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

It is a classical fact that there are bitangents on a smooth plane quartic. In other words, of the theta characteristics, exactly of them are effective. A *bitangent* here will mean a line such that the intersection is a divisor of the form for points, not necessarily distinct. So a line intersecting 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 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 be a smooth curve and on a line bundle. Then, given , there is a -dimensional **jet bundle** , which is a vector bundle on whose fiber over a point consists of -jets of at : equivalently, this is the vector bundle

To make this precise, one way is to use the identification of the symmetric power with the Hilbert scheme of length subschemes of ; one has a natural map

which, in terms of the definition of the Hilbert scheme, is given by the subscheme of which is the diagonal with multiplicity . Now, given , there is a -dimensional vector bundle on which sends a divisor of degree (which is what parametrizes) to the -dimensional vector space

more precisely, if we consider the universal subscheme , then above vector bundle on is given by

for the projections from 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 has a natural filtration whose subquotients are given by the line bundles . These line bundles are precisely : for example, when and is trivial, the line bundle sends

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 ,

While these need not be split, they do (inductively) determine the topological type of , 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 of a given degree . A flex line, by definition, is a line which meets with order of contact at a point.

Let’s try to rephrase the above problem in the language of jet bundles. We have a line bundle , with linearly independent sections , so that a line in is simply a linear combination of these (up to scaling). Now, given a line bundle on , a *global* section of certainly defines global sections of for each ; this operation associates to a global section its Taylor expansion (to some order ) at each point.

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

or equivalently, three global sections of : namely, it sends a global line on to the Taylor expansion up to order 3 at a given point . By definition, is a flex point precisely when there is a line intersecting to order , which means that the line maps to zero in .

In other words, we have a three-dimensional vector bundle on , and three global sections of ; 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 of vanishes, and the number of points in the vanishing locus is the **degree** or **first Chern class** of . So, the number of points where fail to be independent in the fiber of the jet bundle is

Topologically, one has

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

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

so that the degree of this line bundle on , or the number of flex points, is

Taking , 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 , while there is not necessarily a canonical basepoint to make into an elliptic curve, there is a natural space of nine possible choices, corresponding to the flex points.) For , the formula gives 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 , 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 , the multiplicity of the vanishing of the section of is greater than one. In other words, the result is:

Theorem 1If is a smooth curve of degree with no hyperflexes, then has flex points.

To make this theorem non-vacuous, we should claim that the general degree curve has no hyperflexes. To see this, let be the space of degree smooth curves (an open subset in a projective space). We consider the collection of triples where:

- is a degree curve.
- is a point along the line .

This is flat over with fibers given by a flag variety, so has dimension . Now consider the subvariety where we require that , which cuts down the dimension by 1; so . But that’s not quite we want. Impose the stronger condition that has length at least four, to get a subvariety . To compute the dimension of , map to the flag variety, so that the fiber of above consists of degree curves that meet at with contact to order . That is four linear conditions, so that . In particular, the image of is a proper subvariety; that’s equivalent to saying that most degree curves have no hyperflexes.

**ramification**of a linear series, and thus count objects such as Weierstrass points.

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