Is the space of complex structures a loop space?

Question

Define the space of (normalized) complex structures $\mathcal{J}_{2k}$ on $\mathbb{R}^{2k}$ as the orthogonal transformations in $SO(2k)$ that square to minus the identity.

My question is if there exists a space $X$ such that $\Omega (X) \simeq \mathcal{J}_{2k}$?

I suspect this is not the case - if so then how would one prove this?

As far as I am aware, a space can be de-looped if and only if it can be given an $A_\infty$-structure. Since $\mathcal{J}_{2k}$ can be identified with the quotient $SO(2k)/U(k)$ (and $U(k)$ is not a normal subgroup of $SO(2k)$ for all $k\neq 1$*) then there is no obvious well-defined product to put on the quotient, and hence $\mathcal{J}_{2k}$.

Another note is that the space of complex structures can be identified with the space of minimal geodesic paths between two fixed points in $SO(2k)$. So to de-loop $\mathcal{J}_{2k}$ is to de-loop $\Omega^m_{p,q}(SO(2k))$.

A final point of possible interest is that stably (in the colimit) $\mathcal{J}:= SO/U \simeq \Omega SO$** can be de-looped.

*Note: I am identifying $U(k)$ with the subgroup of $SO(2k)$ of those transformations that commute with a fixed complex structure $J \in \mathcal{J}_{2k}$

**Note: This is a part of the story of Bott periodicity - see eg. Milnor's Morse theory.

Answer 1

No, it is not even an H-space.

I prefer slightly different notation, I write $J_k$ to mean $SO(2k)/U(k)$. Topologically your space is the product of this with a two point set.

1. There is a fiber sequence $J_k \to J_{k+1} \to S^{2k}.$

I leave this as an exercise.

2. Euler characteristic is multiplicative under fibrations. It follows from $J_1 = \{*\}$ that $\chi(J_k) = 2^{k-1}$.

3. A finite-dimensional H-space has zero Euler characteristic. In fact, the alternating sum of dimensions in any finite dimensional Hopf algebra is zero. This follows from their basic structure theory; see Hatcher 3C.4.

QED.