SPEAKER: JOHN BERGDALL

November 24, 2014 HODGE TATE AND DE RHAM REPRESENTATIONS SPEAKER: JOHN BERGDALL My goal today is to just go over some results regarding Hodge-Tate and de Rham representations. We always let K/Q p be a finite extension and let C = C K be the completion of an algebraic closure. I m going to go the opposite of Rob Pollack and do some formalism first and then an example at the end. 0.1. Hodge-Tate representations. In his talk, Robert Sayer defined a period ring B HT = C[t, t 1 ] where the Galois action of a local Galois group G K is given by g(t) = χ cycl (g)t. Using Fontaine s formalism we have a period functor A remark on the graded part: we define D HT : Rep Qp (G K ) graded K-vector spaces V (V Qp B HT ) G K. Gr i D HT (V ) = (V Qp t i C) G K The point about saying it this way is that the grading is functorial in V V (as you can obviously see). By Fontaine s formalism of admissible representations we have dim K D HT (V ) dim Qp V and a representation is Hodge-Tate (by definition) if and only if this inequality is an equality. In that case one defines the Hodge-Tate weights of V as the indices i such that Gr i D HT (V ) 0. Example 1. (a) Q p (n) has a unique Hodge-Tate weight n. (b) If G is a p-divisible group over K of height h and dimension d then the Tate-module T (G) has Hodge-Tate weights 1 (with multiplicity d) and 0 with multiplicity h d (see [2, Corollary 2, 4]) (c) As some exercises you can easily compute the Hodge-Tate weights of V W or V if V and W are Hodge-Tate. 0.2. de Rham represenations. There was a slightly more sophisticated period ring, called B dr, that Ben Fischer and David Corwin constructed over the last two lectures. Let me recall the properties: (a) B dr is a complete discretely valued field with residue field C. It has a continuous action of G K. (b) If B + dr B dr is the valuation ring of B dr then the map B + dr C is G K-equivariant and the kernel is generated by an element t such that Z p t Z p (1). (c) B dr is filtered by powers of its maximal ideal t i B + dr B dr. The associated graded of this filtration is Gr B dr = t i B + dr /ti+1 B + dr B HT i Z in a G K -equivariant way. Since B dr is a field, it is automatically a (Q p, G K )-regular ring in the sense of Rob Sayer s talk and thus we have a good notion of B dr -admissible representation, called de Rham representations. A little subtle here is that in order to make things concrete we need to know what B G K dr is. Lemma 2. B G K dr = K. Proof. This follows formally from the property Ben discussed last time that Gr B dr = B HT and B G K HT = K. Now let D dr (V ) = (V B dr ) G K. By the previous lemma this is a K-vector space and using Fontaine s formalism we have the following definition. Definition. V is de Rham if dim K D dr (V ) = dim Qp V. 1

Just as with Hodge-Tate representations there is a special category that D dr takes values. Indeed, B dr is filtered by { t i B + dr} and so if V is a representation we define Fil i D dr (V ) = (V Qp t i B + dr )G K. Note the similarity between Fil i D dr (V ) and Gr i D HT (V ). As one can see easily from the definition, if V V is a morphism of representations then the corresponding morphism D dr (V ) D dr (V ) respects the filtration. Moreover, if i is given then the natural map t i B + dr /ti+1 B + dr C(i) defines a natural sequence What I d like to say is that the diagram of functors 0 Fil i+1 D dr (V ) Fil i D dr (V ) Gr i D HT (V ) Rep Qp (G K ) D dr {filtered K-vector spaces} D HT {graded K-vector spaces} commutes. Is that quite right? I have to show that Fil i D dr (V )/ Fil i+1 D dr (V ) = Gr i D HT (V ). Hmmm...this isn t quite obvious. Before moving on, let me make a remark. Remark. The Fontaine formalism implies that dim K D dr (V ) is finite-dimensional. Further investigation into the definition of Fil D dr (V ) (e.g. the fact that tensor products have only finitely many terms) show that the filtration Fil D dr (V ) is both exhaustive (stably equal to D dr (V ) for i small) and separated (vanishes for i large). Proposition. Suppose that V is de Rham. Then V is Hodge-Tate and Fil i D dr (V )/ Fil i+1 D dr (V ) = Gr i D HT (V ). In particular, the Hodge-Tate weights of V are the indices i such that Fil i D dr (V ) Fil i+1 D dr (V ). Proof. We re going to use the Fontaine formlism. The short exact sequence above implies that Gr D dr (V ) = i Z Fil i D dr (V )/ Fil i+1 D dr (V ) D HT (V ). Since the filtration is exhaustive and separated we get dim K D dr (V ) = dim K Gr D dr (V ) dim K D HT (V ). Now use that V is de Rham. Then the left hand side is dim Qp V and thus dim Qp V dim K D HT (V ). But the other inequality is Fontaine s formalism. Thus V is Hodge-Tate. Moreover, we get an equality of the maps in the first line and so the moreover follows. The converse of the proposition is more subtle. One way to understand this is to use the following easy to remember criterion. Proposition 3. Suppose that K /K is a complete discretely valued extension inside C. Then V is de Rham if and only if V GK is de Rham. To be clear, since maybe K is some infinite extension, such as the maximal unramified extension of K, this means that dim K (V Qp B dr ) G K = dim Qp V dim K (V Qp B dr ) G K = dim Qp V It s really a wonderful result. One way to prove it is to study the canonical map (1) D dr,k (V ) K K D dr,k (V ) where D dr, (V ) is computed with respect to G. I m not going to prove this, it is [1, Proposition 6.3.8] but let me say two things: (a) First, it should be stressed that this is a feature (albeit a bad one) of B dr containing a copy of K and a continuous copy of discretely valued extension K K K. 2

(b) That last bit is really the key (so that you can apply what Brinon and Conrad called unramified descent ) (c) The canonical map (1) is an isomorphism of all V, not just de Rham V. Here is a corollary whose proof is so simple it helps remember the previous proposition. Note: we will see that the corollary is false if de Rham is replaced by crystalline. Corollary 4. Suppose that η : G K Q p is a character. Then η is de Rham if and only if η is Hodge-Tate. Proof. We already showed that de Rham implies Hodge-Tate. We now have to show that Hodge-Tate implies de Rham. Since η is Hodge-Tate there exists some i such that D HT (η) = (C p (i) η) G K is onedimensional. Replace η by χ i cyclη and we may assume that i = 0. Thus the Hodge-Tate assumption becomes C(η) G K = K. By Tate s theorem, η is finitely ramified. Thus there exists a complete disceretely valued extension K such that η GK = 1 (e.g. make a finite extension K /K to make η unramified then pass to K nr ). But this implies that η GK is de Rham (since the trivial character is obviously de Rham). We finish by the proposition then. One can be explicit in the previous example I think. But I don t have the energy right now to explain this; I ll do an explicit example of the Tate curve instead. We ve seen that de Rham implies Hodge-Tate and that the converse holds for dimension one objects. It does not hold in general. Proposition 5. There exists representations which are de Rham but not Hodge-Tate. More specifically any non-split extension 0 Q p (r) V Q p 0 with r < 0 is Hodge-Tate but not de Rham. Remark. More can be said. A computation of Bloch and Kato (I m not sure if this is historically first) says that: If r < 0 then every class in H 1 (G K, Q p (r)) is Hodge-Tate but not de Rham If r 1 then every class in H 1 (G K, Q p (r)) is de Rham If r = 0 then there exists a unique line of classes in H 1 (G K, Q p ) which are de Rham. Actually, it isn t that hard to prove the middle one. See [1, Example 6.3.6]. What I d like to do is be super explicit about it the case where r = 1 and the class is the p-adic representation attached to a curve with split multiplicative reduction. I think Kummer theory can tell you the same argument for every class. 0.3. The Tate curve example. Let E/K be a curve with split multiplicative reduction and use Tate s p-adic uniformization E(K) K /q Z (some q K and q < 1) to describe a p-adic representation T p (E) = ε, ϑ where ε = (1, ζ p, ζ p 2,... ) is a compatible choice of roots of unity and ϑ = (q, q 1/p, q 1/p2,... ) is a compatible choice of p-power roots of q. In a matrix we have that an element g G K acts by a matrix ( ) χ(g) c(g) g 0 1 where c Z p is defined so that g(q 1/pn )/q 1/pn c(g) mod pn = ζp. It s a one-cocycle in n H1 (G K, Q p (1)) (or Z p (1), whatever). If you like, you can say that g(ϑ) = c(g)ε + ϑ as elements of T p (E). This will be important in a moment. I write V p (E) = T p (E) Zp Q p. Notice that ε and ϑ are elements of our perfect local domain R = lim O C. And thus they have Teichmüller lifts [ε] and [ϑ] to W (R). Furthermore, note that when viewed as elements of R we have the equation g(ϑ) = ε c(g) ϑ. Anyways, the ubiquitous element t B + dr was defined as log([ε]) = n+1 ([ε] 1)n ( 1) n n 1 3 B + dr

This was ok, remember, because we checked ahead of time that [ε] 1 ker(θ). Oh yeah, let me remind you that there was this surjective θ map θ : W (R) O C θ([x]) x (0) In particular, the element [ϑ] q B + dr also lands inside the kernel of θ. Let s define the element u q = log([ϑ]). What does this even mean? Here is what I mean: n+1 ([ϑ]/q 1)n log([ϑ]/q) = ( 1) n n=1 converges in the ring B dr because [ϑ] q ker(θ) W (R) and thus [ϑ]/q 1 ker(θ) W (R)[1/p]. And now what we mean by u q is the element Lemma 6. g(u q ) = u q + c(g)tfor all g G K. u q = log(q) + log([ϑ]/q) B dr. Proof. First, g(log q) = log q so that term stays where it is. For the other one we have g(log([ϑ]/q)) = log([g(ϑ)]/q) = log([ε c(g) ϑ]/q) = c(g)t + log([ϑ]/q). Of course, there is the qualification that one has to use carefully the topology of convergence on B dr to justify the formal manipulation (but this was needed even to deal with t...). Theorem 7. V p (E) is de Rham. Proof. We need to show that (V p (E) B dr ) G K is two-dimensional. Consider first the element x = ε t 1 V p (E) B dr. We easily see that g(x) = x for all g G K. The other fixed vector is slightly trickier. Here is what we do. We consider the element u q /t = a q and we compute the Galois action. We get Thus, This is the other fixed vector. g(u q /t) = χ(g) 1 t 1 (u q + c(g)t) = χ(g) 1 (u q /t + c(g)). g(ϑ 1 ε u q /t) = c(g)ε + ϑ 1 χ(g)ε χ(g) 1 (u q /t + c(g)) = ϑ 1 ε u q /t. You might ask about what this u q /t means. I m pretty sure that u q tb + dr and so a q is an element in B + dr. It s image α q = θ(a q ) under θ is an element of C p. And what u q /t is is a lift of the Hodge-Tate period α q (V p (E) C) G K to B dr. 0.4. Closing thoughts. Recall that at the beginning of the seminar we decided that it would be important to understand a p-adic analog of the Néron-Ogg-Shafarevich criterion relating good reduction of abelian varieties to the unramified-ness of local Galois representations away from the residue characteristic. Either the insensitivity of being de Rham to finitely ramified extensions, or the formal calculation in cohomology, or this Tate curve example, show that being de Rham is too coarse of a property to be a good substitute. Compare with [1, p. 81]. The final step in the seminar is to remedy this by the introduction of the crystalline and semi-stable rings. Roughly speaking, a defect of B dr is that is the natural Frobenius operator on W (R) lifting the p-power map on R does not preserve ker(θ). For example ϕ([ε] 1) = [ε p ] 1 = ζ p 1 0. And so we have to introduce a smaller ring B crys B dr which will have this property. In the process the structure gets tightened quite a bit. Unfortunately, B crys misses the Tate curve examples (as it should: 4

those correspond to bad reduction!). To get the ring B st, one that will detect Tate curves, we just add in all the Tate periods I just computed (it turns out you need just one). References [1] O. Brinon and B. Conrad. CMI summer school notes on p-adic hodge theory, 2009. Lectures at CMI Summer School 2009, also available at B Conrad s website. [2] J. T. Tate. p-divisible groups. In Proc. Conf. Local Fields (Driebergen, 1966), pages 158 183. Springer, Berlin, 1967. 5