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!
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 $-1$.