Direct image of vector bundle

Question

Let $f:X\to Y$ be a morphism of projective varieties and $\mathcal{E}$ be a vector bundle on $X$. How can I compute explicitly $f_*\mathcal{E}$ in various situations?

For example, let $Y=\mathbb{P}^1$ and $X$ be a conic. What is $f_*\mathcal{O}_X$?

Answer 1

This is a great question! Hopefully someone more knowledgable than I can give a fulfilling answer. I can only offer a few parlor tricks

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For example, if you have a finite flat morphism $f:X\to\mathbb{P}^1_k$, where $X/k$ is a smooth curve, then $f_\ast\mathcal{E}$ will be a vector bundle. Then, by Grothendieck's theorem you know that

$$f_\ast\mathcal{E}=\mathcal{O}(n_1)\oplus\cdots\oplus\mathcal{O}(n_m)$$

You can then often times deduce what the $n_i$ are from cohomology computations. Namely, since $f$ is affine, you know that

$$H^i\left(\mathbb{P}^1,\mathcal{O}(n_1)\oplus\cdots\oplus\mathcal{O}(n_m)\right)=H^i\left(X,\mathcal{E}\right)$$

the right hand side which you probably know. This will give you a set of equations on the $n_i$ which will usually determine them uniquely (hopefully!).

More generally, if you know what your map $X\to\mathbb{P}^1$ is given by $\mathscr{L}$ then you have the extra data

$$H^i(\mathbb{P}^1,f_\ast\mathcal{E}\otimes\mathcal{O}(n))=H^i(X,\mathcal{E}\otimes\mathscr{L}^{\otimes n})$$

Example: Suppose that $f:E\to\mathbb{P}^1$ is a degree two, non-constant, where $E$ is genus $1$. Then, by degree considerations you know that $f_\ast\mathcal{O}_E$ is $\mathcal{O}(n_1)\oplus\mathcal{O}(n_2)$. But, using the observation above, you have that

$$H^0(\mathbb{P}^1,\mathcal{O}(n_1)\oplus\mathcal{O}(n_2))=H^0(E,\mathcal{O}_E)=1$$

and so clearly we must have, without loss of generality, that $n_1=0$, and $n_2<0$. But, we also have the equation

$$1=H^1(E,\mathcal{O}_E)=H^1(\mathbb{P}^1_k,\mathcal{O}\oplus\mathcal{O}(n_2))=H^1(\mathbb{P}^1,\mathcal{O}(n_2))$$

Looking at the list of values for the cohomology of line bundles, you see that this forces $n_2=-2$. So, $f_\ast\mathcal{O}_E=\mathcal{O}\oplus\mathcal{O}(-2)$.

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In a different vein, if you're map $f:X\to Y$ is finite, then you could leverage the action of $\mathrm{Gal}(X/Y)$ on $f_\ast\mathcal{E}$ and compute its decomposition into irreducibles (assuming nice enough situations).

Example: Let's analyze the same example as above, but under this light. Let's assume, for obvious reasons, that $\mathrm{char}(k)\ne 2$. We have an action of $\mathrm{Gal}(E/\mathbb{P}^1)=\mathbb{Z}/2\mathbb{Z}$. The action of $\mathbb{Z}/2\mathbb{Z}$ on $f_\ast\mathcal{O}_E$ gives us a decomposition into the $\pm 1$-eigenspaces. Thus, we have a decomposition

$$f_\ast\mathcal{O}_E=\mathcal{O}_+\oplus\mathcal{O}_-$$

That said, it's clear that $\mathcal{O}_+=\mathcal{O}$ (why)? That said, since $\mathcal{O}_E=\Omega^1_E$ we have the trace map $$\mathrm{tr}:f_\ast\Omega^1_E=f_\ast\mathcal{O}_E\to \Omega^1_{\mathbb{P}^1}=\mathcal{O}(-2)$$ which is surjective. I think a bit of thought will show that $\ker\mathrm{tr}=\mathcal{O}$, and thus we have an extension $$0\to \mathcal{O}\to f_\ast\mathcal{O}_E\to\mathcal{O}(-2)$$ but of course this splits since $$\mathrm{Ext}^1(\mathcal{O}(-2),\mathcal{O})=\mathrm{Ext}^1(\mathcal{O},\mathcal{O}(2))=0$$