5
$\begingroup$

I suffer from lack of concrete examples in Algebraic Geometry, so I will appreciate it if somebody can help me in understanding a bit better this one:

Let $\mathcal{C}$ be a genus $2$ curve (projective, smooth and complex). Let $K$ be the canonical class and take a couple of different points $p,q\in\mathcal{C}$. Consider the divisor $K+p+q$.

If I'm correct, (using Riemann-Roch theorem) this divisor defines a map taking $\mathcal{C}$ into a plane projective curve with one node. Is this map birational? How can I visualize it in a concrete way?. Many thanks for any useful comments.

$\endgroup$
4
$\begingroup$

This is a very nice example! In fact, provided that $p+q$ is not a canonical divisor, the linear system $K+p+q$ separates "most" points, since $K$ is the only $g^1_2$ on the curve. Thus for any $r$ and $s$ not equal to $p$ and $q$, $\ell(K+p+q-r-s) = 1 = 3-2$, so $r$ and $s$ are separated by the linear system (the dimension drops twice, so you know that $s$ is not in every divisor of the linear system containing $r$, and thus $s$ maps to a different point). The only pair of points for which this doesn't happen is $p$ and $q$ themselves! So the image of $q$ is in every hyperplane ($=$line) containing the image of $p$ (note that imposing that a hyperplane section contain $p$ drops the dimension by $1$, from $3$ to $2$, but further imposing that it contain $q$ leaves the dimension at $2$). This shows that indeed the mapping is birational and that there is exactly one node, which is the image of both $p$ and $q$, in the image curve.

$\endgroup$

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.