Kodaira defines a complex analytic family of compact complex manifolds as the data $(E,B,\pi)$, where $E$ and $B$ are complex manifolds, and $\pi$ is a surjective holomorphic submersion such that the preimage $\pi^{-1}(x)$ of any point $x \in B$ is a compact complex submanifold of $E$. (Here complex manifolds are required to be connected.)

A key property of this definition is that this specifies a differentiable fibre bundle. This fact can almost be obtained from the Ehresmann fibration theorem, which might be stated as follows: "Let $X$, $Y$ be differentiable manifolds, and let $f: X \rightarrow Y$ be a proper surjective submersion. Then $f$ is the projection of a differentiable fibre bundle."

In particular, the map $\pi$ defining a family lacks the properness needed to apply Ehresmann's theorem. It is, however, a corollary of the fact that a family gives a fibre bundle that $\pi$ is indeed proper.

Is there a more elementary way of seeing that $\pi$ must be proper?


I think the following argument answers your question. It reminds me a lot of the exercises in the basic topology course I took during my undergrad. Fun times.

Recall that $\pi$ is a submersion, so for any $x$ in $E$ there exists a neighborhood of the form $U \times V$, such that $\pi$ identifies with the projection $U \times V \to V$. Suppose we only know that $E_t := \pi^{-1}(t)$ is compact for any $t$ in $B$, and let's show that $\pi$ is proper.

Let $K \subset B$ be compact and set $L = \pi^{-1}(K)$. Let $(\mathcal U_\alpha)$ be an covering of $L$ by open subsets of $E$. We will show that there exists a finite covering of $E$ by open subsets of $(\mathcal U_\alpha)$, and thus by sets of the original covering.

Take $t \in K$. Find neighborhoods $\mathcal U_1, \ldots \mathcal U_{n(t)}$ that cover $E_t$. For any point $x \in E_t$, find neighborhoods $U_{t,x} \times V_{t,x} \subset \mathcal U_{j(x)}$ for some $j(x)$ (the $\mathcal U_{j(x)}$ being one of the above), such that the map $\pi$ identifies with the projection $U_{t,x} \times V_{t,x} \to V_{t,x}$.

The compactness of $E_t$ gives finitely many such $U_{t,x_\nu} \times V_{t,x_\nu}$ which cover $E_t$ and such that there is a $V_t \subset B$ contained in the image of any $V_{t,x_\nu}$ by $\pi$ (the intersection of the finitely many such $V$ contains $t$).

Now: $K$ is compact, so it is covered by finitely many of those $V_t$. The corresponding finite collection $(U_{t,x_\nu} \times V_{t,x_\nu})$ covers $L$, and consists of subsets of elements of $(\mathcal U_\alpha)$. Thus finitely many elements of $(\mathcal U_\alpha)$ cover $L$.

  • $\begingroup$ Great proof, Gunnar, but we have to assume that $\pi$ is surjective, else your proof breaks down . And for a vey good reason: the result is false without this assumption of surjectivity! Think of the embedding of $(0,1)$ in $\mathbb R$. But again, kudos: you have shown that essentially all statements of Ehresmann's theorem have a too strong hypothesis, namely that $\pi$ is proper. $\endgroup$ – Georges Elencwajg Mar 19 '11 at 0:46
  • $\begingroup$ Thanks for catching that, we do indeed need $\pi$ to be surjective. I'm glad you like the proof, but I have doubts about its usefulness. We prove that a surjective submersion which has compact fibers is proper - can anyone think of a situation where such a thing pops up without properness being given in advance or being immediate? $\endgroup$ – Gunnar Þór Magnússon Apr 6 '11 at 20:15
  • $\begingroup$ This proof is wrong, when you say that "for any $x$ in $E$ there exists a neighborhood of the form $U \times V$, such that $\pi$ identifies with the projection $U \times V \to V$" you are messing up because there is no natural identification, since $E$ is not a priori a product. $\endgroup$ – Marcos Cossarini Apr 28 '14 at 0:09
  • $\begingroup$ The precise statement of the above identification is like this: "There are open neighborhoods $W_x$ around $x$ in $E$ and $V_x$ around $t$ in $B$, and there is an open set $U_x$ in Euclidean space, and a diffeomorphism $\phi:U_x\times V_x\to W_x$ such that $f\circ\phi=\pi$." (Here I'm reverting to the original notation of the question: $f$ is the submersion that we are studying, and $\pi$ is the projection from $U_x\times V_x$ to $V_x$.) $\endgroup$ – Marcos Cossarini Apr 28 '14 at 0:18
  • $\begingroup$ If you cover the fiber $f^{-1}t$ with finitely many of these neighborhoods $W_{x_i}$ and define $V=\cap_i V_{x_i}$, then there is no guarantee that $f^{-1}V$ will be covered by the sets $W_{x_i}$. $\endgroup$ – Marcos Cossarini Apr 28 '14 at 0:21

Unfortunately, not even a surjective submersion of smooth manifolds whose fibres are compact smooth manifolds is necessarily proper. For instance, consider $\pi: \mathbb{RP}^1 \times (\mathbb{R} \sqcup \mathbb{R}) \rightarrow \mathbb{RP}^1$, where one $\mathbb{RP}^1 \times \mathbb{R}$ is mapped by the inclusion $(a,b) \mapsto b$ and the other is mapped by $(a,b) \mapsto b^{-1}$.

Gunnar's proof is not correct, I am afraid, and the hypothesis of properness is necessary.

  • $\begingroup$ I'm afraid I'm a bit rusty on this stuff, but I don't understand your map $\pi$. How do you consider $b \in \mathbb R$ an element of $\mathbb{RP}^1$, and what makes the fibres of this map connected? Thanks for your answer! $\endgroup$ – Brendan Fong Sep 5 '14 at 18:49

I cannot answer your question but if we take $E$ to be the topologist's sine curve and its projection to $B=[0,1]$ we have a topological (i.e. continuous) family of points (compact subspaces, analogs of the compact complex submanifolds) which however is not proper as a map of topological spaces, since $E=\pi^{-1}(B)$ is not compact while $B$ is.

We may obtain similar nonproper examples of topological families with compact fibers modifying the fibers or the base.

Here the base is a manifold but not the total space/family, and I do not see immediately such counterexamples with families of compact topological manifolds.

Actually I do not understand how Ehresmann's theorem is used in your reasonings, in particular whether you obtain that $\pi$ is proper from it.

  • $\begingroup$ I think this example shows that we need $\pi$ to be a submersion, so the statement is not a topological one, but really depends on the differential structure. $\endgroup$ – Gunnar Þór Magnússon Nov 30 '10 at 19:04

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