$p$-roots of the identity on SO($n$)

Question

Let $S\in R^{n\times n}$ be a skew-symmetric matrix such that $\|S\|_F=1$ ($F$ denotes the Frobenius norm). If $n=2$ or $n=3$, then $\exp(2\sqrt{2}\pi S)=I_n$ (matrix exponential). Are there corresponding constants $\theta_n\in R$ such that $\exp(\theta_n S)=I_n$ for other $n\in N$?

Answer 1

The anwer is no.

Let $U\in M_n(\mathbb{C})$. Note that $e^U=I_n$ iff $U$ is diagonalizable and $spectrum(U)\subset 2i\pi\mathbb{Z}$.

If $S$ is skew symm. and $||S||^2=-tr(S^2)=1$, then $S$ is diagonalizable over $\mathbb{C}$, $spectrum(S)=\{\pm ia_1,\cdots,\pm ia_k,0_{n-2k}\}$ where the $(a_j)$ are real and s.t. $\sum_j a_j^2=1/2$.

When $n=2,3$, in general $k=1$ and $a_1=\pm 1/\sqrt{2}$ and $exp(2\sqrt{2}\pi S)=I$.

When $n\geq 3$, that does not work in general. Choose, for example, $n=4$; then in general $k=2$ and $a_1^2+a_2^2=1/2$. We seek $\lambda\not= 0$ s.t. $\lambda ia_1$ and $\lambda ia_2\in 2i\pi\mathbb{Z}$. In particular $a_1/a_2\in\mathbb{Q}$. Choosing $a_1=1/3,a_2=\sqrt{7/18}$, we obtain a contradiction.