I'm not sure how the →τ matrices are normalized, but we can get quite far without that. The first thing we should do is take the derivative
∂μU=ifπ(→τ⋅∂μ→Φ)U
and
∂μU†=−ifπ(→τ⋅∂μ→Φ†)U†
(I'm making the guess that the
→τ are Hermitian.) Then
∂μU†∂μU=f−2π(→τ⋅∂μ→Φ†)(→τ⋅∂μ→Φ)
Note that the
Us cancel out because they are unitary and commute with the
→τ. We then write
∂μU†∂μU=f−2π∂μΦ†i∂μΦjτiτj
If we know the normalization
tr(τiτj), then we can complete the expression.
This post imported from StackExchange Physics at 2015-01-22 11:33 (UTC), posted by SE-user 0celo7