Intuition: why does a rotation in $3$-dimensional space need $3$ angles to be determined?

Question

We know that $\mathrm{SO}(3)$ is a $3$-dimensional compact manifold. This seems sort of counter intuitive to me. I think that we can travel from any point on a two dimensional sphere to another point by determining only two angles: the change in longitude and the change in latitude. So, why do rotations around the origin need three angles instead of two?

Answer 1

The intuition is that $SO(3)$ comprises rotations of the $2$-dimensional sphere $S^2$. Such a rotation is determined by the axis of the rotation (given by a point on the sphere, so 2-dimensional) and an angle of rotation (1-dimensional). Hence the space $SO(3)$ of rotations has dimension $3 = 2 + 1$. Your argument doesn't work because knowing the effect of a rotation on $1$ point does not determine the axis of the rotation.

If you represent elements of $SO(3)$ as matrices, then you can define a continuous function $h : SO(3) \to S^2$ that maps a matrix $M$ to the point in $\Bbb{R}^3$ given by the 3rd column of $M$. For each $p \in S^2$, $h^{-1}(p)$ is homeomorphic to $S^1$. This mapping (or rather its lift to the double cover of $SO(3)$, called the spin group $\mathrm{Spin}(3)$) is known as the Hopf fibration and displays $SO(3)$ as a kind of product (called a fibration) of the circle $S^1$ and the sphere $S^2$ (in particular each point of $SO(3)$ has a neigbourhood that is isomoporphic to a product $U \times V$ of open sets $U \subseteq S^1$ and $V \subseteq S^2$ so that $SO(3)$ is $3$-dimensional).

Answer 2

You're right that if we pick some point $p\in S^2$, we can specify where $p$ goes under some rotation $R$ with two numbers, the latitude and longitude of $R(p)$. However, $R$ is not determined by $R(p)$ alone! Indeed, any rotation about the axis through $p$ will fix $p$. This gives one extra parameter (for the angle that you rotate about the axis through $p$).

To be more precise, let $G\subset SO(3)$ be the subgroup of rotations that fix $p$. Then $G$ is $1$-dimensional (we can rotate by any angle about the axis through $p$). Your analysis then shows that the quotient space $SO(3)/G$ is $2$-dimensional, since the value of $R(p)$ only determines $R$ up to composition with an element of $G$ (on the right).

Answer 3

One can give another reason for the 3D nature of $SO(3)$ : it is the fact that every element of $SO(3)$ can be written under the form of the exponential of an antisymmetric matrix ; more precisely, the rotation with angle $\theta$ (first degree of freedom (DOF)) around normalized axis defined by $N=(a,b,c)$ (with $a^2+b^2+c^2=1$, thus bringing only two supplementary DOF) can be written under the form :

$$R_{\theta,N}=\exp(\theta N_{\times})$$

where $N_{\times}$ is the antisymetric matrix associated with $N$ :

$$N_{\times}=\begin{pmatrix}0&-c&b\\c&0&-a\\-b&a&0\end{pmatrix}$$

a formula that is a kind of extension of the formula of rotation due to the multiplication by complex number $e^{i\theta}$, because :

$$\exp(\theta i) \ \ \text{can be considered as} \ \ \exp(\theta \frak{I})$$

with

$$\frak{I}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}.$$