forms, then (for ) is potentially semistable (=de Rham). Conversely, if is potentially semistable at and unramified at almost all places, then should be automorphic. From this philosophy it is natural to impose potentially semistable condition at so that theorem has a chance to be true.

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

### The Weil-Deligne functor

When , taking the Weil-Deligne representation gives a functor from all -representations to WD-representations of . When , we only define the Weil-Deligne representation functor for potentially semistable representations. Let be a finite Galois extension and be the maximal unramified subextension over . Take to be sufficient large (containing the Galois closure of ), e.g., . Let be the absolute Frobenius on .

Definition 17 We define to be the category of WD-representations of such that is unramified.
Definition 18 We define to be the category of -modules . Here
1. is a finite rank free -module;
2. is a bijection and -semilinear, i.e., ;
3. is nilpotent and ;
4. The -action on is semilinear and commutes with .
Theorem 6
1. There exists a dimension preserving functor from the category of potentially semistable representations such that is semistable to the category .
2. There exists an equivalence between .

We treat the first functor as a black box (-adic Hodge-Tate theorem). The second functor is purely linear algebraic and can be described more easily. Let be an embedding. Suppose , we define via and , here , is the absolute -adic valuation, and . Then and the isomorphism class does not depend on .

Allowing to be larger and larger, we obtain a functor from potentially semistable -representations to the category of WD-representations of ().

Remark 29 Write . Then is semistable if and only if is unramified; is crystalline if and only if is unramified and ; is potentially crystalline if and only if . Moreover, is semistable if and only if and centralizes .

### Hodge-Tate weights

Given , we will define a collection , where is an unordered multiset of integers. Each in has the following multiplicity It is known that the sum of all these multiplicities is equal to (for any Hodge-Tate representation). These numbers can be read off from some natural filtration (defined after extending coefficients) attached to .

Example 6 The -adic cyclotomic character has Hodge-Tate weight for any .
Example 7 When is an abelian variety and is given by its -adic Tate module. Then with 0 and 1 each occurring times.
Example 8 If is an eigen cuspform of weight . The associated Galois representation has .
Example 9 If has finite image, then all Hodge-Tate weights are 0.
Remark 30 Can one recover potentially semistable from and 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 -side, which is part of the beginning of the -adic Langlands program.

### Potentially semistable local lifting rings

Theorem 7 (Kisin) Suppose we are given , where each consists of distinct integers.
1. There exists a unique -torsion-free, reduced quotient of such that for any closed (geometric) point of , factors through this quotient if and only if and is semistable/crystalline.
2. is equidimensional of dimension . Its generic fiber is generically smooth.
Remark 31 This construction is very difficult and uses serious -adic Hodge theory. The lifting ring may have several irreducible components. Compare the Fontaine-Laffaille condition using relatively easy -adic Hodge theory for crystalline condition and , 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 theorem to be true. See the recent work of Calegari-Geraghty.