Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

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

2 Answers 2

up vote 1 down vote accepted

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 $\mathbb{Z}[1/2]^{2n}$.

share|improve this answer
    
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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.