Let ${C \subset \mathbb{P}^2}$ be a smooth plane quartic, so that ${C}$ is a nonhyperelliptic genus 3 curve imbedded canonically. In the previous post, we saw that bitangent lines to ${C}$ were in natural bijection with effective theta characteristics on ${C}$, or equivalently spin structures (or framings) of the underlying smooth manifold.

It is a classical fact that there are ${28}$ bitangents on a smooth plane quartic. In other words, of the ${64}$ theta characteristics, exactly ${28}$ of them are effective. A bitangent here will mean a line ${L \subset \mathbb{P}^2}$ such that the intersection ${L \cap C}$ is a divisor of the form ${2(p + q)}$ for ${p, q \in C}$ points, not necessarily distinct. So a line intersecting ${C}$ in a single point (with contact necessarily to order four) is counted as a bitangent line. In this post, I’d like to discuss a proof of a closely related claim, that there are ${24}$ flex lines. This is a special case of the Plücker formulas, and this post will describe a couple of the relevant ideas.

1. Jet bundles on curves

Let ${C}$ be a smooth curve and ${\mathcal{L}}$ on ${C}$ a line bundle. Then, given ${k \geq 1}$, there is a ${k}$-dimensional jet bundle ${J_{k-1} L}$, which is a vector bundle on ${C}$ whose fiber over a point ${p \in C}$ consists of ${k-1}$-jets of ${L}$ at ${p}$: equivalently, this is the vector bundle

$\displaystyle p \mapsto H^0( \mathcal{L}/\mathcal{L}(-kp)).$

To make this precise, one way is to use the identification of the symmetric power ${\mathrm{Sym}^k C}$ with the Hilbert scheme of length ${k}$ subschemes of ${C}$; one has a natural map

$\displaystyle C \rightarrow \mathrm{Sym}^k C , \quad p \mapsto (p, p, \dots, p),$

which, in terms of the definition of the Hilbert scheme, is given by the subscheme of ${C \times C}$ which is the diagonal with multiplicity ${k}$. Now, given ${\mathcal{L} \in \mathrm{Pic}(C)}$, there is a ${k}$-dimensional vector bundle ${V_{\mathcal{L}}}$ on ${\mathrm{Sym}^k C}$ which sends a divisor ${D}$ of degree ${k}$ (which is what ${\mathrm{Sym}^k C}$ parametrizes) to the ${k}$-dimensional vector space

$\displaystyle D \mapsto H^0( \mathcal{L}/\mathcal{L}(-D));$

more precisely, if we consider the universal subscheme ${U \subset C \times \mathrm{Sym}^k C}$, then above vector bundle on ${\mathrm{Sym}^k C}$ is given by

$\displaystyle V_{\mathcal{L}} = \pi_{2*} (\pi_1^* \mathcal{L} \otimes \mathcal{O}_U ),$

for ${\pi_1, \pi_2}$ the projections from ${C \times \mathrm{Sym}^k C}$ on each factor. The above definition and discussion are valid only for curves, but the definition of the jet bundles can be extended to any smooth variety.

To compute with the jet bundle, we note that ${J_k L}$ has a natural filtration whose subquotients are given by the line bundles ${p \mapsto H^0( \mathcal{L}(- mp)/\mathcal{L}(-(m+1)p))}$. These line bundles are precisely ${K^{\otimes m} \otimes \mathcal{L}}$: for example, when ${m = 1}$ and ${\mathcal{L}}$ is trivial, the line bundle sends

$\displaystyle p \mapsto \mathcal{O}(-p)/\mathcal{O}(-2p),$

and this is precisely the definition of the cotangent bundle. In other words, from this filtration, we find that there are exact sequences of vector bundles on ${C}$,

$\displaystyle 0 \rightarrow K_C^k \otimes \mathcal{L} \rightarrow J_k \mathcal{L} \rightarrow J_{k-1} \mathcal{L} \rightarrow 0.$

While these need not be split, they do (inductively) determine the topological type of ${J_k \mathcal{L}}$, and enable (for instance) calculation of the Chern classes.

2. Flex lines

The construction of jet bundles plays a fundamental role in solving problems of contact order. As an application, let’s consider (informally) the problem of counting flex points on a general plane curve ${C \subset \mathbb{P}^2}$ of a given degree ${d}$. A flex line, by definition, is a line ${L \subset \mathbb{P}^2}$ which meets ${C}$ with order of contact ${\geq 2}$ at a point.

Let’s try to rephrase the above problem in the language of jet bundles. We have a line bundle ${\mathcal{O}_C(1)}$, with ${3}$ linearly independent sections ${X, Y, Z}$, so that a line in ${\mathbb{P}^2}$ is simply a linear combination of these (up to scaling). Now, given a line bundle ${\mathcal{L}}$ on ${\mathcal{C}}$, a global section of ${\mathcal{L}}$ certainly defines global sections of ${J_k \mathcal{L}}$ for each ${k}$; this operation associates to a global section its Taylor expansion (to some order ${k}$) at each point.

The upshot of this is that we get a map of vector bundles

$\displaystyle \mathbb{C}\left\{X, Y, Z\right\} \otimes \mathcal{O}_C = H^0(C, \mathcal{O}_C(1)) \otimes \mathcal{O}_C \rightarrow J_3 \mathcal{O}(1),$

or equivalently, three global sections of ${J_3 \mathcal{O}(1)}$: namely, it sends a global line on ${\mathbb{P}^2}$ to the Taylor expansion up to order 3 at a given point ${p \in C}$. By definition, ${p}$ is a flex point precisely when there is a line intersecting ${p}$ to order ${\geq 3}$, which means that the line maps to zero in ${J_2 \mathcal{O}(1)}$.

In other words, we have a three-dimensional vector bundle ${J_2 \mathcal{O}(1)}$ on ${C}$, and three global sections ${X, Y, Z}$ of ${J_2 \mathcal{O}(1)}$; we’d like to ask what the locus where they fail to be independent is: that is the locus of flex lines. In fact, that locus is precisely where the section ${X \wedge Y \wedge Z}$ of ${\bigwedge^3 J_2 \mathcal{O}(1)}$ vanishes, and the number of points in the vanishing locus is the degree or first Chern class of ${\bigwedge^3 J_2 \mathcal{O}(1)}$. So, the number of points where ${X, Y, Z}$ fail to be independent in the fiber of the jet bundle ${J_2}$ is

$\displaystyle c_1( \bigwedge^3 J_2 \mathcal{O}(1)) = c_1( J_2 \mathcal{O}(1)).$

Topologically, one has

$\displaystyle J_2 \mathcal{O}_C(1) \sim \mathcal{O}_C(1) \oplus( K_C \otimes \mathcal{O}_C(1)) \oplus (K_C^{2} \otimes \mathcal{O}_C(1)),$

although this need not be true algebraically: the above is true only in the setting of topological bundles, or (better) in the Grothendieck group of algebraic vector bundles. However, using the adjunction relation

$\displaystyle K_C \simeq \mathcal{O}_C(d - 3),$

we now find that (even as algebraic line bundles),

$\displaystyle \bigwedge^3 J_2 \mathcal{O}_C(1) \simeq \mathcal{O}_C(3 + 3(d-3) ),$

so that the degree of this line bundle on ${C}$, or the number of flex points, is

$\displaystyle d( 3d - 6).$

Taking ${d = 3}$, we get the classical nine flex points on a smooth cubic: these correspond to the 3-torsion points of an elliptic curve under the usual imbedding. (This shows that, for an abstract plane cubic ${C}$, while there is not necessarily a canonical basepoint to make ${C}$ into an elliptic curve, there is a natural space of nine possible choices, corresponding to the flex points.) For ${d = 4}$, the formula gives ${24}$ flexes on a smooth quartic curve.

In the above informal argument, there is a serious ignored issue of multiplicities. The argument was that the three-dimensional space of linear forms gave a canonical element of ${\bigwedge^3 J_2 \mathcal{O}(1)}$, which vanished precisely at the flexes. However, we didn’t count the multiplicities. A more detailed local analysis with jet bundles would show that at a hyperflex, where there is a line of order of contact ${\geq 4}$, the multiplicity of the vanishing of the section of ${\bigwedge^3 J_2 \mathcal{O}(1)}$ is greater than one. In other words, the result is:

Theorem 1 If ${C \subset \mathbb{P}^2}$ is a smooth curve of degree ${d}$ with no hyperflexes, then ${C}$ has ${d( 3d - 6)}$ flex points.

To make this theorem non-vacuous, we should claim that the general degree ${d \geq 2}$ curve has no hyperflexes. To see this, let ${X}$ be the space of degree ${d}$ smooth curves (an open subset in a projective space). We consider the collection ${Y}$ of triples ${(C, p, L)}$ where:

• ${C}$ is a degree ${d}$ curve.
• ${p \in \mathbb{P}^2}$ is a point along the line ${L \subset \mathbb{P}^2}$.

This is flat over ${X}$ with fibers given by a flag variety, so ${Y}$ has dimension ${\dim X + 3}$. Now consider the subvariety ${Y_1 \subset Y}$ where we require that ${p \in C}$, which cuts down the dimension by 1; so ${\dim Y_1 = \dim X + 2}$. But that’s not quite we want. Impose the stronger condition that ${C \cap L}$ has length at least four, to get a subvariety ${Y_2 \subset Y}$. To compute the dimension of ${Y_2}$, map ${Y_2}$ to the flag variety, so that the fiber of ${Y_2}$ above ${(L, p)}$ consists of degree ${d}$ curves that meet ${L}$ at ${p}$ with contact to order ${\geq 4}$. That is four linear conditions, so that ${\dim Y_2 = \dim Y - 4 = \dim X - 1}$. In particular, the image of ${Y_2 \rightarrow X}$ is a proper subvariety; that’s equivalent to saying that most degree ${d}$ curves have no hyperflexes.

These ideas can be extended considerably (even for curves); for instance, they can be used to study the notion of ramification of a linear series, and thus count objects such as Weierstrass points.