# If a variety $X$ has finite automorphism group, is the same true for its $n$-fold self-products?

Let $X$ be an algebraic variety over $\mathbb{C}$. Let $n\geq 1$ be an integer and let $X^n$ be the $n$-fold self product of $X$.

Q. Is there an integer $n\geq 1$ and an algebraic variety $X$ over $\mathbb{C}$ such that $\mathrm{Aut}(X)$ is finite and $\mathrm{Aut} (X^n)$ is infinite?

• "self-product" is sometimes called "power" :)
– YCor
Feb 27, 2018 at 22:29

The following result of Biswas and Gómez provides a partial positive answer (for the symmetric product) in the case where $X$ is a curve.

Theorem. Let $X$ be an irreducible smooth projective curve of genus $g > 2$ defined over an algebraically closed field of characteristic different from two, and take an integer $d > 2g-2$. Then the natural group homomorphism $$\mathrm{Aut}(X) \longrightarrow \mathrm{Aut}(\mathrm{Sym}^d X)$$ is an isomorphism.

See

I. Biswas, T. L. Gómez: Automorphisms of a symmetric product of a curve (with an appendix by Najmuddin Fakhruddin). Doc. Math. 22 (2017), 1181–1192. MR3690270.

• Sorry, I misread the question and answered it for the symmetric product. I will leave the answer anyway, maybe it can be useful for someone. Feb 21, 2018 at 11:38
• If $X$ is canonically polarized then $X^d$ is canonically polarized and hence has finite automorphism group. This includes genus $g \geq 2$ curves, so the result is in fact true when $X$ is a curve. Feb 21, 2018 at 11:53
• @Will: you need characteristic 0 for this, right? Anyway, the OP is working over the complex numbers, so it is ok. Feb 21, 2018 at 12:07
• @FrancescoPolizzi The automorphism of a canonically polarized projective variety over a field $k$ is finite. (This is also true in positive characteristic; see mathoverflow.net/questions/152491/… and its answer) Feb 21, 2018 at 14:19
• @AriyanJavanpeykar: ah, I did not know it was known to be true also in positive characteristic. Thanks! Feb 21, 2018 at 14:38

This is partially answered in the answer to this question: - the question itself is different. The answer applies when you have a product of two curves, but as the answer-er (Roberto Pignatelli) points out, the key result (lemma 3.8 of Catanese's paper referenced) does not appear to rely on the variety being a curve).

Catanese, Fabrizio, Fibred surfaces, varieties isogenous to a product and related moduli spaces, Am. J. Math. 122, No. 1, 1-44 (2000). ZBL0983.14013.

• "Two curves": you mean "two smooth projective curves"
– YCor
Feb 27, 2018 at 22:23
• @YCor Isn't that the default? Feb 28, 2018 at 0:30
• Apparently not in the question, and certainly not in most reference books of algebraic geometry such as Hartshorne etc. (And using it as default is not really practical... if a curve is assumed smooth projective, what can be said of an affine curve? or a singular projective curve? a curve which is not really a curve?... I wouldn't have noticed if the question were about smooth projective curves. Actually I'm curious about the question when $X$ is just the affine line minus a finite subset.
– YCor
Feb 28, 2018 at 0:56
• @YCor you confuse me. An affine curve is usually called "an affine curve", and a singular projective curve "singular projective curve" (or "singular curve", being as "projective" tends to be the default) As to the question, where does the OP indicate that this is NOT what he wants? Feb 28, 2018 at 3:53
• Yes but what about a curve which is not supposed singular, not supposed non-singular, not supposed affine, not supposed projective, not supposed non-projective. Standard name is "curve". Of course people tend to add implicit assumptions when they usually need it (irreducible, smooth, projective, etc,)
– YCor
Feb 28, 2018 at 9:42

For $X$ projective, we have that $\mathrm{Aut}(X)$ finite implies $\mathrm{Aut}(X^n)$ finite.

In this case, $\mathrm{Aut}(X)$ is representable by a group scheme of finite type (*Edit. locally of finite type) over $\mathbf{C}$ and its Lie algebra is identified with global tangent fields $H^0(X, T_X)$, after Grothendieck. According to this post, the group scheme $\mathrm{Aut}(X)$ is reduced, so the condition that $\mathrm{Aut}(X)$ being finite is equivalent to that $H^0(X, T_X) = 0$. It then suffices to observe that $H^0\left(X^n, T_{X^n}\right) \cong H^0(X, T_X)^{\otimes n}$.

Edit Yes, a priori $\mathrm{Aut}(X^n)$ may be discrete and infinite. Sorry for that.

• Dear @WilleLiou, the vanishing of $H^0(X,T_X)$ is not equivalent to the finiteness of Aut(X). It is equivalent to the automorphism group scheme being zero-dimensional. However, the automorphism group scheme is not necessarily of finite type, only locally of finite type over $\mathbb{C}$. Thus, it could a priori happen that $Aut(X)$ is finite, and that $Aut(X^n)$ is an infinite countable discrete group. Feb 27, 2018 at 21:25