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Maybe this question is trivial, but I was not able to find an answer. The question is this: Consider the algebraic K-theory of smooth complex projective varieties (such that the K-theory and the G-theory agree). Does it have the Mayer-Vietoris property for the analytic topology? I know it does for the Zarisky topology. If the answer is negative is it possible to have an example?

Update

In view of the comments I felt the need to elaborate my question. Let $X$ be a smooth projective complex variety. Let $G(X)$ be the $K-$theory space of the category of coherent sheaves over $X$. If $U$ and $V$ are two Zariski open subschemes one has the following homotopy cartesian square:$\require{AMScd}$ \begin{CD} G(U \cup V) @>>> G(U)\\ @VVV @VVV \\ G(V) @>>> G(U \cap V) \end{CD}

Does it remain true if $U$ and $V$ are open subsets in the usual (complex) topology of $X$?

Here an open subset $U$ is seen as a ringed space with structure sheaf $\mathcal{O}_X|_U$.

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    $\begingroup$ I'm not sure the question makes sense -- if I take an analytic but not algebraic open set, what is its algebraic K-theory supposed to be? $\endgroup$
    – Allen Knutson
    Commented Apr 9, 2015 at 15:31
  • $\begingroup$ I think that by GAGA an analytic subset will be also algebraic. $\endgroup$
    – Andrei Halanay
    Commented Apr 9, 2015 at 16:44
  • $\begingroup$ @AndreiHalanay I usually understand "analytic topology" to mean the classical topology where we consider $X$ as a real manifold. So, if $X = \mathbb{C}$ with the standard algebraic structure, then the open disc of radius $1$ is open for the analytic topology, but doesn't come with an algebraic structure. If I am confused, could you spell out your definitions? $\endgroup$
    – David E Speyer
    Commented Apr 9, 2015 at 17:29
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    $\begingroup$ I would be doubtful of the thesis, since algebraic K-theory does not satisfy étale descent, which should be weaker than descent for the analytic topology. $\endgroup$
    – Denis Nardin
    Commented Apr 9, 2015 at 17:56
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    $\begingroup$ But then isn't this just asking the question in the Zariski topology? $\endgroup$
    – David E Speyer
    Commented Apr 10, 2015 at 18:18

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