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Let $g\geq 3$. Following the reference below, the locus of curves in $M_g$ with an effective even theta characteristic has codimension $1$. (Those are the curves $C$ with an effective line bundle $L$ such that $L\otimes L=K$ and $h^0(L)\equiv 0\mod 2$)

(QUESTION) Are there complete curves in $M_g$ that avoid this locus? I know that this is not the case when $g=3$ because the curves with an effective even theta characteristic are hyper-elliptic and any complete curve in $M_3$ must intersect the hyper-elliptic locus.

The main reason I ask is because I'd like to construct surfaces in $M_{2g}$ as follows: Take a family $F\rightarrow B$ of genus $g$ curves that avoids all theta-null divisors, i.e. all half-canonical bundles have $0$ or $1$ section. Since the parity of $h^0(K^{1/2})$ is constant in families, the ineffective theta characteristics are locally constant and form an etale cover of $B$. There is a map $F\times_BF\rightarrow Jac_BF$ sending $(x,y)\mapsto O(x-y)$ which contracts the diagonal to the zero section. After a suitable base change, one can compose with a map $$Jac_BF\rightarrow Pic_B^{g-1}F$$ sending $L\rightarrow L\otimes K^{1/2}$ for an ineffective $K^{1/2}$. Thus, under the map $$\phi:F\times_BF\rightarrow Pic^{g-1}_BF$$ $\phi(\Delta)$ does not intersect the theta divisor $\Theta\subset Pic^{g-1}_BF$ (excuse the overuse of the word theta!)!

Hence $\phi^{-1}(\Theta)$ parameterizes a surface of distinct pairs of points in $F_b$. After another base change, one can take the double branched cover of $F_b$ branched at these two points and get a surface inside $M_{2g}$. Will this technique ever work to construct surfaces in $M_{2g}$? That, by the way, is the motivation for the question...


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The Picard group of $M_g$ is known to be isomorphic to $\mathbb{Z}$ for $g \geq 3$ by a theorem of Harer and it is also known (you can search papers of Farkas for a precise reference, I don't recall one right now) that the class of the locus of effective even theta characteristics is non-zero in this group. It follows that there can be no complete curves in the complement of this locus. –  ulrich May 1 '13 at 13:10
Thanks for the reference, this is exactly what I was looking for. –  Philip Engel May 1 '13 at 21:14
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