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If $k=\mathbb{C}$, $X$ is a projective (or even proper) scheme over $k$ and $F$ a coherent sheaf on $X$, then there is an isomorphism

$$ H^p(X,F)\rightarrow H^p(X^{an}, F^{an}) $$

(and there is also a relative version of this, for proper morphisms between $k$-schemes locally of finite type).

If take $k$ to be $\mathbb{R}$, there still is an analytification functor, so I wonder wether this result still holds and if you could provide a reference for this.

Remarks: (i) I must admit that I haven't studied the details of the proof in the complex setting (yet), so from my current knowledge it would be possible that it carries over without major modifications. If this is not the case, what goes wrong?

(ii) In case there are additional assumptions to be made, I would also be happy with a statement for smooth varieties.

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    $\begingroup$ There are lots of nontrivial $X$'s such that $X(\mathbb{R})$ is empty. $\endgroup$
    – Laurent Moret-Bailly
    Commented Dec 21, 2015 at 17:00
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    $\begingroup$ So the only way to make the statement non trivially false would be to change the definition of $X^{an}$. A natural definition would be the quotient of the analytic space $X(\mathbf C)$ by the action of complex conjugation (this is what Berkovich's definition gives, in any case). Then the theorem is a trivial consequence of the complex GAGA. $\endgroup$
    – ACL
    Commented Dec 21, 2015 at 22:11
  • $\begingroup$ thank you both for the very useful comments. In fact I had a situation in mind where the scheme "comes from" a scheme over $\mathbb{C}$ via Weil restriction. In this case the problem that the analytification has no points will not arise and the approach proposed by ACL seems very natural indeed. Could you specify what you mean by Berkovich's definition? $\endgroup$
    – jorst
    Commented Dec 22, 2015 at 14:47

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