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I am studying algebraic groups and we are dealing with homogeneous $G$ space at the moment. Let $X$ be a homogeneous $G$ space with stabilizer $H$. I was wondering if someone could explain me why the dimension of $X$ is equal to $\dim G - \dim H$?

Thank you very much!

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    $H$ is the stabilizer of...a distinguished point $x_0 \in X$? Or something else? – 2017-02-28
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    I don't know much about homogeneous spaces, but assuming $H = \mathrm{Stab}_G(x_0)$, since $G$ acts transitively on $X$, we have an isomorphism of $G$-sets: $X \cong G/H$. The maps are: given a coset $gH$, map it to $gx_0$, and given a point $x \in X$, take $g \in G$ such that $gx_0 = x$ and map $x$ to $gH$. (You can show that these maps are well-defined and mutually inverse.) – 2017-02-28

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Suppose that $X$ is a $G$-space. If $O^\circ_x$ is the orbit of $x\in X$ under the action of $G^\circ$, then it turns out that $\dim\overline{O}_x=\dim\overline{O^\circ_x}$, $\dim\operatorname{Stab}_G(x)=\dim\operatorname{Stab}_{G^\circ}(x)$, and finally $$ \dim(G)=\dim\overline{O}_x+\dim\operatorname{Stab}_G(x). $$ So in particular, if $X$ is homogeneous $G$-space, then $X=O_x$, and the result follows.

Fix $x\in X$, and let $H=\operatorname{Stab}_G(x)$. You can decompose $G=\bigcup g_iG^\circ$ as finite disjoint union of the cosets of the identity component. Then $O_x=\bigcup g_iO^\circ_x$, so that $\overline{O}_x=\bigcup\overline{g_iO_x^\circ}$. But $\overline{g_iO^\circ_x}$ is a closed, irreducible subset for each $i$, since translation is an isomorphism of affine varieties, and thus $\dim\overline{g_iO^\circ_x}=\dim\overline{O^\circ_x}$ for all $i$. Hence we have $\dim\overline{O^\circ_x}=\dim\overline{O_x}$.

Since $G^\circ\unlhd G$, $G^\circ H$ is a subgroup of $G$, hence $$ H/\operatorname{Stab}_{G^\circ}(x)=H/(G^\circ\cap H)\simeq G^\circ H/G^\circ, $$ the isomorphism being one of abstract groups. But it's known $[G:G^\circ]$ is finite, so that $[H:\operatorname{Stab}_{G^\circ}(x)]$ is finite, hence $\dim H=\dim\operatorname{Stab}_{G^\circ}(x)$.

This shows it's essentially fine to just look at how $G^\circ$ acts, so we can assume $G$ is connected.

The orbit map $\varphi_x\colon G\to\overline{O}_x$ is a dominant map of irreducible affine varieties, so standard facts in algebraic geometry give a nonempty open $U\subseteq O_x$ such that $$ \dim\varphi_x^{-1}(y)=\dim G-\dim\overline{O}_x,\quad\forall y\in U. $$ It's well known that $G$-orbits are open in their closure, and so $O_x$ meets $U$ nontrivially. So we can pick $y\in O_x\cap U$, and so $y=gx$ for some $g\in G$. Thus $$ \varphi_x^{-1}(y)=\{h\in G:gx=y=hx\}=\{h\in G:g^{-1}hx=x\}=\{h\in G:g^{-1}h\in H\}=gH. $$

Since $gH\simeq H$ as varieties, $\dim\varphi_x^{-1}(y)=\dim H$. So we get $$ \dim H=\dim G-\dim\overline{O}_x. $$ The equation for when $X$ is a homogeneous $G$-variety now follows.