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Let $G$ be a real semisimple Lie group (say $SL(2,\mathbb{R})$) and $H$ be its Cartan subgroup (say torus or diagonal subgroup of $SL(2,\mathbb{R})$).

My questions is: it is always true that we have a natural symplectic structure on the quotient space $G/H$?

If it is not true, could we consider this weaker version: Let $\mathfrak{g}$ and $\mathfrak{h}$ be complexified Lie algebras of $G$ and $H$ respectively. Is there a natural symplectic structure on the quotient space $\mathfrak{g}/\mathfrak{h}$?

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A plausible answer is that $G/H$ is the cotangent bundle of the full flag variety $G/B$ where $B$ is a Borel subgroup. A story like this works out over the complex numbers, not sure about over the reals. $\mathfrak{g}/\mathfrak{h}$ is an even dimensional vector space and I believe the killing form gives it a symplectic structure. –  solbap Aug 31 '12 at 5:50
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This morning Killing form was symmetric, not skew-symmetric... –  Bugs Bunny Aug 31 '12 at 10:07
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@solbap: One needs to be a bit careful here - the cotangent bundle of $G/B$ is not a homogeneous $G$-space and hence will never be of the form $G/H$. –  Chuck Hague Aug 31 '12 at 13:19
    
@Chuck you make a good point, what I had in mind in the complex case was not that algebraic cotangent bundle. What I've heard is that if $G$ is a compact connected lie group with maximal torus $T \cong (S^1)^n$ then one can look at the cotangent bundle of $G/T$ in the category of smooth manifolds: $T^* G/T$; I've heard that $G_\mathbb{C}/T_\mathbb{C}$ is diffeomorphic to $T^* G/T$. –  solbap Aug 31 '12 at 16:11

2 Answers 2

up vote 11 down vote accepted

To add to Robert Bryant's answer, the coadjoint orbits of a Lie group always have a natural symplectic structure, the (Lie-)Kirillov-Kostant(-Souriau) form. These are the symplectic leaves of the natrual Lie-Poisson structure on $\mathfrak{g}^\ast$. For $x \in \mathfrak{g}^\ast$, we can identify the coadjoint orbit $\mathcal{O}_x$ with $G/H_x$, where $H_x$ is the stabilizer of $x$ under the coadjoint action. For $x$ such that $H_x$ is Cartan (which should happen generically), this endows $G/H$ with a symplectic structure. However, this depends on the choice of $x$. For example, even in the absolute simplest case of $G = SU(2)$, the coadjoint orbits are spheres of different radii, so the orbits $\mathcal{O}_x$ and $\mathcal{O}_y$ are not symplectomorphic unless $y$ is conjugate to $x$.

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Yes, Jonathan is absolutely correct. The essential point is that, when the rank of $G$ is $r$ there will be an $r$-parameter family of $\xi\in\frak{g}^\ast$ that have the same stabilizer subgroup (which is a Cartan subgroup) under the coadjoint action. Thus, you will have an $r$-parameter family of 'natural' symplectic structures on $G/T$. –  Robert Bryant Aug 31 '12 at 14:50
    
A word of caution: a non-compact semisimple Lie group always has more than one conjugacy class of Cartan subgroups (e.g. maximally split and maximally compact), so "generically" here does not have the usual meaning. –  Victor Protsak Aug 31 '12 at 19:39
    
It may be worth pointing out that for semisimple Lie groups the adjoint and the coadjoint orbits are the "same": the Killing form gives an equivariant identification of the Lie algebra with its dual. –  Eugene Lerman Aug 31 '12 at 21:51
    
Yes it is very natural from the viewpoint of coadjoint orbits. Maybe I should think more carefully before asking this question. Nevertheless, thank you all! –  Zhaoting Wei Sep 1 '12 at 6:41

Actually, the issue is that there is usually more than one 'natural' symplectic structure on $G/T$, where $T$ is a Cartan subgroup. The space of $G$-invariant closed $2$-forms on $G/T$ has dimension equal to the rank of $G$ (which is the dimension of $T$), and the generic one is nonsingular, so you might have trouble picking out a 'natural' one from among this class when the rank of $G$ is greater than $1$.

For example, when $G = \mathrm{SO}(4)$, which has rank $2$, the quotient $G/T$ is $S^2\times S^2$ and you can make the symplectic form take arbitrary nonzero values on each $S^2$-factor.

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