Let $G$ be an almost-simple classical real algebraic group, and $H$ a maximal proper closed subgroup. Assume that the $n$ in $G := A_n,B_n,C_n,D_n$ is sufficiently large.
Is it true that dim$(G)$ - dim$(H)$ $\geq$ rank($H$)?
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$\begingroup$ I was first confused by your notation. $\ge O(f)$ means $\ge Cf$ for some $C$? this is very non-standard. $\endgroup$– YCorCommented Mar 24, 2017 at 5:37
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$\begingroup$ No I erased them because I first misinterpreted your question (because it's not big O notation, it's some invention of yours). big O notation is like $g=O(f)$ and is an asympotic upper bound on $f$. The standard notation for what you mean is "$=\Omega(\mathrm{rank}(H))$". $\endgroup$– YCorCommented Mar 24, 2017 at 5:43
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2 Answers
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If is very easy to see that if $G$ is simple and $H$ is a proper subgroup then $$\dim G-\dim H\ge \mathrm{rank}\,G.$$ Proof: Since $G$ is simple and $H$ is proper, the action of $G$ on $G/H$ is (locally) effective. Hence, the action of the maximal torus $T$ of $G$ is also effective. For a torus this means that the generic orbits have dimension $\dim T$. Thus, $\dim G-\dim H=\dim G/H\ge\dim T=\mathrm{rank}\,G$.
answered Mar 24, 2017 at 10:45
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1$\begingroup$ To show the claim that generic $T$-orbits (in the variety $X$) have dimension $\dim(T)$: we can assume to work with the complex numbers and that $\dim(T)\ge 1$. For every 1-dimensional subtorus $K$ let $X_K$ the points fixed by $K$. This is a closed subset, distinct of $X$. Since there are only countably many $K$, $\bigcup_KX_K$ is distinct of $X$; hence there is a point with finite stabilizer. [In contrast the unipotent abelian 2-dimensional group can act faithfully with $\le 1$-dimensional orbits: $(u,v).(x,y)=(x+u+yv,y)$.] $\endgroup$– YCorCommented Mar 24, 2017 at 16:51
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Yes. The table p45 of this article by Cantat-Xie (see the references therein) shows that it is indeed true that for any proper subgroup $H$ of $G$, we have $\mathrm{codim}(H)=\Omega(\mathrm{rank}(G))$ (more precisely for all $G$ and $H$ we have $\mathrm{codim}(H)\ge\mathrm{rank}(G)$ with possible equality only for $G$ of type $A_n$). Hence it's also $\ge\mathrm{rank}(H)$. I guess it's originally due to Dynkin.
