Topology

Fact: $SU(n)$ is simply connected. So $SU(2)$ is the universal covering space of $SO(3)$. The covering map is 2-1. It follows from standard results in topology that the fundamental group of $SO(3)$ is ${\bf Z}_2$.

Recall that elements of $SU(2)$ all take the form $a{\bf 1}+b{\bf i}+c{\bf j}+d{\bf k}$ with $a^2+b^2+c^2+d^2=1$. Therefore $SU(2)$ is topologically $S^3$, the 3-dimensional hypersphere. (This is enough to show that $SU(2)$ is simply connected.) Because the kernel is $\{\pm{\bf 1}\}$ (verify!), antipodal points are identified on mapping into $SO(3)$, so $SO(3)$ is homeomorphic to real projective 3-space.

This can be seen another way. A rotation can be specified by a vector along the axis of the rotation, with magnitude giving the angle of the rotation. This serves to identify elements of $SO(3)$ with points inside or on a ball of radius $\pi$. However, antipodal points on the surface of the ball represent the same rotation. The resulting space (a three-dimensional ball with antipodal points on the surface identified) is well-known to be homeomorphic to real projective 3-space. (If you think about this argument for a bit, you should see an implicit use of the exponential mapping from $so(3)$ into $SO(3)$.)

A loop in the topological space $SO(3)$ can be visualized as a continous ``trajectory'' of rotations: we take a rigid object and turn it around in some fashion, finally restoring it to its original orientation. The following fact can be deduced from this: if a solid object is connected by threads to a surrounding room, and the object is turned through $720^\circ$, then the threads can be untangled without turning the object any more. However, if the object is turned through $360^\circ$, then the threads cannot be untangled. (The two-thread version of this is known as ``Dirac's string trick''.) In this sense, a continuous turn through $360^\circ$ is not the same as no turn at all (but a $720^\circ$ turn is.)

© 2001 Michael Weiss

home