Do there exist countably many proper holomorphic submersions of complex manifolds $\mathcal{X}_n \to B_n$ such that every compact complex manifold appears as a fiber in at least one of the families?

(Conventions: Assume that each $B_n$ is Hausdorff, connected, and paracompact; these conditions imply that $B_n$ is also second countable. For the question at hand, one could equivalently require each $B_n$ to be an open subset of $\mathbf{C}^m$ for some $m$ depending on $n$.)

  • $\begingroup$ More generally, one could ask whether there exist countably many proper morphisms of complex analytic spaces such that every compact complex analytic space appears as a fiber in at least one of the families? $\endgroup$ May 1, 2017 at 22:20
  • $\begingroup$ In the algebraic context, the answer is yes, even without properness. For projective varieties, one could use the theory of the Hilbert scheme. But there is an easier argument that applies to all schemes of finite type over $\mathbf{C}$: enumerate finitely generated field extensions of $\mathbf{Q}$ up to isomorphism, enumerate all schemes of finite type over these, and spread out each one to a family over an integral $\mathbf{Q}$-variety. $\endgroup$ May 1, 2017 at 22:22
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    $\begingroup$ You must want the $B_n$'s to be connected. $\endgroup$
    – nfdc23
    May 1, 2017 at 22:56
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    $\begingroup$ @nfcdc23: A paracompact, second countable manifold only has countably many connected components, so I don't think it actually matters. $\endgroup$ May 1, 2017 at 23:02
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    $\begingroup$ @MichaelAlbanese: A second-countable manifold has only countably many connected components without need for paracompactness, and an arbitrary disjoint union of paracompact spaces is paracompact. There is nothing in the hypotheses that is requiring the $B_n$'s to be second countable (could be discrete with uncountably many points), so depending on one's implicit conventions for the definition of "complex manifold" the connectedness would matter (but of course I am just nitpicking, since none of this addresses the real issues in the question). $\endgroup$
    – nfdc23
    May 2, 2017 at 2:26

2 Answers 2


Edit. I spoke with some of the experts in my department. I added one or two references. I reorganized some of the arguments.

This is true. This follows from a couple of big theorems and then a separability / second countability argument that ultimately reduces to the Stone-Weierstrass approximation theorem.

Step 1. Countably many closed oriented manifolds. Dennis Sullivan has a general theory with the following goal: find explicit topological invariants that bound differentiable manifolds $M$ up to finite indeterminacy and diffeomorphism.

MR0646078 (58 #31119)
Sullivan, Dennis
Infinitesimal computations in topology.
IHES. Publ. Math. No. 47 (1977), 269–331 (1978).

To be honest, I do not completely understand the definitions of models and lattices. There is a different proof that follows from Cheeger's finiteness theorem. Every closed manifold $M$ admits a Riemannian metric $d_M$. Cheeger's original finiteness theorem was only for simply connected Riemannian manifolds, and the proof required stronger hypotheses in dimension $4$. That theorem was generalized by Cheeger and Ebin. The most general form of the theorem that I have found is due to Stefan Peters and independently to Gromov.

MR0743966 (85j:53046)
Peters, Stefan(D-DORT)
Cheeger's finiteness theorem for diffeomorphism classes of Riemannian manifolds.
J. Reine Angew. Math. 349 (1984), 77–82.

Generalized Cheeger Finiteness Theorem. For fixed dimension $n$, for fixed diameter $\delta$, for fixed volume $V$, and for fixed bound on the norm of the sectional curvature $\Lambda$, there are only finitely many diffeomorphism types of closed $n$-dimensional manifolds $M$ that admit a Riemannian metric $d_M$ such that $\text{diam}(M)\leq \delta$, $\text{vol}(M)\geq V$, and the norm of the sectional curvature satisfies $|K_M|\leq \Lambda^2$.

Considering $\delta$ an integer, $V$ of the form $1/m$ for $m$ a positive integer, and $\Lambda$ an integer, it follows that there are at most countably many diffeomorphism types of closed manifolds. Sullivan's theorem proves a similar finiteness theorem, but fixing topological data rather than Riemannian data. Because of these finiteness theorems, it suffices to prove the result for complex manifolds whose underlying differentiable manifold is diffeomorphic to a fixed closed, oriented, differentiable manifold.

Edit. As Bjorn Poonen points out, countability of closed manifolds up to diffeomorphism also follows from Nash's theorem that every closed manifold is diffeomorphic to one connected component of the set of real points in $\mathbb{R}^n$ of a smooth, real algebraic variety (with the differentiable structure inherited from $\mathbb{R}^n$ via the Implicit Function Theorem). The reason I refer to the theorem of Sullivan a is that this theorem (and similar theorems) is essential in important boundedness theorems for compact complex manifolds of particular classes (e.g., hyper-Kähler manifolds). I learned of these boundedness results from Ljudmila Kamenova and Misha Verbitsky.

Step 2. The Fréchet manifold of almost complex structures. Let $M$ be a fixed closed, oriented, differentiable manifold. There exists a smooth embedding of $M$ in a Euclidean space $\mathbb{R}^e$. Let $d_M$ be the restriction to $M$ of the Euclidean Riemannian metric on $\mathbb{R}^e$. There is an induced Riemannian metric on the differentiable vector bundle $T_{\mathbb{R}^e}|_M$, the restriction to $M$ of the tangent bundle of $\mathbb{R}^e$, and this metric is induced by a smooth family of inner products on the fibers. These inner products determine a smooth and orthogonal direct sum decomposition of differentiable vector bundles, $T_{\mathbb{R}^e}|_M = T_M\oplus N_{M/\mathbb{R}^e}$. This in turn defines a smooth and orthogonal direct sum decomposition of the differentiable vector bundle $\text{Hom}_M(T_{\mathbb{R}^e}|_M,T_{\mathbb{R}^e}|_M)$ with one summand equal to $\text{Hom}_M(T_M,T_M)$. There is a closed submanifold $AC_M$ of this summand bundle whose fiber over every $t\in M$ is the set of orientation-preserving linear automorphisms $J:T_{M,p}\to T_{M,p}$ such that $J\circ J$ equals $-\text{Id}$. This is a differentiable fiber bundle over $M$. The direct sum decompositions above determine a smooth and isometric embedding of $AC_M$ into the vector bundle $T_{\mathbb{R}^e}$, which in turn is isometrically diffeomorphic to $\mathbb{R}^{2e}$ with its standard inner product.

Denote by $\mathcal{J}$ the set of all smooth sections $J$ of the fiber bundle $AC_M\to M$, i.e., the set of all almost complex structures on $M$. This is a subset of the set of all smooth functions $C^\infty(M,AC_M)$. Via the embedding above, this is a subset of the set of all smooth function $C^\infty(M,\mathbb{R}^{2e})$. This set is a Fréchet topological vector space topologized by using the countably many metrics $(d_n)_{n\in \mathbb{N}}$ arising from the uniform metrics for partial derivatives of all orders. (Because $M$ is compact, and thus has finite volume, the uniform topology is finer than any $L^p$-topology.) Since it does not change the metric topology, replace each $d_n$ by $d_n'$, the pointwise minimum of $d_n$ and the constant $1$. For the associated metric $d=\sum_n (1/2^n) d'_n$, the metric topology of $d$ on $C^\infty(M,\mathbb{R}^{2e})$ is the Fréchet topology.

By the Stone-Weierstrass approximation theorem, and the generalization due to Nachbin, there is a countable dense subset of this Fréchet space, e.g., it suffices to take the restrictions to $M\subset \mathbb{R}^e$ of the countable set of polynomial functions $\mathbb{R}^e\to \mathbb{R}^{2e}$ that have rational coefficients. Thus, this Fréchet space with the metric $d$ is separable, or equivalently, it is second countable.

The subset $\mathcal{J}$ with the induced metric from $d$ is a metric space. As a susbspace of a second countable space, also $\mathcal{J}$ is second countable. In fact, $\mathcal{J}$ is a Fréchet submanifold of $C^\infty(M,\mathbb{R}^{2e})$.

Step 3. Newlander-Nirenberg and Kuranishi's theorem. By the Newlander-Nirenberg theorem, for the Fréchet manifold $\mathcal{J}$ of smooth sections of $AC_M \to M$, there is a closed subspace $\text{Comp}_M$ of almost complex structures that are integrable -- this turns out to be those for which the Nijenhuis tensor is zero (which is defined in terms of smooth sections and finitely many derivatives, thus is a continuous function on this Fréchet manifold). Kuranishi showed how to reduce to a fixed Sobolev manifold. Moreover, Kuranishi proved a completeness theorem at one point for his families of complex structures; what algebraic geometers would call "versality". Finally, Kuranishi proved completeness for all points in an open neighborhood; what algebraic geometers would call "openness of versality."

MR0141139 (25 #4550)
Kuranishi, M.
On the locally complete families of complex analytic structures.
Ann. of Math. (2) 75 1962 536–577.

Define a Kuranishi patch to be a triple $(P,\pi:\mathcal{X}\to S,g:P\to S)$ of an open subset $P$ of $\text{Comp}_M$ (with its subspace topology), a proper, holomorphic submersion of complex analytic spaces $\pi:\mathcal{X}\to S$ such that $S$ is connected, complex analytic subspace of an affine space, and a surjective function $g:P\to S$ such that (i) for every $J\in P$ with image $s=g(J)$, there exists a diffeomorphism of $M$ with the fiber $\mathcal{X}_s=\pi^{-1}(\{s\})$ such that $J$ is identified with the natural complex structure on $\mathcal{X}_s$, and (ii) for every $s\in S$, the family is complete at $s$. Precisely, for every proper holomorphic submersion of complex analytic spaces, $\pi':\mathcal{X}'\to S'$, for every point $s'\in S'$ and a biholomorphism $\mathcal{X}'_{s'}\cong \mathcal{X}_s$, up to replacing $S'$ by an open neighborhood of $s$ and replacing $\mathcal{X}$ by the inverse image of this open neighborhood, there exists a holomorphic morphism $h:S'\to S$ mapping $s'$ to $s$ and such that $\mathcal{X}'$ is biholomorphic, over $S'$, to the pullback by $h$ of $\mathcal{X}$.

Kuranishi's Theorem. Every integrable almost complex structure $J$ is contained in a Kuranishi patch. Moreover, there exists such a Kuranishi patch with $S$ a complex analytic subspace of $H^1(M,T^{1,0}_{M,J})$.

By the way, I learned all of this from Misha Verbitsky and Ljudmila Kamenova. Verbitsky has a quick summary of Kuranishi theory at the beginning of the following article. The Bourbaki seminar by Adrien Douady reporting on Kuranishi's theorem is also very readable.

Teichmueller spaces, ergodic theory and global Torelli theorem
Misha Verbitsky

Douady, Adrien
Le problème des modules pour les variétés analytiques complexes (d'après Masatake Kuranishi).
Séminaire Bourbaki, Vol. 9, Exp. No. 277, 7–13, Soc. Math. France, Paris, 1995.

Step 4. Second countability. The Fréchet space $C^\infty(M,\mathbb{R}^{2e})$ is second countable by the Stone-Weierstrass approximation theorem / Nachbin's theorem. Thus the subspaces $\mathcal{J}$ and $\text{Comp}_M$ are also second countable. Let $\{U_n\}_{n\in \mathbb{Z}}$ be a countable basis for the topology of $\text{Comp}_M$. Let $I\subset \mathbb{Z}$ be the subset of $n$ such that $U_n$ is contained in a Kuranishi patch. As a subset of a countable set, also $I$ is countable. For every $n\in I$, let $P_n$ be a Kuranishi patch that contains $n$.

By Kuranishi's Theorem, for every $J$ in $\text{Comp}_M$, there exists a Kuranishi patch that contains $J$. Since $\{U_n\}_n$ is a basis for the topology, there exists $n$ such that $J$ is contained in in $U_n$, and $U_n$ is contained in this Kuranishi patch. Thus, $n$ is an element of $I$, and $J$ is contained in $P_n$, since $P_n$ contains $U_n$. Thus, the countably many Kuranishi patches $(P_n)_{n\in I}$ form a countable open covering of $\text{Comp}_M$. So the countably many associated families $\pi_n:\mathcal{X}_n \to S_n$ have the desired property: every complex manifold that is diffeomorphic to $M$ is biholomorphic to a fiber of one of the proper, holomorphic submersions $S_n$.

If you really insist, you can form a resolution of singularities $\widetilde{S}_{n}\to S_{n}$. More sensibly, as nfdc23 suggests, you can partition $S_{n}$ into a union of finitely many locally closed analytic subspaces, each of which is smooth. Personally, I am happy to work with singular complex analytic spaces $S_n$, provided that they are a little bit closer to a "universal family".

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    $\begingroup$ Actually, the request for $B$ to be smooth doesn't require going through resolution since the demands on $B$ are so mild (just a request on fibers over $B$, no stronger mapping property): if we had some $B$ which does the job but is a non-smooth complex-analytic space then we can pass to the members of a locally finite stratification of $B_{\rm{red}}$ for the analytic Zariski topology by smooth strata. How do $p$-adic numbers potentially arise in giving further invariants? $\endgroup$
    – nfdc23
    May 2, 2017 at 12:06
  • $\begingroup$ @nfdc23. That is true. I was actually trying to see whether specifying a basis for $H^1(M,T^{1,0}_M)$ and perhaps some "finite" data about obstruction maps might reduce to a parameter problem that is "countable up to deformation". However, that does not seem to help. A priori, there might be uncountably many non-diffeomorphic complex structures that each have $H^1(M,T^{1,0}_M)$ equal to zero. $\endgroup$ May 2, 2017 at 12:09
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    $\begingroup$ In Step 3 you make an open subset $\cup_{\sigma} U_{\sigma}$ containing a dense subset of the Frechet manifold $\mathcal{J}$ and then claim this open subset must therefore be the entire space. But this already fails quite strikingly for open sets of arbitrarily small measure containing a countable dense subset of $\mathbf{R}^n$ ($n>0$), so the argument that one has covered $\mathcal{J}$ seems to have a gap. Or am I missing something specific to the above situation? Maybe you can dispense with $\Sigma$ and use that any (yuuuge) open cover of $\mathcal{J}$ has a countable subcover? $\endgroup$
    – nfdc23
    May 3, 2017 at 17:30
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    $\begingroup$ I think Step 1 must have been known long before the papers you cite. For instance, Nash, "Real algebraic manifolds", Annals of Math. in 1952 proved that every compact differentiable real manifold is diffeomorphic to a component of a real algebraic variety; there are only countably many families of real algebraic varieties; and the diffeomorphism type is locally constant within each family, by Ehresmann's theorem proved in 1950. $\endgroup$ May 3, 2017 at 20:44
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    $\begingroup$ @BjornPoonen. Yes, those results of Sullivan and Cheeger are trying to classify manifolds with specified topological, resp. Riemannian, invariants, rather than trying to prove that there are only countably many closed manifolds up to diffeomorphism. $\endgroup$ May 3, 2017 at 21:09

This is an attempt to express a variant of Jason's answer in slightly more down-to-earth terms, so that maybe less has to be referred to his experts. (This answer is too long for a comment, so I made it a community wiki.) It seems to me that there is really only one difficult point in the argument beyond the existence of the Kuranishi deformation space, namely the "Theorem" below.

Step 1: Nash's article "Real algebraic manifolds" in Annals of Math. (1952) proved that every connected compact $C^\infty$ real manifold is diffeomorphic to a component of a real algebraic variety.
There are only countably many families of real algebraic varieties, and Ehresmann's fibration theorem (1950) shows that the diffeomorphism type is locally constant within each family. Thus there are only countably many diffeomorphism types. Therefore we now restrict attention to one.

Step 2: Let $M$ be a compact $C^\infty$ real manifold, let $T$ be its tangent bundle, and let $E$ be the bundle with fibers $E_x := \operatorname{End} T_x$. Fix $r$ not too small, and let $C^r(M,E)$ be $\{\text{$C^r$ functions $M \to E$}\}$ with the $C^r$ compact-open topology (Hirsch, Differential topology, 2.1), which is second countable. The space $\operatorname{Comp}$ of complex structures (i.e., integrable almost complex structures) on $M$ is a subset of $C^r(M,E)$; give it the subspace topology, so it too is second countable. Explicitly, complex structures $I$ and $I'$ are close if uniformly on $M$ they are within $\epsilon$ (with respect to some metric) and the same holds for their derivatives up to order $r$.

Step 3: On the other hand, let $\mathcal{M}$ be the set of isomorphism classes of compact complex manifolds whose underlying real manifold is diffeomorphic to $M$. Kuranishi proved that each compact complex manifold $X$ admits a versal deformation over a pointed connected complex analytic space $(B,0)$. Each such versal deformation gives rise to a subset of $\mathcal{M}$, namely the set of isomorphism classes of the fibers; define a topology on $\mathcal{M}$ in which these subsets form a basis of neighborhoods of the point $[X] \in \mathcal{M}$ determined by $X$.

Step 4: Let $\Phi \colon \operatorname{Comp} \to \mathcal{M}$ be the map sending $(M,I)$ to its isomorphism class as a complex manifold. Then $\Phi$ is open because for any open subset $U \subset \operatorname{Comp}$ and $[X] \in \Phi(U)$, a choice of $C^\infty$ trivialization of a versal deformation over $B$ defines a continuous lifting $B \to \operatorname{Comp}$ of $B \to \mathcal{M}$.

The key point is now the following (which seems to be a consequence of what Jason's experts claimed):

Theorem: $\Phi$ is continuous. (In other words, given a versal deformation of $X = (M,I)$, any complex structure $I'$ sufficiently close to $I$ in the $C^r$ sense defines a complex manifold $(M,I')$ isomorphic to a fiber of the deformation.)

Given this, $\mathcal{M}$ is second countable because it is the image of a second countable space under an open continuous map. This means that $\mathcal{M}$ is covered by countably many subsets each arising from a family over a complex analytic subspace in $\mathbf{C}^n$ for some $n$. Each such base can be stratified into complex manifolds, so we are done.


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