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.


2 Answers 2


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 $O(2)$.

  • 1
    $\begingroup$ 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, 2015 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}


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