I have a question similar to this one, but that question is not answered. The question is to show that $SO(3)/SO(2)$ is isomorphic to the 2-sphere: $$ SO(3)/SO(2)\cong S^2 $$ How does one establish the isomorphism? Similarly, how do I show that the following is also an isomorphism: $$ SO(3)/O(2)\cong \mathbb{R}P^2 $$ Thank you very much in advance.


Consider $SO(R^3)$ which acts on $R^3$ by rotations, but restricts to an action of $S^2$. For every point $x\in S^2$ we have a unique orthogonal plane $V$, hence $SO(V)\subset SO(R^3)$ will fix $x$. It is easy to see that in fact $Stab(x)=SO(V) \cong SO(2)$. Hence we have a fiber bundle $$ SO(3) \to S^2 $$

with fiber being $SO(2)$. The map is basically just fixing a point in $S^2$ e.g. $(0,0,1)$ and consider its image under the group action. More fancy: a group action is a map $G\times S^2 \to S^2$ which you can restrict to $G\times \{*\}$. Since the bundle and its fiber are lie groups, this induces an isomorphism $$SO(2) \to SO(3) \stackrel \cong \to S^2 \cong SO(3)/SO(2) $$

Now we compose $SO(3) \to S^2 \to S^2/\mathbb Z_2 = RP^2$. The fiber will be twice as much as before. It is easy and a nice exercise to fill in the details, that the fiber is $O(2)$.

  • $\begingroup$ Isn't the bundle with base space $S^2$ and fiber $SO(2)$ a double cover of $SO(3)$? By taking antipodal points on the sphere as fixed points and opposie rotations in $SO(2)$ we get the same action on $S^2$, don't we? $\endgroup$ – Blazej Aug 29 '15 at 10:25

$\bullet \space \mathbf{SO(3) / SO(2) \simeq S^2}:$

Consider a fundamental representation of the Lie group $G := SO(3)$. Any element $M$ of $G$ can be written as a linear map $M : \mathbb{R}^3 \rightarrow \mathbb{R}^3$ such that $M^{-1} = M^T$ and $\det(M) = 1$. We can easily restrict to $M : S^2 \rightarrow S^2$. For any arbitrary $x \in S^2 \subset \mathbb{R}^3$ we write $x = (x_1,x_2,x_3)$ and $-x = (-x_1,-x_2,-x_3)$, so that $x_1^2 + x_2^2 + x_3^2 = 1$.

Let now $\iota : SO(2) \rightarrow SO(3)$ be some embedding such that $\iota(SO(2))$ is a subgroup of $SO(3)$. Note that there is some $x \in S^2$ such that $\iota(SO(2)) (x) = x$ and $\iota(SO(2)) (-x) = -x$. Thus $\iota(SO(2))$ is a stabilizer $G_x = G_{-x} \subset G$, so that $G_x x = x$ and $G_{-x} (-x) = -x$. Let now $g \in G - G_x$, thus $g \in SO(3)$ but $g \not \in \iota(SO(2))$. Then $g G_x \subset G$ is a left coset of $G_x$, so that $g G_x \cap G_x = \emptyset$. Then \begin{equation} y := g x = g G_x x = (g G_x g^{-1}) g x = G_y y. \end{equation}

Note that $g G_x$ is a subset of $G$ but not a subgroup. But it should be clear that $G_y$ is some conjugate of $G_x$. Then $G_y \simeq G_x$ and $G_x \cap G_y = e$ if $y \not \in \{x,-x\}$, where $e$ is the identity element of $SO(3)$. Also note that $g G_x g^{-1} = G_{-x} = G_x$ for any $g \in SO(3)$ such that $g x = -x$. Then $g^2 = e$, so that $g^{-1} = g$ and $g h g^{-1} = g h g = h^{-1}$ for any $h \in G_x$.

For any $y \in S^2$ there exists an element $g \in G$ such that $y = g x$. Now it should be clear that the left coset space (i.e. the smooth set of left cosets of $G_x$) is isomorphic to $S^2$. Then we can say that there is a principal fiber bundle $(SO(3),S^2,\pi,SO(2))$ with surjective map $\pi : SO(3) \rightarrow S^2$, with a short exact sequence: \begin{equation} 1 \rightarrow SO(2) \rightarrow SO(3) \rightarrow SO(3) / \iota(SO(2)) \simeq S^2 \rightarrow 0. \end{equation}

(This is similar to the principal fiber bundle $(SU(2),S^2,\pi,U(1))$.) Now note that any $x \in S^2$ induces a pair $\{x,-x\} \subset S^2$, so that $\{x,-x\} \in \mathbb{R} P^2$. Then it is straight forward to see that \begin{equation} SO(3) = G = \cup_{\{x,-x\} \in \mathbb{R} P^2} G_x. \end{equation}

$\bullet \space \mathbf{SO(3) / O(2) \simeq \mathbb{R} P^2}:$

There is no proper embedding of $O(2)$ into $SO(3)$ with a fundamental representation. Consider a projective representation $SO(3) : \mathbb{R} P^2 \rightarrow \mathbb{R} P^2$.

Let $L \in O(2)$ and $l := \det(L)$, so that $l \in \{1,-1\}$. Let now $\iota : O(2) \rightarrow SO(3)$ be some embedding such that $\iota(O(2))$ is a subgroup of $SO(3)$. Now define $M := \iota(L) \in \iota(O(2))$ such that $\det(M) = 1$: \begin{equation} L = \left( \begin{array}{cc} L_{1 1} & L_{1 2} \\ L_{2 1} & L_{2 2} \end{array} \right) \Rightarrow M = \left( \begin{array}{ccc} L_{1 1} & L_{1 2} & 0 \\ L_{2 1} & L_{2 2} & 0 \\ 0 & 0 & l \end{array} \right) . \end{equation} Note that this is just an arbitrary embedding; there is no canonical one. As discussed: for any $x \in S^2$ there exists an element $g$ such that $g x = -x$, thus also $g (-x) = x$. There is a projection $S^2 \rightarrow \mathbb{R} P^2$ so that this action turns into $g \{x,-x\} = \{-x,x\} = \{x,-x\}$. This shows that in this case $g$ is also an element of the stabilizer. All these $g$ generate an extension to the stabilizer we already constructed, related to the fundamental representation. This extended stabilizer can really be regarded as a proper embedding from $O(2)$ to $SO(3)$. Thus: \begin{equation} \iota(O(2)) = G_{\{x,-x\}} = G_{\{-x,x\}} \simeq O(2). \end{equation}

It should be clear that if $l = 1$, then $M$ acts like $SO(2)$ and we may assume that $g = e$. If $l = -1$, then $g$ generates an axis $a(g) \in \mathbb{R} P^2$ which is perpendicular to the ${\{x,-x\}}$ axis. This axis $a(g)$ generates the direction of a mirror. There is a principal fiber bundle $(SO(3),\mathbb{R} P^2,\pi,O(2))$ with surjective map $\pi : SO(3) \rightarrow \mathbb{R} P^2$, with a short exact sequence: \begin{equation} 1 \rightarrow O(2) \rightarrow SO(3) \rightarrow SO(3) / \iota(O(2)) \simeq \mathbb{R} P^2 \rightarrow 0. \end{equation}


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.