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Is there a known way to construct flat families of smooth curves $\mathcal{C}/\mathcal{B}$ which are fiberwise embedded in a family of projective varieties $\mathcal{X}/\mathcal{B}$ and which have general moduli (i.e., the forgetful map $\mathcal{C} \to \mathcal{M}_g$ which discards the embedding and takes every curve to its isomorphism type is dominant)?

When we take all fibers of $\mathcal{X}$ to be projective spaces this can be done by Brill-Noether theory. I'm looking for alternative constructions which either don't use Brill-Noether, or consider different target spaces.

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  • $\begingroup$ For every smooth projective variety $X$ of dimension at least $3$ that is separably rationally connected, for every $g\geq 0$, for every $r\geq 0$, for all sufficiently positive curve classes $\beta$, for the stack $\mathcal{M}_{g,r}(X,\beta)^o$ parameterizing embeddings of $r$-pointed, genus $g$ curves with class $\beta$, for the associated morphism $(\text{ev},\Phi):\mathcal{M}_{g,r}(X,\beta) \to (X^r)\times \mathcal{M}_{g,r}$, the open subset of the domain where the morphism is smooth is nonempty . . . $\endgroup$
    – Jason Starr
    Mar 18, 2017 at 13:40
  • $\begingroup$ . . . Thus, for any $Y$ that contains such an $X$, again $\mathcal{M}_{g,r}(Y,\beta)\to \mathcal{M}_{g,r}$ is surjective for infinitely many $\beta$, e.g., for the classes of curves in $X$. Thus, fix an integer $d$, and consider a $d$-fold $Y$ containing no rationally connected subvariety. Thus $Y$ contains no uniruled $Z$. For all $g$ with $3g-3>d$, by Bend-and-Break, if there exists a family of curves in $Z$ dominating $\mathcal{M}_g$ and such that a general point of $Z$ is contained in one of these curves, then $Z$ is uniruled. Thus, a general curve of genus $g$ does not embed in $Y$. $\endgroup$
    – Jason Starr
    Mar 18, 2017 at 13:49
  • $\begingroup$ @Jason: thanks! If I understand your answer correctly, you're saying that a necessary and sufficient condition for the problem to be meaningful with a constant target space $X$ is to have it contain a rationally connected subvariety? ... is the answer different if I want to let $X$ be a flat family over the base? $\endgroup$
    – Nati
    Mar 19, 2017 at 1:09
  • $\begingroup$ I do not understand what you are asking. What is the "base" in your question? $\endgroup$
    – Jason Starr
    Mar 19, 2017 at 3:37
  • $\begingroup$ The base $B$ parametrizes both the family of algebraic curves, and the family of varieties they can be embedded in. The map $B \to M_g$ which sends each curve to its isomorphism type should be surjective. $\endgroup$
    – Nati
    Mar 19, 2017 at 4:49

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