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 *S*^{3} and **RP**^{3}. 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).