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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 something of an exercise is unwinding the definitions, but there's an interesting twist to it as well, so here's an outline of an answer: Let $G$ be a connected $3$-dimensional Lie group (not …

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 …

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 …

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 …

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 …

Note that the answer depends on the pair $(G,K)$. For example, if $K=\{e\}$, then one is asking whether the ring of left-invariant forms on $G$ consists only of closed forms. This only happens when $ …

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 …

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 …

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 …

Actually, though this may seem pedantic, there are two almost-complex structures on $\mathbb{CP}^m$ that are invariant under $\mathrm{SU}(m{+}1)$, namely the 'standard' one and its conjugate. Of cour …

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 …

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 …

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 …

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 …