I am a physics student with only a rudimentary knowledge of differential geometry, so please feel free to point out if I miss something elementary / trivial.
According to https://arxiv.org/abs/1408.2760, $ SO(2n+1)/U(n) $ is not a symmetric space because it does not have the right Cartan decomposition of the Lie algebra. That is, suppose that $ \mathfrak{g} $ is the Lie algebra of $ SO(2n+1)$ and $ \mathfrak{h} $ is the Lie algebra of $ U(n) $. There is a decomposition $$ \mathfrak{g} = \mathfrak{h} + \mathfrak{p} $$ for some $ \mathfrak{p} $ such that $ [\mathfrak{h},\mathfrak{h}] \subset \mathfrak{h}$ and $[\mathfrak{h},\mathfrak{p}] \subset \mathfrak{p}$. But $ [\mathfrak{p},\mathfrak{p}] $ is not in $ \mathfrak{h} $, so we do not have a Cartan involution on this space.
I'm wondering if it is that simple. I'll be grateful if someone can clear up my confusion below.
I think that $SO(2n+1)/U(n)$ and $SO(2n+2)/U(n+1)$ are diffeomorphic. For instance, this book shows that the two are the same homogeneous spaces by showing that
- Any element $SO(2n+1)$ can be written as an ordered product of two elements, one in $SO(2n+1)$ and another in $U(n+1)$. They write this as $SO(2n+2)=SO(2n+1)\cdot U(n+1)$
- The quotient $X = SO(2n+2)/U(n+1) = SO(2n+1)\cdot U(n+1) /
U(n+1)$ can be thought of as $ SO(2n+1)/U(n)$ because the
$SO(2n+1)$-action on $X$ is transitive and the stabilizer of the identity
$eU(n+1)$ of $X$ are the elements of $SO(2n+1)$ that are also in
$U(n+1)$: $SO(2n+1) \cap U(n+1) = U(n)$.
An example in low dimensions is $ SO(5)/U(2) = SO(6)/U(3) = \mathbb{C}P^3$, a complex projective space.
But $SO(2n+2)/U(n+1)$ is a symmetric space. So why is $SO(2n+1)/U(n)$ not?
Best Answer
Let me complement Claudio's answer. There is indeed a definition of symmetric space which works for any Riemannian manifold $M$: For any point $p\in M$ there is an involutive isometry $\iota_p$ of $M$ such that $p$ is an isolated fixed point. This involution is easy to describe: Take any geodesic $\gamma(t)$ with $\gamma(0)=p$. Then $\iota_p$ sends $\gamma(t)$ to $\gamma(-t)$. In this sense, both of your spaces are symmetric since they are isometric.
In practice, it is difficult to work with the $\iota_p$ directly. Instead let $\tilde G\subseteq\text{Aut}(M)$ be the group which is generated by all the $\iota_p$ and let $G$ be its connected component. Let $H=G_p$ be the stabilizer of a point. Then one shows that $G$ is a Lie group, that $G$ acts transitively on $M$, i.e., $M=G/H$ and that there is an involutory automorphism $\iota$ of $G$ having $H$ as the set of fixed points. Conversely, given $G$, $\iota$ and $H=G^\iota$ then $G/H$ is a symmetric space in the sense above, provided that $H$ is compact. Therefore, it is common to take this as a definition of a symmetric space. One usually even requires $G$ to be semisimple which excludes the Euclidean space as symmetric.
In your situation it happens that a smaller group $G_0$ still acts transitively on $G/H$. Thus $G_0/H_0=G/H$ where $H_0=G_0\cap H$. Then $G_0/H_0$ is not symmetric with respect to the second definition since the group acting is too small.
The difference between the two definitions is not too serious since the phenomenon that there is a smaller group $G_0\subset G$ acting transitively on a homogeneous space $G/H$ is extremely rare. In fact, it is so exceptional that Onishchik was able to classify all triples $(G,H,G_0)$ in a famous paper "Inclusion relations among transitive compact transformation groups" under the condition that $G$ is compact and simple. In his list you'll find more examples of the kind above, e.g. $SU(2n+1)/Sp(2n)=SU(2n+2)/Sp(2n+2)$.