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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 …

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 …

The same technique Allen mentioned also shows that $\mathbb{H}P^{2n}$ doesn't admit any orientation reversing diffeomorphisms. However, it's also true that $\mathbb{H}P^{2n+1}$ doesn't admit any orie …

Even if you ask that $f$ induces trivial maps on all (singular) homology and cohomology groups, there are still easy manifold examples. (This actually arises as an exercise in Hatcher's AT). For ins …

Yes, $M$ must be compact. In fact, if $M$ is non-compact, it admits a non-surjective self embedding $f:M\rightarrow M$. When $n=1$, the only non-compact manifold is $\mathbb{R}$, which obviously admi …

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 …

First, one can clearly assume $M$ is connected by simply applying the argument to each componenet of $M$. The key fact is a generalization of your argument for $M=\mathbb{R}^n$: that if $f:M\rightarr …

If $M$ and $N$ are both compact, then the submersion $F$ can be thought of as a fiber bundle map with fiber $F^{-1}(p)$ for any $p\in N$. Then one can apply the long exact sequence of homotopy groups …

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 …

My question, in its most general form is this: Given a fiber bundle $F\rightarrow E\rightarrow B$, when is there a fiber bundle $B\rightarrow E\rightarrow F$? Here, F,E, and B can lie in whichev …

As Anton writes, this is unknown. The main issue is that as of now, the only method we have of creating positively curved closed manifolds with non-trivial fundamental group is to start with a simply …

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} = …

$\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\Sp{Sp}$A maximum symmetry metric on $M:=\mathbb{O}P^2$ must be equivalent to the one you described. Here's one way to see it. Suppose $G$ is the is …

The equality $D_p = C_p$ holds for compact symmetric spaces (CROSSes) of rank $1$, but not in general for higher rank symmetric spaces. A rank $1$ CROSS is isometric to a round sphere or projective sp …

This question is motivated by What manifolds are bounded by RP^odd? (as well as a question a fellow grad student asked me) but I can't seem to generalize any of the provided answers to this setting. …