3
$\begingroup$

Let $f: X \to S$ be a smooth morphism of schemes of relative dimension $1$. Then $K=\Omega^1_{X/S}$ is a line bundle on $X$. I am interested in the following question:

When does $\Omega_{X/S}$ have a square root $K^{1/2}$ on $X$?

Example: start with a rank $2$ vector bundle $E$ over $S$, we could consider the projective bundle $\mathbb P(E) \to S$. Then we could compute $K$ via the relative Euler sequence (see Canonical sheaf projective bundle).

Motivation: Assume $f: X \to S$ is a family of smooth projective curves of genus $g>0$ over $\mathbb C$ (in particular, the fiber of $f$ at any closed point is a smooth projective curve over $\mathbb C$). Then fiberwise on $S$ the answer is yes.

For example, we could consider the universal curve over $M_g$ the moduli space of curves and consider good locus of $M_g$ where $K^{1/2}$ exists.

The case I am most interested in is when $S$ is also a smooth projective curve.

$\endgroup$
5
  • 2
    $\begingroup$ I don't think this has a simple answer. Take your example: $S$ a curve, $X=\mathbb{P}(E)$ with $E$ a rank 2 vector bundle; then $K_{X/S}$ has a square root if and only if $\deg E$ is even. $\endgroup$
    – abx
    Commented Dec 8 at 5:12
  • 2
    $\begingroup$ The "good locus" is a finite etale double cover (outside characteristic 2). $\endgroup$ Commented Dec 8 at 8:01
  • 2
    $\begingroup$ @Piotr Achinger: double?? I would say of degree $2^{2g}$ (or $2^{g-1}(2^g\pm 1)$, to be more precise). $\endgroup$
    – abx
    Commented Dec 8 at 11:29
  • $\begingroup$ As alluded by user @abx, these are classically known as "theta characteristics." They can be either "even" or "odd" depending on the parity of the dimension of the vector space of global sections. $\endgroup$ Commented Dec 8 at 14:26
  • $\begingroup$ Yes, $2^{2g}$ of course. $\endgroup$ Commented Dec 8 at 19:16

1 Answer 1

6
$\begingroup$

Let's ignore characteristic $2$.

If $f$ is smooth and proper, then we have a map $S \to \mathcal M_g$. There exists a finite étale cover of $\mathcal M_g$ of degree $2^{2g}$, with two components, parameterizing theta characteristics, as explained by Piotry Achinger, abx, and Jason Starr in the comments. A necessary condition for such a square root to exist is that the map $S \to \mathcal M_g$ lifts to this double cover.

In particular, since the degree of each component is at least $1$ as soon as $g>1$, such a square root does not exist over an open subset of $\mathcal M_g$ (and hence not over an open subset of $M_g$).

This necessary condition is not sufficient, as shown by abx in the genus $0$ case. A section of the étale cover means we have chosen, locally on $S$, an isomorphism class of a square root of the canonical bundle, but these may not glue as to glue them we need to choose, in intersections of two open sets in $S$, an isomorphism between square roots, and these choices must be compatible on intersections of three open sets. The obstruction to doing this is a cohomology class in $H^2_{\textrm{et}}(S, \mathbb Z/2)$, which pulls back from a universal class on the finite étale cover parameterizing theta characteristics, and a square root exists if and only if that class vanishes.

$\endgroup$
2
  • $\begingroup$ Thank you! If S is a Dedekind scheme say Spec O for an order O in number field, do we still have an obstruction theory (without referring to the moduli space)? $\endgroup$
    – Zhiyu
    Commented Dec 8 at 19:48
  • 1
    $\begingroup$ @Zhiyu The obstruction from the finite étale cover can also be expressed as $H^1$ of $S$ with coefficients in the $2$-torsion group scheme of the family of Jacobians of $X$ (since once you have a theta characteristic, the other theta characteristics are parameterized by $2$-torsion). $\endgroup$
    – Will Sawin
    Commented Dec 8 at 20:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .