1
$\begingroup$

Let $\overline{M}_{0,n}$ be the usual Deligne-Mumford compactification of $M_{0,n}$ the moduli space of smooth $n$-pointed rational curves.

The canonical divisor $K_{\overline{M}_{0,n}}$ can be written as a combination of the irreducible components of the boundary of $\overline{M}_{0,n}$.

If $f:\overline{M}_{0,n}\rightarrow\overline{M}_{0,n}$ is an automorphisms then $f^{*}K_{\overline{M}_{0,n}} = K_{\overline{M}_{0,n}}$.

If $n\geq 5$, is this enough to conclude that $f$ has to map the boundary $\partial\overline{M}_{0,n}$ to itself?

$\endgroup$
1
  • 3
    $\begingroup$ The argument would not be correct as the canonical divisor is automatically preserved as a line bundle, but not as a divisor. Moreover, the statement is false in general. When $n=4$, $\bar{M}_{0,4}=\mathbb{P}^1$ has many automorphisms ! $\endgroup$ Sep 24, 2014 at 20:17

1 Answer 1

3
$\begingroup$

If $n\geq 5$ then $Aut(\overline{M}_{0,n})\cong S_n$ (http://arxiv.org/abs/1006.0987). For instance $\overline{M}_{0,5}$ is a Del Pezzo surface of degree five. Its automorphism group is well known to be $S_5$ (http://arxiv.org/abs/math/0610595).

In particular, any automorphism of $\overline{M}_{0,5}$ preserves the boundary.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy