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In the below problem I want to answer second part:

Show that the curve in $\mathbb{P}^3$ defined by the two equations $x_0 x_3 = x_1 x_2$ and $x_0^2 + x_1^2 + x_2^2 + x_3^2 = 0$ is a smooth complete intersection curve. What is its topological genus?

but I don't know how can I calculate it.

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If you are over $\mathbb C$, since the curve is smooth, you can use Corollary 2 of this paper; because arithmetic and geometric genus coincide in that case. If you have to change base, this approach seems more complicated... –  Jesko Hüttenhain Jan 16 '13 at 8:24
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Related mathoverflow.net/questions/55312/… –  Ehsan M. Kermani Jan 17 '13 at 8:18

1 Answer 1

up vote 6 down vote accepted

The jacobian $2\times 4$ matrix $Jac_P(F_1,F_2)=(\frac {\partial F_i}{\partial x_j}(P))$ has rank $2$ at every point $P\in C$ of the intersection curve $C=V(F_1,F_2)$.
This follows from an easy calculation and has as a consequence that $C$ is smooth.
The topological genus of $C$ is then equal to its arithmetic genus.
The arithmetic genus of the complete intersection of two surfaces of degrees $a,b$ in $\mathbb P^3_\mathbb C$ is given by the formula $$p_a(C)=\frac {1}{2}ab(a+b-4)+1$$ [cf. Hartshorne, Chapter I, exercise 7.2 (d), page 54]
In your case $a=b=2$ and the formula shows that the required genus of your curve is $1$.

The above reasoning shows that more generally the smooth complete intersection of two quadrics in projective $3$-space always has genus one i.e. is an elliptic curve : this a well-known and fundamental result in the theory of space curves.

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