Algebraic atlas on smooth manifolds

A real/complex rational atlas on a smooth closed manifold $$M$$ is an atlas with charts homeomorphic to Euclidean open sets in $$\Bbb{R}^n$$/$$\Bbb{C}^n$$ covering $$M$$ and real/complex rational transition maps. A real/complex rational structure is a maximal collection of compatible mentioned real/complex atlases. Two rational structures are called isomorphic if there is a topological automorphism $$\sigma$$ of $$M$$ such that each coordinate representation of $$\sigma$$ under a rational chart pair (from each structure) is rational. The existence of a rational structure does not imply being algebraic because every complex torus admits one (the transition maps are translations, thus polynomial), so I want to ask about how rigid these structures are.

Q$$1$$: Are there topological invariants classifying whether a smooth manifold admits a real/complex rational structure? Or on the contrary every smooth manifold admits one?

Q$$2$$: Obviously a real/complex rational structure belongs to a smooth/holomorphic structure. Consider the converse problem: what is the moduli space of real/complex rational structures compatible with the smooth/holomorphic structure on $$M$$? Specifically, is it true that the number of complex rational structures belonging to the holomorphic structure of $$M$$ is at most $$1$$? Counterexamples are welcome.

Q$$3$$: A related question exists on this website, and the first answer gives a sheaf-sense definition of rational structures. However its conclusion -- the algebraic structure of $$\Bbb{C}^n$$ can be pullbacked onto $$M$$ -- seems to contradict my complex torus example. I hope someone can help.

• It seems a strong condition that each chart have image homeomorphic to $\mathbb{C}^n$, and not just to a Zariski open subset of $\mathbb{C}^n$. Why do you want that? Dec 4 '21 at 8:58
• Zerox, you probably want to have charts being open subsets of $\mathbf{K}^n$ (in the Euclidean topology). Otherwise tori $\mathbf{K}^n/\Lambda$ would have no charts as desired, and you would also miss most affine/projective manifolds.
– YCor
Dec 4 '21 at 9:55
• @YCor You are right. I changed my description. Dec 4 '21 at 13:47

The answers to your question in the case $$n=1$$ are well-known. In higher dimensions, the answers are less complete, but something is known.

For example, in the real case when $$n=1$$, there is only one smooth, connected compact $$1$$-manifold, the circle, and, for each natural number $$k\ge1$$, there is a rational structure $$\mathcal{R}_k$$, which is the rational structure induced on the $$k$$-fold connected cover of $$\mathbb{RP}^1\simeq S^1$$ (which is smoothly diffeomorphic to $$S^1$$, of course). Two rational structures $$\mathcal{R}_i$$ and $$\mathcal{R}_j$$ on the circle are isomorphic if and only if $$i=j$$, and every rational structure on the circle is isomorphic to some $$\mathcal{R}_i$$.

Meanwhile, in the complex case when $$n=1$$, specifying a rational structure on a compact Riemann surface of genus $$g$$ (i.e., a topological surface of genus $$g$$ with a fixed underlying holomorphic structure) is easily seen to be equivalent to specifying a projective structure on the surface. For $$g>1$$, it is known that the moduli of projective structures on a given compact Riemann surface is equivalent to the moduli of quadratic holomorphic differentials, a complex vector space of dimension $$3g{-}3$$. For example, see R. C. Gunning's On uniformization of complex manifolds: The role of connections. In particular, these provide counterexamples sought by the OP.

In higher dimensions the situation is more complicated because the pseudogroup of rational maps with a rational inverse, i.e., the so-called birational pseudogroup, in either $$\mathbb{R}^n$$ or $$\mathbb{C}^n$$ is not well-understood when $$n>1$$.

However, when $$n=2$$, the question of when a given compact complex surface has a rational structure is well-understood since we have a classification of compact complex surfaces, by the work of Kodaira. Not all such surfaces have a rational structure; for example, a K3 surface does not.

• For the classification of holomorphic projective connections on complex surfaces, see Bruno Klingler, Structures affines et projectives sur les surfaces complexes, Ann. Inst. Fourier (Grenoble) 48 (1998), no. 2, 441–477. MR MR1625606 (99c:32038). But these are not the same as rational structures, since the birational automorphisms of the complex affine plane are more complicated. Dec 4 '21 at 13:32
• Are there results about the well-undertood case of real $2$-manifolds and geometrizable real $3$-manifolds? Dec 4 '21 at 13:53
• Every primary Kodaira surface admits infinitely many holomorphic rational structures not derived from any projective connection: McKay, Benjamin Exotic geometric structures on Kodaira surfaces. Indiana Univ. Math. J. 62 (2013), no. 2, 643–670. Dec 4 '21 at 14:01
• @Zerox: Any connected compact real smooth 2-manifold has a metric of constant curvature $K$, so it has an atlas whose transition maps are isometries between contractible open sets in the simply-connected surface of constant curvature $K$. Since these latter spaces can be presented in such a way that their isometries are rational maps, every compact real smooth 2-manifold has at least one rational structure. I'm not aware of any classification up to rational equivalence of such real rational structures on 2-manifolds, though. The case of geometrizable 3-manifolds is probably similar. Dec 5 '21 at 10:39