4
$\begingroup$

I had previously asked: Help motivating log-structures

I now have some more questions regarding the role of log structures in moduli problems (you can assume that the moduli problem is the compactification of $n$-marked genus $g$ smooth projective curves for simplicity):

  1. It seems that one of the mantras of the subject is that outside of the boundary, the objects have a unique log-structure. In terms of the example moduli problem I gave, that $n$-marked smooth projective curves of genus $g$ have unique log-structures. In what sense is this true? It doesn't seem literally true to me. Surely they must mean that they have unique log-structures such that they satisfy some property, right? If you can enlighten me about the essence of this mantra, please do!

  2. One of the strengths of log-structures, evidently, is that in the degenerations, they give a unique deformation. So in the example, if we had a stable $n$-marked curve of genus $g$ with a log-structure, there there would be a unique way to extend it to a complete DVR. My question is: what is the virtue of log-structures as opposed to deformation data? Why not instead of a log-structure attached to each (possibly semi-stable) curve, just add some data that will say how it deforms over a complete DVR? Would it be fair to say that log-structures is the natural way to encode this deformation data? Or perhaps there is an extra virtue? I'm confused about this.

Any help would be much appreciated. I've zigzagging between various texts about log-structures, and it is still difficult to get the gist of how to think about them!

P.S. I put this question under community wiki also, but I wasn't sure this time that it was merited. If you have objections, let me know.

$\endgroup$
1
  • $\begingroup$ 1. log stuructures with some universal property is unique. In section 2 of the following paper, the existence of special log structure is proved in a more general setting. Martin Olsson, Universal log structures on semi-stable varieties. Tohoku Math Journal 55 (2003) 397--438 $\endgroup$
    – Naturalmap
    Jul 13, 2011 at 1:02

2 Answers 2

2
$\begingroup$

If you're compactifying a moduli space by choosing degenerations, you typically assign the trivial log structure to the uncompactified space. Given a point $x$, the size of the characteristic $M_{X,x}/\alpha^{-1}\mathscr{O}_{X,x}^\times$ roughly describes how degenerate the object over $x$ is, and in a place where the log structure is trivial, the characteristic is the trivial monoid.

When someone says that the log structures on the moduli space of marked curves and the tautological curve over it are unique, that is relative to some condition that needs to be specified, e.g., being an essentially semistable morphism. If that condition is assumed, then the log structure is unique. In the case of marked curves, the locus of schematically smooth curves is then given the trivial log structure.

I don't know what you mean by unique deformation. The tangent and jet spaces of a smooth compactified moduli space are just as big on the boundary as they are elsewhere.

$\endgroup$
1
$\begingroup$
  1. if $f: X \rightarrow S $ is a proper, log smooth, integral and vertical morphism with semistable geometric fibers, then there is a special log structure on $f' : X' \rightarrow S'$ with same underlying scheme. This log structure is minimal: $X$ is fibered product of $X'$ and $S$ over $S'$. Look at section 2 of

    Martin Olsson, Universal log structures on semi-stable varieties. Tohoku Math Journal 55 (2003) 397--438

For example, let $X= \textrm{Spec} k[x,y]/xy$ and $S = \textrm{Spec} k$. Basic log structures on $X \rightarrow S$ will be given by monoids: $\mathbb{N}^2 \rightarrow k[x,y]/xy$, $(1,0) \mapsto x$ and $(0,1) \mapsto y$. For $S$, $\mathbb{N} \rightarrow k$, $1 \mapsto 0$. You can put other log structures using different monoids on $X$ and $S$, but as long as it satisfies log smothness and so on, it is pull back from this basic log structure.

By the way, does anyone know how to draw a commutative diagram on mathoverflow?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.