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Let $k$ be a field, and $K$ its separable closure. Consider two different $k$-schemes, $X$ and $Y$, which become isomorphic upon extension of scalars to $K$: $X_K \cong Y_K$. Then the etale cohomologies (and indeed, etale homotopy types) of $X_K$ and $Y_K$ will be equivalent, but may differ at $k$. I'd like to know some sample computations in which $H^*_{et}(X_k)$ and $H^*_{et}(Y_k)$ differ, or agree. I'm happy to take any sort of coefficients here, but being an algebraic topologist, I'm lazy, so I would prefer that they're constant sheaves.

To give some focus to the question, let's consider the case $k=\mathbb{R}$, and $K=\mathbb{C}$. Take $X = GL_2(\mathbb{R})$, and let $Y$ be the nonzero quaternions, $\mathbb{H}^{\times}$. These become isomorphic (to $GL_2(\mathbb{C})$) upon extension of scalars to $\mathbb{C}$. Do they have the same etale cohomology over $\mathbb{R}$?

Let's take coefficients such that $2$ is invertible; first note that by a comparison to the analytic topology, $H^*_{et}(GL_2(\mathbb{C}))$ is an exterior algebra on two generators in dimensions 1 and 3, $\Lambda[x_1, x_3]$. There are actions of $\mathbb{Z} / 2 = Gal(\mathbb{C} / \mathbb{R})$ on $H^*_{et}(X_{\mathbb{C}})$ and $H^*_{et}(Y_{\mathbb{C}})$, and the invariants are $H^*_{et}(GL_2(\mathbb{R}))$ and $H^*_{et}(\mathbb{H}^{\times})$, respectively. I presume that the action of $\mathbb{Z} / 2$ on $\Lambda[x_1, x_3]$ must differ for the two examples, but I'm not really sure how to compute these actions.

Lastly, are there any criteria on a scheme under which the etale cohomology over $\mathbb{R}$ bears any resemblance to the singular cohomology of the real points? I have read that this is related to the Sullivan conjecture in homotopy theory, but don't know much more than that slogan.

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Let $\Gamma=Gal(K/k)$. For a variety $X$ over $k$ and a finite abelian group $F$ say, there is a spectral sequence $H^{r}(\Gamma,H^{s}(X_{K},F))\implies H^{r+s}(X,F)$. In general, if you take a different form of $X$, you will get a different action of $\Gamma$ on $H^{s}(X_{K},F)$, and hence you will expect to get different cohomology. It shouldn't be too hard to work everything out in the particular case you mention (but I also am lazy). –  abz Jul 9 '13 at 1:37
Just an obvious comment--if you take $X=\operatorname{Spec} \mathbb{R}\sqcup \operatorname{Spec} \mathbb{R}$ and $Y=\mathbb{C}$, viewed as $\mathbb{R}$-schemes. The homotopy type of the former is $B(\mathbb{Z}/2\mathbb{Z})\sqcup B(\mathbb{Z}/2\mathbb{Z})$ and the latter is that of a point. If you'd like an group example where the same phenomenon happens, consider $\mu_4$ and the constant group scheme $\mathbb{Z}/4\mathbb{Z}$. –  Daniel Litt Jul 9 '13 at 2:04

1 Answer 1

up vote 13 down vote accepted

For your first question: the two forms of $GL_2$ differ by modifying the Galois action by an automorphism of $GL_2$. The outer automorphism group of $GL_2$ is cyclic of order $2$: an automorphism is inner if and only if it is trivial on the center of $GL_2$. Since the form of $GL_2$ you're considering has the same center as the standard form, it follows that it's an inner twist of $GL_2$. Now $GL_2$ is connected, so any inner automorphism acts trivially on its cohomology. It follows that the two Galois actions are the same (both act by a sign on the class in degree $1$, and preserve the class in degree $3$).

As for your second question: the Sullivan conjecture says that if $X$ is a finite $\mathbf{Z}/2 \mathbf{Z}$-complex, then the $2$-adic completion of the fixed set of $X$ is homotopy equivalent to the homotopy fixed set of the $2$-adic completion of $X$ (here you need to be careful about the meaning of "$2$-adic completion" if the fundamental group is nontrivial: you should really take a $2$-profinite completion, rather than a Bousfield localization). If $Y$ is an algebraic variety defined over $\mathbf{R}$, then $Y(\mathbf{C})$ is a finite $\mathbf{Z}/2\mathbf{Z}$-complex having fixed set $Y(\mathbf{R})$. It follows that the $2$-adic completion of $Y( \mathbf{R})$ can be recovered as the homotopy fixed points of complex conjugation on the $2$-adic completion of $Y(\mathbf{C})$. And the latter can be recovered in a "purely algebraic" way using etale homotopy theory. So in this sense, etale homotopy theory "knows" the $2$-adic completion of $Y(\mathbf{R})$ (and, in particular, invariants like the $\mathbf{F}_2$-cohomology of $Y(\mathbf{R})$).

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Thanks a lot, Jacob! –  Craig Westerland Jul 9 '13 at 4:10

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