Is SO(2n+1)/U(n) a symmetric space?

Question

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

1. 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 $SO(2n+2)=SO(2n+1)\cdot U(n+1)$
2. 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?

Answer 1

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. $SU(2n+1)/Sp(2n)=SU(2n+2)/Sp(2n+2)$.

Answer 2

I think you are right to say that both homogeneous spaces are diffeomorphic as smooth manifolds. Therefore if one of them is a symmetric space, then one can transfer this structure to the other one.

However it is common usage and an abuse of language to say that $G/K$ is a symmetric space to mean that $(G,K)$ is a symmetric pair, that is, $G$ is a connected Lie group and $K$ is an open subgroup in the fixed point set of an involutive automorphism of $G$. This is essentially equivalent to your Lie algebraic formulation. In the strict latter sense, $SO(2n+1)/U(n)$ is not a symmetric space.

Answer 3

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).