Consider a holomoprhic map from a Riemann surface $$ f: \Sigma_g \to \mathbb{CP}^n. $$ This is given by some homogeneous polynomials in some variables.

How can we show that the homogeneous degree $d$ of these polynomials coincides with the definition of degree from fundamental homology class, namely $d = f_* ([\Sigma_g]) \in H_2(\mathbb{CP}^n;\mathbb{Z})$?


1 Answer 1


Let me rename $C=\Sigma_g$, as the genus is unimportant here. To give a map from a smooth complex projective curve $C$ to $\mathbb P^n$ is the same as to give a line bundle $L$ on $C$. This line bundle can be recovered up to isomorphism as $$L=f^\ast(\mathscr O_{\mathbb P^n}(1))=g^\ast(\mathscr O_{\mathbb P^n}(1)|_D),$$ where I have denoted by $g$ the same map $f$, but restricted to the image $D=f(C)$; in other words we factor $f$ as follows: $$C\overset{g}{\longrightarrow}D\hookrightarrow\mathbb P^n.$$ On $\mathbb P^n$, the one-cycle $f_\ast[C]$ has degree $$\deg f_\ast[C]=\deg g_\ast[C]=(\deg g)\cdot(\deg D).$$ On the other hand, the degree of the polynomials defining $f$ is the degree of $L$, which is $$\deg L=(\deg g)\cdot \deg \mathscr O_{\mathbb P^n}(1)|_D=(\deg g)\cdot(\deg D).$$

  • $\begingroup$ Thanks for the answer. You mean two-cycle, right? So, requiring that $f$ maps to a general quintic in $\mathbb{CP}^4$ would amount to take $L=f^*(\mathcal{O}_{\mathbb{CP}^4}(5))$, correct? Moreover, is it possible to understand what is the effect, at the level of line bundles (or divisors) of making $f$ an embedding (->very ample?), or making $f$ equivariant with respect to some involutions on $C$ and $\mathbb{CP}^n$, or again deforming one $f$ into another, e.g. when smoothing a node? $\endgroup$
    – jj_p
    Mar 18, 2014 at 12:15
  • 1
    $\begingroup$ Sorry for the confusion but I called "curve" what you called "surface"! so yes, two-cycles is fine. And yes, your $L$ looks fine (it has the correct degree). "Very ample" is the key word. Sorry I do not feel able to say more about your comment! Maybe post a new question and someone will! :) $\endgroup$
    – Brenin
    Mar 18, 2014 at 22:13

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