# Quotient of a compact Lie group by maximal Torus

I start with a noncompact connected semisimple Lie group with finite center $G$ and fix a maximal compact subgroup $K$ of $G$. I am considering these compact groups $K$. If $\mathbb T$ is the maximal torus in $K$, I take the quotient $K/\mathbb T$. I read that they are Kahler manifold. I am interested to know if there is a natural way to find a dense set in $K/\mathbb T$ which is parametrized by $(j_1/2^{n_1}, \dots, j_m/2^{n_m})$, where $j_i\in \mathbb Z$ and $n_i\in \mathbb N$.

Something like a dense grid, as can be constructed on $\mathbb R^d$ and some other groups/spaces like in a Heisenberg group, is what I have in mind.

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The group $G$ seems to play no role in the question. Weird. –  Alain Valette Dec 22 '12 at 21:39

The answer is yes. As Aakumadala mentioned in a comment, the quotient $K/\mathbb{T}$ is isomorphic to the flag variety of $K_{\mathbb{C}}$. If we let $N = N_{\mathbb{C}}$ denote the nilpotent radical of a Borel subgroup of $K_{\mathbb{C}}$, the flag variety has a dense $N$-orbit with a simply transitive $N$ action. Note that when $K = SU(3)$, we find that $N$ is a Heisenberg group, as you anticipated.

The underlying manifold of the dense $N$-orbit is just an affine space, isomorphic to $\mathbb{C}^n = \mathbb{R}^{2n}$ for some $n$. If you make this identification, then the dense embedding $\mathbb{Z}[1/2] \subset \mathbb{R}$ gives you a dense set parametrized by $\mathbb{Z}[1/2]^{2n}$.

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Ah nice. Use the large cell in the Bruhat decomposition which is dense. –  Michael Murray Dec 23 '12 at 13:13
@Michael Murray Could you elaborate a bit? –  spr Dec 24 '12 at 4:46

Maybe try first with $G=SL(2, \mathbb{C})$, $K = SU(2)$ and $T = U(1)$, the diagonal matrices with determinant $1$. Then $K/T = S^2$.

Edit: I was thinking of a discrete lattice so this answer which I thought was a counter example isn't.

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In $S^1$ my grid is roots of unity $e^{i\alpha\pi 2^{-n}}$ where $\alpha\in \mathbb Z$ and $n\in \mathbb N$. I guess similar construction will work for $S^2$. Precisely any point on $S^2$ has latitude and longitude, I will approximate them separately by roots of unity. Will not that work? –  spr Dec 22 '12 at 12:35
Perhaps the word grid is misleading. My grid in $\mathbb R^d$ is dense. The grid in $\mathbb R^d$: $(j_1/2^{n_1},\dots, j_d/2^{n_d})$ where $j_i\in mathbb Z, n_i \in \mathbb N$. –  spr Dec 22 '12 at 12:41
Sorry I missed the word dense and was thinking discrete lattice! Your latitude and longitude idea should work. There might be a general approach using a dense grid in the Lie algebra mapped to the group with the exponential map. –  Michael Murray Dec 22 '12 at 14:36
As Alain Valette says, $G$ plays no role in this. If $K=SU(2)$ take the subgroup of $K$ with entries in ${\mathbb Q}(i)$ the field of fractions of Gaussian integers. This gives a dense subset of $K/T$ which is fairly natural. You can get a similar dense set in any compact connected Lie group $K$. Incidentally, $K/T$ is an algebraic variety, the flag variety of the reductive group $K_{\mathbb C}$ (the complexification of $K$). –  Venkataramana Dec 23 '12 at 2:14
I wanted to say that I was interested in those $K$ which are coming as maximal compact subgroups of such $G$. I thought that would restrict the class of compact Lie groups $K$ which might be useful. It appears from your answers/comments that we can deal with any compact Lie group. –  spr Dec 24 '12 at 4:44