Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

[EDIT]: After getting a very nice answer by Kevin Buzzard I realize that my question was a little bit too vague and I try to restate it more precisely.

As the title says, I would like to understand an isomorphism of Hida from a more geometric perspective than what I normally read. What bothers me is that there are two construction of the universal (ordinary) Hida-Hecke algebra and they turn out to give isomorphic objects: fix a prime $p\geq 5$, and a tame level $N$ prime to $p$.

  1. Take the projective limit over the level $r$ of the Hecke algebra acting on $S_k(\Gamma_1(Np^r),\mathbb{Z}_p)$ where $k$ is any weight. By applying the usual idempotent, one gets the Hida-Hecke ordinary algebra $h_k^0(Np^\infty;\mathbb{Z}_p)$, where I adopt notations as in Hida's paper in Inventiones, 1986, "Galois representations into $\mathrm{GL}(2,\mathbb{Z}_p[[X]])$...".
  2. Consider now the injective limit over the weight of the spaces of cusp forms $S_k(Np;\mathbb{Z}_p)$. By taking a suitable completion of this injective limit, one sees that the projective limit (over the weight, now) of Hecke algebras acts on the above completion. Applying again the idempotent, we get the Hida-Hecke ordinary algebra $h^0(N,\mathbb{Z}_p)$.

Theorem 1.1 in the quoted paper by Hida shows that these two algebras are isomorphic (in the most compatible way one can dream of, in particular inducing the same Hecke action on spaces of cusp forms) but his proof is entirely algebraic.

My question is: is there a reasonable way to prove this isomorphism geometrically?As Kevin Buzzard suggested, several papers of Katz (and successive work by Coleman-Mazur, Buzzard himself et al.) discuss geometric interpretation of $p$-adic modular forms and $p$-adic families of modular forms. Still, I do not understand how Hida's isomorphism comparing the Hecke algebra as acting on the projective limit over the level (so ''at the top of the modular tower'') or on the inductive limit over the weight (so, ''over the first curve $X_1(Np)$'') can be given a geometric interpretation.

share|improve this question
    
Dear Filippo, Here is a more geometric formulation of what is happening: fix a tame level $N$. Then over $X_1(N)$ there is a line bundle $\omega$; weight $k$ modular forms are global sections of $\omega^{\otimes k}$. Think of this in the old-fashioned way as a sheaf with structure group $\mathbb G_m$. The fact that we can raise it to integral powers reflets the fact that the character group of $\mathbb G_m$ is precisely $\mathbb Z$. Now if we restrict to $X_1(N)_{ord}$ (the ordinary locus, an affinoid open subspace of $X_1(N)_{/ \mathbb Q_p}$, Katz shows that we may write ... –  Emerton Apr 21 '12 at 12:11
    
... $\omega = \mathcal O \otimes \mathcal L$, where $\mathcal L$ is an etale local system of rank one over $\mathbb Z_p$. (More precisely, the fibre of $\mathcal L$ over a point $x$ --- which corresponds to an ordinary elliptic curve $E_x$ --- is the etale part of the $p$-divisible group of $E_x$, which is rank one because we are over the ordinary locus. [And maybe I should replace $\mathcal L$ by $\mathcal L^{-1}$ here; I didn't think this through right now.]) This means that we have reduced the structure group from $\mathbb G_m$ to $\mathbb Z_p^{\times}$ --- which is why we can define ... –  Emerton Apr 21 '12 at 12:14
    
... $p$-adic modular forms of arbitrary weight $\kappa \in Hom(\mathbb Z_p^{\times}, \overline{\mathbb Q}_p^{\times})$: because we are allowed to raise $\mathcal L$ (and hence $\mathbb \omega$ restricted to the ordinary locus) to powers that are arbitrary characters of the structure group $\mathbb Z_p^{\times}$. Now if we pass from $X_1(N)_{ord}$ to $X_1(Np^r)_{ord}$, we reduce the structure group even more, from $\mathbb Z_p^{\times}$ to $1 + p^r \mathbb Z_p$. In the limit over all levels $p^r$, we thus make the structure group trivial, and so $\omega$ as become the trivial bundle. –  Emerton Apr 21 '12 at 12:18
    
The conclusion is that if we add all $p$-power levels, the notion of weight has disappeared, because $\omega$ (and hence all its powers) have gone away. This is the source of what you call Hida's isomorphism. To see this discussed in the literature you should look at Katz's LNM 350 article and the subsequent articles he wrote; a good summary is provided by his paper "$p$-adic $L$-functions via moduli" in the Arcata proceedings. You can also look at Gross's "Tameness criterion" paper, where he works mod $p$, and shows that just by going to $X_1(Np)$ the distinction between weight and ... –  Emerton Apr 21 '12 at 12:20
    
... disappears for mod $p$ forms. (This is because the structure group $1 + p\mathbb Z_p^{\times}$ is already trivial once we are looking at $\omega$ on the mod $p$ ordinary locus.) None of these sources use the language of "reduction of the structure group" that I am using here, although I find it a convenient point-of-view. Instead, you will find discussion of the Hasse invariant, of a $p-1$st root of the Hasse invariant (this is in Gross), and of lifts of these mod powers of $p$ (this is in Katz). These are just the particular objects which, in my phrasing, achieve the reduction of ... –  Emerton Apr 21 '12 at 12:24
show 2 more comments

2 Answers 2

up vote 3 down vote accepted

I am not sure what your criteria would be for a proof to be given a geometric interpretation, but the reason why weights "disappear" when we take the inverse limit on the level stems from the contraction property of Hecke operators (at $p$), or informally from the fact that Hecke operators at $p$ diminish the level.

As you know, the proof of the isomorphism between the two different Hecke algebras requires the definition of a map between Hecke algebras acting on forms of weight 2 and forms of weight $k$. Because theses two algebras are sub-algebras of endomorphisms generated by the same abstract elements but acting on different objects, this amounts to constructing a map between the cohomology of (one of the level of) the modular tower with coefficients in the constant sheaf and (one of the level of) the modular tower with coefficients in a sheaf of weight $k$ (or the same thing with the modular tower replaced by the Igusa tower, as in Kevin's answer). This last map is really no big deal: if memory serves, on the sheaves it is just projection on the last component. The remarkable fact is that the map on cohomology then is surjective with a finite kernel (and is an isomorphism in the ordinary case); the proof of this assertion being exactly the contraction property. Note that the proof necessarily requires the choice of a level at some point; how else would you even state the result?

Note for instance that for a tower of more general Shimura varieties, it is not at all obvious how the contraction property will play out: group-theoretic properties of $\operatorname{GL}_{2}(\mathbb Q_{p})$ really do play an important role in the proof. See the reference below though, for an answer to these questions.

So in the end, the isomorphism between the two Hecke algebras seems to me to come from the interplay between the cohomology of modular varieties and group-theoretic properties of the Hecke algebra. A very general formulation of this fact can be found in D.Mauger Algèbres de Hecke quasi-ordinaires universelles. Ann. Sci. École Norm. Sup. (4) 37 (2004) (section 2.4 to be precise)

share|improve this answer
    
Dear Olivier, this sounds good – and sad. I guess you are saying that you expect the isomorphism only to show up at level of (Betti?) realizations of our motives and not to be motivic itself. May be that is what I would have called "geometric". Or, in more down-to-earth terms that I myself can understand, I would love the existence of a sheaf (?) $\omega_\infty(k)$ on the pro-scheme $\varprojlim_r X_1(Np^r)$ that for ''some reason'' is isomorphic to a ind-sheaf $\varinjlim_k \omega^{\otimes k}$ over $X_1(N)$. Beside the nonsense, the ''some reason'' might spoil all the fun... Filippo –  Filippo Alberto Edoardo Apr 11 '12 at 10:28
    
Dear Filippo, I am not sure that this is what I am saying honestly, because I am not sure I understand what you are looking for (through no fault of yours). My (very shallow) understanding of the situation is that a motivic isomorphism would be an isomorphism between the towers (for the level) of Jacobians of modular curves and the "tower with respect to $k$" (that is to say the simplicial scheme) of canonical desingularizations of the $k$-fold fibre product of the universal generalized elliptic curve over the compact modular curve. –  Olivier Apr 11 '12 at 11:55
    
It may be easy to cook up a proof by combining the classical proof I outlined and Scholl's result on motives for modular forms. In fact, I think Scholl does everything needed at the end of his Invent. Math. 100 article (well, strictly speaking you would need to work with finite coefficients instead of $\mathbb Q$ coefficients but this does not look too serious). –  Olivier Apr 11 '12 at 12:02
    
Dear Olivier, this seems very nice and very close to what I was looking for. I'll look at School's paper, merci! Filippo –  Filippo Alberto Edoardo Apr 11 '12 at 12:20
add comment

As you've spotted, there are two ways to do $p$-adic modular forms. The point is that at some point you take a limit of classical modular forms, and there are a whole host of modular forms which may have e.g. very big weight but whose $q$-expansions are $p$-adically very close 1. Multiplication by such a gadget can completely change the "classical data" (weight, level) attached to a modular form but might not change the $p$-adic properties of its $q$-expansion much at all. I'm being a little informal, but the upshot is that as long as you let something tend to infinity you might well have enough space to get a decent theory. Oh by the way you speak of Hida theory but there is no ordinarity assumption built in at this point -- you may as well just think of the entire huge space of $p$-adic modular forms when formulating your question.

A precise way of formulating an answer to your question can be found in work of Katz from the 1970s (several papers). I have paper copies of all the relevant papers in my office but I am not in my office, and am too lazy to google for the exact references, so here is a cheat: find Fernando Gouvea's thesis, which was published in Springer Lecture Notes in Mathematics, where he gives a summary of Katz' approach, and then follow up the references there. You'll find that in some sense you're "nearly" right: but you can't work with $X_1(Np^r)$ over $\mathbf{Q}_p$ or $\mathbf{Z}_p$ immediately -- the trick is to work with a model of $X_1(Np^r)$ over $\mathbf{Z}/p^n\mathbf{Z}$, look at differentials or functions on such spaces, and then take a direct limit over $r$ (corresponding to a projective limit of the schemes but with base $\mathbf{Z}/p^n\mathbf{Z}$), and then take a projective limit over $n$. The two limits don't commute so you have to be careful, but I think that Katz' approach is the approach that you're looking for. Note that these limits do not give you just Hida's ordinary modular forms, but all $p$-adic modular forms; you can then cut out the ordinary subspace using the $U_p$ operator.

But as you already know, there is more than one approach to the theory, and here is another one that I find quite geometric; the catch is that it relies on the theory of rigid spaces, i.e. (in this setting) a $p$-adic version of Riemann surfaces. If you're happy with this theory then let me briefly explain how to use it to give another geometric interpretation. Take the modular curve $X_1(Np)$ and consider it as a $p$-adic Riemann surface. Now remove all the points corresponding to elliptic curves with good supersingular reduction. The curve falls into two pieces -- take the piece containing infinity. This is an affinoid rigid space and it has a sheaf $\omega$ on it (the pushforward of the differentials on the universal generalised elliptic curve). We could call this space the ordinary locus. Now because we are in a $p$-adic setting we can make sense of $\omega^\kappa$ for $\kappa$ in a much more general space than the integers -- basically $\kappa$ can be any continuous group homomorphism $\mathbf{Z}_p^\times\to\mathbf{Z}_p^\times$ (and even more, in fact). It's a bit like the fact that for $p>2$ you can raise $(1+p)$ to the power $k$ for $k$ in $\mathbf{Z}_p$ using the binomial theorem, but you can't raise 2 to the power $k$; the point is that $1+p$ is sufficiently close to 1. Similarly the ordinary locus is sufficiently close to the cusp for things to work. You could define a $p$-adic modular form of weight $\kappa$ to be a section of $\omega^\kappa$ and you could define a family to be a continuous, or analytic, family of such things, as $\kappa$ varies.

share|improve this answer
    
@Kevin. Thanks, I will try to look at Katz' paper(s) – I just read his Antwerp one and thought that the gadget I was looking for did not show up. By the way, I agree that I needed no ordinarity assumption, but thought that if something geometric happens over the eigencurve it should also be clear on the $0$-slope subspace, that's why I restricted to that case. But I am very happy with that theory, too. Anyhow, my point was to try to understand "informally" why we do not expect a weight anymore: your sheaves $\omega^\kappa$ over the ordinary locus do depend on $\kappa$... –  Filippo Alberto Edoardo Apr 9 '12 at 9:30
    
I don't really know what you mean by "we do not expect a weight anymore". These Katz space of generalised $p$-adic modular functions are basically a natural candidate for a space for which the topology is governed by the $p$-adic properties of the $q$-expansion, and such that we can see classical forms of all weights. So you can take a form of weight $2$ and a form of weight $4$ and add them up, and the resulting form will not have a weight. But many of the forms in the space have weights. Using the "let the level go to infinity" model, the forms with weights are exactly the forms which are... –  Kevin Buzzard Apr 11 '12 at 11:57
    
...eigenvalues for the diamond operators (which induce an action of $\mathbf{Z}_p^\times$ on the space). Is this a sufficient informal explanation of why some forms have weights and others don't or are you looking for something else? –  Kevin Buzzard Apr 11 '12 at 11:58
    
I think this explanation of ''do not having a weight anymore'' is nice and convincing - and pretty much what I was looking for: thanks! In the meanwhile, I realized (thanks to your answer) that my question was a little bit too vague, and I edited it. I apologize for the first misleading formulation - although it turned out to be fruitful in understanding this weight issue ;-). Filippo –  Filippo Alberto Edoardo Apr 11 '12 at 12:23
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.