Let $X\subset\mathbb{P}^{n+c}$ be a smooth complete intersection of dimension $n$.

Is it known when $Aut(X)$ is finite ?

Does there exist a formula for the dimension of the tangent space to the space of first order infinitesimal deformations $H^1(X,T_X)$ ?

I am especially interested in the case of a codimension two complete intersection of two quadrics.

  • $\begingroup$ I think Bjorn Poonen has an article about this . . . $\endgroup$ Nov 16 '15 at 18:58
  • $\begingroup$ Poonen's article is about writing down explicit hypersurfaces where the only automorphism is the identity. Here is the reference, in case it is useful. Poonen, B. "Varieties without extra automorphisms. III. Hypersurfaces." Finite Fields Appl. 11 (2005), no. 2, 230–268. $\endgroup$ Nov 16 '15 at 19:15

It is known that there exists complete intersection of given type $T=(d_1,\ldots,d_c;n)$ with infinite automorphism group if and only if the type satisfies one of the following: $$T \in \{(2;n), (3;1), (2,2;1), (4;2), (2,3;2), (2,2,2;2) \}.$$ I.e. quadrics, curves of genus $1$ and K3 surfaces. In all other cases, the automorphism group is always finite.

See for example Theorem 3.1 of:

Olivier Benoist - Séparation et propriété de Deligne-Mumford des champs de modules d'intersections complètes lisses.

As for your other question; yes there are formulae for $\mathrm{H}^1(X, T_X)$ in terms of the degrees of its defining equations. They get quite messy though. I'm not sure where to find these; perhaps in SGAVII or Hirzebruch's book "Topological methods in algebraic geometry".

In the case of intersections of two quadrics however it is quite easy to work out. First you can use the Euler sequence and standard cohomology computations as in this question Deformations of smooth projective hypersurfaces and the Jacobian ring , but there is also a nice heuristic argument. Namely, a complete intersections of two quadrics of dimension $n$ is determined by the discriminant of its associated pencil of quadrics, which is a closed subscheme of $\mathbb{P}^1$ of degree $n+3$. These have $n+3-3=n$ moduli. Hence we find that $$\mathrm{H}^1(X, T_X) = n.$$

Many useful facts about intersections of two quadrics can be found in Miles Reid's PhD thesis: http://homepages.warwick.ac.uk/~masda/3folds/qu.pdf


I can suggest you a shorthand for your second question, in case you need a practical answer. First, if $X$ is a smooth complete intersection, let me denote by $A_X$ the affine cone over it, by $U_X= A_X \smallsetminus v$ the punctured affine cone and $\pi: U_X \to X$ the natural projection. We have then a standard short exact sequence (see for example Schlessinger's paper "On Rigid Singularities") $$ 0 \to \mathcal{O}_{U_X}\to T_{U_X} \to \pi^* T_X \to 0 $$ since we have a trivialization of the relative tangent by the Euler vector field. Now, for a smooth complete intersection of dimension $\geq 2$ we have $$H^1(U_X, T_X) \cong T^1_{A_X}$$ where the object on the right hand side is the (graded) module of first order deformations of the affine cone $A_X$. Now, passing in cohomology and since in general $$H^q(U_X, \pi^* \mathcal F)= \bigoplus_{k \in \mathbb Z} H^q(X, \mathcal F(k))$$ we have (when we restrict to the degree 0) $$ \cdots \to H^1(X, \mathcal{O}_X) \to (T^1_{A_X})_0 \to H^1(X, T_X) \to H^2(X, \mathcal O_X) \to \cdots $$

For a complete intersection of dimension $>2$ $H^i(X, \mathcal O_X)=0$, $i=1,2$, and thus we have $$(T^1_{A_X})_0 \cong H^1(X, T_X).$$

Now, the point is that the dimension of the graded component of the $T^1_{A_X}$ are super-easily computable using a bit of computer algebra: for example SINGULAR is extremely efficient in doing this (see for example https://www.singular.uni-kl.de/Manual/latest/sing_874.htm#SEC925), and basically you can get your answer with a couple of 'one-line command'.


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