Dual curves, bitangents, and jet bundles

Let ${C \subset \mathbb{P}^2}$ be a smooth degree ${d}$ curve. Then there is a dual curve

$\displaystyle C \rightarrow (\mathbb{P}^2)^*,$

which sends ${p \in C \mapsto \mathbb{T}_p C}$ , to the (projectivized) tangent line at ${p \in C}$ . Such lines live in the dual projective space ${(\mathbb{P}^2)^*}$ of lines in ${\mathbb{P}^2}$ . We will denote the image by ${C^* \subset \mathbb{P}^2}$ ; it is another irreducible curve, birational to ${C}$ .

This map is naturally of interest to us, because, for example, it lets us count bitangents. A bitangent to ${C}$ will correspond to a node of the image of the dual curve, or equivalently it will be a point in ${(\mathbb{P}^2)^*}$ where the dual map ${C \rightarrow (\mathbb{P}^2)^*}$ fails to be one-to-one. In fact, if ${C}$ is general, then ${C^*}$ will have only nodal and cuspidal singularities, and we we will be able to work out the degree of ${C^*}$ . By the genus formula, this will determine the number of nodes in ${C^*}$ and let us count bitangents.

The purpose of this post is to describe this, and to discuss this map from the point of view of jet bundles, discussed in the previous post.

1. Jet bundles and the dual map

Let ${J_1( \mathcal{O}_C(1))}$ be the first jet bundle of the hyperplane bundle ${\mathcal{O}_C(1)}$ : ${J_C(\mathcal{O}_C(1))}$ is a two-dimensional vector bundle on ${C}$ whose fibers over a point ${p \in C}$ record not only sections of ${\mathcal{O}_C(1)}$ , but their “derivatives” at ${p}$ : in other words, 1-jets. To compute with ${J_1( \mathcal{O}_C(1))}$ , we can use the exact sequence

$\displaystyle 0 \rightarrow K_C(1) \rightarrow J_1( \mathcal{O}_C(1)) \rightarrow \mathcal{O}_C(1) \rightarrow 0, \ \ \ \ \ (1)$

where the last map sends a 1-jet to its “value.” Moreover, given a global section of ${\mathcal{O}_C(1)}$ , we have (by “Taylor expansion”) a global section of ${J_1( \mathcal{O}_C(1))}$ .

Recall from the previous post that we have a map

$\displaystyle \mathcal{O}_C^3 \twoheadrightarrow J_1 \mathcal{O}_C(1) \rightarrow 0,$

where the three global sections of the jet bundle ${J_1 \mathcal{O}(1)}$ come from the global sections of ${\mathcal{O}_C(1)}$ , as before. The kernel ${\mathcal{L}}$ of this map is a one-dimensional subbundle of ${\mathcal{O}_C^3}$ whose fiber above a point is the tangent line.

This gives a description of the dual curve: the dual curve is the map ${C \rightarrow \mathbb{P}^2}$ corresponding to the line subbundle ${\mathcal{L} \subset \mathcal{O}_C^3}$ . In other words, we use the universal property of ${\mathbb{P}^2}$ : a map into ${\mathbb{P}^2}$ is equivalent to giving a line subbundle of ${\mathcal{O}_C^3}$ . (One could equivalently use line quotients; it is here that the “duality” appears.)

Proposition 1 The dual curve map ${C \rightarrow( \mathbb{P}^2)^*}$ has degree ${ d(d-1) }$ .

Proof: The dual curve map (or Gauss map) ${ C \rightarrow (\mathbb{P}^2)^*}$ has the property that ${\mathcal{O}(-1)}$ pulls back to the line bundle ${\mathcal{L}}$ on ${C}$ , which was the kernel of the surjection ${\mathcal{O}_C^3 \twoheadrightarrow J_1 \mathcal{O}_C(1)}$ . It thus suffices to compute the degree of the first Chern class of ${\mathcal{L}}$ , which is minus the first Chern class of ${J_1 \mathcal{O}_C(1)}$ .

To do so, observe that from the exact sequence (1), the degree of ${J_1( \mathcal{O}_C(1))}$ is

$\displaystyle \deg K_C(1) + d = (2g_C - 2) + 2d = (d-1)(d-2) - 2 + 2d,$

using the genus formula. This implies the claim. $\Box$

2. General properties of the dual map

In the previous section, we gave a definition of the dual map ${C \rightarrow ( \mathbb{P}^2)^*}$ in terms of jet bundles, and showed that the map had degree ${d(d-1)}$ . However, that in itself doesn’t determine the degree of the image: we don’t know that the map is birational onto its image, let alone what the singularities of its image ${C^* \subset ( \mathbb{P}^2)^*}$ might look like.

So we should start with the following result, which requires characteristic zero:

Proposition 2 The dual map ${C \rightarrow (\mathbb{P}^2)^*}$ is birational onto its image.

Equivalently, it suffices to show that the general tangent line to ${C}$ is not a bitangent.

Proof: Here is a rough geometric argument, which is based upon the result that the bidual of a smooth curve ${C \subset \mathbb{P}^2}$ is ${C}$ again. (The dual ${C^*}$ is not necessarily smooth, but one can still define a Gauss map ${C^* \rightarrow \mathbb{P}^2}$ away from the singular locus.)

To define the tangent line to ${C}$ at a point ${p \in C}$ , take a point ${q \in C}$ near ${p}$ , and consider the secant line ${\overline{pq}}$ : as ${q \rightarrow p}$ , this will approach the tangent line. Thus, to define the tangent line to ${C^*}$ at a point ${\ell \in ( \mathbb{P}^2)^*}$ , which is interpreted as a line ${\ell \subset \mathbb{P}^2}$ , take lines ${\ell' \in (\mathbb{P}^2)^*}$ near ${\ell}$ (which are in ${C^*}$ , so are tangent lines to ${C}$ at some point), and “draw the line through ${\ell}$ and ${\ell'}$ .” In ${( \mathbb{P}^2)^{**} = \mathbb{P}^2}$ , that corresponds to intersecting ${\ell \cap \ell'}$ .

So if ${\ell}$ was the projectivized tangent line ${\mathbb{T}_p C}$ for ${p \in C}$ , then ${\ell}$ will map to, in the bidual, the intersection of ${\mathbb{T}_p C}$ and ${\mathbb{T}_q C}$ for ${q \in C}$ close to ${p}$ . As ${q \rightarrow p}$ , this intersection tends to ${p}$ , so the bidual of ${C}$ is ${C^{**}}$ .

$\Box$

This already tells us something: now that we know the degree of ${C^*}$ , it tells us that the intersection of ${C^*}$ with a general line in ${( \mathbb{P}^2)^*}$ consists of ${d( d-1)}$ points. This means that if ${p \in \mathbb{P}^2}$ is a general point, there are ${d(d-1)}$ tangent lines to ${C}$ that pass through ${p}$ . We could see this (assuming birationality but without using Chern classes) as follows: if ${C}$ is given by the degree ${d}$ polynomial equation ${P(x, y, z) = 0}$ , then the line through ${p = [1: 0: 0]}$ and a point ${q \in C}$ is tangent to ${C}$ at ${q}$ if and only if

$\displaystyle \frac{\partial P}{\partial x}(q) = 0.$

In other words, the condition on ${q}$ that the tangent line through ${q}$ pass through ${p}$ is that a certain degree ${d-1}$ polynomial vanish on ${q}$ . So the collection of such ${q}$ is the intersection of ${P}$ with ${\frac{\partial P}{\partial x}}$ , which by Bezout’s theorem gives ${d( d-1)}$ .

To understand the singularities of the dual curve, we use the following result, which is a local calculation that we omit.

Proposition 3 If ${p \in C}$ is not a flex point, then the Gauss map ${C \rightarrow ( \mathbb{P}^2)^*}$ is an immersion at ${p}$ . If ${p \in C}$ is a flex but not a hyperflex, then the dual curve ${C^*}$ has an ordinary cusp at the image of ${p}$ .

3. The Plücker formulas

Let ${C \subset \mathbb{P}^2}$ be a smooth curve. In the previous section, we showed that the dual ${C^* \subset \mathbb{P}^2}$ , and stated that if ${C}$ was general (no hyperflexes), then ${C^*}$ was not too singular: it had only nodes and cusps, with the nodes occurring at bitangents and cusps at flex lines.

We know now that the degree of ${C^*}$ is ${d( d-1)}$ , and that ${C^*}$ is birational to ${C}$ , so the normalization has genus ${g}$ . In other words, ${C^*}$ is a plane curve of degree ${d(d-1)}$ with ${b}$ nodes and ${f }$ cusps, if ${C}$ has ${b}$ bitangents and ${f}$ flexes. It follows that we have the Plücker formula

$\displaystyle \frac{(d-1)(d-2)}{2} = g(C) = g(C^*) = \frac{(d^2 - d - 1)(d^2 - d - 2)}{2} - b - f,$

because each node and each cusp reduces the genus of the normalization of a plane curve by one from the “expected” one.

However, in the previous post, we showed that for a general plane curve of degree ${d}$ ,

$\displaystyle f = 3d( d - 2),$

so that this formula enables us to work out the number of bitangents.

For a plane quartic, we have ${d = 4}$ and the genus is three; the degree of the dual curve is ${12}$ , which gives

$\displaystyle 3 = 55 - b - f,$

and we showed in the previous post that ${f = 24}$ , which gives ${b = 28}$ as desired.

Jason Starr Says:

June 28, 2013 at 2:26 pm

You can also compute 28 = 27 + 1, i.e., the 27 lines on a cubic surface together with one more (an exceptional divisor from blowing up one more point on the cubic surface). For a smooth cubic surface X and a general point p on X, projection of X away from p realizes the blowing up of X at p as a degree 2 cover of P^2 branched on a plane quartic. The -1 curves on the blowing up map isomorphically to bitangent lines of the plane quartic.

Akhil Mathew Says:

June 29, 2013 at 8:34 am
Very interesting! I was curious about the connection with the 27 lines.

Climbing Mount Bourbaki