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?


  1. If $SU(k)/SO(k)$ is a manifold for every $k$, how does this manifold behave?
  2. 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?

  • 3
    $\begingroup$ The quotient of a Lie group by a Lie subgroup is always a manifold (in a natural way). Please use Math.stackexchange for this kind of general questions. $\endgroup$
    – abx
    Aug 11, 2021 at 16:23
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    $\begingroup$ I would be interested in knowing simple descriptions of the manifold SU(n)/SO(n). Regarding that it is a manifold, see math.stackexchange.com/questions/2221222 $\endgroup$ Aug 11, 2021 at 16:34
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    $\begingroup$ @МаринаMarinaS: the quotient $G/H$ of a Lie group by a closed subgroup is a smooth manifold, with a unique $G$-invariant smooth structure; see any textbook on Lie groups. $\endgroup$
    – Ben McKay
    Aug 11, 2021 at 16:50
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    $\begingroup$ This particular manifold is the collection of all choices of a totally real linear subspace of a complex vector space, as $SU_n$ acts transitively on such, while $SO_n$ is the subgroup fixing one such. $\endgroup$
    – Ben McKay
    Aug 11, 2021 at 16:51
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    $\begingroup$ In fact $SLag_k=\mathrm{SU}(k)/\mathrm{SO}(k)$ is not only homogeneous, it is an irreducible symmetric space on Cartan's list, where the defining involution $\sigma:\mathrm{SU}(k)\to \mathrm{SU}(k)$ is simply $\sigma(A)=\overline{A}$. In calibrated geometry, $SLag_k$ is known as the 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 $$SLag_4\simeq\mathrm{Gr}(3,6)\simeq\mathrm{SO}(6)/\bigl(\mathrm{SO}(3)\times \mathrm{SO}(3)\bigr)$$ $\endgroup$ Aug 12, 2021 at 0:39

1 Answer 1


$SU(k)/SO(k)$ is the set of real structures on $\mathbb C^k$, i.e. $k$-dimensional real subspaces of $\mathbb C^k$ which generate it as a complex vector space, satisfying two conditions:

(1) The Hermitian form $((z_1,\dots, z_n) , (w_1,\dots, w_n)) \mapsto z_1\overline{w}_1+ z_2 \overline{w}_2 + \dots + z_n \overline{w}_n$ takes real values on each pair of vectors in the subspace.

(2) Given a basis for the $k$-dimensional real subspace, the determinant of the corresponding $k \times k$ complex matrix is real.

To prove this is $SU(k)$, it suffices to show that $SU(k)$ acts transitively on it, with stabilizer $SO(k)$. The action of $SU(k)$ is by its action on $\mathbb C^k$.

Restricting the Hermitian form to any such real subspace gives a symmetric form. We can choose a basis which is orthonormal respect to this form, and we can further choose the basis to have positive determinant (by negating the last vector if necessary). That will give an orthonormal basis for $\mathbb C$. So the determinant of the corresponding matrix is a complex number of unit norm, and, because it is a positive real, must be $1$. So there is a matrix in $SU(k)$ transferring the standard basis to this basis, thus transferring the standard real subspace to this real subspace.

Finally, for the standard real subspace, an element of the stabilizer is a $k\times k$ real matrix that also lies in $SU(k)$, i.e. an element of $SO(k)$.

Wikipedia claims that we should look at real structures only satisfying condition (2), but this seems to me to give the wrong dimension $2k^2 - k^2 -1 = k^2-1$. Possibly (1) is suppsoed to be assumed in the definition of a real structure.

  • $\begingroup$ I'll point out that condition (1) can be thought of as saying that your $k$-plane is Lagrangian for the symplectic form $\langle z, w \rangle = \mathrm{Im}(\sum z_j \overline{w_j})$. So this is some closed subset of the Lagrangian Grassmannian $G(k,2k)$. $\endgroup$ Aug 12, 2021 at 0:02
  • $\begingroup$ It will be great if you can add a few words on 𝑘=3, 𝑆𝑈(3)/𝑆𝑂(3) = Wu manifold case. I will accept the answer for now. $\endgroup$ Aug 12, 2021 at 0:32