Automorphism groups of general type varieties

Question

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!

Answer 1

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

Answer 2

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