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.



Etale cohomology of and forms of algebraic groups
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 $Y(\mathbf{R})$).


