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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 …

$\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\Sp{Sp}$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 …

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} = …

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 …

Here is another obstruction. Suppose $M^n$ is a closed simply connected manifold which admits only trivial vector bundles. Then $M$ cannot be a $\mathbb{Z}/2\mathbb{Z}$-homology sphere, unless $n=3$ …

The only if direction fails. That is, there are $K$-equivariant vector bundles which are not homogeneous. For example, the Mobius band has the form $O(2)\times_{O(1)} \mathbb{R}$, and is not Riemann …

I call such bundles "homogeneous bundles", but it's not a totally standard terminology. It is true that the map $G/H\rightarrow G/H'$ is a fiber bundle map with fiber $H/H'$. One way to see this is t …

My understanding is that this is generally unknown. Of course, a few of the total spaces (e.g., $S^7\times S^8$, $S^{15}$, and the unit tangent bundle of $S^8$) are homogeneous spaces, so admit a non …

Suppose $M = S^n$ is a sphere with $n$ odd and at least $5$. Pick your favorite Lie group $G$ for which $\pi_{n-1}(G)$ is non-trivial. (Many examples may be found here. For example, for any $n > 3$ …

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 …

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 …

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 …

Given a compact smooth manifold $M$, it's relatively well known that $C^\infty(M)$ determines $M$ up to diffeomorphism. That is, if $M$ and $N$ are two smooth manifolds and there is an $\mathbb{R}$-a …

The Kazdan-Warner theorem goes a long way toward answering the first and second questions. (For notes typed up by Kazdan, see http://www.math.upenn.edu/~kazdan/japan/japan.pdf.) Here's what is says …