I want to prove the following statement

Let $H$ be a 1-dimensional closed subgroup of $SO(3)$. Then $H$ has exactly one connected component or two connected components.

I have note that:

(1) The lie algebra $\mathfrak h$ is isomorphic to $\mathbb R$.

(2) $SO(3)$ is compact and so the exponential map $\exp:\mathfrak{so}(3)\to SO(3)$ is surjective.

But I have no idea how to prove the statement. Any idea?


Since $H$ is a subgroup of $\mathrm{SO}(3)$, it acts on the sphere $S^2\subseteq\mathbb{R}^3$ in a natural way. Moreover, $H$ is $1$-dimensional so this action generates a vector field $X$ on $S^2$. By the Hairy Ball Theorem, $X$ vanishes at some $p\in S^2$.

Lemma. $H$ preserves the set $\{\pm p\}$.

Proof. Since $X_p=0$, the subgroup $H_p=\{h\in H:h\cdot p=p\}$ is $1$-dimensional. But $\mathrm{SO}(3)_p$ is a $1$-dimensional connected group (isomorphic to circle $S^1$) and $H_p$ is a $1$-dimensional closed subgroup of $\mathrm{SO}(3)_p$, so $H_p=\mathrm{SO}(3)_p$. Now, for all $h\in H$ we have $H_{h\cdot p}=hH_ph^{-1}=h\mathrm{SO}(3)_ph^{-1}=\mathrm{SO}(3)_{h\cdot p}$. Thus, $\mathrm{SO}(3)_{h\cdot p}\subseteq H$ for all $h\in H$. Since $\mathrm{SO}(3)_{x}$ and $\mathrm{SO}(3)_{y}$ generate $\mathrm{SO}(3)$ whenever $x,y\in S^2$ are not collinear, we conclude that $h\cdot p=\pm p$ for all $h\in H$. $\square$

The subgroup of $\mathrm{SO}(3)$ which preserves $\{\pm p\}$ has exactly two connected components. It consists of rotations around the line spanned by $p$ together with flipping the line (or in other words, it is the symmetry group of a unit circle in $\mathbb{R}^3$). Explicitly, when $p=(0,0,1)$, this is the group of matrices of the form $$ \begin{pmatrix} \cos\theta & \mp\sin\theta & 0 \\ \sin\theta & \pm\cos\theta & 0 \\ 0 & 0 & \pm 1\end{pmatrix}.$$

Since $H$ is a one-dimensional closed subgroup of that group, it has at most two connected components. It is either $\mathrm{SO}(3)_{\pm p}$ itself, or the $S^1$-subgroup which preserves $p$ (i.e. with $\pm1=1$ above).

  • $\begingroup$ Thank you so much for your answer!. Do you know the bibliography in which your answer is? $\endgroup$ – FUUNK1000 Feb 5 '17 at 19:08
  • $\begingroup$ @FUUNK1000 I just made it up. I don't know any reference for this problem. Where did you hear about it? $\endgroup$ – Spenser Feb 5 '17 at 21:31
  • 1
    $\begingroup$ @FUUNK1000 If you have trouble understanding some parts, two important facts that I am using implicitly in this answer are the following: (1) If $M$ is a smooth manifold and $S$ a submanifold of $M$ of the same dimension as $M$, then $S$ is open in $M$. (2) If a subset $N$ of $M$ is both open and closed, then it is a union of connected components of $M$. I hope that helps. $\endgroup$ – Spenser Feb 5 '17 at 21:38
  • $\begingroup$ I am studying Lorentzian manifolds with spherical symmetry. I have found this statement in the book: Geometria diferencial i relativitat by J. Girbau (teacher of the university of Barcelona, Spain). I think I understood your answer. Thank you very much $\endgroup$ – FUUNK1000 Feb 5 '17 at 21:43

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