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For simplicity let's work with smooth projective varieties over $\mathbb C.$ I'm wondering how to compute the pushforward of a sheaf by a finite map $f:X\to Y.$ [The actual example I care about is with the double cover of $\mathbb P^2$ branched over a smooth sextic, but I would rather figure this out on my own. My intuition says that the pushforward of the structure sheaf in this case is $\mathcal O_{\mathbb P^2}\oplus\mathcal O_{\mathbb P^2}(3),$ but the cohomology is not of the correct rank if this were true.]

Let's take for example the map $t:\mathbb P^1\to \mathbb P^1$ given by $t(Z:W)=(Z^2:W^2).$ Then I can show that $t_*\mathcal O_{\mathbb P^2}$ is locally free of rank 2 (I know this is a theorem but it's not one I have seen yet). However I'm having trouble figuring out the global picture. For example, near the point $(0:1)$ with $z=Z/W$ we have the induced map $t^*:\mathbb C[z]\to \mathbb C[z]$ which sends $z\mapsto z^2$ and as a module, the codomain splits as $1t^*(\mathbb C[z])\oplus zt^*(\mathbb C[z])$ and hence has rank 2. But now I want to compute how the section $z$ transforms into the other affine piece of $\mathbb P^1$ and I'm at a loss. How do you compute the transition functions in this case?

When I told my advisor that I'm trying to look at this simple example to get an understanding for how to do this he looked at it and instantly said that it was $\mathcal O_{\mathbb P^1}\oplus\mathcal O_{\mathbb P^1}(-1).$ I will ask him at our next meeting, but I'm also wondering if there are any good heuristics for computing these pushforwards. Right now I have no intuition for how this should go.

I have no knowledge of schemes so if any explanation could avoid that language I would appreciate it.

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  • $\begingroup$ I remember answering a similar question in the past! math.stackexchange.com/questions/3063864/… Does this help at all? $\endgroup$ – Kenny Wong Aug 6 at 20:03
  • $\begingroup$ Yes actually this is almost exactly the solution a friend and I came up with after discussing for a while. You've removed any doubt I had in my thinking. Thanks a lot! $\endgroup$ – D. Brogan Aug 6 at 21:46
  • $\begingroup$ Awesome! Glad it was useful. $\endgroup$ – Kenny Wong Aug 6 at 23:18
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Some simple facts. Let $f:X\to Y$ be a double cover of smooth varieties over complex numbers. Then $f_*O_X$ is a rank two vector bundle and you have the natural map $O_Y\to f_*O_X$ which splits using trace. So, $f_*O_X=O_Y\oplus L$ for a line bundle $L$ on $Y$. Now, let us specialize to the case when $Y$ is a projective space and then $L=O_Y(n)$ for some $n$. If $Y=\mathbb{P}^m$, then $H^m(X, O_X)=H^m(Y, O_Y\oplus O_Y(n))=H^m(Y, O_Y(n))$. There can be at most one $n$ as a solution since the dimension of these vector spaces are distinct most of the time.

In your $m=1$ case, since $H^1(X,O_X)=0$, we must have $H^1(O_Y, O_Y(n))=0$ and thus $n\geq -1$. But $n\leq -1$ since $H^0(X,O_X)$ is one dimensional.

Similarly in your $m=2$ case, $n\leq -1$ is forced since $H^0(X,O_X)$ is one dimensional. Since ramification locus is a sextic, you can check that $H^2(X,O_X)$ is one dimensional and thus $H^2(O_Y(n))$ is one dimensional. The only $n$ with that property is $n=-3$.

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  • $\begingroup$ This is some good news because after talking with a friend for a while we came up with something that gives $\mathcal O\oplus \mathcal O(-1)$ and $\mathcal O\oplus \mathcal O(-3)$ respectively. Can I ask some clarifying questions? First, what do you mean when you say that the natural map $\mathcal O_Y\to f_*\mathcal O_X$ splits using trace? Why does that imply $f_*O_X=O_Y\oplus L$ instead of $f_*O_X=L\oplus L'?$ Also, what does the ramification locus have to do with $H^2(X,\mathcal O_X)$ a priori, before knowing what the pushforward looks like? $\endgroup$ – D. Brogan Aug 6 at 21:38
  • $\begingroup$ There is always a trace map which splits the above map if characteristic is not two. Once you have some unknown $L$ as above, to determine it you have to use some information about $X$. In the two dimensional case, the fact that it it is ramified along a sextic implies the canonical bundle of $X$ is trivial and then by Serre duality, you have the $H^2(K_X)$ is one dimensional. $\endgroup$ – Mohan Aug 7 at 0:39
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    $\begingroup$ I don't want to be obtuse, but this comment didn't actually answer any of my questions. I'll try and be more specific. 1.) I don't know what "trace map" means in this context. In my mind trace is reserved for linear maps between vector spaces and I don't see any of those here 2.) How do you know that $f_*\mathcal O_X$ has $\mathcal O_Y$ as a summand? Why can't it be the sum of two general line bundles? Why is it a sum of line bundles in the first place? 3.) My goal is to show that the branched cover of the plane is a $K3,$ so it seems like you're assuming what I want to show here. $\endgroup$ – D. Brogan Aug 7 at 14:52

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