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?
Intuition: why does a rotation in $3$-dimensional space need $3$ angles to be determined
geometryintuitionrotations
Best Answer
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).