TITLE: Is SO(2n+1)/U(n) a symmetric space?
QUESTION [15 upvotes]: 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?

REPLY [3 votes]: The homogeneous space $M=G/K=SO(2n+1)/U(n)$ is not a symmetric space but a   generalized flag manifold of the compact simple Lie group $SO(2n+1)$.
It arises from the more general family 
$$
M_{n, p}:=SO(2n+1)/(U(p) \times SO(2(n-p)+1) )
$$
for $p=n$. 
Notice that for any $2\leq p\leq n$ the isotropy representation of $M_{n, p}$ decomposes intro two isotropy summands:
$${\frak m}\cong T_{o}M_{n, p}=\frak{m}_{1}\oplus\frak{m}_{2}.$$
This is true, since in this case you paint black in the Dynkin diagram of $G=SO(2n+1)$ a simple root of Dynkin mark  equal to 2.
It is easy to see that $[\frak{m}, \frak{m}]\neq \frak{k}$, in particular one computes
$$
[\frak{m}_{1}, \frak{m}_{1}]\subset\frak{k}\oplus\frak{m}_{2}, \quad [\frak{m}_{1}, \frak{m}_{2}]\subset\frak{m}_{1},\quad  [\frak{m}_{2}, \frak{m}_{2}]\subset\frak{k}
$$
 For $p=1$ one gets the Hermitian symmetric space $SO(2n+1)/SO(2)\times SO(2n-1)$, which of course is isotropy irreducible (in this case, in the Dynkin diagram of $G=SO(2n+1)$ we paint black  the first simple root, which has Dynkin mark equal to 1, recall that the highest root of $SO(2n+1)$ is given by $\tilde{\alpha}=\alpha_1+2\alpha_2+\cdots+2\alpha_{n}$ where $\{\alpha_1, \ldots, \alpha_{n}\}$ is a basis of simple roots. The coefficients $\{1, 2, \ldots, 2\}$ are the so called Dynkin marks).
In fact, the only generalized flag manifolds which are the same time symmetric spaces are  the (compact) Hermitian symmetric spaces and these are the unique flag manifolds which  are isotropy irreducible. 
(see Which Kahler Manifolds are also Einstein Manifolds? for more details)
Finally notice that a smooth manifold can has more than one expressions as a homogeneous space, $M=G/K=G'/K'$.
For example 
$$S^{6}=SO(7)/SO(6)=G_2/SU(3), \quad S^{7}=SO(8)/SO(7)=Sp(2)/Sp(1)=Spin(7)/G_2.$$
 However, only the pairs $(SO(7), SO(6))$ and $(SO(8), SO(7))$ are symmetric pairs (and so passing to the double coverings you get simply-connected symmetric spaces). And similar in your example: the first complex projective space $$SO(5)/U(2)=ℂP^{3}_{{\frak m}_{1}\oplus{\frak m}_{2}}$$ is not a symmetric space but a flag manifold with two isotropy summands, but $SU(4)/U(3)=ℂP^{3}$ is a Hermitian symmetric space (irreducible).