Let $M$ be a complex manifold admitting an atlas with each chart biholomorphic to $\mathbb{C}^n$ and transition maps being rational functions.

Is it true that there exists a smooth integral separated scheme $X$ of finite type over $\mathbb{C}$ such that its set of complex points $X(\mathbb{C})$ endowed with analytic topology is biholomorphic to $M$? Is the converse true (i.e. if such a scheme exists, then there has to exist an atlas as in the first sentence)?


2 Answers 2


I'm a bit confused by your question. I'll try to interpret what you are after: you want to put a condition on complex manifolds (defined via holomorphic atlases) so that the resulting class of manifolds $M$ is exactly that of algebraic ones (i.e. $M=X_{an}$ where $X_{an}$ is the analytification of a smooth complex algebraic variety $X$).

The condition that charts locally resemble $\mathbb{C}^n$ algebraically is too stringent. Indeed, most smooth algebraic varieties are not locally isomorphic to each other and/or to affine space. Take an elliptic curve $X$ over $\mathbb{C}$ for example: $X_{an}$ is certainly the complex manifold associated to a smooth algebraic variety, but $X$ is not locally isomoprhic to $\mathbb{A}^1_{\mathbb{C}}$ (so $X_{an}$ does not have holomorphic charts of the form you require).

Also, I think your atlas condition on manifolds, though too stringent to get all the smooth algebraic varieties, does get you only (complex manifolds that are the analytification of) algebraic varieties. Why? Because if you keep only the datum of the images of the charts, and the transition functions, you're essentially presenting your manifold as a gluing of opens of $\mathbb{C}^n$ by holomorphic identifications. Your requirement forces such identifications to be regular isomorphisms in the sense of algebraic geometry, so you are gluing schemes in the standard way and you will get a scheme (and I have the impression the conditions on the scheme of being reduced, finite type, smooth, separated, etc should follow easily).

Added. I think even if you allow charts to be open balls (but impose the transition functions to be rational maps with no indeterminacies on the considered domain) you still won't get all smooth varieties for the following reason.

Consider the transition functions $$\varphi_{ij}:V_{i,ij}:=\varphi_i(U_{ij})\to V_{j,ij}:=\varphi_j(U_{ij})$$ where now the $V_i$s are open balls in $\mathbb{C}^n$. Each $\varphi_{ij}$, since after all it's given by ratios of polynomials, has domain and codomain that can be extended to a Zariski open $W_i$, with $V_i\subseteq W_i \subseteq\mathbb{C}^n$ and, up to further shrinking the $W_i\;$'s, $\varphi_{ij}:W_i\to W_j$ will still be a regular isomorphism. Now you're back to the case in which local charts are valued in Zariski opens of affine space (and transition maps are regular).

Remark. Taking Zariski opens of affine space as algebraic local models for smooth varieties is too strong a condition. Think of this: if two varieties have isomorphic local rings at a single point, then they are necessarily birational to each other. If you impose the charts to be valued in Zariski opens of affine space, then your variety has to be rational (https://en.wikipedia.org/wiki/Rational_variety) - There are plenty of non-rational smooth varieties.

Analogy. You want to specify a class of atlases on smooth (in the sense of real, differentiable) manifolds to get something equivalent to Riemannian manifolds. Then you say: let's take charts valued in opens of $\mathbb{R}^n$ with transition maps which are restrictions of isometries for the standard Euclidean Riemannian structure on $\mathbb{R}^n$. Will you get all Riemannian manifolds this way? Of course not: you will get only the flat ones. Why? Because Riemannian manifolds have a lot of possible local behaviors: very few of them locally resemble flat space.

Added later:

Let $M$ be a topological manifold locally homeomorphic to $\mathbb{C}^n$, and let $\mathcal{C}^0_M$ be its sheaf of continuous $\mathbb C$-valued functions. Let $\mathcal R\subseteq \mathcal{C}^0_{\mathbb{C}^n}$ be a subsheaf of $\mathbb{C}$-algebras on $\mathbb{C}^n$ (in the following I'll say "sheaf of rings" for simplicity).

Definition. A map $\varphi: U\to V$, with $U,V\subseteq\mathbb{C}^n$ opens, is $\mathcal R$-compatible if the map of ringed spaces $$(\varphi,\varphi^{\sharp}):(U,\mathcal{R}|_U)\to(V,\mathcal{R}|_V)$$ is an isomorphism, where $\varphi^{\sharp}$ is the natural map induced from that coming from $\mathcal{C}^0_{\mathbb{C}^n}$. An atlas $\mathcal A =\{(U_i,\varphi_i)\}$ for $M$ is said $\mathcal R$-compatible if each transition map $\varphi_{ij}:=\varphi_j\circ\varphi_i^{-1}$ is $\mathcal R$-compatible.

Lemma. Let $\varphi:U\to V$ be a homeomorphism ($U,V$ open in $\mathbb{C}^n$). Then

1) $\varphi$ is holomorphic iff it is $\mathcal{O}_{\mathbb{C}^n}^{an}$-compatible.

2) $\varphi$ is algebraic iff it is $\mathcal{O}_{\mathbb{C}^n}^{alg}$-compatible.

Corollary. An atlas on $M$ is

1) holomorphic iff it is $\mathcal{O}_{\mathbb{C}^n}^{an}$-compatible

2) algebraic iff it is $\mathcal{O}_{\mathbb{C}^n}^{alg}$-compatible.

Proposition. Let $\mathcal R\subseteq \mathcal{C}^0_{\mathbb{C}^n}$ be a subsheaf of rings, and $\mathcal A$ an $\mathcal R$-compatible atlas on $M$. Then there is a unique subsheaf of rings $\mathcal R_M^{\mathcal A}\subseteq \mathcal{C}^0_M$ ($\mathcal{R}_M$ for brevity) such that

1) $\mathcal{R}_M|_{U_i}=\varphi_i^{-1}(\mathcal{R}|_{\varphi_i(U_i)})$ for every $i$.

2) $\varphi_i:(U_i,\mathcal{R}_M|_{U_i})\to(\varphi_i(U_i),\mathcal{R}|_{\varphi_i(U_i)})$ is an isomorphism of ringed spaces.

3) For every $x\in M$, the stalk $\mathcal{R}_{M,x}\simeq \mathcal{R}_x$.

Proof: you define $\mathcal{R}_i:=\varphi_i^{-1}(\mathcal{R}|_{\varphi_i(U_i)})$ on $U_i$, and verify that $\mathcal{R}_i|_{U_{ij}}=\mathcal{R}_j|_{U_{ij}}$.

If you take $\mathcal{R}=\mathcal{O}_{\mathbb{C}^n}^{an}$ and $\mathcal A$ is a holomorphic atlas on $M$, then $\mathcal{R}_M$ is $\mathcal{O}_M^{an}$, the sheaf of holomorphic functions on the complex manifold $M$ (with respect to $\mathcal A$). If $\mathcal A$ is even algebraic, then $\mathcal{R}_M$ gives you the sheaf of regular functions on $M$ in the sense of algebraic geometry. Notice that you are putting further structure on $M$, and that there are algebraic manifolds with the same underlying holomorphic manifold.

By point 3) of the Proposition, the local ring at a point $x\in M$ has to be that of affine space, so an algebraic manifold constructed as above is necessarily rational.

  • $\begingroup$ but what if I demand charts to be open balls? I was thinking of larger charts because of Wolfram article (mathworld.wolfram.com/AlgebraicManifold.html). Do I get all smooth varieties then? If this also does not work, what does? $\endgroup$
    – rori
    Dec 8, 2018 at 5:21
  • $\begingroup$ (Added a paragraph in response to your comment) $\endgroup$
    – Qfwfq
    Dec 8, 2018 at 15:09
  • $\begingroup$ do I understand correctly that your remark is not directly relevant to the question? In the analytification of a variety, we only have access to the sheaf of holomorphic functions, whose stalks at different points are isomorphic, right? $\endgroup$
    – rori
    Dec 8, 2018 at 18:19
  • $\begingroup$ I have added an explanation of how you pass from atlas to sheaf of rings. I hope this clarifies the situation. $\endgroup$
    – Qfwfq
    Dec 8, 2018 at 21:43
  • $\begingroup$ (By the way, this is a question -I mean the OP question- more suited for MSE than MO) $\endgroup$
    – Qfwfq
    Dec 8, 2018 at 22:04

As @Qfwfq pointed out, if the transition maps are rational functions, then the open charts are (can be extended to) Zariski opens of affine space, so in particular $X$ must be rational. Such varieties are called uniformly rational. It was apparently a fairly-long open question due to Gromov whether every rational variety is uniformly rational, but Karzhemanov recently proved this is false, i.e. not every rational variety has such a system of charts. See here for more info: Local structure of rational varieties


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