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Well, the identity component $H^0$ of $H$ would have to be abelian, since, otherwise, it would have a compact simple component, and the induced map on $\pi_3$ would then be nontrivial. If $H^0$ has p …

Edit: (21 May 2017) I have modified my answer to cover the case that the OP meant to ask, i.e., the assumption is that the closure of an orbit has nonempty interior. Now that you have added the assu …

Because $\mathrm{SU}(N{+}1)$ acts transitively on the projectivized tangent space of $\mathbb{CP}^N$ (a fact that has nothing to do with metrics on $\mathbb{CP}^N$), any $\mathrm{SU}(N{+}1)$-invariant …

Well, here is what I can say. Perhaps this will answer some of your questions about $S^{2n+1}$ at least. Suppose that $G/H = S^{2n+1}$ where $n>0$ and that the action of $G$ on $S^{2n+1}$ is effecti …

I'm sure this can be found in the literature, though I don't know exactly where to look. On the other hand, it is easy to calculate the automorphism group directly from the following observations: A …

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 …

I'm afraid that you are mistranslating Cartan's notation. The first 3-dimensional example that Cartan gives is this: Every diffeomorphism of the line, written in the form $X = f(x)$, can be 'lifted' …

The quotient is a cone on $\mathbb{CP}^m$. When $m$ is even, $\mathbb{CP}^m$ is not the boundary of any compact smooth $(2m{+}1)$-manifold, so you can't smooth the singularity at the tip of the cone b …

The various different ways that $\mathrm{U}(1)\simeq S^1$ can appear as a subgroup of $\mathrm{SU}(3)$ are indexed by a lattice of rank $2$, and the $7$-dimensional quotients are now known as Aloff-Wa …

I don't have the book handy, but I seem to remember that this formula is written out explicitly in S.-s Chern's Complex manifolds without potential theory (Second Edition), in the chapter "The Grassma …

Here is an algorithm to compute the Lie algebra of the group of isometries of a homogeneous space $G/H$ endowed with a $G$-invariant (pseudo-)Riemannian metric $g$. It is phrased in terms of essentia …

Remark: I assume that you want $A$ to be a non-trivial bundle. Otherwise, of course, any parallelizable compact manifold would be an example. In particular, any compact Lie group would be an exampl …

This is true. One can use a few facts from Helgason's Differential Geometry, Lie Groups, and Symmetric Spaces to show that, indeed, $K = \mathrm{Stab}_G(Z)$. Since $M=G/K$ is an irreducible Hermitian …

If I understand your question correctly, the answer is that the zero Poisson structure on $S^4$ is the only homogeneous one. In fact, this is the only Poisson structure on $S^4$ of constant rank. To …

Perhaps the OP really wants to know why quaternionic Grassmannians other than the quaternionic projective spaces are not considered to be 'quaternion-Kähler'. The reason goes back to Berger's classifi …