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Suppose f:*X*→Y 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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1 Answer 1

up vote 16 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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2  
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
2  
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

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