One reason I call this site a “blag” and not a “blog” is that I’m always late.
For example, I’m finally typing up the group-theory homework assignment which Ben gave last Monday (and which will be due next Monday). During our seminar over in BU territory, we discussed the relations among the Lie groups SU(2) and SO(3) and the manifolds S3 and RP3. Problems will be given below the fold.
Also, Eric will be discussing statistical physics this afternoon at NECSI.
Problem 1. To understand the relation between SU(2) and SO(3), we need to identify a 2×2 Hermitian matrix with a rotation in 3D Euclidean space. We do this by pulling out the Pauli matrices. Define a vector space V as
[tex]V = \hbox{span}\left\{\left(\begin{array}{cc} i & 0 \\ 0 & -i\end{array}\right), \left(\begin{array}{cc} 0 & 1 \\ -1 & 0\end{array}\right), \left(\begin{array}{cc} 0 & i \\ i & 0\end{array}\right)\right\}.[/tex]
Next, define an action for A ∈ V, P ∈ SU(2),
[tex]P \cdot A \equiv P A P^\dag.[/tex]
Prove that P acting on A always falls within V (closure), and that the action is linear, i.e.,
[tex] P\cdot(A + kB) = P\cdot A + kP\cdot B[/tex]
for A and B in V, k a real number.
Assuming the results of homework 1, a map P from V to itself can be represented as a 3×3 matrix, which we denote as ψ(P):
[tex]\psi(P)A = P\cdot A.[/tex]
Problem 2. Prove that
[tex]\psi:\ SU(2) \rightarrow GL(3,R)[/tex]
is a homomorphism. That is, show
[tex]\psi(PQ) = \psi(P)\psi(Q)[/tex]
and that (this is slightly tricker) ψ(P) is invertible.
Problem 3. Test the relation
[tex]PAP^\dag = A[/tex]
on the basis of V, the Pauli matrices, and show that the matrix P must be plus-or-minus the identity. This implies that the preimage of the identity in SO(3) is two points in SU(2).