Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a divisor $D$ on an algebraic curve $X$, there is a corresponding map $\phi_D$ from $X$ to the projective space (of dimension $\dim L(D)-1$). In particular, we know that if $D$ is a very ample divisor, then this is a holomorphic embedding into projective space. When $D$ is not very ample, how does one read off information about the map?

Here is a specific problem from Miranda's textbook to provide context:

Let $X$ be an algebraic curve of genus $2$. Let $K$ be a canonical divisor on $X$, and let $\phi_{2K}$ be the associated map to projective space. Show that the image of $\phi_{2K}$ is a smooth projective plane conic, and that the map has degree $2$.

Here are my thoughts so far. We know that $\deg(2K) = 4g-4 = 4$. By Riemann-Roch, we also have $\deg L(2K) = \deg(2K) + 1 - g = 3$. This would suggest that the image is in $\mathbb(P)^2$, and by the degree it could be a double covering of a conic, an embedding as a degree $4$ curve, or perhaps a quadruple covering of a line?

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.