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Let $\bar M_g$ be the Deligne-Mumford compactifiction of the moduli space of complex genus $g$ curves $M_g$. Is this correct that through every point of the boundary $\bar M_g\setminus M_g$ passes a rational curve that lies in the boundary $\bar M_g\setminus M_g$?

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up vote 10 down vote accepted

It seems unlikely. Say $g \geq 24$ (or so). There's a divisor $\Delta_g$ on the boundary $M_{2g+1}$ corresponding to curves with a single node consisting of two genus $g$ curves glued at a point. This component is birational to $M_{g,1} \times M_{g,1}$, which is of general type by the assumption on $g$. Thus through a general point of $\Delta_g$ there is no rational curve. For that matter there is no rational curve on $\bar{M}_g$ through this point at all.

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Thank you, I will think of this answer – Dmitri Nov 6 '12 at 15:43
I see now my mistake (that led me to the question). I was assuming (without any reason) that Satake compactification of $M_g$ is a smooth space. And since Deligne-Mumford is birational to Stake compactification I was thinking that boundary divisors in $\bar M_g$ should be ruled since they have larger dimension than in Satke. This sounded strange to me. But now I see that Satake is singular and so not surprisingly its blow up does not need to contain additional rational curves. Thank you for the answer again. – Dmitri Nov 6 '12 at 22:00

The irreducible components of the normalization $\mathcal{B}^{\nu}$ of the boundary $\mathcal{B}\subset\overline{\mathcal{M}}_{g,n}$ are finite images of the moduli spaces:

  • $\overline{\mathcal{M}}_{g_{1},S_{1}\cup\{n_{1}+1\}}\times\overline{\mathcal{M}}_{g_{2},S_{2}\cup\{n_{2}+1\}}$, where $g_{1}+g_{2} = g$ and $S_{1},S_{2}$ is a partition of $\{1,...,n\}$,
  • $\overline{\mathcal{M}}_{g-1,n+2}$,

Therefore any component of the boundary can be interpred as a moduli space of curves and you question can be rewritten in this way: For which $(g,n)$ is $\overline{\mathcal{M}}_{g,n}$ uniruled?

It is well known that $\overline{\mathcal{M}}_{g,n}$ is of general type for $g\geq 24, n\geq 0$. Then it is not uniruled. In lower genus, at the best of my knowledge:

  • $\overline{\mathcal{M}}_{0,n}$ is rational for any $n\geq 3$.
  • $\overline{\mathcal{M}}_{1,n}$ is rational for any $1\leq n\leq 10$, P. Belorousski, Chow rings of moduli spaces of pointed elliptic curves, Ph.D. thesis, Chicago, 1998.
  • $\overline{\mathcal{M}}_{g,n}$ for $g = 2$ and $1\leq n\leq 12$, $g = 3$ and $1\leq n\leq 14$, $g = 4$ and $1\leq n\leq 15$, $g = 5$ and $1\leq n\leq 12$,
  • $\overline{\mathcal{M}}_{g,n}$ is rational for $g=6$ and $1\leq n\leq 18$, and it is unirational for $g=8$ and $1\leq n\leq 11$, $g=10$ and $1\leq n\leq 3$, $g=12$ and $n=1$,
  • In Section $7$ of A. Logan, The Kodaira dimension of moduli spaces of curves with marked points, Am. J. Math. 125 (2003), 105–138, the author determines for $g = 2, . . . , 9, 11$ an integer $f(g)$ such that $\overline{\mathcal{M}}_{g,n}$ is unirational for $n \leq f(g)$.
  • $\overline{\mathcal{M}}_{g}$ is unirational for $g\leq 14$, $\overline{\mathcal{M}}_{15}$ is rationally connected, $\overline{\mathcal{M}}_{16}$ is uniruled,
  • $\overline{\mathcal{M}}_{22}$ is of general type. The Kodaira dimension of $\overline{\mathcal{M}}_{g}$ for $17\leq g\leq 21$ and $g = 23$ is not know,
  • $\overline{\mathcal{M}}_{g,n}$ is uniruled for $g = 12$ and $n \leq 5$, $g = 13$ and $n \leq 3$, $g = 15$ and $n \leq 2$,
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