Geometric interpretation of the Riemann-Roch for curves

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 $3$. 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?

• 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 '14 at 11:12
• Thanks, that was a bad typo! – Abramo Feb 20 '14 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 '14 at 12:31
• What a beautiful, large, coloured picture: just like a picture in algebraic geometry should be ! +1 – Georges Elencwajg Feb 20 '14 at 22:10
• In case you don't know where to find the flag responses: Questions older than 60 days cannot be migrated. You could however re-ask. After not getting an answer for a long enough time, that's legitimate. One year is more than long enough. – Daniel Fischer Mar 11 '15 at 23:53

The answer to your first question is yes. As for the second question I would make a minor adjustment but the idea is correct. Please be patient with me as I set things up first.

Notation

Let $\omega_C$ be the sheaf of holomorphic differential forms on $C$. For any vector space $V$ I will denote by $|V|$ the projective space of lines in $V$ and by $\mathbb{P}(V)$ the projective space of codimension 1 planes in $V$, henceforth referred to as hyperplanes.

Throughout I will let $V = H^0(C,\omega_C)$ stand for the global holomorphic differentials on $C$. For a divisor $D$ on $C$ let $V(-D) = H^0(C,\omega_C(-D))$ which is to be viewed as a subspace of $V$ using the natural injection $\omega_C(-D) \hookrightarrow \omega_C$.

Let me write the map $\varphi : C \to \mathbb{P}^{g-1}$ with the proper notation, because as things stand the notation suggests that generators for $V$ have been chosen. The preferred version, as you put it, can be written as $\varphi: C \to \mathbb{P}(V)$ where $$p \mapsto V_p := \mathrm{ker}(V \to \omega_C \to \omega_C|_p).$$ It is clear that $V(-p) \subset V_p$ but Riemann-Roch says that the dimensions match so we get $V(-p) = V_p$.

Basic observations

Given $q \in \mathbb{P}(V)$ we get a hyperplane $H_q \subset |V|$ by duality. Here, $H_q$ parametrizes hyperplanes in $\mathbb{P}(V)$ containing $q$. Given two points $q_1, q_2$ the line between $q_1$ and $q_2$ can be viewed as the intersection of all hyperplanes containing $q_i$'s. This is the dual of $H_{q_1}\cap H_{q_2}$. And similarly for more points. So far this is linear algebra, let us apply it to our specific setting.

Our description of the map $\varphi$ ensures that $H_{\varphi(p)} = |V(-p)|$. Given $p_1, p_2 \in C$ the line between $\varphi(p_1)$ and $\varphi(p_2)$ has dual $|V(-p_1)|\cap|V(-p_2)|$ which is easily seen to be $|V(-p_1-p_2)|$ (Check!). And if $D$ is reduced, this argument is sufficient to conclude that $|V(-D)| \subset |V|$ parametrizes the hyperplanes through $\varphi(D)$.

It is possible, but difficult, to describe the locus of hyperplanes having high order contact at a point $p \in C$ using just geometry. So we go back to doing a little more tautology.

If you give me any divisor $D \in |\omega_C| = |V|$ then this gives a line in $H^0(C,\omega_C)$. Pick a section $\sigma$ generating this line. By the standard relations $D = (\sigma)_0$. Now this point $D$ has a dual hyperplane $W_D \subset \mathbb{P}(V)$ consisting of all hyperplanes in $|V|$ containing $D$. I claim that $W_D \cdot \varphi(C) = D$.
Indeed, $W_D$ corresponds to a divisor in the linear system of the tautological line bundle $\mathcal{O}(1)$ which by construction has global sections canonically isomorphic to $V$. Furthermore, the divisor $W_D$ corresponds precisely to the (line generated by) $\sigma \in V$. Now the intersection of $W_D$ with $C$ can be obtained by pulling back $(\mathcal{O}(1),\sigma)$ to $C$. However, the pullback of $\mathcal{O}(1)$ is of course (canonically isomorphic to) $\omega_C$ and the section $\sigma$ maps to $\sigma$ again by canonical identifications. This proves the desired statement.
So far what we have been doing was largely exploratory. Now let's tackle your question head on. The inclusion $V(-D) \hookrightarrow V$ identifies sections of $\omega_C(-D)$ with sections of $\omega_C$ that vanish on $D$ (this follows from the exact sequence obtained from $\omega_C(-D) \hookrightarrow \omega_C$). Therefore if we take $\sigma \in V(-D)$ then the corresponding hyperplane $H \subset \mathbb{P}(V)$ satisfies $H \cdot C \ge D$. Conversely, any hyperplane that satisfies $H \cdot C \ge D$ corresponds to (the line generated by) a section of $\omega_C$ vanishing on $D$, hence to a section of $\omega_C(-D)$. This answers your first question.
As for the second question, if $d' = \mathrm{deg} D'$ and $d' < g$ then we expect $\varphi(D')$ to span a $d'-1$ dimensional projective subspace. Then your calculation shows that $r(D')$ in fact measures the failure of our expectation. More precisely, $r(D') = d'-1 - \mathrm{dim}(\mathrm{span}(\varphi(D')))$. The space of hyperplanes in $\mathbb{P}^{g-1}$ passing through a $k$-dimensional projective space has dimension $(g-2) - k.$ Then putting these together we get: $$r(D) = (g-2) - (d'-1 - r(D')) = (g-1) - (d' - r(D'))$$