Higher direct images of canonical extensions

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:

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

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.