The quotient space $SU(k)/SO(k)$ is also a homogeneous space constructed out of the Lie groups (special unitary $SU(k)$ and special orthogonal $SO(k)$).

Because the $SO(k)$ may not be a normal subgroup of $SU(k)$, so $SU(k)/SO(k)$ may not be a quotient group, or may not be any Lie group. However $SU(k)/SO(k)$ may be a manifold?

Question:

- If $SU(k)/SO(k)$ is a manifold for every $k$, how does this manifold behave?
- Are the different ways to specify the quotient so we may obtain different results? (see $k=2$ below)

Note that

$k=1$, $SU(1)/SO(1)= $ a point.

$k=2$, $SU(2)/SO(2)=SU(2)/U(1)= S^3/S^1= S^2$. However, there are different ways to have $S^1$ fibered over $S^2$, to get $S^3/\mathbf{Z}_k$. So I am interested in knowing $S^3/S^1$ can be something else other than $S^2$?

$k=3$, $SU(3)/SO(3)$ = Wu manifold as a 5 real dimensional manifold. But how exactly this is a manifold? How is this $SU(3)/SO(3)$ related to a Dold manifold as a $\mathbf{CP}^2$ fibered over $U(1)$? (correct me if I said the fibration the other way around.)

$k=4,\dots$, do we have a general understanding for this manifold?

special Lagrangian Grassmannian. In addition to the cases you have looked at already, it may be worth noting that when $k=4$ the exceptional isomorphism $\mathrm{SU}(4)=\mathrm{Spin}(6)$ gives the identification $$SLag_4\simeq\mathrm{Gr}(3,6)\simeq\mathrm{SO}(6)/\bigl(\mathrm{SO}(3)\times \mathrm{SO}(3)\bigr)$$ $\endgroup$5more comments