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Suppose f:XY is a morphism of schemes. We can define a function on the underlying topological space Y by sending yY to the dimension of the fiber of f over y. When is this function upper semi-continuous?

I have a "concrete" application in mind. If an algebraic group G acts on a scheme X, I'm pretty sure the dimensions of the stabilizers is upper semi-continuous (i.e. it can jump up on closed subschemes), but I don't know a proof. The stabilizers of points are the fibers of the map StabX in the following cartesean square:

Stab ---> GxX
  |         |
  |  cart   |
  v         v
  X ----> XxX

The map GxX→XxX is given by (g,x)→(gx,x), and the map X→XxX is the diagonal x→(x,x). So it would be nice to have a condition satisfied by GxX→XxX which guarantees upper semi-continuity of the fiber dimension.

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Dear @Transcendental, if your edit suggestion is rejected, please do not resubmit it. Thank you. –  Ricardo Andrade 42 mins ago

2 Answers 2

up vote 19 down vote accepted

Theorem (EGA IV 13.1.3): Let $f \colon X \to Y$ be a morphism of schemes, locally of finite type. Then $$x \mapsto \dim_x(X_{f(x)})$$ is upper semi-continuous.

Corollary (Chevalley's upper semi-continuous theorem, EGA IV 13.1.5): Let $f \colon X \to Y$ be proper, then: $$y \mapsto \dim(X_y)$$ is upper semi-continuous.

Corollary (SGA3, ??): Let $X/Y$ be a group scheme, locally of finite type. Then $$y \mapsto \dim(X_y)$$ is upper semi-continuous.

Proof: The dimension of a group scheme over a field is the same as the dimension at the identity. Thus the function $$y \mapsto \dim(X_y)$$ is the composition of the continuous function $y \to e(y)$ and the upper semi-continuous function $x \mapsto \dim_x(X_{f(x)})$.

Concerning your application: The fiber dimensions of the stabilizer group scheme Stab/X is upper semi-continuous, but the "diagonal" $G \times X \to X \times X$ does not always have this property (unless it is proper, i.e., "$G$ acts properly").

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Oh, I see; you get semi-continuity on X much more generally than semi-continuity on Y. I got a little confused in the last Corollary, so I'll add a comment to clarify to my future self. X/S is a group scheme with an S-morphism f:X-->Y. The map y-->e(y) is the composition of the structure morphism Y-->S and the identity S-->X. To get the application from this Corollary, take X=Stab and Y=X. –  Anton Geraschenko Oct 8 '09 at 17:53
Well, there was a typo: It should be X/Y and not X/S in the last Corollary (now corrected). As stated before, it was wrong. The reason is that the dimension of the fiber of X->Y at the identity cannot be related to the dimension at other points unless X->Y is equivariant (which is the case if Y=S) so that the fiber acts transitively on itself. –  David Rydh Oct 9 '09 at 4:57
Stupid example: Let $Y = {\mathbb A}^1$, and $X$ the disjoint union of $(Y\setminus 0) \times {\mathbb P}^1$ and $0$. Then the evident map $X\to Y$ violates Chevalley's theorem, even though it has projective fibers (but is not actually projective). –  Allen Knutson Mar 14 '13 at 0:35

I just discovered that Shavarevich (second edition) has a wrong answer to this question. In Section I.6.3, after Theorem 7 (which is correct), he gives the following Corollary. This quotation combines the Theorem and the Corollary.

Let $f: X \to Y$ be a regular map between irreducible varieties. Suppose that $f$ is surjective ... The sets $Y_k := \{ y \in Y: \dim f^{-1}(y) \geq k \}$ are closed in $Y$.

Note that this differs from the true EGA IV 13.1.5 by replacing "proper" with "surjective". I figured I'd record a counterexample here, which is slightly more public than just creating a handout for my class.

Our map is a composition $X \subset \mathbb{A}^3 \to \mathbb{A}^3 \to Y \subset \mathbb{A}^4$. We'll call the two $\mathbb{A}^3$'s $A$ and $B$ respectively.

$X$ is the quasi-affine variety $A \setminus \{ (0,\ast,0) \}$. We map $A \to B$ by $(x,y,z) \mapsto (x, xy, z)$. We map the $B$ to $\mathbb{A}^4$ by $(p,q,r) \mapsto (p(p-1), p^2(p-1), q,r)$. $Y$ is the affine variety $\{ (a,b,c,d) : a^3 = b(b-a) \}$. In other words, $Y$ is the product of a nodal cubic with $\mathbb{A}^2$.

To see surjectivity, note that $X$ hits every point of $B$ where $p$ is nonzero. The points of $B$ where $p \neq 0$ map to the points of $Y$ where $(a,b) \neq (0,0)$. The points $(0,0,c,d)$ in $Y$ are the images of $(1,c,d) \in B$, which are in turn the images of $(1,c,d)$ in $X$.

Now, let's look at $\dim f^{-1}(0,0,0,r)$. When $r \neq 0$, this is the union of $(1,0,r)$ and $(0, \ast, r)$, so one dimensional. When $r = 0$, the line $(0, \ast, 0)$ is deleted, so the preimage is only a point.

This suggests that something nicer may happen if we insist that fibers are irreducible, or that $Y$ is normal (perhaps Zariski's Main Theorem gets involved?) but I don't have a proposed statement to make.

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