Concise version of the question

On a generic curve of genus $g$, the odd theta characteristics will have exactly one global section. Therefore each odd theta characteristic will correspond to a unique effective divisor. The support of this effective divisor, that is the set of points appearing in the sum, consists of $g-1$ points. We will call this set the support of the odd theta characteristic.

So we ask:

Given two odd theta characteristics on a generic curve, is the support of the two disjoint?

Geometry and Motivation

Take a canonically embedded generic curve of genus $g$ and denote it by $C \hookrightarrow \mathbb{P}^{g-1}$. There are precisely $2^{g-1} (2^g -1) $ hyperplanes in the ambient space which have contact of order 2 at every point of intersection with the curve $C$. Such a hyperplane is often called a bitangent. Lets call a point of intersection of a bitangent with the curve, a point of contact.

One way to formulate the question is:

Can two bitangents share a common point of contanct?

A note on genericity of the curve: If the curve is not general, the statement is false. Moreover, on certain curves what we want to call a 'bitangent' can have higher order contact, and it could move in families. We avoid all that.

In this generic situation, odd theta characteristics and bitangents correspond to one another. Given a bitangent, take the reduced sum of points of contact to get the odd theta characteristic.

The set points which appear in a divisor is called the support of that divisor. The unique hyperplane passing through the points in the support of an odd theta characteristic will give us the bitangent corresponding to it.

With this point of view we can extend the question so that it would apply to genera 1 and 2, but is essentially equivalent:

Is the support of two odd theta characteristics disjoint?

Low genera

The statement is true on generic curves of genera 1,2,3 and I can show that it is almost true on genera 4 and 5. The arguments are easy so I give them below:

Genus 1: The trivial bundle is the only odd theta characteristic, hence the statement is void and thus true.

Genus 2: The odd theta characteristics (otc) are in correspondence with the Weierstrass points. The support of an otc being one of the Weierstrass points. These are disjoint.

Genus 3: The geometric picture helps here. Of course a bitangent line is determined by any one of its points of contact, the tangent line at that point happens to be the bitangent.

Genus 4: A Brill-Noether general curve will not allow any two of its points to share a tangent line in the canonical embedding. Thus, if $H$ is tangent to the canonical curve $C \subset \mathbb{P}^3$ at two points $p$ and $q$ then the tangent lines at those two points determine $H$. Which means that no two bitangents could share two points of contact. But I don't know how to eliminate a single point in common.

Genus 5: Similar to the argument in Genus 4. Two points of contact will determine the tangent 3-space because the tangent lines at those points will be skew. Hence no two distinct bitangents could share 2 points of contact. Again I don't know how to extend this to a single point.

This argument can be extended to higher genera, but the result is successively less impressive. Hoping that one single classical idea can be an inspiration to solve it for all cases, I propose the

Baby question: Is there a classical solution for the main question in genus 4?


1 Answer 1


On the general curve (for $g$ large), i.e. the curve defined over the function field $K$ of $M_g$, the only divisors defined over $K$ are the multiples of the canonical divisor. This is a theorem of Hain, I think. So you cannot have a distinguished point (or a distinguished divisor of low degree) on the general curve. This may be overkill but proves what you want.

  • $\begingroup$ I think the question makes sense on the geometric generic curve, i.e. defined over $\bar{K}$. By the way, the result you use is actually due to John Harer, see Arbarello et al., ch. XIX for a very detailed discussion. $\endgroup$
    – abx
    Commented Sep 17, 2014 at 20:16
  • $\begingroup$ @abx If the point is unique, it's defined over $K$. If there is more than one, then I guess you are right but then it's even more unlikely. Thanks for the correct attribution. $\endgroup$ Commented Sep 18, 2014 at 0:58
  • $\begingroup$ Since there are $2^{g-1}(2^g-1)$ odd theta characteristics, each with $g-1$ points, one can not rule out the possibility that in fact quite a lot of points end up being in the common support of two odd theta characteristics. In any case, a point defined over $K$ corresponds to a global section of stack $U_g \to M_g$ (R. Hain - Rational points of universal curves, p. 7). It should be impossible to realize this distinguished point as a section, since on the hyperelliptic curves all the fixed points of the involution are common to many odd theta characteristics. Could you clarify further please? $\endgroup$
    – Emre
    Commented Sep 18, 2014 at 11:17
  • $\begingroup$ @Torquemada I don't understand your final objection. I can't rule out lots of points, yes. But as you show in genus four, ruling out lots of points should be easier. You should reconsider your username, BTW. $\endgroup$ Commented Sep 18, 2014 at 13:51

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