# Smooth curve of genus $1$ in $\mathbb{P}_{\mathbb{C}}^1\times \mathbb{P}_{\mathbb{C}}^1$.

This question comes from Gathmann's notes of Algebraic Geometry:

Show that $$\{((x_0:x_1),(y_0:y_1)): (x_0^2+x_1^2)(y_0^2+y_1^2)=x_0x_1y_0y_1\}\subset \mathbb{P}_{\mathbb{C}}^1\times \mathbb{P}_{\mathbb{C}}^1$$ is a smooth curve of genus $1$.

The author briefly talked about the concept genus in this section without assuming much knowledge in topology. One proposition said that the genus of smooth curve of degree $d$ in $\mathbb{P}_{\mathbb{C}}^2$ is $\binom{d-1}{2}$.

Then in one exercise I showed that the arithmetic genus ($(-1)^{\dim X}\cdot (\chi_X(0)-1)$) of a smooth curve in $\mathbb{P}_{\mathbb{C}}^2$ agrees with the geometric genus, where $\chi_X$ is the Hilbert polynomial of the projective variety $X$. The author then stated that this can be generalized to any smooth projective curve over $\mathbb{C}$.

So I think I should use that formula for this exercise. The smoothness can be shown by the Segre embedding. Now the formula says that the genus is $$1-\chi_X(0).$$ And $\chi_X(d)=\binom{3+d}{3}-\binom{3+d-4}{3}$. But I cannot get $1$. I must have made some mistake when calculating $\chi_X(d)$. But I couldn't figure out where.

Thank you for any help!

## 1 Answer

Using the Segre embedding $\mathbb P^1 \times \mathbb P^1 \hookrightarrow \mathbb P^3$ given by $(x_0:x_1)\times(y_0:y_1) \mapsto (x_0y_0,x_0y_1,x_1y_0,x_1y_1)$, we can rewrite the curve as an intersection of two hypersurfaces in $\mathbb P^3$. Explicitly, these hypersurfaces will be $xw=yz$ and $x^2+y^2+z^2+w^2=xw$. These are both quadrics, so that we have a free resolution of $\mathscr O_X$ ($X$ is the curve): $$0 \to \mathscr O_{\mathbb P^3}(-4) \to \mathscr O_{\mathbb P^3}(-2)^2 \to \mathscr O_{\mathbb P^3} \to \mathscr O_X \to 0$$

Twisting by $d >>0$, we get $$0 \to \mathscr O_{\mathbb P^3}(-4+d) \to \mathscr O_{\mathbb P^3}(-2+d)^2 \to \mathscr O_{\mathbb P^3}(d) \to \mathscr O_X(d) \to 0$$ By additivity of Euler characteristics, the Hilbert polynomial of $X$ is given as the alternating sum $$\binom{3+d}{3}-2\binom{3+d-2}{3}+\binom{3+d-4}{3}=4d.$$ Putting $d=0$ gives $g=1$ by your formula.

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Here is another way to compute the genus, using Hurwitz' formula: Let $X$ be the curve again, thought of as sitting inside $\mathbb P^1 \times \mathbb P^1$. Then consider the projection to $\mathbb P^1$ This gives a $2:1$ map to $\mathbb P^1$, and by Hurwitz' formula we have $$2g-2=2(0-2) + \deg R,$$ where $\deg R$ is the degree of the ramification divisor, that is, degree of the defining equation of the points of $\mathbb P^1 \times \mathbb P^1$ where the map fails to be $2:1$. One can compute that this degree is $4$ (by the abc-formula). Hence $2g-2=-4+4$, which implies that $g=1$.