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

In the following article : "H. Matsumura, P. Monsky, On the automorphisms of hypersurfaces, J. Math. Kyoto Univ. 3 (1964) 347-361", it is shown that in finite characteristic, automorphism groups of smooth hypersurfaces of $\mathbb{P}^N$ are finite (with known exceptions such as quadrics, elliptic curves, K3 surfaces). However, the question of their reducedness is left open. Does anyone know something about it ?

In fact, we need to show that $H^0(X,T_X)=0$. In characteristic $0$, you can use Bott's theorem to do that. What can you do in finite characteristic ?

share|improve this question
1  
Could you clarify what you mean by "known exceptions"? So far as I know, a necessary and sufficient condition for the automorphism group of every smooth degree $d$ hypersurface in $\mathbb{P}^N$ [over a fixed, say, algebraically closed field] to be finite is $d > N+1$, equivalently for the variety to be of general type. I also thought that it was a general theorem that varieties of general type had finite automorphism group: is this harder to prove in characteristic $p$? (Is it false?!?) –  Pete L. Clark Jan 4 '10 at 23:13
1  
The projective automorphism group of a hypersurface is finite, except for quadrics. The whole automorphism group coincides with the projective one except in the cases mentioned by Olivier. But this happens for different reasons if $N>2,d\neq N+1$ and if $N>3$. Namely, if $N>2$, then we can use the fact that the Picard is torsion free and the canonical class is $n+1$ times the hyperplane section. If $N>3$, then the Picard is $\mathbf{Z}$ and the effective cone is spanned by the hyperplane section. –  algori Jan 4 '10 at 23:53
    
Olivier -- I'm not 100% sure, but you could try cubic curves over a field of characteristic 3 to get a non-reduced example. But in any characteristic the projective automorphism groups are finite. –  algori Jan 4 '10 at 23:58
    
@Pete : It is true that varieties of general type have finite automorphism groups, but the converse is not (think of a blow-up of the plane in more than four points !). And for hypersurfaces, this happens most of the time (example : cubic surfaces which are blow-ups of the plane in six points). However, as long as I know, varietes of general type could very well have nonreduced automorphism groups (?). I don't have any example at hand. –  Olivier Benoist Jan 5 '10 at 0:07
1  
@algori : In fact, we were both right : the automorphism group of elliptic curves in char $3$ is reduced, but their projective automorphism group isn't. It is well explained in the book of Katz and Sarnak to which Bjorn Poonen refers in his answer. –  Olivier Benoist Jan 5 '10 at 12:21
show 8 more comments

1 Answer

up vote 12 down vote accepted

If $X$ is a smooth hypersurface in $\mathbf{P}^{n+1}$ of degree $d$, where $n \ge 1$, $d \ge 3$, and $(n,d) \ne (1,3)$, then $H^0(X,T_X)=0$ by Theorem 11.5.2 in Katz and Sarnak, Random matrices, Frobenius eigenvalues, and monodromy, AMS Colloquium Publications, vol. 45, 1999. So the connected component of the identity of the automorphism group scheme is trivial in these cases. See also Theorem 11.1 in http://www-math.mit.edu/~poonen/papers/projaut.pdf , which is Poonen, Varieties without extra automorphisms III: hypersurfaces, Finite fields and their applications 11 (2005), 230-268.

share|improve this answer
    
Thank you so much for the reference ! –  Olivier Benoist Jan 5 '10 at 1:38
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.