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Let $X$ be a scheme defined over $\mathbb{C}$ with an involution $\sigma$. How to get a $X_{\mathbb{R}}$ scheme defined over $\mathbb{R}$ such that $X_{\mathbb{R}}\times_\mathbb{R} \mathbb{C} = X$ ? How are the real algebraic bundle on $X_{\mathbb{R}}$ and complex bundle on $X$ related?

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You need a semi-linear $\sigma$, i.e., it covers complex conjugation on $\mathbb C$. In that case the $\mathbb R$-scheme is obtained by descent theory as the quotient by $X$ under $\sigma$ (though I guess in general this quotient will only be an algebraic space). –  Torsten Ekedahl Aug 4 '10 at 7:54
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As an illustration of Torsten's parenthetical remark, on p. 210 of the book "Neron models" there's a nice example (due to Mumford) of a scheme loc. of finite type over $\mathbf{C}[[t]]$ equipped with descent data relative to $\mathbf{R}[[t]]$ so that the descent is an algebraic space and not a scheme. The same example works using $\mathbf{C}[t]$ and $\mathbf{R}[t]$, so it provides schemes of finite type over $\mathbf{C}$ which descend to non-scheme alg. spaces over $\mathbf{R}$. Unless one assumes $X$ is q-projective, it seems hard to determine when the descent as an alg. space is a scheme. –  BCnrd Aug 4 '10 at 11:28
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The paper by Atiyah on K-theory and reality Quart. J. Math. Oxford Ser. (2) 17 1966 367--386. MR0206940 discusses the topological analogue of this question, showing how to relate complex vector bundles over a space with vector bundles over the fixed points of an involution. The idea is to define a sort of intermediate K-group KR of vector bundles with involution, and compare it to the two other K-groups.

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