# Automorphism groups of general type varieties

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!

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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)? –  YCor 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

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
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 $$(\lambda_1,\dots,\lambda_n)\cdot ([x_0,\dots,x_n],[y_0,\dots,y_n]) = ([x_0,\dots,x_n],[y_0, \lambda_1y_1,\dots,\lambda_ny_n]).$$