Let $C$ be a smooth projective curve. Is it true that
$$\textrm{Aut}(C\times C)\cong S_2 \ltimes (\textrm{Aut}(C)\times \textrm{Aut}(C))$$
and in case, what would be a reference for this? Thanks.
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3 Answers
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This is a particular case of a more general rigidity result, whose proof (similar to the one given in abx's answer) can be found in Lemma 3.8 of
F. Catanese, Fibred surfaces, varieties isogenous to a product and related moduli spaces, Amer. J. Math. 122 (2000), no. 1, 1-44.
Proposition. Let $f \colon C_1 \times C_2 \longrightarrow B_1 \times B_2$ be a surjective holomorphic map between products of curves. Assume that both $B_1$ and $B_2$ have genus $\geq 2$. Then, after possibly exchanging $B_1$ with $B_2$, there are holomorphic maps $f_i \colon C_i \to B_i$ such that $$f(x, \, y) = (f_1(x), \, f_2(y)).$$
answered Feb 29, 2016 at 10:33
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$\begingroup$ Thanks a lot. Do you know if this is true more generally for $C_1\times ...\times C_n$? Perhaps we could prove it by induction on $n$ starting from the Proposition. $\endgroup$– user82886Commented Apr 13, 2016 at 20:55
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That is certainly not true. Consider the case that $C$ is an elliptic curve. Then $\text{Aut}(C\times C)$ contains $\text{GL}(2,\mathbb{Z})$ as a subgroup.
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$\begingroup$ Thanks. What if $g(C)\neq 1$? For instance if $g(C)=0$ it is true. $\endgroup$– user82886Commented Feb 28, 2016 at 21:26
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$\begingroup$ Just a question. How do you see $GL(2,\mathbb{Z})$ as a subgroup of $Aut(C\times C)$? Are you thinking $C\times C$ as the Jacobian of a genus two curve? $\endgroup$– user82886Commented Mar 21, 2016 at 17:15
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1$\begingroup$ For a triple of integers $(a,b,c,d)$ such that $ad-bc$ equals $\pm 1$, form the map $f:C\times C \to C\times C$ by $f_{a,b,c,d}(x,y) = (ax+by,cx+dy)$, where $+$ is the group operation on the elliptic curve, and (as usual) $nx$ means $x+\dots +x$ ($n$ times) for nonnegative integers $n$. There exist integers $(s,t,u,v)$ such that $f_{a,b,c,d}\circ f_{s,t,u,v} = f_{s,t,u,v}\circ f_{a,b,c,d} = \text{Id}_{C\times C}$. $\endgroup$ Commented Mar 21, 2016 at 17:18
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$\begingroup$ I am fine with this. But in this way you get an action of $SL(2,\mathbb{Z})$. If $ad-bc\neq \pm 1$ we can not invert the matrix. $\endgroup$– user82886Commented Mar 21, 2016 at 20:56
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$\begingroup$ "But in this way you get an action of $\text{SL}(2,\mathbb{Z})$. That is incorrect. For instance, $(a,b,c,d) = (1,0,0,-1)$ is not in $\text{SL}(2,\mathbb{Z})$. $\endgroup$ Commented Mar 21, 2016 at 21:19
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It is true if $g(C)\geq 2$. The point is that if you map nontrivially $C$ to $C\times C$, the projections of the image on each factor have degree $0$ or $1$ (say, by Hurwitz formula). Thus the image is either $C\times \{p\} $ or $\{q\}\times C $ for some $p,q\in C$, or it is the graph of an automorphism. Since $\mathrm{Aut}(C)$ is finite for $g(C)\geq 2$, your statement follows easily.
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$\begingroup$ Sorry, but I do not see where you use that the automorphism group of $C$ is finite. $\endgroup$– user82886Commented Mar 6, 2016 at 19:06
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$\begingroup$ This implies that the numer of graph of automorphisms is finite, hence the curves $C\times \{p\}$ cannot map to such a graph (which happens in Jason Starr's example). $\endgroup$– abxCommented Mar 7, 2016 at 0:32