Higher direct images of canonical extensions

Posted on January 4, 2015 by Persiflage

I like Kai-Wen’s talks; he gives lots of examples, writes big with big chalk, and clearly explains the key points of the argument. I’m not sure I would classify his thesis as light reading material, but if he produced a video series explaining all the details in lecture format, I would buy the DVD. Speaking of different ideas for disseminating mathematics, I have some thoughts on that, but they will have to wait for another time. For now, I just wanted to make the smallest remark concerning Kai-Wen’s lecture at the Harris conference.

As all my readers surely know (this is code for I am not going to explain why), a key ingredient in the Harris-Lan-Taylor-Thorne argument is the fact the the higher direct images of the of subcanonical automorphic vector bundles under the projection from the toroidal compactification to the minimal compactification of quite general classes of Shimura varieties vanish. In contrast, this does not hold for the higher direct images of the canonical extensions, and when this was first being discussed, it was not entirely clear (at least to me) what was going on. But Kai-Wen’s talk actually does make the situation very clear! That is what I want to talk about.

Let \(X\) be the open Shimura variety, let \(Y\) be a minimal compactification, and let \(Z\) be a toroidal compactification. To avoid silliness, assume that \(Y \setminus X\) has codimension at least two. Let \(W\) be an automorphic vector bundle on \(X\), and let \({W^{\mathrm{can}}}\) and \({W^{\mathrm{sub}}}\) denote the canonical and subcanonical extensions of \(W\) to \(Z\). There’s a short exact sequence

\(0 \rightarrow {W^{\mathrm{sub}}} \rightarrow {W^{\mathrm{can}}} \rightarrow Q \rightarrow 0.\)

Take the pushforward of this to \(Y\). We know that the higher direct images of the first sheaf vanish, and so we obtain an exact sequence

\(0 \rightarrow \pi_*{W^{\mathrm{sub}}} \rightarrow \pi_*{W^{\mathrm{can}}} \rightarrow \pi_*Q \rightarrow 0.\)

The last sheaf is supported on \(Y \setminus Z\), which has fairly small dimension, so its cohomology groups vanish in high degree by Grothendieck. Now let us assume that the higher direct images also vanish for \({W^{\mathrm{can}}}\). It follows that the Leray spectral sequence degenerates (for both \({W^{\mathrm{sub}}}\) and \({W^{\mathrm{can}}}\)), and so we obtain isomorphisms

\(H^*(Z,{W^{\mathrm{sub}}}) = H^*(Z,{W^{\mathrm{can}}})\)

in sufficiently high degree. Now the canonical bundle on \(Z\) is also an automorphic vector bundle, and so Serre duality relates the cohomology of \({W^{\mathrm{sub}}}\) to the cohomology of \({V^{\mathrm{can}}}\) for another automorphic vector bundle \(V\), and relates the cohomology of \({W^{\mathrm{can}}}\) to \({V^{\mathrm{sub}}}\). For example, for modular curves, the Serre dual of \(\omega^k\) is \(\omega^{2-k}(\infty),\) because the canonical sheaf of the modular curve is \(\Omega^1 \simeq \omega^2(\infty)\). Hence (using the assumption on codimensions made above so the numerology works out) we end up with the isomorphism

\(H^0(Z,{V^{\mathrm{sub}}}) = H^0(Z,{V^{\mathrm{can}}}).\)

But this formula says that all cusp forms modular forms of weight \(V\) are cuspidal! So this gives an easy proof of:

Lemma If there exists at least one form of weight \(V\) which is not cuspidal, then at least one of \({W^{\mathrm{sub}}}\) or \({W^{\mathrm{can}}}\) has non-trivial higher direct images under \(\pi\).

Of course, we know from HLTT that it will be the second (because the higher direct images of the first vanish), but we didn’t prove that. Now I just chatted with Kai-Wen, who did one better than this lemma. First of all, remember that there is an automorphic line bundle \(\omega\) on \(X\) (corresponding to “parallel weight”) which is ample, and the corresponding canonical extension to \(Z\) descends to an ample on \(Y\), which we also call \(\omega\). What’s nice about this is that, using the projection formula, one can replace the question about the vanishing of the higher direct images of \(W\) by the vanishing of \(W\) under twists by powers of this bundle. But that means one can translate the problem of asking whether there exists a non-cusp form in the dual weight \(V\) to whether there exists a non-cusp form in weight \(V \otimes \omega^n\) for some arbitrarily large \(n.\) Now as before, we have an exact sequence:

$latex 0 \rightarrow \pi_*{V^{\mathrm{sub}}} \otimes \omega^n \rightarrow \pi_*{V^{\mathrm{can}}}
\otimes \omega^n \rightarrow \pi_*R \otimes \omega^n \rightarrow 0.$

twisted by some arbitrarily high power of \(\omega\), where we have used the vanishing of \(R^1 \pi_* {V^{\mathrm{sub}}}\) and the projection formula. Here \(R\) is just \({V^{\mathrm{can}}}/{V^{\mathrm{sub}}}.\) On the other hand, because \(\omega\) is ample on \(Y\), we know that

  1. \(H^1(Y,\pi_*{V^{\mathrm{sub}}} \otimes \omega^n)\) vanishes for sufficiently large \(n,\)
  2. \(\pi_* R \otimes \omega^n\) is generated by global sections for sufficiently large \(n,\) and so, for such \(n,\) we have \(H^0(Y,\pi_* R \otimes \omega^n) \ne 0\) as long as \(\pi_* R \ne 0\).

So if one shows that \(\pi_* R\) is non-zero then one is done. Certainly \(R\) is non-zero, but analyzing \(\pi_* R\) is a bit more subtle (I jumped the gun a little on the first version of this post, but Kai-Wen told me I needed to be a little more careful). On the other hand, there are many classical examples where one can explicitly construct non-cuspidal forms. For example, one can take \(X = \mathcal{A}_g\) with \(g \ge 2\) to be the Siegel moduli space, and take \(W\) to be the line bundle \(\omega^k.\) Then Siegel himself constructed the so-called Siegel Eisenstein series for high enough \(k\). Kai-Wen also tells me the non-vanishing of \(\pi_* R\) can be proved more generally for \(X = \mathcal{A}_g,\) and so one has:

Lemma [Kai-Wen] Let \(g \ge 2,\) let \(X =\mathcal{A}_g,\) and let \(W\) be an automorphic bundle. Then at least one of higher direct images \(R^i \pi_* {W^{\mathrm{can}}}\) with \(i > 0\) must be non-zero.

In fact, Kai-Wen also tells me he had a proof of (a more general version of) this last result even before HLTT knew about the vanishing of \(R^i \pi_*{V^{\mathrm{sub}}},\) but this argument gives a completely transparent proof of why they can’t both vanish.

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Derived Langlands

Posted on December 22, 2014 by Persiflage

Although it has been in the air for some time, it seems as though ideas from derived algebraic geometry have begun to inform developments in the Langlands program. (A necessary qualifier: I am talking about reciprocity in the classical arithmetic Langlands program here.)

I want to describe a very simple instance of this which came up in Akshay’s MSRI talk which I linked to last time.

Start by fixing a global residual (GL-)odd Galois representation:

\(\displaystyle{\overline{\rho}: G_{\mathbf{Q}} \rightarrow \mathrm{GL}_N(\mathbf{F}_p)}\)

Let us suppose that \(\overline{\rho}\) is surjective. Associated to this representation is a fixed determinant unrestricted global deformation ring \(R\), and a fixed determinant unrestricted local deformation ring which I will call \(S.\) (Apologies for the notation, but wordpress is not great with lots of subscripts.) I assume that the reader can make the appropriate adjustments to these definitions if the local representation is not irreducible by adding framings. One knows, at least if \(p > 2N+1\), that the map

\(\mathrm{Spec}(R) \rightarrow \mathrm{Spec}(S)\)

is finite. Let us now suppose that \(\overline{\rho}\) restricted to \(G_{\mathbf{Q}_p}\) admits a crystalline lift of some regular weight; associated to this weight is a Kisin local deformation ring which I shall call \(T\). If you like, you can even imagine that we are working in small weight so that \(T\) has nice properties; perhaps it is even smooth. Barry Mazur has made a conjecture for what the (relative) dimension of \(R\) should be over \(\mathbf{Z}_p.\) Namely, it should be given by the Euler characteristic of the adjoint representation, which is equal (see Def.2.1 and the subsequent comment) to

\(\dim B – \ell_0,\)

where \(B\) is a Borel of \(\mathrm{SL}_N(\mathbf{R})\), and \(\ell_0\) is the difference between the rank of \(\mathrm{SL}_N(\mathbf{R})\) and \(\mathrm{SO}_N(\mathbf{R})\). Of course, these quantities can easily be calculated explicity in this (or any) case; for \(\mathrm{SL}(N)/\mathbf{Q},\) \(\ell_0\) is the integer part of \((N-1)/2.\) On the other hand, we may also compute the (relative) dimensions of the rings \(S\) and \(T\), and we find that

\(\mathrm{dim}(S/\mathbf{Z}_p) = N^2 – 1, \quad \mathrm{dim}(T/\mathbf{Z}_p) = N(N-1)/2.\)

(The notation here means the relative dimension.) The Fontaine-Mazur sanity check is to see that, on the associated rigid analytic spaces, the assumption that \(R\) and \(T\) meet transversally inside \(S\) should imply that their intersection only has finitely many points. Indeed, we can compute that the expected dimension of the intersection is exactly:

\( N(N-1)/2 + \dim(B) – \ell_0 – (N^2 – 1) = – \ell_0.\)

When \(N = 2\), we have \(\ell_0 = 0\), and everything is as expected. However, as soon as \(N > 2\), we have \(\ell_0 > 0\), and so the expected dimension is negative. This says that regular algebraic automorphic forms for such \(N\) are much rarer beasts than their counterparts for \(N = 2\), where modular forms are abundant. For example, it is not known if there exists a regular algebraic cusp form for \(\mathrm{GL}(N)/\mathbf{Q}\) giving rise to a \(\overline{\rho}\) as above for any \(N \ge 5\). (Note that forms from smaller groups coming via functorial lifts will fail to give rise to representations with such large image.) Now all of this is a philosophy that has been known and exploited for some time. But suppose we actually try to interpret this heuristic a little more literally. For a start, we do expect that forms of characteristic zero do exist. This means that, in general, there are unlikely intersections of \(\mathrm{Spf}(R)\) and \(\mathrm{Spf}(T)\) inside \(\mathrm{Spf}(S).\) That is, \(R\) and \(T\) will not, in general, be transverse! This is exactly a context in where, to understand the intersection, it makes sense to introduce the derived world (see, for example, the introduction to DAG-V).

In the classical picture, to recover the usual minimal deformation ring, one considers the intersection \(X:= R \otimes_S T.\) However, science now tells us it is more natural to consider the derived tensor product

\(Y = R \otimes^{\mathbf{L}}_{S} T.\)

If \(\ell_0 = 0\), then the cohomology of \(Y\) should exactly recover \(X\) in degree zero and be zero otherwise — that is, the classical context should be related to a completely transverse intersection, and we are still in the usual word of schemes (or even complete local Noetherian rings). However, this will (essentially) never happen when \(\ell_0 > 0.\) Classically, the ring \(X\) may be identified with the ring of endomorphisms generated by Hecke operators on a single extremal degree of cohomology. More generally, the cohomology of \(Y\) should be identified with the ring of Hecke operators acting on the cohomology now in degrees \(q_0, \ldots q_0 + \ell_0\), where the notation is as in Borel-Wallach and is fixed for all time. In particular, the only context in which one should expect the intersection to be transverse (beyond \(\ell_0 = 0\)) is the case when \(\ell_0 = 1\) and \(X\) is a finite ring (which can happen). Indeed, in such contexts, the cohomology over \(\mathbf{Z}_p\) also occurs exactly in one degree. It might be worth noting here that the ring \(S\) is not in general regular, and so \(Y\), a priori, is not even bounded.

On the other hand, science also tells us that the complex \(Y\) has more information than its cohomology; and so one should really think of \(Y\) as the correct object. Unfortunately, I don’t have anything clever to say about derived arguments, but let me use the Fontaine–Mazur heuristic to extract some tiny amount of juice. Since I am not so DAGgy, I will only use algebra that goes back 30 years or more.

Instead of looking at the intersection of \(R\) and \(T\) inside the formal spectrum of \(S\), let us look at their intersection over \(\mathbf{Z}_p\). In this case, all the dimensions have been shifted by one, so, when \(\ell_0 = 0\), their intersection should be infinite (that is, the length over \(\mathbf{Z}_p\)). This is obviously the case, because \(X\) (which by assumption exists) is flat over \(\mathbf{Z}_p\) and so automatically infinite. Well, obvious modulo Serre’s conjecture and Fontaine-Mazur for \(\mathrm{GL}(2)/\mathbf{Q}\), at least. On the other hand, when \(\ell_0 > 1\), then the dimensions don’t add up and the intersection multiplicity should be zero. Thus, assuming the dimensions of the rings are correct, we should have:

  1. If \(\ell_0 = 1\), then the intersection multiplicity is finite and non-zero;
  2. If \(\ell_0 > 1\), then the intersection multiplicity is zero.

In the latter case, we are invoking the conjecture of Serre (proved by Roberts and Gillet-Soulé) that the intersection multiplicity is zero when the dimensions are too small. (Heuristically, one can “move around” classes whose codimension is sufficiently large so they don’t intersect at all, although one cannot literally do this in a local ring!) Actually, even this is a lie, because \(S\) is not necessarily regular, as mentioned above, but pretend for this paragraph that it is. Serre’s fomula tells us that the intersection multiplicity is given by the Euler characteristic of the complex. Let me now suppose that \(Y\) has no characteristic zero cohomology. For example, we could be working in a weight corresponding to a (strongly) acyclic local system (as long as \(\ell_0 \ne 0\)). What we want to compute is the alternating product of the cohomology groups, which should be equal to the alternating product of the integral cohomology groups (everything localized at the appropriate maximal ideal) of the corresponding congruence subgroup of \(\mathrm{GL}_N(\mathbf{Z}).\) Yet this product (without localizing at a maximal ideal) is basically equal to the Reidemeister torsion, which is equal to the analytic torsion, which always vanishes when \(\ell_0 > 1\)! Under our finiteness conditions, when \(\ell_0 = 1\), all the cohomology occurs in just a single degree, and so the multiplicity is just the length of \(X.\) But when \(\ell_0 > 1\), this gives (the expected) refinement of the vanishing of analytic torsion after localizing at a non-Eisenstein maximal ideal, which was one of the questions implicitly raised in the last blog post. To be more precise about our assumptions and conclusions, we have:

Proposition: Let \(\mathfrak{m}\) be a non-Eisenstein maximal ideal of the cohomology of a congruence subgroup of \(\mathrm{SL}_N(\mathcal{O}_F)\) for a CM number field \(F\), and let \(\overline{\rho}\) be the corresponding Galois representation. Assume that:

  1. The required assumptions in C-G hold: (vanishing outside degrees \(q_0, \ldots, q_0 + \ell_0\) after localization at \(\mathfrak{m}\), the representation \(\overline{\rho}\) has big image, local-global compatibility, etc.);
  2. The cohomology localized at \(\mathfrak{m}\) vanishes in characteristic zero.

Then the alternating product of the orders of the cohomology groups localized at \(\mathfrak{m}\) is non-zero if and only if \(\ell_0 = 1\).

One issue with proving this directly using the above argument is that we don’t actually know the dimension of \(R\) in general. So, instead of working with \(Y\), we work instead with the output of the Taylor-Wiles method as in C-G, namely:

\(\displaystyle{Y_{\infty} = R_{\infty} \otimes^{\mathbf{L}}_{S_{\infty}} S_{\infty}/\mathfrak{a}}\)

Here \(S_{\infty}\) is an Iwasawa algebra of diamond operators of dimension \(q + \ell_0\), the ideal \(\mathfrak{a}\) is the augmentation ideal with \(S_{\infty}/\mathfrak{a} = \mathbf{Z}_p,\) and \(R_{\infty}\) is a patched minimal deformation ring of dimension \(q.\) These are relative dimensions, so the transverse case over \(\mathbf{Z}_p\) corresponds exactly to the case when \(\ell_0 = 1\). We note here that the ring \(S_{\infty}\) is regular, so we are in the appropriate context of Serre’s multiplicity formula, and the result follows. (Exercise: we are only using a very special case of the vanishing claim for intersection multiplicities when one component is \(\mathbf{Z}_p\) inside \(S_{\infty};\) the vanishing should be easy to prove directly in this case.) This is good, because it gives a purely Galois theoretic proof (well, really only a heuristic because of all the conjectures one needs to assume) of a result (vanishing of analytic torsion) which is not at all obvious. (Well, not quite; the result is localized at a maximal ideal — which one can’t do analytically — but it only applies to non-Eisenstein maximal ideals.) At any rate, thinking through this example after Akshay’s talk has convinced me that this derived perspective is a very good one.

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Harris 60

Posted on December 15, 2014 by Persiflage

I’ve just returned from the excellent MSRI workshop which honored Michael Harris’ 60th birthday, and here is a brief summary of some of the gossip and mathematics I picked up when I was there.

First, let me take note of the nicely designed posters for the conference:

One thing that can be said about the original poster (on the left) is that it does not contain a single mistake. No word on whether a third poster is in the works.

Next up, celebrity fashion. Of course, my main source for celebrity gossip is Richard Taylor, fresh from hobnobbing on the red carpet at the recent breakthrough prize awards.


Richard Taylor and Kate Bekinsdale at the Breakthough Award ceremony

Apparently the hot news from RLT is that flip phones are now back in style. You won’t see Richard or Kate brandish a smartphone; both now proudly sport the retro flip phones as if it’s 2007. Of course, all this comes months after I finally got around to getting my first iPhone.

Finally, something not entirely silly, Akshay gave a very intriguing talk on integral structures in cohomology. It reminded me of a question that we discussed a long time ago near Washington Square Park in NYC. Recall that, for tempered automorphic forms contributing to the cohomology of \(\mathrm{GL}(n)/\mathbf{Q}\), a computation with \((\mathfrak{g},K)\) cohomology shows that each such form occurs in cohomology in degrees \(q_0, \ldots, q_0 + \ell_0\) and contributes (a multiple) of

\(\displaystyle{\binom{\ell_0}{k}}\)

dimensions in degree \(q_0 + k\). Is there any analogue of this for torsion classes or in characteristic p? Assume here that the residual representation also occurs only inside the relevant ranges of cohomology, which should be the case as long as the corresponding residual representation is not Eisenstein. If \(\ell_0 = 0\), there is nothing to say. If \(\ell_0 = 1\), then the result follows from an Euler characteristic argument; that is, the cohomology in each of the two non-zero degrees over \(\mathbf{F}_p\) will have the same dimension. If \(\ell_0 = 2,\) then Poincaré duality shows that the cohomology groups in the two outer degrees will have the same dimension (again we are working with non-Eisenstein classes and trivial coefficients, so from this point of view things look compact), and then an Euler characteristic argument shows that the space appears in some multiplicities of dimensions \((1,2,1),\) as in characteristic zero. Now suppose that \(\ell_0\) is arbitrary. I will assume we are in a multiplicity one situtation and that all the local deformation rings are smooth. The CG-method (under suitable hypotheses) produces a resolution \(P^{\bullet}\) of \(R_{\infty}\) consisting of finite free \(S_{\infty}\)-modules, where \(R_{\infty}\) is smooth of relative dimension \(q\) and \(S_{\infty}\) is free of relative dimension \(n:=q + \ell_0\). To recover the cohomology over a finite field, one takes the quotient of this resolution by the maximal ideal of \(S_{\infty}\) and then takes cohomology. In other words, what we are really computing is

\(\displaystyle{\mathrm{Ext}^*_{S_{\infty}}(R_{\infty},k)}.\)

This answer depends only on the rings involved and not on the resolution. For example, suppose that \(q =0\), which is the same as saying that there are no non-trivial local infinitesimal deformations. Then \(R_{\infty} = \mathbf{Z}_p\), and one can compute the cohomology of this ring using the Koszul complex, and one gets the expected dimensions. Note that if \(R = \mathbf{Z}_p\), then this recovers the automorphic computation, but this is already slightly interesting if \(R = \mathbf{F}_p\) or even \(\mathbf{Z}/p^k \mathbf{Z}.\) However, it seems a little optimistic to expect this pattern to hold in general. I spent some time trying to prove it using commutative algebra, but one problem is that it is not true in that generality. For example, suppose that \(\ell_0 = 3\), and that

$latex \displaystyle{R_{\infty} = \mathbf{Z}_p[[y_1,y_2,y_3]], \quad
S_{\infty} = \mathbf{Z}_p[[x_1,x_2,x_3,x_4,x_5,x_6]],}$

where the map from \(S_{\infty}\) sends the generators to the six possibile monomials of degree two. Then the appropriate dimensions of the ext groups are \(4,14,14,4\). (Thanks to Daniel Erman for this example.) Now this example actually can’t occur globally, because the same computation implies that one the Betti numbers over \(\mathbf{Q}_p\) are also given by these numbers, which violates the previously referenced computation with \((\mathfrak{g},K)\)-cohomology. However, one should easily be able to deform it very slightly to kill off any cohomology in characteristic zero, for example replacing \(y_i\) by \(y_i – \epsilon_i\) for small constants \(\epsilon_i\). Of course this doesn’t disprove anything, but it does strongly suggest that the dimensions over \(\mathbf{F}_p\) could be all over the place, subject to the Poincaré and Euler characteristic conditions. Akshay has also pointed out that the case \(\ell_0 = 3\) is interesting from a different but related perspective: the analytic torsion will vanish in this case, which implies that, at least morally (since we have localized at a maximal ideal), that the alternating product of the the orders of the cohomology groups over \(\mathbf{Z}_p\) should equal one. Is this a consequence of the TW-method? I just thought of this question right now and it may have an obvious answer which Akshay knows, I’ll ask him today and report back. A second obvious question is what happens if one looks only at \(\mathfrak{m}\)-torsion rather than \(\mathbf{F}_p\)-torsion; perhaps that is the more sensible generalization of the characteristic zero question anyway.

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The Modular Forms Database is Annoying

Posted on November 23, 2014 by Persiflage

It used to be the case, 10 or so years ago, that William Stein maintained a website with extensive tables of \(q\)-expansions of modular forms, computed using magma. However, as the edifice of civilization begins to crumble, this website no longer works, and it no longer seems possible to easily access q-expansions online. The only current online resource I can find appears to be the modular forms database. Unfortunately, it appears that the designers of that website have invested more time into aesthetics than usability. First, some minor grumbles. The choice of how to represent Dirichlet characters seems somewhat cumbersome, at best. This is somewhat excusable, because there isn’t any obvious elegant solution, although the best choice will surely be one which is easier for the viewer to use. It’s also a little unclear, certainly in comparison to William’s old site, what range of levels and weights have been computed. More seriously, after spending five minutes chasing down the right Dirichlet character for a particular space of modular forms I wanted to compute, I clicked on the appropriate link only for it to misdirect me to a page with the same level and a different character. Indeed, this character even had the wrong sign, and so all the web page did was to unhelpfully inform me that the space of forms was zero. I then started clicking randomly on other characters in the hope that they would send me the the web page with the character I actually wanted, but no such luck. Another very annoying “feature” of the website is the format in which the answers are returned: they are given in (compiled) LaTeX! This is a terrible design decision. Given that I (and possibly other users) actually want to do something with the answer (for example, compute the reduction modulo a prime using pari), a pretty formula is completely useless, because it is impossible to cut and paste it in any useful way. At one point I did find a way to ask for the answer to be given in text format, but looking at the page again now I can’t even remember what I did. Even that was not so useful, because it didn’t actually work on my particular example and instead and caused the website to crash. In fact, the only way I was ever able to extract the answer in some useful way was to tell Safari to give me the source code of the web page. I should note that there is a “feedback” button at the top right of the website. On both occasions when these problems occurred, I dutifully submitted my questions/requests with a detailed explanation of what I did, but neither time did I get any response, nor any indication that those messages were even read. In fact, the website does not seem to contain information any about who maintains it (beyond that it is funded by the NSF), so I can’t even email someone directly to complain. So here I am complaining, in the hope that one of my readers has more influence with the designers of this web site than I do.

Update: This post has already been successful — various people have already contacted me to say they are working to address my concerns. Of course, this is what I expected, since the people I know who do these computations are genuinely interested in their data being useful to others…

Update II: Instead of recommending how to deal with characters, let me just mention what I often end up wanting to compute, namely, a space of weight one modular forms of level \(\Gamma_1(N)\) and character \(\chi\) in characteristic \(p.\) In practice, this means that I compute in weight two, level \(\Gamma_1(Np)\), and character \(\chi \cdot \epsilon\), where \(\epsilon\) is the mod-p cyclotomic factor. This ultimately requires making a number of choices of embeddings \(\overline{\mathbf{Q}} \hookrightarrow \overline{\mathbf{Q}}_p\) and \(\overline{\mathbf{Q}} \hookrightarrow \mathbf{C}\) which have to be compatible in some way but ultimately are irrelevant. So even though (for example) if \(\chi\) is a quadratic character (say) then \(\chi \epsilon\) is easy to describe in words, it becomes a little bit of a mess to get the LMFDB to give me the information I need…

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Mysterious Formulae

Posted on November 15, 2014 by Persiflage

I’m not one of those mathematicians who is in love with abstraction for its own sake (not that there’s anything wrong with that). I can still be seduced by an explicit example, or even — quell horreur — a definite integral. When I was younger, however, those tendencies were certainly more pronounced than they are now. Still, who can fail to appreciate an identity like the following:

\(\displaystyle{e^{-2 \pi} \prod_{n=1}^{\infty} (1 – e^{-2 \pi n})^{24} = \frac{\Gamma(1/4)^{24}}{2^{24} \pi^{18}}}.\)

But man can not live on identities alone, and ultimately one’s efforts turn in other directions. So it’s always nice when the old and new words coincide, and an identity is revealed to have a deeper meaning. The formula above is a special case of the Chowla-Selberg formula, which is, possible typos in transcription aside,

$latex \displaystyle{\sum_{CM(K)} \log \left( y^{6} |\Delta(\tau)| \right) + 6h \log(4 \pi \sqrt{\Delta_K})
= 3 w_K \sum \chi(r) \log \Gamma(r/\Delta_K).}$

Here the notation is as you might guess — \(y\) is the imaginary part of \(\tau\), which is ranging over the equivalence classes of CM points for a fixed ring of integers in an imaginary quadratic field (there is presumably a version for orders as well). The existence of this identity (and a vague sense that it was related to the Kronecker limit formula) was basically all that I new about this identity, but Tonghai Yang gave a beautiful number theory seminar this week explaining the geometric ideas behind this formula, and some generalizations (the latter being the new work). So, just as in the Gross-Zagier paper on the special values of \(j\) at CM points, one now has *two* proofs of this result which complement each other, one analytic, and one geometric. (I apologize in advance for not being able to attribute all [or really any] of the ideas, Tonghai certainly mentioned many names but I never take notes and this was 5 days ago.) The first remark is that the RHS is essentially the logarithmic derivative of the corresponding Artin L-function. On the other hand, it turns out (non-obviously) that the left hand side can be related to the Faltings height(s) of the corresponding Elliptic curves with CM by \(\mathcal{O}_K\). I think this relation was discovered by Colmez in his ’93 Annals paper. The Faltings height has always been a slippery concept to me, and in fact the theory of heights in general has always struck me as being connected to the dark arts. In particular, various definitions depend on certain choices of height function, although they actually don’t depend on that choice in the end. So when actually doing a calculation, it’s always nice if you can magically produce some choice which makes calculation possible. And of course, when making a choice of function on some (tensor power of) \(\omega\) over the modular curve, what better choice is there (if one wants to control the zeros and poles) than \(\Delta\). (Tonghai mentioned another version of the formula where one instead used certain forms which are Borcherds products — of which \(\Delta\) is a highly degenerate example. I had the sense that this formulation was more generalizable to other Shimura varieties, but I never understood Borcherds products so I shall say no more.) Key difficulties in understanding generalizations of these formulas involve ruling out certain vertical components in certain arithmetic divisors on Shimura varieties, which I guess must ultimately be related to understanding the mod-p reduction of these varieties in recalcitrant characteristics (blech).

Colmez also formulated a conjectural generalization of the CS-formula, which is what Tonghai was talking about, and on which he (and now he together with his co-authors) have made some progress. The viewpoint in the talk was to re-interpret these identities in terms of arithmetic intersection numbers of arithmetic divisors on Shimura varieties. Of course, this is intimately related to the ideas of Gross-Zagier and its subsequent developments, especially in the work of Kudla, Rapoport, Brunier, Ben Howard, and Tonghai himself (and surely others… see caveat above). In light of this, one can start to see how special values of L-functions and their derivatives might appear. I can’t possibly begin to do this topic justice in a blog post, but I will at least strongly recommend watching Ben Howard talk about this at MSRI in a few weeks (Harris-fest, Tuesday Dec 2 at 11:00). I’ll be there to watch in person, but for those of you playing at home, the video will certainly be posted online. Ben is talking about exactly this problem. Since he is an excellent lecturer, I can safely promise this will be a great talk.

Added: Dick Gross emailed me the following (which also gives me the chance to say that Tonghai did indeed mention Greg Anderson during his talk):

************

…if you want to read a nice analytic treatment of the Chowla-Selberg formula, using Kronecker’s first limit formula, you can find it in the last chapter of Weil’s book “Eisenstein and Kronecker”.

I found an algebraic proof of C-S when I was a graduate student, using the moduli of abelian varieties with multiplication by an imaginary quadratic field (what we would now call unitary Shimura varieties). Deligne figured out what I was actually doing, and generalized it to prove his wonderful theorem that Hodge cycles on abelian varieties are absolutely Hodge.

Greg Anderson formulated a generalization of C-S for the periods of abelian varieties with complex multiplication. This was refined by Colmez, and we know how to prove all the refinements when the CM field is abelian over Q. Tonghai and Ben have been making progress in some non-abelian cases.

Dick

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In Brief

Posted on October 21, 2014 by Persiflage

The start of the academic year has a habit of bringing forth distractions, not least of all to someone as disorganized as me. So here are a few remarks in brief.

The class number of \(\mathbf{Q}(\zeta_{151})^{+}\) is one.

John Miller, a student of Iwaniec at Rutgers, wrote the following nice paper, which improves upon a previous result of Schoof. One technique that is useful in computing the class numbers of fields with small discriminant is to make use of the Odlyzko bounds. Here’s a typical example. If \(K =\mathbf{Q}(\zeta_{37})^{+}\), then the root discriminant of \(K\) is \(30\) or so. However, by consulting Odlzyko, one sees that any totally real field with this root discriminant has degree at most \(40\). Hence the class number of \(K\) is either one or two, and it is easy to rule out the second possibility by using genus theory. More generally, whenever one has an a priori bound on \(h^{+}\), one can compute \(h^{+}\) by relating \(h^{+}\) to the index of the circular units (Schoof did this in a previous paper.) This trick only works if the root discriminant of the totally real field is at most \(60\) (or so), which seems to prevent one from applying this to real cyclotomic fields for \(p > 67\). (There’s always a bound on the class group by Minkowski, but that is a terrible bound.) The idea behind this paper is that Odlyzko’s bound can be improved if one in addition knows that certain primes of small norm are principal. And since one has explicit fields, it is possible to show that the relevant ideals are principal by exhibiting explicit elements with the appropriate norm. I can’t quite tell how lucky the author was to find such elements (he searches for cyclotomic elements expressible as a small number of roots of unity), but it works! Perhaps, a postiori, it is useful that these fields do actually turn out to have class number one.

Matisse cut-outs

The NYT reports on an exhibit of Matisse cut-outs at MoMA. I have a particular soft spot for these works. I was particularly struck by a cut-out I once saw at an exhibit at the Centre Pompidou, so much so that I painted a replica (almost full sized) on the wall of my rental apartment in Cambridge:

My version

Original

I’m not sure if this particular cut-out is at MoMA, though.

256

Thanks to DS, I was hooked on 2048 for far too long. I eventually got bored trying to get the 16384 tile, and moved on to the more compact 256 instead. The latter game is slightly less random in that only 2s are created on each turn. The highest possible score is 7172, which is obtained when one ends up with (in any configuration) the powers of \(2\) from \(2\) to \(512\). Recently, I finally managed to complete the game:

Notice that the \(512,256,128\) tiles are not along the same edge (I think it must be theoretically possible to finish in that way, but it would be harder). Unfortunately, having reached this point, it has not cured me of my addiction. Curse you, Savitt!

Stickelberger’s Theorem.

I proved Stickelberger’s theorem in class the other day — well, with one caveat. I proved that all the ideals \(\mathfrak{q}\) of prime norm are annihilated by the Stickelberger ideal. This certainly implies the result, because the class group is generated by such ideals. This follows, for example, by the Cebotarev density theorem applied to the Hilbert class field (which was my argument in class). But then I worried that this was an anachronistic argument, and indeed Stickelberger’s theorem was a solidly 19th century result. So what did Stickelberger do?

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iTunes top Ten

Posted on October 6, 2014 by Persiflage

tl;dr: lots of Bach, if you’re not into that sort of thing, at least check out Mel Brooks. And if you’re not into that *either*, well then I don’t know what’s wrong with you.

Following Jordan, here is a list of my top 10 iTunes tracks by play count. In reality, it’s more like the top 10 *albums*, because most of the works are spread out over multiple tracks.

Technically speaking, my most-played “song” on iTunes is “Ambient Waterfall Sounds for Ultimate Bedtime Relaxation, Deeply Lucid Dreams,” with 188869 plays. But that’s a little bit misleading. The track basically consists of white noise, but (slightly irritatingly to me) it varies slightly in tone and pitch over the 4 minutes 44 seconds of the track. So I rigged it to run on a 2-second loop, which i play overnight when trying to sleep at conferences.

#1. Musical Offering, 296 plays [the number of plays from each track is not constant, so I will just go by the highest number for each album]. My recording (by the Ensemble Sonnerie) is arranged for Oboe, Violin, Viola, Harpsichord, Flute, and Viol. There comes a certain point in the evening where the only possible music one wants to listen to the Musical Offering. And that time is 3AM. As you can see from the play count, I am up a lot at 3AM.

#2. The French Suites, Glenn Gould 215 plays. The French Suites are easy to play and even easier to listen to.

Then again, perhaps you would prefer the French suites Mel Brooks style:

#3. The Cello Suites, Yo Yo Ma (his second recording), 207 plays. OK, this is also something to listen to at 3AM.

#4. The Art of Fugue (Juilliard String Quartet) 200 plays. They built their own custom-made viola in order to avoid having to transpose the score up a fifth (which is what the Delme string quartet do in one of my other recordings of this piece). I highly recommend this recording from 1982. Unfortunately, I couldn’t find a clip on youtube.

#5. The Art of Fugue (Glenn Gould) 196 plays. My only regret is that it is not complete, as Gould only recorded the fugues I,II,IV,IX,XI,XIII and, of course, XIV. (There’s also a version by Gould on the organ on this recording which I don’t listen to). Instead of linking to this recording, let me link instead to a fascinating interview between Gould and Bruno Monsaingeon with Gould at the piano (I’ve linked to the the video at the end of the first (fairly ordinary and early) fugue:

http://www.youtube.com/watch?v=AirAT7gN6A0&t=1m33s

#6. Inventions and Sinfonias (Andras Schiff) 151 plays. There’s a common theme to this list so far, and it is Bach. This CD holds a special place for me as it was my first piano Bach recording. Glenn Gould has an even more than usually idiosyncratic recording of these, so I more often turn to Schiff on this one.

#7. Ave Verum Corpus (Byrd, King’s Singers) 150 plays. We’ve finally broken out of Bach! The intimacy of this recording (for so few voices rather than a choir) is what appeals to me.

#8. Arias “Erbarme dich, mein Gott” and “Aus Liebe will mein Heiland sterben” from St Matthew Passion, 129 plays (Michael Chance/Ann Monoyios). I don’t usually have three hours to listen to the entire recording, but these Arias are certainly some of the highlights. Here’s Michael Chance in a different recording of the same aria:

http://www.youtube.com/watch?v=dHbOOe8n2gY

#9. Gnossiennes #1-#3 Jean-Yves Thibaudet, 121 plays. Perhaps you would like to see (actors portraying) Wittgenstein, John Maynard Keynes, and Lydia Lopokova pretend to be the solar system? I actually saw this Derek Jarman film in the theatre with Patrick Emerton…

#10. An Die Musik (Schwarzkopf) 119 plays. Whenever I sing/play lieder, I always finish by singing this.
This isn’t my favourite recording, but it’s the oldest one I had. In fact, when it comes to Schwarzkopf singer lieder, my favourite performance is her singing of Litanei auf das Fest Allerseelen here:

I should note that these numbers are all coming from iTunes on my laptop. My listening habits are somewhat different on my iPod, where I’m much more likely to listen to (say) Beethoven or Mahler. As an indication of how long a time period these numbers represents, I gave the analogous numbers on Jordan’s Blog http://quomodocumque.wordpress.com/2010/07/01/real-life-rock-top-10/ in 2010. Interpolating between these two sources, it suggests I listen (at home) to either the Art of Fugue or the Musical Offering about once a week, which seems about right.

Added: Please list your own top 10/top 5 in the comments! Already TG has introduced me to something interesting.

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Applying for an NSF grant

Posted on September 26, 2014 by Persiflage

It’s not easy to write a good grant proposal. But it can be even harder to write one for the first time, especially if you’re not quite sure who will be reading your proposal. So today, I want to say a little bit about how an NSF mathematics panel is run, and give you some idea of who your target audience should be.

Before I start, I want to include a pseudo-legal disclaimer. For fairly obvious reasons, you are not supposed to reveal that you served on any particular panel. But I am allowed to say that I have served on *some* panels, and there is enough uniformity in the process to make me confident that what I say should resemble your reality if you decide to apply. (Let me also mention that I had some help on this post from a friend [whom I shall refer to as the Hawk] who is much more of an NSF pro than I am. He made various corrections and suggestions on a first draft of this blog, and I even included a few of his remarks verbatim in the text.)

The NSF administers many different types of grants. I’m not just talking about graduate fellowships, postgraduate fellowships and research grants here. There are FRG grants, RTG grants, CAREER NSF grants, REUs, conference grants, and so on. However, for the purpose of this email, I want to concentrate on research grants.

The Mechanics: The panel is comprised of approximately 10 or so mathematicians, who consider approximately 40-50 or so proposals. About six weeks before the panel takes place, each panelist is given the list of proposals and asked to rank the proposals 1,2,3,C based on the following criteria:

1 = I feel comfortable reviewing this proposal

2 = I could review this proposal if necessary

3 = It would be very difficult for me to review this proposal

C = I have a conflict of interest with this proposal

Here “conflict of interest” is defined in a fairly precise way. It includes some obvious things (recent co-authors, people at your institution, family members) and some non-obvious ones (people with whom you serve with on an editorial board, people at institutions that have paid you an honorarium for giving a recent talk). You are also free to declare a conflict of interest which is not on that list. About a month before the panel meets, each panelist is given 12 or so files to read (all the files are online, of course). It is not unusual for a panelist to be given (in their suite of proposals) one or two grants they graded as a “3” above — it depends on how parsimonious they were in their initial grading. For each of these files, the panelist is asked to grade the proposal on both intellectual merit and broader impact. Many panelists also unofficially rank the proposals that they read. In addition to grading the proposals, the panelist writes a brief summary indicating what they feel are the strengths and merits of each proposal. A panelist can, if they wish, also read other proposals.

The next step is that the panel meets at the NSF headquarters in Virginia, sometime between November and March. A typical panel may last 2.5 days. The panel is chaired by the relevant program officer and three or so other NSF employees (usually professional mathematicians who have taken a leave of absence for a two year position at the NSF), so there will typically be 14-15 people in a conference room, each with either their laptop or a supplied computer. The first 1.5 days of the panel consist of going through the files one by one. For each file, the three (or so) panelists who were assigned the proposal read out their review of the proposal. During this time, other panelists (especially those with some expertise) will also offer their opinions. During this period, anyone who is conflicted with the proposal has to leave the room. At the end of each discussion (which takes about 10 minutes), a yellow sticky sheet with the PI’s name has to be placed on a white board with three columns. The columns are officially designated as “strongly recommended for funding,” “recommended for funding if possible,” and “not recommended for funding.” The desired outcome is to have 10% of proposals in the first column, 30% in the second, and 60% in the third. Within the first two columns the names are ordered, although, during the process, certain proposals can float up or down as they are re-evaluated in light of other proposals. During each discussion, a panelist who was not assigned to read the proposal is assigned to be a scribe and record the highlights of each discussion. Each panelist is a scribe on 3-4 proposals.

The final step is for each panelist to write up a summary of the panel discussions for which they were a scribe, highlighting what the panel thought were the strengths and weaknesses of the proposal, indicating “which column” the panel placed the name, and reflecting the extent to which there was uniform agreement or not. Everyone then goes over these summaries to confirm that the summary does reflect the panel discussion. If you ever apply for a grant, you will be able to read this summary, together with the evaluation of the three members who read your proposal in depth. (The panelists assigned to your proposal have an opportunity to modify their evaluations during the meeting if they change their minds in light of the discussion.)

Then the panel ends; the panel has given the program officer a (roughly) ordered set of names, and it is up to the NSF to decide whom to fund. I’m not sure the extent to which the recommendations of the panel exactly mirror the actual results, although I suspect that it is quite close. I can imagine, however, that a programme officer feels that a certain proposal suffered because nobody on the particular panel was an expert in that area, and they may decide to send that proposal off for further review. The Hawk says: The actual results can deviate significantly from the advice of the panel.  I think it’s safe to say that the ‘highly recommended’ proposals always get funded. After that, there are various other objectives that the program officers are trying to achieve — gender diversity, racial diversity, support for young PIs, support for worthy PIs at undergraduate-only institutions. The panel list is typically the default in cases where none of those other objectives apply, though you can imagine reasons to deviate from it (e.g. you might not let the same person suffer the bad luck of being the first person after the cutoff two years running, you might support a proposal in a subdiscipline that has otherwise been shut out, etc). So in the ‘recommended’ zone there are certainly some inversions. It’s also not unheard of for a ‘not recommended’ proposal to end up being funded. One way this can happen is for the proposal to be looked at by a second (perhaps more appropriate) panel that likes the proposal much better. But also, the program officers can simply decide that the panel’s conclusions about a proposal were unjust for some reason, and raise the proposal up in the rankings.

How narrow is the focus of each panel? As I mentioned, there are approximately 40-50 proposals for each panel, of which maybe 15 are funded. So take the 80 or so people who are research active and applying for grants who are closest to you mathematically, and that gives you a rough idea. If you study Galois representations and modular forms, or Iwasawa theory, or the arithmetic of Shimura varieties, or arithmetic geometry of some kind, your proposal may well end up in the same panel as mine was (it can happen — as it did to me last year — that your proposal ends up being evaluated by *two* panels — this is possibly done in order to normalize the orderings in some way. Because I wasn’t there, I can’t quite tell what the difference was between the two panels). The Hawk says: This is the first time I’ve heard a suggestion that normalization is the reason that some proposals are looked at by two panels. I think this happens either because the program officers feel that the proposal straddles two panels to such an extent that they feel both opinions could be useful; or because the proposal has two very different parts that genuinely fit in separate panels; or because the assigned panel decided that there were parts of a proposal that they didn’t have the expertise to comment on, and so they suggest getting the input of another panel. On the other hand, I’m pretty sure that my proposal would not be on the same panel as someone like Ken Ono or Soundararajan. Could my proposal be on the same panel as Akshay’s? I’m not sure. I probably would have said no if my proposal didn’t end up on two panels last time. And Akshay is a collaborator of mine! So it’s pretty focused. On the other hand, there are certainly areas in each field which are smaller than others, and if you work in such a sub-field, then it’s more likely that the panelists will not be experts in your area.

Who serves on the panel? First, there are a few NSF rules which apply to panels. (update: this information was wrong — it turns out there are no formal NSF requirements for the constitution of any panel.) Beyond this formal requirement, who is a typical member of the panel? Well, of course, one goal of the program officer is to make the panel is not *too* uniform. But, for example, I would expect that there would always be at least one person on the committee who knows as much (say) about modular forms and Galois representations as I do. So if that is what you do, then you can be pretty sure that whomever that person is will be reading your file. But you can also be sure that someone who is *not* an expert will also be reading your file, perhaps someone in Iwasawa theory, say. And this already should give you a pretty good idea of your target audience. In other words, you have to do two things:

  • You have to explain to Iwasawa theory person why the modularity theorems you are going to prove are interesting. When is math interesting? Well, there are plenty of ways it can be interesting. You may have an idea of how to apply previous machinery in a novel way. You may have an interesting application in mind. You may have a completely new approach to an old theorem. You may have a completely new idea on how to solve an open problem. This is what you want to get across when you are talking to Iwasawa theory person — to give a sense of why the general problem you are studying is interesting, and how you are going to make a contribution to that field.
  • You have an easier job convincing me (or equivalent) why your modularity theorems are broadly interesting, but you still have to conveince me that you particular proposal is interesting. More importantly, you have to convince me that you can carry out your proposal successfully, or at least to the point of producing interesting mathematics. The Hawk says: I think it would be worth mentioning here the fine line between saying enough about how you intend to carry out your plans that the panel is convinced you can do it, and saying so much that they think you’ve done it already. I think new proposers often struggle on this point.

Of course, if you do something other than what I do, then replace “Iwasawa theory person” above by me or equivalent and “me” by someone with expertise in your field.

What should I take away from this? First up, I think that an NSF grant proposal is probably the most technical audience you will write for in a context that is not one of your research papers. So you don’t need (beyond a cursory mention) to say how modular forms played a role in the proof of Fermat’s Last Theorem which you might do (say) in a job application. Nor do you need to define the class group of a number field, or explain what a modular curve is. But, at the same time, and this is very important, it can still be incredibly useful to place your work in a broader context. For example, on my last NSF proposal, I started out by reminding the reader briefly how there are very general conjectures linking Galois representations coming from geometry to automorphic L-functions. I reminded the reader that special degenerate cases of this conjecture correspond to very classical objects like the Riemann zeta function. I then mention how the work of Wiles addresses the case when the representation comes from the cohomology of an elliptic curve over \(\mathbf{Q}.\) Then I explain how all the generalizations of Wiles’ theorem share a common assumption, namely, that the Galois representations over \(\mathbf{Q}\) that one can study by this method have the property that they are, up to a twist, self-dual. So already, in perhaps not much more than a half a page, I have given the context to explain how proving that a non-self-dual Galois representation is modular is “interesting.” Of course, then I have to go on an explain *how* I am going to say anything interesting about non-self-dual representations.

Do fat cats just get their grants without trying? Every proposal is evaluated on its merits, but of course “prior success” is taken into account when judging future chances of success, and so it should be. But if Peter Scholze (say, to take someone who is not in the US so I can use his name) sends in an application consisting solely of “I am working on several projects that I decline to disclose but that I expect to be of the same importance as my prior results,” he would not be funded. More realistically, I have heard that it has been the case that fields medalists have been turned down for grants, but because all grants that are turned down are never officially acknowledged, this is just hearsay. My feeling is that, on the whole, the panels do a pretty good job, and (apart from the occasional controversial case) there is more of a uniform agreement than you might guess. The Hawk brings up the key point that this opinion only concerns number theory panels. It may be the case (and I occasionally here rumours to this effect) that other areas are not run as well. I would also say that the fat cats (on the whole) seem to put as much effort into writing their NSF proposals as everyone else.

How can I compete with the fat cats given I’m only just starting out? This is taken into account. If you are at most 6 years from your PhD, your proposal is evaluated in that context; an effort is made to fund promising young people, and also people who have never received prior NSF support. That said, it’s not easy to get a grant the first time you apply coming straight out of a postdoctoral position.

What about broader impact? This is hard for younger people. But everyone on the panel realizes this and so the expectations are lower. You probably don’t have any grad students yet, so what can you say? Perhaps you have given expository talks at a workshop? Perhaps you have written up detailed notes on otherwise hard to access topics? Perhaps you have gone into the public schools in some hardscrabble inner surburban neighbourhood and and taught calculus? (Not the last one? Then don’t suggest that you might if there’s no reason to suspect that you have any previous inclination to do so.)

Don’t Imagine that you are going to be held account for what you say you are going to prove in future proposals. Future proposals will be evaluated on their own merits (as well as prior research), and nobody is going to know or remember what you said in your previous NSF grants. It’s expected that some of problems you are working on might not work out, and that you will have new ideas while working on the proposal.

Two further suggestions from the Hawk:

When will I hear back? Answer: who the hell knows. Usually within six months from the deadline, but not always, especially these days when the federal government is funded from continuing resolution to continuing resolution. If you hear in January, either you are Peter Scholze or it’s bad news. By May, no news is good news: you probably weren’t in the ‘not recommended’ pile, and they’re waiting to see how far they can stretch the money in the ‘recommended’ pile.

If I get the grant, how much money will I get? Answer: probably less than what you asked for in your budget, and if not, you probably didn’t budget enough. Less glib answer: the program officers do adjust the award sizes in order to hit their target funding rates. You shouldn’t fret that if you ask for too much and the person who’s next on the list asks for a lower number, that could hurt your chances. The natural followup: “If program officers have that kind of discretion, wouldn’t it be better if they gave smaller awards to more people?” You can certainly argue that in theory that might be better, but in practice the answer is emphatically no. DMS’s (DMS = division of mathematical sciences at the NSF) funding rate is already much higher than that of other divisions, as high as can politically be sustained within NSF. If the funding rate went up, DMS’s budget would be cut, and the rate would go back down again.

Do you have any other thoughts? The fact that approximately 30% percent of proposals get accepted is a fairly immutable law of nature. It is no doubt depressing to be continually rejected by the NSF, and good people simply stop applying, in some sense making it then harder for everyone else. If, for some reason, the number of applications suddenly doubled, it wouldn’t be the case that the success rate would halve, but more proposals would be awarded. So, there is a real sense in which the more people who apply the more grants are awarded.

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The nearly ordinary deformation ring is (usually) torsion over weight space

Posted on September 19, 2014 by Persiflage

Let \(F/{\mathbf{Q}}\) be an arbitrary number field. Let \(p\) be a prime which splits completely in \(F\), and consider an absolutely irreducible representation:

\(\rho: G_{F} \rightarrow {\mathrm{GL}}_2({\overline{\mathbf{Q}}}_p)\)

which is unramified outside finitely many primes.

If one assumes that \(\rho\) is geometric, then the Fontaine–Mazur conjecture predicts that \(\rho\) should be motivic, and the Langlands reciprocity conjecture predicts that \(\rho\) should be automorphic. This is probably difficult, so let’s make our lives easier by adding some hypotheses. For example, let us assume that:

  • A: For all \(v|p\), the representation \(\rho|G_{v}\) is crystalline and nearly ordinary,
  • B: The residual representation \({\overline{\rho}}\) has suitably big image (Taylor–Wiles type condition.)

Proving the modularity of \(\rho\) under these hypotheses is still too ambitious — it still includes even icosahedral representations and Elliptic curves over arbitrary number fields. Natural further hypotheses to make include conditions on the Hodge–Tate weights and conditions on complex conjugation.

We prove the following:

Theorem I: Assume, in addition to conditions A and B+, that

  • C: The Hodge-Tate weights \([a_v, b_v]\) at each \(v|p\) are sufficiently generic,
  • D: If \(F\) is totally real, then there exists at least one infinite place such that \(\rho\) is even.

Then \(\rho\) does not exist.

The condition B+ (which will be defined during the proof) is more restrictive than the usual Taylor–Wiles condition — we shall see from the proof exactly what it entails. Condition C will also be explained — but let us note that, for any suitable method of counting, almost all choices of integers are generic, even after imposing some condition on the determinant (say \(a_v + b_v\) is constant) to rule out stupidities.

One should think of this theorem as follows. If \(F\) is totally real, then condition D should be sufficient to rule out the existence of any automorphic \(\rho\) in regular weight, because (for motivic reasons) such representations should be totally odd. On the other hand, if \(F\) is not totally real, then the weights of any motive (with coefficients) should satisfy a certain non-trivial symmetry property with respect to the action of complex conjugation. So, for example, if \(F\) has signature \((1,2)\), then either condition C or D should be sufficient, but we will require both. In fact, even condition C is stronger than what should be necessary. In addition to assuming regularity at all primes, it amounts to (on the representation theoretic side) insisting that none of the \(\mathrm{GL}_2(\mathbf{C})\) weights are fixed by any conjugate of complex conjugation, whereas a single such example should be enough for a contradiction.

Perhaps a useful way to think about Theorem I is to make the following comparison. Hida proves the following theorem:

Theorem [Hida]: The nearly ordinary Hida family for \(\mathrm{SL}(2)/F\) is finite over weight space and has positive rank if and only if \(F\) is totally real and the corresponding \({\overline{\rho}}\) is odd at all infinite places.

On the other hand, a consequence of Theorem I is:

Theorem II: The fixed determinant nearly ordinary deformation ring of a residual representation \({\overline{\rho}}\) satisfying condition B+ is finite over weight space and has positive rank if and only if \(F\) is totally real and the corresponding \({\overline{\rho}}\) is odd at all infinite places.

In both cases, I am only considering the deformation rings up to twist — the deformation ring of the character is torsion over the corresponding weight space whenever \(\mathcal{O}_F\) has infinitely many units. Also in both cases, it is of interest to determine the exact co-dimension of the ordinary family — this is a difficult problem, because strong enough results would allow you do deduce Leopoldt by considering induced representations.

OK, so what is the argument? If you have read some of my papers, you can probably guess.

Assume that \(\rho\) exists. Let \(U\) be the representation corresponding to \(\rho\). Now replace \(U\) by \(V = {\mathrm{Sym}}^2(U)\). Now replace \(V\) by the tensor induction:

\(\displaystyle{ W = \bigotimes_{G_F/G_{{\mathbf{Q}}}} V}\)

of dimension \(3^{[F:{\mathbf{Q}}]}\). We now let C be the condition that \(W\) has distinct Hodge–Tate weights. To see that this is generic, it really suffices to show that there is at least one choice of weights for which this is true. But one can let the weights of \(U\) up to translation consist of the 2-uples \([0,1]\), \([0,3]\), \([0,9]\), etc. and then the weights of \(W\) are, again up to translation, \([0,1,2,\ldots,3^{[F:{\mathbf{Q}}]} – 1].\) We now let B+ be the condition that the residual representation is absolutely irreducible, and that the prime \(p > 2 \cdot 3^{[F:{\mathbf{Q}}]} + 1\). This is generically true, and amounts to saying that the conjugates of \({\overline{\rho}}\) under \(G_{{\mathbf{Q}}}\) are sufficiently distinct. Since the dimension of \(W\) is odd, and because it is essentially self-dual (exercise), orthogonal (obvious), nearly ordinary (by assumption), has distinct Hodge–Tate weights (by construction), satisfies the required sign condition (automatic in odd dimension), we deduce that it is potentially modular by [BLGGT]. In order to win, it suffices to show, by a theorem I made Richard prove, that the action of complex conjugation on \(W\) has trace \(\pm 1.\) However, this is equivalent to condition D (see below). QED.

One can relax condition B+ slightly by only inducing down to the largest totally real subfield of \(F\). On the other hand, there are plenty of examples to which Theorem II applies. I think one can take any elliptic curve \(E/F\) without CM and such that \(j_E \in F\) does not lie in any subfield of \(F\), and then take \(p\) to be any sufficiently large ordinary prime which splits completely in \(F\) (caveat emptor, I didn’t check this). Of course, the condition that \(p\) splits isn’t really necessary either, I guess…

The second theorem follows along the exact same lines — the conditions are strong enough to ensure, using results of Thorne, that the nearly ordinary deformation ring of (the now residual) representation \(W\) is finite over weight space, which translates back into finiteness of deformations of \(U\) over weight space. The result is obvious if \(F\) is totally real and \({\overline{\rho}}\) is odd. Otherwise, we choose a sufficiently generic point in weight space (in the sense of C), and then, by Theorem I, we see that the specialization of the nearly ordinary deformation ring at that point must be torsion.

It remains to compute the sign of \(W\). This is an exercise in finite group theory, we only recall enough of the details for our purposes. Let \(V\) be a representation of \(H\) of dimension \(d\). Consider the tensor induction:

\(\displaystyle{\bigotimes_{G/H} \sigma V}.\)

Let \(T\) denote a set of representatives of right cosets of \(H\) in \(G\). Let \(t g \in T\) denote the corresponding choice for the coset \(Htg\). For \(g \in G\), let \(n(t)\) denote the size of the \(\langle g \rangle\)-orbit which contains \(T\). If \(g = c\) has order \(2\), then either \(n(t) = 1\) or \(n(t) = 2\) Certainly

\(t c^{n(t)} t^{-1} \in H, \quad t \in T.\)

Let \(T_0\) be a set of representatives for the \(\langle g \rangle\) orbits on \(T_0\). Then (proof omitted)

\(\displaystyle{\phi^{\otimes G}(c) = \prod_{t \in T_0} \phi(t c^{n(t)} t^{-1})}.\)

We observe that:

  1. If \(n(t) = 2\), then \(\phi(t c^{n(t)} t^{-1}) = \phi(t t^{-1}) = \phi(1) = d\).
  2. If \(n(t) = 1\), then \(tct^{-1} \in H\) and \(\phi(t c t^{-1})\) is what it is. For example, it is \(0,\pm 1\) if and only if \(V\) is \({\mathrm{GL}}\)-odd with respect to \(tct^{-1}\).

Now suppose that \(G = G_{{\mathbf{Q}}}\) and \(H = G_{F}\). The elements \(tct^{-1}\) are exactly the different complex conjugations of the representations of the conjugates of \(H\). We deduce:

  1. If \(\dim(V)\) is even, then \(W\) is \({\mathrm{GL}}\)-odd if and only if there exists at least one real place of \(F\) such that \(V\) is \({\mathrm{GL}}\)-odd.
  2. If \(\dim(V)\) is odd, then \(W\) is \({\mathrm{GL}}\)-odd if and only if \(F\) is totally real and \(V\) is \({\mathrm{GL}}\)-odd at every real place.

Equivalently, a product of even integers can equal zero only if at least one of them is zero, and a product of odd integers can equal \(\pm 1\) if and only if all of them are \(\pm 1\).

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Tricky Fingers

Posted on September 18, 2014 by Persiflage

How is one supposed to play this exactly?

Contrapuctus XIII

One can neither can play a 14th in the right hand (my hands are not that big) nor play legato parallel 10ths in the left; hence some sort of arpeggiation is required. But I can’t quite seem to reproduce how Glenn Gould plays this measure. Then again, that phenomenon is not unique to this passage.

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