The following question arises from the proof of "bend-and-break" lemma:

Let $X$ be a projective variety over $\mathbb{C}$ and $C$ be an irrational smooth curve. Let $c \in C$ be a fixed closed point. Let $f: C \to X$ be a nonconstant morphism such that $f(c)=x$.

Suppose there exists an irreducible one dimensional variety $T \subset Mor(C,X;f|_{c})$ passing through $[f]$ (we use $[f]$ to denote the point corresponding to $f$ in the moduli space), where by $Mor(C,X;f|_{c})$, we mean the moduli space of morphisms from $C$ to $X$ such that any morphism maps the point $c$ to $x$. Let $e$ be the evaluation map restricted to $C \times T$, that is

$$e: C \times T \to C \times Mor(C,X;f|_{c}) \to X.$$

My questions is, why $\dim(e(C \times T)) >1$?

I understand when $g(C)>0$, with one point fixed, $C$ only has finite automorphism. But I don't know how to use this fact to show the claim.

  • $\begingroup$ Without any additional hypotheses, this is not true. You can take $e(z,t)=f(z),\forall z,t$ and the image will be one-dimensional. $\endgroup$ – Felipe Voloch May 31 '14 at 15:31
  • $\begingroup$ Yes, I had edited the post, hopefully, this will make the question clear. $\endgroup$ – Li Yutong May 31 '14 at 15:45

You may assume that $T$ is irreducible. You have $f(C)\subseteq e(C\times T)$. Then $dim(e(C\times T)) \geq 1$.

If $dim(e(C\times T)) = 1$ then $e(C\times T) = f(C)$. This implies that for any $t\in T$ the image of the morphism $f_t$ corresponding to $t$ is $f(C)$. Now, two morphisms $f_t:C\rightarrow f(C)$ and $f_{t^{'}}:C\rightarrow f(C)$ such that $f_{t}(c) = f_{t^{'}}(c)$ differ by an automorphism of the pointed curve $(C,c)$. That is, there exists an automorphism $\alpha:C\rightarrow C$ such that $\alpha(c) = c$ and $ f_{t^{'}} = f_{t}\circ\alpha$.

Since $C$ is smooth and $g\geq 1$ we have that $Aut(C,c)$ is finite. A contradiction because we assumed that $dim(e(C\times T)) = 1$. Therefore $dim(e(C\times T)) \geq 2$.

  • $\begingroup$ I don't understand your claim that "$f_{t'},f_{t}$ differ by an autumorphism". Suppose $C$ is an elliptic curve, with $c$ to be identity element under the group law. Suppose $f_t = id$, while $f_n$ is the multiply by $n$ morhpism(under the group law), they are not factor through an automorphism. $\endgroup$ – Li Yutong Jun 1 '14 at 1:47
  • $\begingroup$ It isn't true that any two pointed finite maps from $C$ onto a fixed target curve are related through an automorphism of the source: at the very least one needs to bring in degree bounds (which certainly hold in the case of interest) but one also has to rule out other things such as automorphisms of the target which preserve the base point (of which there can be infinitely many when the target curve is a rational curve). $\endgroup$ – user76758 Jun 1 '14 at 1:49
  • $\begingroup$ @LiYutong: In the case of interest one can see that the degree of the $f_t$'s is bounded, so that can avoid your proposed counterexamples (if we allow ourselves to drop some $t$'s). But degree bound alone is insufficient, as my previous comment indicates. $\endgroup$ – user76758 Jun 1 '14 at 1:50
  • $\begingroup$ @user76758 I see what you mean. However, where I got wrong in the statement? I checked two books on MMP (Debarre and Matsuki), book proves seems rely on that claim $\endgroup$ – Li Yutong Jun 1 '14 at 1:55

Your Answer

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

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