6
$\begingroup$

This might be a trivial question and I might be overlooking something:

Suppose $k$ is a field with algebraic closure $\overline k$ and absolute Galois group $\Gamma$. Let $X,Y$ be two distinct varieties over $k$ that are isomorphic over $\overline k$. Consider their automorphisms groups $Aut_{\overline k}(X)$ and $Aut_{\overline k}(Y)$. These groups are isomorphic as abstract groups but they each also have a Galois action of $\Gamma$ acting on the groups by conjugation.

Are these two automorphism groups isomorphic as "groups with a $\Gamma$ action"? Note that $H^1(\Gamma,Aut_{\overline k}(X)) \cong H^1(\Gamma,Aut_{\overline k}(X))$ because both groups classify twists of $X$ (or equivalently of $Y$).

If these two "groups with $\Gamma$ action" are distinct, are their higher cohomologies distinct too in any examples? How are these two related otherwise?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
12
$\begingroup$

They are not. $H^0(\Gamma, \text{Aut}(X_{\bar{k}}))$ computes the subgroup of automorphisms which are Galois-invariant, which is equivalently the automorphism group of $X$, and similarly for $Y$, so to find a counterexample it suffices to find varieties $X, Y$ which are isomorphic over $\bar{k}$ but which have non-isomorphic automorphism groups over $k$.

Let $k = \mathbb{R}$, let $X = \mathbb{P}^1$, and let $Y$ be the conic $\{ X^2 + Y^2 + Z^2 = 0 \}$ in $\mathbb{P}^2$. Then $\text{Aut}(X) \cong PGL_2(\mathbb{R})$ but $\text{Aut}(Y) \cong PO(3) \cong SO(3)$ (actually I'm not entirely confident I know how to prove this, at least not without passing to the complexification). These two groups can be distinguished abstractly by their abelianizations: I believe (but haven't checked carefully) that the abelianization of $PGL_2(\mathbb{R})$ is $\{ \pm 1 \}$ (given by the sign of the determinant) but the abelianization of $SO(3)$ is trivial.

Improve this answer
$\endgroup$
7
  • 5
    $\begingroup$ You can see (at least intuitively) that twisting can kill (or introduce) automorphisms over the base field: by twisting the Fermat cubic $x^3+y^3+z^3=0$ you obtain e.g. $ax^3+by^3+cz^3=0$, where all of the symmetry is broken. $\endgroup$
    – R.P.
    Commented Oct 14, 2020 at 8:07
  • 4
    $\begingroup$ I think you can even get a hands-on $0$-dimensional counterexample: Let $X = \operatorname{Spec}\mathbb{R}[x]/(x^2 + 1)x$ and $Y=\operatorname{Spec}\mathbb{R}[x]/(x^2-1)x$. Then I'd think that over $\mathbb{C}$ both have automorphism group $\Sigma_3$, but the action is trivial for $Y$ and conjugation with a transposition for $X$. (Am I messing anything up here?) $\endgroup$
    – Achim Krause
    Commented Oct 14, 2020 at 8:14
  • 2
    $\begingroup$ @AchimKrause I think that is correct. Another way to see it is to write out the automorphisms of $X$ over $\mathbb{C}$ in terms of their corresponding ring isomorphisms: they are given by $x \mapsto x$, $x \mapsto -x$, $x \mapsto (2x+i)(x-i)$, $x \mapsto (2x-i)(x+i)$, $x \mapsto (-2x-i)(x-i)$, and $x \mapsto (2x+i)(x+i)$ if I haven't made any mistakes. Now it is easy to see that only the first two are stable under complex conjugation. For me, the Galois action is always easier to understand on the ring level, since the action there is just given by the action on the coefficients. $\endgroup$
    – R.P.
    Commented Oct 14, 2020 at 10:39
  • $\begingroup$ That's great, thanks! I thought of the twisted $\mathbb P^1$ example too but I too did not know how to compute the automorphism group of the conic over the reals. It's a little strange to me to be having two different groups with $\Gamma$ actions that nevertheless give the same $H^1$. Is there a purely group theoretic explanation of what's going on here? For instance, is there some classification of what sort of different $\Gamma$ actions can you get by twisting? $\endgroup$
    – Asvin
    Commented Oct 14, 2020 at 18:15
  • 1
    $\begingroup$ @Asvin: when you stipulate that $X$ and $Y$ are isomorphic over $\bar{k}$ you're imposing exactly that they have the same forms but that's all you're imposing. If you work through the proof that $H^1$ classifies forms what you'll see is that different $1$-cocycles are used to twist the Galois action (starting from a base; in other words the bijection requires picking a $k$-form) and taking fixed points of that action is what produces the different forms. $\endgroup$
    – Qiaochu Yuan
    Commented Oct 14, 2020 at 18:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .