Edit 1: I think that the question was not stated clearly enough so modified it a little
Edit 2:
I thought over the physics that lies behind this question which led me to reformulation of the original problem. Orbit itself in non physically significant. What really counts is its image in projective space!
Edit 3:
I changed the title and added new part of the question (see below). It is directly related to the original problem ( which was answered by David Bar Moshe) so I decided not to make it a separate question.
Edit 4: I erased the second part of the question and made it into a separate one – see here. Sorry for this mess…
Setting:
Let $G_0$ be a compact, simply connected Lie group giving rise to a semi-simple Lie Group $G$ (its Lie algebra I denote by $\mathfrak{g}$). Let $V_{\lambda}$ be a finite dimensional complex vector space on which $\mathfrak{g}$ is irreducibly represented (with the highest weight $\lambda$ and highest weight vector $v_\lambda$).
Context:
In the article by Lichteinstein:
http://www.ams.org/journals/proc/1982-084-04/S0002-9939-1982-0643758-8/S0002-9939-1982-0643758-8.pdf
there is a quadratic criterion that enables one to say whether a given vector $v\in V_\lambda$ belongs to the orbit of $G$ through the highest-weight vector $v_\lambda$. It says that a given $v$ is on the orbit of our interest iff:
$\Omega (v\otimes v) =\langle 2\lambda+2\delta,2\lambda\rangle (v\otimes v)$ (that's the equation (1) in this article)
$\Omega$ is the representation of the second order Casimir operator (treated as a member of universal enveloping algebra of $\mathfrak{g}$) in $V_{\lambda}\otimes V_{\lambda}$.
$\delta=\frac{1}{2}\sum_{\alpha>0}\alpha$ (summation over all possitive roots of $\mathfrak{g}$)
$\langle\cdot ,\cdot \rangle $ – a standard inner product on Cartan algebra dual $\mathfrak{h}^*$ .
Question:
Let $\mathbb{P}V_\lambda$ be a complex projective space of $V_\lambda$ and let $\pi:V_\lambda\rightarrow \mathbb{P}V_\lambda$ be a canonical map to projective space. Define:
$O_{v_{\lambda}} $ – orbit of $G$ through $v_\lambda$
$O_{v_\lambda}^0 $ – orbit of $G_0$ through $v_\lambda$
Is it true that $\pi(O_{v_\lambda}) = \pi(O_{v_\lambda}^0)$ ?
In physics articles I came across criterion stated in the "context" part is interpreted as a necessary and sufficient condition for a given $v$ to be precisely on the orbit of $G_0$ through $v_\lambda$. In the quantum mechanical context what really matters are images of vectors from $V_\lambda$ in associated projective space (phase and normalization do not play a role ) – that's why physically interesting object is associated with the image of actual orbit in projective space.
Best Answer
Let $H$ be the isotropy group of the highest weight ray $\pi(v_\lambda)$ in $G_0$ and $P$ the highest weight ray isotropy group in $G$. It is easy to see that $H$ is the centralizer of the torus generated by the coroots corresponding to the nonvanishing components of $\lambda$ in the weight basis, which is a consequence of the fact that $v_\lambda$ is annihilated by the all positive roots generators and by the negative root generators corresponding to the vanishing weight components of $\lambda$. By the same reasoning $P$ is a parabolic subgroup of $G$ whose Lie algebra is the union of the complexification of the $H$ Lie algebra and the Borel subalgebra of the positive root generators.
Thus, what we are really looking for is the standard isomorphism between $G_0/H$ and $G/P$ as real homogeneous spaces. A sketch of the proof is given nicely for the nondegenerate case by Hogreve,Muller,Potthoff,Schrader in Commun. Math. Phys. 131, 465-494 (1990) , (section 2.3). I'll try to repeat it here for completeness:
On one hand the real span of the Lie algebras of $G_0$ and $P$ is the whole of $G$, thus the $G_0$ orbit is open in the $G$ orbit, on the other hand it is closed by being compact, thus the orbits are the same.