Does every path in $SO(2k)$ from $1$ to $-1$ pass through the space of complex structures?

Recall the space of (normalised) complex structures $$\mathcal{J}_{2k} : = \{J \in SO(2k) \mid J^2 = -1\}$$ on $$\mathbb{R}^{2k}$$. I am curious to know if every path from $$1$$ to $$-1$$ in $$SO(2k)$$ must intersect the space of complex structures $$\mathcal{J}_{2k}$$.

For $$k=1$$ we clearly have that any path from $$1$$ to $$-1$$ must pass through $$\pm i$$.

For $$k=2$$ I believe it is also true. Recall that $$SO(4)$$ has a unique double cover by $$S^3_L \times S^3 _R$$ where $$S^3_L$$ can be identified with the elements in $$SO(4)$$ which represent left multiplication by a (unit) quaternion, and similarily $$S^3_R$$ as right multiplication by a quaternion. These two $$S^3$$'s are not disjoint - but they only share $$1$$ and $$-1$$ in common. Hence, I am, heuristically at least, seeing $$SO(4)$$ as two $$S^3$$'s that are connected at the north ($$1$$) and south ($$-1$$) poles. I also know that each path component of the space of complex structures on $$\mathbb{R}^4$$ is homotopy equivalent to $$S^2$$ (seen as the space of all purely imaginary unit quaternions). It is in this way that I see that any path from $$1$$ to $$-1$$ in $$SO(4)$$ must necessarily pass through either the left equatorial sphere $$S^2_L$$ of $$S^3_L$$, or the right equatorial sphere $$S^2_R$$ of $$S^3_R$$, and hence a complex structure.

A general proof or known result would be great!

• Just a note for anyone reading this later: it turns out that the space of minimal geodesics from $1$ to $-1$ in $SO(2k)$ is homeomorphic to the space of complex structures (see Milnor's Morse Theory Lemma 24.2). Feb 2, 2021 at 23:48

No, this is not true for any $$k>1$$. Indeed, inside the subgroup $$U(k)\subseteq SO(2k)$$, we can move from $$1$$ to $$-1$$ by one-by-one moving the diagonal entries around the unit circle, so that at any given time at most one of the entries is different from $$\pm 1$$. In particular, then, at all times there is some entry that is $$\pm 1$$, so the square will not be $$-1$$.