4
$\begingroup$

Let $f : S \to X$ be a dominant morphism of smooth complex compact surfaces. Let $C \subset S$ be a smooth curve such that $df$, seen as a map from $T_S \to T_X$, is generically of rank $1$ along $C$ and that in a neighborhood of $C$, $df_x$ is an isomorphism outside $C$. Suppose that $C$ is not contracted by $f$, is it true that $f_{|C}$ is an immersion?

One might want to decompose it into two sub-questions:

  1. Can one show that the rank of $df$ is exactly $1$ along $C$?
  2. Suppose the rank of $df$ is exactly $1$ along $C$, can we show that $f_{|C}$ is an immersion?
$\endgroup$
4
  • $\begingroup$ your surfaces are compact? $\endgroup$
    – pinaki
    Jun 12, 2013 at 10:08
  • $\begingroup$ Yes, they are compact. Thanks for asking the question. $\endgroup$
    – JacobI
    Jun 12, 2013 at 10:24
  • $\begingroup$ What do you mean by generically rank 1 along $C$? Do you mean that on all but a finite number of points of $C$ the differential of $f$ (seen as a map from $S\to X$ has rank one or that the restriction of $f$ to $C$ (thus seen as a map from $C$ to $X$) has injective differential at all but finitely many points of $C$? $\endgroup$
    – diverietti
    Jun 12, 2013 at 12:27
  • $\begingroup$ It means on all but a finite number of points of $C$, the rank of $df_x$ is 1, seen as map from $T_xS \to T_xX$. $\endgroup$
    – JacobI
    Jun 12, 2013 at 12:49

1 Answer 1

6
$\begingroup$

The question in essentially local on $X$, so the compacteness assumption is irrelevant. The answer to both questions is no: consider $S\subset \mathbb C^3$ defined by $z^3+xz+y=0$ and the map to $X=\mathbb C^2$ given by $(x,y,z)\mapsto (x,y)$.

The ramification curve $C$ is smooth (given by $z^3+xz+y=3z^2+x=0$) while its image has a cusp at $x=y=0$. The differential of the map has rank $\ge 1$ everywhere

$\endgroup$
1
  • $\begingroup$ Great! Thanks for your counter-ex Rita. $\endgroup$
    – JacobI
    Jun 13, 2013 at 14:25

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.