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Here is Asaf's agrument expanded a bit. It has the advantage of working for all Lie groups simultaneously. Given a Lie group $G$ with Lie algebra $\mathfrak{g}$, consider the exponential map $\exp:\ …

The short answer is that many (most? all?) homogeneous spaces do NOT have such a nice description. In particular, at any point $p\in M$, the set of the $G$ or $H$ invariant vectors is a strict subset …

Berger showed the homogeneous space $SU(5)/Sp(2)\cdot S^1$ has a metric of positive sectional curvature. Here, $Sp(2)\cdot S^1 = Sp(2)\times_{\mathbb{Z}/2\mathbb{Z}} S^1$. More generally, Bazaikin f …

To add a bit, There are also many examples of compact manifolds with multiple group structures. As a quick example, first recall that $SU(2)$ is the collection of all $A \in M_2(\mathbb{C})$ with $A …

I've been studying isometric quotients (by compact Lie groups) of compact simply connected homogeneous spaces $G/H$ and their inherited curvature. One of the issues that continually arises is the fol …

For ease of writing, I'll say a compact connected Lie group $G$ has the Krull–Schmidt property if, up to order, it has a unique decomposition as a direct product where each factor itself cannot be dec …

I don't know of anything as bare hands as the proof that $\pi_1(G)$ must be abelian, but here's a sketch proof I know (which can be found in Milnor's Morse Theory book. Plus, as an added bonus, one l …

Yes, there is always such an $M$. To see this, first note that saying two representations of $G$ on a vector space $V$ are equivalent is the same as saying the two images of $G$ in $Gl(V)$ are conjug …

I just wanted to add that there is a fairly easy proof for your final question: Is every continuous homomorphism between Lie groups actually smooth? The theorem we need is the closed subgroup theore …

I just wanted to add two points: A bi-invariant metric on a compact Lie group $G$ does not always induced the maximum symmetric metric on $G/H$. The most familiar examples are spheres: $S^{2n+1} = …

As Igor shows, every endomorphism of a simple Lie group $G$ has degree $\in\{0,\pm 1\}$. On the other hand, every compact Lie group admits self maps of other degrees. Namely, the $k$-th power map $g\ …

Consider the usual $G = S^1$ action on $S^2$ given by rotations. This action respects the antipodal map, so descends to a $G$ action on $M = \mathbb{R}P^2$. Let $p\in M$ be any point on the "equator …

(In some sense, this is just a restatement of what Eric said above....) For compact groups, quite a lot can be said. Every compact group H' has a finite cover H which is Lie group isomorphic to $T^{ …

First, here's a rough outline of how the classification works: Prove that if G and H are simply connected and have the same Lie algebra, then G and H are isomorphic as Lie groups. Prove that if G is …

The following result is, for example, exercise 3 on pg. 104 of Do Carmo's Riemannian Geometry book. Suppose $X$ is a Killing field on a compact even dimensional Riemannian manifold of positive cur …