Do all orbits have the same dimension? Well, I've already asked this question at math.SE — but no-one's answered or commented. So now I'm posting it here (it's about the research paper — I think that it isn't an off-topic for this forum) — if it's breaking the rules, tell me about that, please.
Let $G$ be an algebraic group and let $X$ be a $G$-variety. It's stated in the paper Dufresne and Kraft - Invariants and Separating Morphisms for Algebraic Group Actions (p.13) that all orbits are closed and have the same dimension if the graph
$$\Gamma_{X}\mathrel{:=}\{(g x, x) \mid g \in G, x \in X\}=\bigcup_{x \in X} G x \times G x \subset X \times X$$
is closed.
They say that the statement about the dimensions is true because "$G x \times\{x\}=p_{2}^{-1}(x)$ where $p_{2}: \Gamma_{X} \rightarrow X$ is the second projection."
But I don't understand this reasoning (the "dimension of the fiber" argument is legal only for "general fibers", isn't it?). Can someone explain it, please?
 A: I think you can use upper semicontinuity of fiber dimension, i.e. that if $f:X\to Y$ is a morphism of varieties then $\dim_x(f^{-1}(f(x)))$ is an upper semicontinuous function on $X$.
The assumption that $\Gamma_X$ is closed implies that it is a variety and that orbits are closed.
There are then two morphisms of varieties:
$$\pi_2:\Gamma_X \to X $$
and
$$G\times X \to X\times X$$
$$(g,y)\mapsto (g.y,y).$$
Take $x \in X$. The fiber of $x$ under the first morphism is its orbit, while the fiber of $(x,x)$ under the second morphism is isomorphic to the stabilizer. Hence the sum of the dimensions of these two fibers is $\dim(G)$. Both of these dimensions can only jump up at a limit point of a curve. Hence if $G$ and $X$ are connected they are constant.

It may be useful to be a bit more explicit.

*

*It may seem as though we applied the semicontinuity theorem to limits in $X$ and not in the domain of each of the two morphisms under consideration. But both the morphism $\Gamma_X\to X$ and the composition $G\times X \to X\times X \overset{p_2}{\to} X$ here have right inverses $\phi:X\to \Gamma_X$ and $\psi:X \to G\times X$, given by $\phi(x)=(x,x)$ and $\psi(x)=(e,x)$. We are really applying semicontinuity to limits in the image of $X$ in $\Gamma_X$ and in $G\times X$ under these right inverses.

*This argument works in case $X$ is irreducible. If it is not, since we assume $X$ is connected, we can apply it to each irreducible component separately. Whenever two irreducible components intersect it is then clear that their orbit dimensions are equal; connectivity yields the result.

(I get confused between left and right inverses, but I think I have it right this time.)
