Are the 1-parameters subgroups of $SO(3)$ closed?

Question

I'm trying to solve the following question

Question: Prove that all $1$-parameters subgroup of $SO(3)$ are closed. Does this statement holds for $SO(n),$ $n>3$?

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Some comments

The $1$-parameters subgroups of $SO(n)$ are the groups

$$P_A := \{e^{tA};\ A \in M_n(\mathbb{R})\ \text{and}\ A+A^T = 0\}. $$

Since $SO(n)$ is compact the exponential map is surjective, and the map $\mathbb{R}\to P_A\ (t\mapsto e^{tA})$ is an isomorphism of groups (it is easy to see that $0$ is the unique element of the kernel, sinse $e^B = \text{Id}$ $\Leftrightarrow B =0$) homomorphism of groups.

Being honest I can't believe that $P_A$ is closed when $n=3$, the fact of $P_A \cong \mathbb{R}$ as a group makes me have no idea of what is going on, once $P_A$ closed would imply $P_A$ compact.

EDIT: After Reuns' help I realised that I wroted thing that does not make any sense.

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Can anyone help me?

Answer 1

Hint One can compute the $1$-parameter subgroups rather explicitly.

Consider the one-parameter subgroup $t \mapsto \exp t A$ determined by the any $n \times n$ skew-symmetric (and hence diagonalizable) matrix $A$. By skew-symmetry its eigenvalues are imaginary and (since $A$ is real) the set of eigenvalues is closed under conjugation.

For $n = 3$, the eigenvalues are $+ \lambda i, -\lambda i, 0$ for some $\lambda \in \Bbb R$. Diagonalizing $A$ then gives $$A = P D P^{-1}, \qquad D := \operatorname{diag}(\lambda i, -\lambda i, 0) ,$$ for some matrix $P$.

Then, $$\exp t A = \exp t (P D P^{-1}) = P \exp (tD) P^{-1} = P \operatorname{diag}(\exp(\lambda i t),\exp(-\lambda i t),1) P^{-1}. $$ Is $t \mapsto \exp t A$ periodic?

For $n \geq 4$ the situation is qualitatively different, and the case $n = 4$ illustrates the general phenomenon. The eigenvalues are $\pm \lambda i, \pm \mu i$ for some $\lambda, \mu \in \Bbb R$, and so $D$ in the above statement is replaced by $$D := \operatorname{diag}(\lambda i, -\lambda i, \mu i, -\mu i) .$$

Proceeding as before gives $$\exp tA = P \operatorname{diag}(\exp (\lambda i t), \exp(-\lambda i t), \exp(\mu i t), \exp(-\mu i t)) P^{-1}.$$ Under what condition on $\lambda, \mu$ is $t \mapsto \exp t A$ periodic? Show that when this map is not periodic, the image of $\exp tA$ is dense in $SO(4)$ but does not contain $-I$.