Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The question Are automorphism groups of hypersurfaces reduced ? reminded me of the following related question that I have not seen discussed.

If $X$ is a smooth projective variety over an algebraically closed field $k$ of characteristic $p$ such that the canonical bundle $K_X$ is ample, then is the automorphism group of $X$ reduced?

Some observations:

  1. It is equivalent to ask if $H^0(X,T_X) = 0$.
  2. When $X$ is a hypersurface in $\mathbb{P}^n$, this is exactly the question answered in the link above.
  3. If $d := \mathrm{dim}(X) < p$ and $X$ lifts flatly to $W_2(k)$, then the answer is "yes" as $H^0(X,T_X) = (H^d(X, \Omega^1_X \otimes K_X))^\vee = 0$ by Kodaira vanishing (which holds because of Deligne-Illusie-Raynaud under these hypotheses).

Arguments or references or counterexamples would be very appreciated!

share|improve this question
    
What does "reduced" mean (for the automorphism group)? Is it always a group scheme in a natural way (in which case I understand what "reduced" means)? –  Yves Cornulier Jan 21 '13 at 12:38
    
I think the automorphism functor is representable by a finite type group scheme since $K_X$ is ample (look at the action on sections of $K_X^n$). Also, $\mathrm{Aut}(X)$ is always represented by a locally finitely presented group scheme for $X$ projective (look inside the Hilbert scheme of $X \times X$), so "reducedness" makes sense. However, I don't want to worry about these issues too much here, so I'm happy to take reducedness to mean that $H^0(X,T_X) = 0$. –  anon Jan 21 '13 at 13:51
add comment

2 Answers

up vote 7 down vote accepted

William Lang produced examples of surfaces of general type in positive characteristic with non-zero vector fields. Since surfaces of general type must have finite automorphism groups, this gives examples. See William Lang, "Examples of surfaces of general type with vector fields", Arithmetic and geometry, Vol. II, Progr. Math., 36, Birkhäuser, pp. 167–173

share|improve this answer
    
Ah, thanks. I don't have access to Lang's paper, so I just wanted to check: do Lang's examples also have $K_X$ ample (as opposed to say big)? –  anon Jan 21 '13 at 18:43
    
If I remember correctly, Lang's examples are fibrations $X \to C$, where $C$ is a curve of genus at least 2, $X$ is a smooth surface of general type, and the geometric fibers of $X \to C$ are irreducible rational curves with a single higher-order cusp. I believe that for these surfaces the canonical has to be ample. –  Angelo Jan 21 '13 at 21:59
    
Perfect, thank you! –  anon Jan 22 '13 at 6:26
add comment

I suspect that there are examples. Here is an example, relative to the existence of a smooth, projective morphism $\pi:X\to \mathbb{P}^n$ with $\omega_X\otimes \pi^*\mathcal{O}(1)$ ample, for some integer $n\geq 1$. There are examples due to Moret-Bailly of smooth, projective morphisms to $\mathbb{P}^n$ that might do the trick -- I need to double-check the canonical bundle. Anyway, given such a morphism, consider the induced morphism $$ \pi \times \text{Id}_{\mathbb{P}^n}: X\times \mathbb{P}^n \to \mathbb{P}^n\times \mathbb{P}^n.$$ This is also a smooth, projective morphism. Let $d\geq 1$ be an integer such that $pd > n+1$, and consider the closed subscheme, $$ Z = \{ ([x_0,\dots,x_n],[y_0,\dots,y_n])\in \mathbb{P}^n\times \mathbb{P}^n | x_0y_0^{pd} + \dots + x_ny_n^{pd} = 0\}. $$ Define $Y$ to be the inverse image of $Z$ under $\pi\times \text{Id}_{\mathbb{P}^n}$. Of course $Z$ is everywhere smooth. Since $\pi$ is smooth, also $Y$ is smooth. By adjunction, $\omega_Y$ is the restriction to $Y$ of $$ \text{pr}_X^*(\omega_X \otimes \pi^* \mathcal{O}(1)) \otimes \text{pr}_{\mathbb{P}^n}^* \mathcal{O}(dp-n-1), $$ which is ample by hypothesis. On the other hand, there is an action of $\mathbf{\mu}_{pd}^n$ by $$(\lambda_1,\dots,\lambda_n)\cdot ([x_0,\dots,x_n],[y_0,\dots,y_n]) = ([x_0,\dots,x_n],[y_0, \lambda_1y_1,\dots,\lambda_ny_n]).$$

share|improve this answer
    
Thanks, this is a nice construction! –  anon Jan 21 '13 at 18:44
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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