Let $M_n \subseteq SO(2n)$ be the set of real $2n \times 2n$ matrices $J$ satisfying $J + J^{T} = 0$ and $J J^T = I$. Equivalently, these are the linear transformations such that, for all $x \in \mathbb{R}^{2n}$, we have $\langle Jx, Jx \rangle = \langle x, x \rangle$ and $\langle Jx, x \rangle = 0$. They can also be viewed as the linear complex structures on $\mathbb{R}^{2n}$ which preserve the inner product.

I'd like to understand $M_n$ better as a topological space, namely an $(n^2-n)$-manifold.

$M_1$ is just a discrete space consisting of two matrices: the anticlockwise and clockwise rotations by $\pi/2$.

For $n \geq 2$, we can see that $M_n$ is an $M_{n-1}$-bundle over $S^{2n-2}$. Specifically, given an arbitrary unit vector $x$, the image $y := Jx$ must lie in the intersection $S^{2n-2}$ of the orthogonal complement of $x$ with the unit sphere $S^{2n-1}$. Then the orthogonal complement of the space spanned by $x$ and $y$ is isomorphic to $\mathbb{R}^{2n-2}$, and the restriction of $J$ to this space can be any element of $M_{n-1}$.

Since the even-dimensional spheres are all simply-connected, it follows (by induction) that $M_n$ has two connected components for all $n \in \mathbb{N}$, each of which is simply-connected. For instance, $M_2$ is the union of two disjoint 2-spheres: the left- and right-isoclinic rotations by $\pi/2$. The two connected components of $M_n$ are two conjugacy classes in $SO(2n)$; they are interchanged by conjugating with an arbitrary reflection in $O(2n)$.

Is [each connected component of] $M_n$ homeomorphic to a known well-studied space? They're each an:

$S^2$-bundle over an $S^4$-bundle over $\dots$ an $S^{2n-4}$-bundle over $S^{2n-2}$

but that's not really very much information; can we say anything more specific about their topology?

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    $\begingroup$ Your $M_n$ is the Riemannian symmetric space $\mathrm{SO}(2n)/\mathrm{U}(n)$. I believe that its topology is quite well studied from that point of view. $\endgroup$ – Robert Bryant Sep 18 at 17:43
  • $\begingroup$ Thanks! Yes, it looks like the space is called DIII in Cartan's classification of compact Riemannian symmetric spaces. (I'll accept this if you post it as an answer.) $\endgroup$ – Adam P. Goucher Sep 18 at 18:35
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    $\begingroup$ yes, this is class DIII, the two connected components are distinguished by the Pfaffian, which equals $\pm 1$; in the physics context this matrix $M_n$ is the scattering matrix of a superconductor with preserved time-reversal symmetry but broken spin-rotation symmetry. The sign of the Pfaffian then distinguishes topologically trivial from topologically nontrivial superconductors. $\endgroup$ – Carlo Beenakker Sep 18 at 19:30
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    $\begingroup$ In Morse Theory, Milnor has proved that $\mathrm{O}(2n)/\mathrm{U}(n)$ can be identified with orthogonal linear transformations $J$ with $J^2=-I$ i.e. complex structure. $\endgroup$ – C.F.G Sep 18 at 19:34

Your $M_n$ is (two copies of) the Riemannian symmetric space $\mathrm{SO}(2n)/\mathrm{U}(n)$ (which is $DI\!I\!I$ in Cartan's nomenclature). Its topology is well-studied from that point of view.

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