Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be a smooth projective curve of genus $g\geq2$ over an algebraically closed field $k$ and denote by $K$ a canonical divisor.

I have some clues about the geometrical interpretation of the Riemann-Roch Theorem for smooth algebraic curves, but also some doubts which I would like to clarify. Recall that the RR formula is $$ h^0(X,\,D)-h^0(X,\,K-D) = d-g+1\,. $$

Assume that $X$ is not hyperelliptic, so that the canonical map is actually a canonical embedding $$ \phi_K : X \to \mathbb{P}^{g-1} \qquad P\mapsto\{ \; s\in H^0(X,\,K) \mid s(P)=0 \; \} $$ giving a preferred realization of the curve inside a $(g-1)$-dimensional projective space.

The key feature of such an embedding is that there is a bijective correspondence between hyperplanes $W\subset \mathbb{P}^{g-1}$ and effective divisors in the linear system $ |K| \cong \mathbb{P}H^0(X,\,K) $.

The picture shows the canonical embedding in $\mathbb{P}^2$ of a non hyperelliptic curve of genus $3$.

The canonical embedding in the plane of a non hyperelliptic curve of genus 3

Let $D=\sum_{i=1}^d P_i$ be an effective divisor consisting of $d<g$ distinct points of $X$. We define $$ \phi_K(D) := \operatorname{span}\{\phi_K(P_1), \dots, \phi_K(P_d)\}. $$

The vector space $H^0(X,\,K-D)$ can be interpreted as the space of canonical divisors containing $D$, and here comes my first question:

(1) Is it correct to identify $\mathbb{P}H^0(X,\,K-D)$ with the set of hyperplanes of $\mathbb{P}^{g-1}$ passing through $\phi_K(D)$ ? If so, how can one see it formally?

Let $r(D) := \dim |D|$ denote the dimension of the complete linear series associated to $D$. Further, denote by $D'=K-D$ the residual divisor of degree $d'=2g-2-d$.

If (1) is correct, then it follows that $r(D)$ equals the number of hyperplanes of $\mathbb{P}^{g-1}$ passing through $\phi_K(D')$. Now, notice that the RR can be rewritten as

$$ r(D)=[g-1]-[d' - r(D')] $$

so that we deduce that $r(D')$ counts the number of independent linear relations on the points of $D'$ and we can give the following geometrical interpretation of the Riemann-Roch:

The integer $r(D)$ is the number of hyperplanes passing though $\phi_K(D')$, hence it equals the difference between the dimension $g-1$ of the ambient space and the dimension of the space spanned by the points of $\phi_K(D')$.

Of course my second question is:

(2) Do you agree with this geometrical interpretation?

share|improve this question
Nice question. A side remark: your picture shows a plane quartic (hence of genus $3$, not $4$)! on such a curve a canonical divisor is exactly what you drew: $4$ points on a line. –  Brenin Feb 20 at 11:12
Thanks, that was a bad typo! –  Abramo Feb 20 at 11:27
Btw, I agree with your interpretation. I do not know if this is precise enough, but a way to see it would be to just observe that $H^0(X,K-D)$ is the subspace of $H^0(X,K)$ consisting of sections vanishing along (the support of) $D$ according to the multiplicities of the (supporting) points in $D$. What you wrote is the projectivized ($\mathbb P$) version of this. –  Brenin Feb 20 at 12:31
What a beautiful, large, coloured picture: just like a picture in algebraic geometry should be ! +1 –  Georges Elencwajg Feb 20 at 22:10
Thanks Georges! Sometimes I feel like the best way to give my contribution to the field could be by just producing pictures! –  Abramo Feb 21 at 14:21

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.