The space of minimal geodesics on $SU(2m)$

221 Views Asked by At

In the proof of Bott periodicity for the unitary group in Milnor's Morse theory (Lemma 23.1, page 128), it is asserted that the space of minimal geodesics from $I$ to $-I$ in the special unitary group $SU(2m)$ is homeomorphic to the complex Grassmannian of all $m$-dimensional vector subspaces of $\mathbb{C}^{2m}$.

Prior to this, Milnor explained how every minimal geodesic comes from an initial tangent vector $A \in \mathfrak{su}(2m)$, which, as a linear endomorphism splits $\mathbb{C}^{2m}$ as the orthogonal direct sum of two eigenspaces corresponding to the eigenvalues $+i\pi$ and $-i\pi$. Each eigenspace is a $m$-dimensional subspace, so we can identify the minimal geodesic with the $+i\pi$-eigenspace, and thus obtain a correspondence between the set of minimal geodesics and the complex Grassmannian.

However, it is not obvious to me that this bijection is a homeomorphism. So I would appreciate any arguments or ideas about why this is so.

Firstly, we are equipping the space of minimal geodesics with compact-open topology, which in this case is metrizable with metric $d(\gamma,\gamma') = \max_{0 \leq t \leq 1} \rho(\gamma(t), \gamma'(t))$, where $\rho$ is the distance function induced by the Riemannian metric on $SU(2m)$, which in turn is given by $\langle A, B \rangle = \operatorname{Re} \operatorname{trace}(AB^*)$.

On the other hand, the Grassmannian has the topological structure inherited as an orbit space, e.g., as $U(2m)/U(m) \times U(m)$.

I don't have very much intuition with these topologies, but perhaps it is plausible that if two $m$-dimensional subspaces are "close", then the matrices in $\mathfrak{su}(2m)$ to which they correspond are also close, and so the geodesics are close, and vice-versa. How I can make this rigorous?

(Later, Milnor gives a similar lemma (Lemma 24.1) for the space of minimal geodesics on the orthogonal group $O(n)$ and the space of complex structures on $\mathbb{R}^n$, but as far as I can tell he gives no argument on their topological equivalence either.)