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One of the first theorems encountered in algebraic geometry is the upper semicontinuity of fiber dimension:

Let $ f : X \to Y $ be a surjective regular map between irreducible varieties with irreducible fibers. Then $ {\rm dim} \; f^{-1}(y) \geq {\rm dim} \; X - {\rm dim} \; Y $ and the equation ${\rm dim} \; f^{-1}(f(x)) \leq b $ defines a zariski open subset of $X$.

This theorem is not true in the category of smooth manifolds. Indeed, consider the height function $ h : S^2 \to \mathbb{R}$. Moreover, the proof uses commutative algebra.

Question 1: Is there a more geometric proof? Maybe one that works for holomorphic manifolds?

Here is my idea for how a proof might go: If $f : M \to N $ is a surjective map of smooth manifolds, then ${\rm Null} \; f'(m) \leq b $ defines an open subset of $M$. Somehow relate this nullspace to the fiber dimension in the complex manifold case. I have no idea if this will work...

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  • $\begingroup$ Data point: at this point I've taken three courses called "algebraic geometry" and this theorem was never mentioned in any of them. $\endgroup$ – Qiaochu Yuan Aug 21 '15 at 6:47
  • $\begingroup$ @QiaochuYuan: Semicontinuity of fiber dimension is one of the most important basic facts in the dimension theory of algebraic varieties (after setting up the theory of codimension). Your instructors did you a disservice if they taught you about algebraic varieties without ever discussing and illustrating this theorem. $\endgroup$ – grghxy Aug 21 '15 at 14:04
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Let $X$ and $Y$ be complex manifolds and $f:X \to Y$ a holomorphic map.

If $f$ is surjective then, by Sard's theorem, the generic fiber of $f$ has dimension $\dim X - \dim Y$. So, once we prove upper semicontinuity, we will know that all fibers have dimension at least $\dim X - \dim Y$.

To show semicontinuity, it is enough to show $\{ x : \dim f^{-1}(f(x))=0 \}$ is open. Once we have done this, suppose that $x$ lies on a $b$-dimensional component of $f^{-1}(f(x))$. Then we can choose a neighborhood $U$ of $x$ and a map $g: U \to \mathbb{C}^b$ such that the fiber of $f \times g : U \to Y \times \mathbb{C}^b$ though $x$ is just the singleton $\{ x \}$. There will then be some open neighborhood $V$ of $x$ where the fibers of $f \times g$ are finite, and hence the fibers of $f$ have dimension $\leq b$.

So, suppose that $f^{-1}(f(x))$ is finite. We must build an open neighborhood of $x$ where all fibers are finite. Shrinking $X$ around $x$, we may assume that $f^{-1}(f(x)) = \{ x \}$ and that $X$ is open in $\mathbb{C}^{d}$ for $d = \dim X$. Let $B$ be a closed ball in $X$ around $x$. Let $B^{\circ}$ be the interior and $\partial B$ the boundary. Write $y = f(x)$.

Now, $\partial B$ is compact, so $f(\partial B)$ is closed in $Y$. Since $f^{-1}(y) = \{ x \}$, we see that $y \not \in f(\partial B)$, so we can take an open set $V$ around $y$ disjoint from $f(\partial B)$. Take $U = f^{-1}(V) \cap B^{\circ}$. I claim that, for any $x' \in U$, the fiber $f^{-1}(f(x')) \cap U$ is finite.

Let $y' = f(x')$. Note that $y' \in V$ and hence $f^{-1}(y') \cap U = f^{-1}(y') \cap B^{\circ} = f^{-1}(y') \cap B$. The last is a closed subset of the compact set $B$, hence is compact. So far, we have not used complex geometry in any way.

Let $z$ be any holomorphic function on $U$. Then the restriction of $z$ to $f^{-1}(y') \cap U$ is a holomorphic function on a compact complex manifold, so it is locally constant by the maximum modulus principle. (I am glossing over the fact that $f^{-1}(y') \cap U$ could have singularities.) Applying this fact with $z_1$, $z_2$, ..., $z_d$ the coordinate functions on $U \subset \mathbb{C}^d$, we see that $f^{-1}(y') \cap U$ is finite.

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    $\begingroup$ The analytic Zariski topology is not transitive, so this method does not prove the desired Zariski-openness (just openness for the classical topology). Also, the method of reduction to fiber-dimension 0 uses non-trivial input from commutative algebra and complex analysis, namely the existence of a system of parameters in local noetherian rings, applied to local rings on complex-analytic spaces (though I can't imagine how to deal with $x$ not smooth in its fiber without some algebra!). Hence, it uses non-trivial results in the algebraic theory of coherent analytic sheaves. $\endgroup$ – grghxy Aug 21 '15 at 1:00
  • $\begingroup$ Agreed that this only gives classical openness. I am stumped as to how you are going to get Zariski openness without algebraic methods. I don't object so much to use of the use of system of parameters -- I don't think I am cheating any more than the OP is cheating by referring to the dimension of a singular fiber in the first place. $\endgroup$ – David E Speyer Aug 21 '15 at 1:18
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    $\begingroup$ @DavidSpeyer: I completely agree, the use of such kind of algebraic methods is a basic tool in any serious complex-analytic geometry that confronts singularities (and was indeed invented exactly for such purposes, to use apply algebro-geometric ideas in an analytic setting). That being said, even admitting the full force of coherent analytic spaces, the proof of Zariski-semicontinuity of fiber dimension remains a serious theorem. This should surely be proved in the book by Banica and Stanasila with the apt title "Algebraic methods in the global theory of complex spaces"! $\endgroup$ – grghxy Aug 21 '15 at 1:34
  • $\begingroup$ @DanielBarter You're welcome! I will say, I originally was planning to write this just as a suggestive argument, and found it firmer than I expected when I wrote it out. Actually defining dimension for a singular space, building the system of parameters to reduce to dimension $0$ and proving the "maximum modulus principle" for singular fibers would all take additional work, which would make this answer a lot longer. $\endgroup$ – David E Speyer Aug 21 '15 at 2:02
  • $\begingroup$ @DavidSpeyer: Logically prior to defining dimension for general (not necessarily smooth ) complex-analytic spaces comes the local structure of morphisms around points isolated in their fibers, which is shown to be open for the classical topology early in the theory (without any role for maximum modulus principle or consideration of compact analytic spaces): see the 3.1/2 in the book "Coherent Analytic Sheaves" by Grauert & Remmert (which rests on coherent sheaf theory from Chapter 2, where the noetherian property of local rings is also proved, whereas dimension theory is in Chapter 5). $\endgroup$ – grghxy Aug 21 '15 at 2:28

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