2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.