Dimension of a $G$-variety $X$ that is a finite union of $G$-orbits

Question

Suppose that $G$ is an algebraic group acting on a variety $X$, and $X$ is a finite disjoint union of $G$-orbits $\mathcal{O}_i$, $i=1,\ldots,n$, under this action. Is it true that the dimension of $X$ is the maximum of the dimensions of the $\mathcal{O}_i$? It seems like this should be true, and is probably a result of something more general. Thanks in advance for any help with this.

Answer 1

Disclaimer: I only know the following for a fact if we are over an algebraically closed field of characteristic zero.

Let $X=X_1\cup\cdots\cup X_r$ is the decomposition of $X$ into irreducible components and assume without loss of generality that $X_1$ is of maximal dimension.

It is a well-known and basic fact that any $G$-orbit is open in its closure of $G.x$ is the union of $G.x$ and orbits of strictly smaller dimension. See these very nice notes by Michel Brion. Now by assumption, there are finitely many $x_1,\ldots,x_n\in X$ such that $X=\bigsqcup_{i=1}^n \overline{G.x_i}$ (disjoint union). Thus, $X_1 = \bigsqcup_{i=1}^n (\overline{G.x_i}\cap X_1)$. Since $X_1$ is irreducible, we have (again wlog) $X_1 = \overline{G.x_1}\cap X_1$. Thus, $$\dim(X)=\dim(X_1)=\dim(\overline{G.x_1}\cap X_1)\le \dim(\overline{G.x_1})=\dim(G.x_1)\le \dim(X).$$

Answer 2

Yes since the relation between the orbits and the variety is finite to one maximum dimension of the orbits is the same with the dimension of X.