2
$\begingroup$

Fix a field $k$ of characteristic $0$, and let $X\subset\mathbb{P}_k^1\times_k\mathbb{P}_k^1$ be defined by the type $(2,6)$-equation $x_0^2 y_1^6 - x_0 x_1 y_0^6 + x_1^2 y_0^3 y_1^3 + x_1^2 y_1^6\in k[x_0,x_1,y_0,y_1]$, so it can be shown that $X$ is a nonsingular curve in $X$, and basic computations of homology using the invertible sheaves $\mathscr{O}_{\mathbb{P}^1\times\mathbb{P}^1}(d,e)$ give us that $h^1(\mathscr{O}_X)=5$, so that $X$ has arithmetic genus $5$.

I wish to prove by computation that in this case, the geometric genus $h^0(\omega_X)$ is also $5$, where $\omega_X=\Omega_{X/k}^1$. However, I've hit a snag when I try to gain any explicit information about the sheaf $\Omega_{X/k}$ – quite simply, I don't know ay quick and easy way to calculate that sheaf. My first thought was to try and use the closed immersion $i:X\hookrightarrow\mathbb{P}^1\times\mathbb{P}^1$ and the exact sequence $$i^*\Omega_{\mathbb{P}_k^1\times\mathbb{P}_k^1/k}\to\Omega_{X/k}\to 0$$ since it is not so difficult to calculate that $$\Omega_{\mathbb{P}_k^1\times\mathbb{P}_k^1/k}\cong\mathscr{O}_{\mathbb{P}^1\times\mathbb{P}^1}(-2,0)\oplus\mathscr{O}_{\mathbb{P}^1\times\mathbb{P}^1}(0,-2)$$ but this only tells me that $\Omega_{X/k}$ can be thought of as a quotient sheaf, it doesn't give me any hard information about the sheaf itself. What is the best way to calculate $h^0(\omega_X)$ here?

$\endgroup$
5
$\begingroup$

Adjunction formula says $$ \omega_X = \omega_{\mathbb{P}^1 \times \mathbb{P}^1}([X])\vert_X. $$ So, $\omega_X = (\mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(-2,-2) \otimes_{\mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}} \mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(2,6))\vert_X = \mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(0,4)\vert_X$, You can compute its cohomology by the Koszul complex $$ 0 \to \mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(-2,-6) \to \mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1} \to \mathscr{O}_X \to 0. $$ When tensored by $\mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(0,4)$, it gives $$ 0 \to \mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(-2,-2) \to \mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(0,4) \to \mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(0,4)\vert_X \to 0. $$ The cohomology exact sequence gives $$ H^0(X,\omega_X) = H^0(X,\mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(0,4)\vert_X) = H^0(\mathbb{P}^1\times \mathbb{P}^1,\mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(0,4)) = H^0(\mathbb{P}^1,\mathscr{O}_{ \mathbb{P}^1 \times \mathbb{P}^1}(4)). $$ This is a 5-dimensional space.

$\endgroup$
  • $\begingroup$ @MonstrousMoonshine: I have reluctantly approved your edit on the representation that you are correcting an error introduced by your earlier edit. If I may suggest a way to avoid this kind of situation, instead of making an extensive edit to an Answer composed (for your Question) by someone else, put the proposed edit in a Comment for the other user to review. $\endgroup$ – hardmath Apr 22 '17 at 20:15

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.