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 '19 at 3:33
  • 1
    $\begingroup$ Welcome new contributor! Flatness is not a miracle in this case, it is simply the classification of f.g. modules over a PID. $\endgroup$ Apr 24 '19 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 '19 at 4:23
  • $\begingroup$ @Sasha they are not quasi-finite though (if you are over an algebraically closed field). $\endgroup$
    – user138661
    Apr 24 '19 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 '19 at 5:07
  • 3
    $\begingroup$ @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 $\mathcal{O}(-1)$. $\endgroup$ Apr 24 '19 at 19:07
  • $\begingroup$ Ah perfect, thank you very much! $\endgroup$
    – cgodfrey
    Apr 24 '19 at 22:00

Here is another way to view this.

Let's look at local sections of $\mathbb{P}^1_w$. Cover $\mathbb{P}^1_w$ by affine charts $U_0$ and $U_1$. Then $\mathcal{O}_{\mathbb{P}^1_w}$ and $\mathcal{O}_{\mathbb{P}^1_w}(-1)$ can be viewed as following pictures:

$$ \rlap{\underbrace{\phantom{\cdots w^{-3}, w^{-2}, w^{-1}}}_{U_0}} w^{-3}, w^{-2}, w^{-1}, \overbrace{1, w , w^2, w^3, \cdots}^{U_1} $$ $$ \underbrace{\cdots w^{-3}, w^{-2}, w^{-1}, 1\,}_{U_0} , \, \overbrace{ w , w^2, w^3, \cdots}^{U_1} $$

For example, the local sections of $\mathcal{O}_{\mathbb{P}^1_w}$ over Spec$k[w]$ are $1, w, w^2,\cdots$ and over Spec$k[w^{-1}]$ are $1, w^{-1}, w^{-2},\cdots$. So the global section is only $k$.

Now suppose we have the morphism given by $z=w^3$. Then we can decompose $\mathcal{O}_{\mathbb{P}^1_w}$ as following three:

$$ \rlap{\underbrace{\phantom{\cdots w^{-9}, w^{-6}, w^{-3}}}_{U_0}} w^{-9}, w^{-6}, w^{-3}, \overbrace{1, w^3 , w^6, w^9, \cdots}^{U_1} $$ $$ \underbrace{\cdots w^{-8}, w^{-5}, w^{-2}}_{U_0}, \, \overbrace{ w^{1}, w^4 , w^7, \cdots}^{U_1} $$ $$ \underbrace{\cdots w^{-7}, w^{-4}, w^{-1}}_{U_0}, \, \overbrace{ w^{2}, w^5 , w^8, \cdots}^{U_1} $$ Hence $$f_* \mathcal{O}_{\mathbb{P}^1_w} = \mathcal{O}_{\mathbb{P}^1_z} \oplus \mathcal{O}_{\mathbb{P}^1_z}(-1)^{2}.$$

  • $\begingroup$ I didn't understand completely what the braces signify, but I think that this is quite similar to the argument on p. 2 of J. F. Thomsen "Frobenius Direct Images of Line Bundles on Toric Varieties" (J. Algebra 226 (2000)). $\endgroup$ Apr 22 at 5:26
  • $\begingroup$ I'm going to echo Piotr's comment and add that the first two displays are identical, which doesn't help. $\endgroup$
    – user347489
    Apr 22 at 6:15
  • 2
    $\begingroup$ Okay, I just learned how to use phantom to get overlapped brackets in Latex. $\endgroup$
    – WWK
    Apr 22 at 9:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.