Let $k$ be an algebraically closed field. Any finite $k$-morphism $P^1_k\rightarrow P^1_k$ is flat (miracle flatness) and surjective on the underlying spaces. Therefore, the pushforward of a coherent locally free sheaf is coherent locally free (on $P^1_k$, such sheaves can be described by a finite sequence of integers using the fact that Picard rank is 1 and there is Birkhoff--Grothendieck splitting).

Assume we have a finite $k$-morphism $P^1_k\rightarrow P^1_k$ such that the inverse image of the generic point has cardinality $n\geq 2$. Which sheaves can we get as the pushforward of a locally free sheaf of rank 1?

  • $\begingroup$ I believe that a finite morphism sends closed points to closed points. Any point on a curve is either closed or generic. The question does not make much sense right now. Maybe you mean that the maximum cardinality of a fiber of the morphism is $n\geq 2$? $\endgroup$ – user138661 Apr 24 at 3:33
  • $\begingroup$ Welcome new contributor! Flatness is not a miracle in this case, it is simply the classification of f.g. modules over a PID. $\endgroup$ – Piotr Achinger Apr 24 at 4:03
  • $\begingroup$ There are non-flat maps from $P^1$ to $P^1$, namely compositions $P^1 \to Spec(k) \to P^1$. $\endgroup$ – Sasha Apr 24 at 4:23
  • $\begingroup$ @Sasha they are not quasi-finite though (if you are over an algebraically closed field). $\endgroup$ – user138661 Apr 24 at 4:31

Write $$ f_* \mathcal{O}(m) = \bigoplus_{k\in\mathbb{Z}} \mathcal{O}(k)^{\alpha(m, k)}. $$ We want to compute the multiplicities $\alpha(m,k)$. We have $f^* \mathcal{O}(k) = \mathcal{O}(nk)$ where $n = \deg f$, so the projection formula gives $$ (f_* \mathcal{O}(m)) \otimes \mathcal{O}(-k) = f_* \mathcal{O}(m-nk),$$ and hence $\alpha(m,k) = \alpha(m-nk, 0)$. If $S(x) = \sum_m (m+1)x^m = 1/(1-x)^2$ and $A(x) = \alpha(m,0) x^m$, then applying $h^0(-)$ to both sides of the first displayed formula, multiplying by $x^m$ and summing over $m\in\mathbb{Z}$ gives $$ S(x) = A(x)\cdot S(x^n).$$ Thus $$ \alpha(m,k) = \left(\text{coefficient of }x^{m-nk}\text{ in } S(x)/S(x^n) = (1+x+\cdots + x^{n-1})^2\right). $$ For example, $$f_* \mathcal{O} = \mathcal{O} \oplus \mathcal{O}(-1)^{n-1}.$$

(The above argument appears in my paper "Frobenius Push-Forwards on Quadrics", and works similarly for $\mathbb{P}^N$. The first place I know where these pushforwards are computed is the paper "Frobenius direct images of line bundles on toric varieties" by J. F. Thomsen)

  • $\begingroup$ I will never get sick of this calculation. Follow up (please don't feel obligated to reply!): is there a perspective (or even a calculation for a particular $f$) from which $f_* \mathcal{O}_X = \mathcal{O}_X \oplus \mathcal{O}(-1)^{n-1}$ is "obvious"? $\endgroup$ – cgodfrey Apr 24 at 5:07
  • 2
    $\begingroup$ @cgodfrey Yes: if $f$ is the map $(x^n\colon y^n)$ (and say if $n$ is prime to the characteristic, otherwise you need to be more careful), then $f$ is a cyclic covering with group $\mu_n$ ramified along $D = 0+\infty$. The usual theory of cyclic coverings gives $f_* \mathcal{O} = \bigoplus_{i=0}^{n-1} \mathcal{O}(\lfloor -\frac{iD}{n} \rfloor)$. Now the $i=0$ summand is $\mathcal{O}$, and the other summands are $\mathcal{O}(-1)$. $\endgroup$ – Piotr Achinger Apr 24 at 19:07
  • $\begingroup$ Ah perfect, thank you very much! $\endgroup$ – cgodfrey Apr 24 at 22:00

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