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If we consider complex projective varieties, to be defined over $\mathbb{Q}$ means that there is a projective embedding whose image is the vanishing locus of a polynomial system with coefficients in $\mathbb{Q}$.

If we consider closed complex manifolds there is no obvious ambient space.

However, we could require that there be a positive integer $n$ and a choice of

  • for any $1\leq i\leq n$, open sets $U_i\subset M$
  • for any $1\leq i<j\leq n$ such that $U_i\cap U_j\neq \emptyset$, points $p_{i, j}\in U_i\cap U_j$
  • for any $1\leq i \leq n$, holomorphic embeddings $\phi_i:U_i\to \mathbb{C}^{d}$ such that for any $i< j \leq n$ with $U_i\cap U_j\neq \emptyset$ the transition maps $\phi_i(U_i\cap U_j)\to \phi_j(U_i\cap U_j)$ are ratios of two holomorphic functions $\phi_i(U_i\cap U_j)\to \phi_j(U_i\cap U_j)$ each having Taylor series with coefficients in $\mathbb{Q}$ around $p_{i, j}$ that converge on all of $\phi_i(U_i\cap U_j)$?

Has this notion been studied? Is there a closed complex manifold not in Fujiki class $\mathcal{C}$ satisfying this condition?

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    $\begingroup$ Any open subset of $\mathbb{C}^n$ satisfies your condition. You may need compactness to hope for something more interesting. $\endgroup$
    – Chris
    Commented Sep 29, 2020 at 16:46
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    $\begingroup$ Moishezon manifolds (compact complex manifolds whose field of meromorphic functions has the expected transcendence degree) are equivalent to proper algebraic spaces, so for them one could ask that the associated algebraic space is defined over $\mathbb Q$. $\endgroup$
    – Will Sawin
    Commented Sep 29, 2020 at 16:51
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    $\begingroup$ Taylor series with respect to what coordinates? The Taylor expansion of the constant polynomial $z$ about the point $z=\pi$ does not have rational coefficients. $\endgroup$ Commented Sep 29, 2020 at 16:53
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    $\begingroup$ A projective manifold defined over $\mathbb{Q}$ may have no "rational point", i.e., it may have no (complex) point whose coordinates on the charts of a "rational atlas" are all rational coordinates. $\endgroup$ Commented Sep 29, 2020 at 16:56
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    $\begingroup$ @FrançoisBrunault then a generic complex torus is defined over $\mathbb{Q}$. $\endgroup$
    – user164740
    Commented Sep 29, 2020 at 19:20

1 Answer 1

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Note that this is a correct answer to the original question, so I will leave it here, even though the question has now been changed. (The original question is recoverable by going back to the previous versions.)

In fact, every complex manifold has such an atlas.

Let $(M,J)$ be a (finite-dimensional) complex $n$-manifold and let $\mathscr{U}$ be an open cover of $M$ with the properties that (i) for each $U\in\mathscr{U}$, there is a $J$-holomorphic chart $\zeta:U\to\mathbb{C}^n$, and (ii) For each $U\in\mathscr{U}$ there is a point $p\in U$ that does not lie in any $V\in\mathscr{U}$ other than $U$. (Using paracompactness, it is not difficult to construct such a chart.) Then by choosing one such 'reference point' $p_U\in U$ with $p_U\not\in V\in\mathscr{U}$ for $V\not=U$ and one $J$-holomorphic chart $\zeta_U:U\to\mathbb{C}^n$ so that $\zeta_U(p_U) = 0\in\mathbb{C}^n$, we arrive at a 'pointed atlas' $$ \widehat{\mathscr{U}} = \{ (U,\zeta_U,p_U)\ |\ U\in \mathscr{U}\ \} $$ with all the stated properties. The reason is that the only time the point $p_U$ is in the domain of a transition function for the pointed atlas $ \widehat{\mathscr{U}}$ is when one is 'transitioning' from $U$ to $V=U$, and, in that case, the only transition function is the identity mapping on $\zeta_U(U)\subset\mathbb{C}^n$, whose Taylor series at $\zeta_U(p_U) = 0\in\mathbb{C}^n$ clearly has all coefficients in $\mathbb{Q}$ (in fact, all the coefficients are in $\mathbb{Z}$).

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    $\begingroup$ It also reveals a gap in convenientness of established terminology: ambiguity for "projective" (given closed subvariety of projective space / embeddable as such a closed subvariety), ambiguity for "defined over $\mathbf{Q}$" (used both for "defined over $\mathbf{Q}$" [endowed with a $\mathbf{Q}$-structure] and "definable over $\mathbf{Q}$")... $\endgroup$
    – YCor
    Commented Oct 6, 2020 at 12:30

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