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This question concerns a seemingly folk lore result, which states that automorphism groups of generic complete intersections are trivial, under certain assumptions.

To state the question, let $r \geq 1$ and let $d_1,\ldots,d_r, n \in \mathbb{N}$ be such that

  1. $n \geq 3$.
  2. $d_i \geq 2$ for each $i$.
  3. $(d_1,\ldots,d_r;n) \neq (2;n)$ nor $(2,2;n)$.

Let $X$ be a generic smooth complete intersection of dimension $n$ in projective space $\mathbb{P}^{n+r}$ over $\mathbb{C}$ with equations of degrees $d_1,\ldots,d_r$. Is $\mathrm{Aut}(X)$ trivial?

Remarks:

  • The conditions (1), (2), (3) imply that $\mathrm{Aut}(X)$ is finite and preserves the hyperplane class, for each such smooth complete intersection $X$ (not necessarily generic).
  • The result is not true without condition (3), as here the generic such complete intersection has a non-trivial automorphism group (hopefully I did not miss any other ''bad'' cases).
  • The answer is well-known to be yes when $r=1$, i.e. for generic hypersurfaces.

I need a version of this result in a paper I am currently writing. We are able to use this to show that the action of $\mathrm{Aut}(X)$ on $H^n(X,\mathbb{C})$ is faithful for any such smooth complete intersection (not necessarily generic). I would be most interested if anyone has any ideas on alternative approaches to this problem as well.

We are able to handle many special cases, such as when the degrees $d_i$ are all different or $X$ is of general type. A critical case is for example when all the degrees are the same and $X$ is Fano of high codimension.

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    $\begingroup$ I think you did miss a "bad" case, viz. $(d;n)=(4;3)$. Here, of course, the generic group is trivial, but your first remark does not hold: automorphisms to not need to preserve $h$, which is a big problem sometimes :) $\endgroup$
    – Alex Degtyarev
    Commented Feb 8, 2015 at 14:15
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    $\begingroup$ Are you referring to the case of quartic surfaces? I have avoided these by assuming that $n \geq 3$ (my $n$ is the dimension of $X$, not the dimension of the ambient projective space). $\endgroup$
    – Daniel Loughran
    Commented Feb 8, 2015 at 14:26
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    $\begingroup$ Oops! Sorry, I missed the meaning of $n$! $\endgroup$
    – Alex Degtyarev
    Commented Feb 8, 2015 at 14:42

2 Answers 2

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You may try the following:

I believe (I did not check the details) that the monodromy representation on the primitive part of $H^n(X,\mathbb{C})_{prim}$ is irreducible, as in the case of hypersurfaces.

If this is true then for a very general $X$ the Hodge structure $H^n(X,\mathbb{C})_{prim}$ does not have any nontrivial sub-Hodge structures.

Let $\sigma$ be an automorphism of $X$. As you mentioned $\sigma^*$ respects the hyperplane class and hence it induces an automorphism of $H^n(X,\mathbb{C})_{prim}$. The 1-eigenspace of $\sigma^*$ is a sub-Hodge structure and there either all of $H^n(X,\mathbb{C})_{prim}$ or this eigenspace is the zero-space. In the first case you have that $\sigma^*$ is the identity and you claimed that you then can deduce that $\sigma$ is trivial. In the second case you obtain that the quotient $X/\langle \sigma \rangle$ has the same Betti numbers as $\mathbb{P}^n$. I do not see how you can exclude this case, but it seems very weird that this happens when the $d_i$ are large.

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    $\begingroup$ Thanks for the answer, but I think you have misunderstood the problem. I'm ultimately trying to show that $\sigma^*$ can't be the identity when $\sigma$ is non-trivial. So I have no idea how to handle this first case! The second case you mention is fine for me. $\endgroup$
    – Daniel Loughran
    Commented Feb 18, 2015 at 21:12
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    $\begingroup$ The argument you have in mind (I think) also appears in the book of Katz-Sarnak when they show that the generic autom group of a smooth curve of genus $g>1$ is trivial. They do this by combining 1) the bigness of the monodromy group with 2) the injectivity of the map $Aut(X) \to Aut(H^1(X))$ and 3) the fact that a generic curve $X$ is not hyperelliptic (so that no automorphism of $X$ induces $-1$ on $H^1$). In the case of complete intersections, the bigness still holds (see the book by Peters-Steenbrink for a reference).... $\endgroup$
    – Ariyan Javanpeykar
    Commented Feb 18, 2015 at 21:23
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    $\begingroup$ ...the problem therefore becomes showing the analogue of 2). Namely, showing the injectivity of the canonical map $Aut(X) \to Aut(H^n_{prim}(X)$ (which is easy in fact if $\sum d_i > n+1$). As you note, 1) and 2) only show that the generic autom group is then trivial or $\mathbb Z/2\mathbb Z$. To exclude that there are autom's acting as $-1$ on cohomology (ie. prove 3) might be a bit difficult because as you say the (highly singular) quotient will be a fake projective space. The only way to do this I know of is using Lefschetz trace formula. $\endgroup$
    – Ariyan Javanpeykar
    Commented Feb 18, 2015 at 21:24
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I think the result you are looking for is worked out in Theorem 1.3 of Chen-Pan-Zhang "Automorphism and Cohomology II: Complete intersections" (+ Remark 1.4), see: https://arxiv.org/pdf/1511.07906.pdf

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