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A Riemannian manifold $M$ is said to be homogeneous if the group of isometries $Isom(M)$ acts transitively on $M$.

A Riemannian manifold is said to be symmetric if it is connected, homogenuous, and in addition, there exists a point $p\in M$ and an involution $\phi\in Isom(M)$ such that $p$ is an isolated fixed point of $\phi$ (that is, there exists an open neighborhood $V$ of $p$ where $p$ is the only fixed point of $\phi$ among all elements of $V$)

It thus follows that even though a homogeneous spaces are very nice and "symmetric" in the sense that all points look geometrically the same, they are nevertheless not considered to be a symmetric space by the definition above (at least a priori).

The inevitable question is thus:

Is there a relatively simple example of a connected and homogeneous space which is not a symmetric space?

stag
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1 Answers1

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Here is my favorite example. Consider the flag space $$G(k,\ell,n) = \{(L,H)\in G(k,n)\times G(\ell,n): L\subset H\}.$$ Here $G(k,n)$ is the Grassmannian of $k$-planes in $n$-space, and $G(\ell,n)$ is the Grassmannian of $\ell$-planes in $n$-space, and we take $1\le k<\ell<n$. Depending on whether we're doing real or complex subspaces, this will be a homogeneous space with group $O(n)$ or $U(n)$.

You can check Lie-algebraically that this is not a symmetric space, or — my preferred way — you can find invariant differential forms on this space that are not closed (on a symmetric space $G/K$ with $G$ compact, every invariant form is closed).

Ted Shifrin
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