I'm not sure how the $\vec\tau$ matrices are normalized, but we can get quite far without that. The first thing we should do is take the derivative
$$\partial^\mu U=\frac{i}{f_\pi}(\vec\tau\cdot\partial^\mu\vec\Phi)U$$
and
$$\partial_\mu U^\dagger=-\frac{i}{f_\pi}(\vec\tau\cdot\partial^\mu\vec\Phi^\dagger)U^\dagger$$
(I'm making the guess that the $\vec\tau$ are Hermitian.) Then
$$\partial_\mu U^\dagger \partial^\mu U=f^{-2}_\pi(\vec\tau\cdot\partial^\mu\vec\Phi^\dagger)(\vec\tau\cdot\partial^\mu\vec\Phi)$$
Note that the $U$s cancel out because they are unitary and commute with the $\vec\tau$. We then write
$$\partial_\mu U^\dagger \partial^\mu U=f^{-2}_\pi \partial_\mu\Phi^\dagger_i\partial^\mu\Phi_j\tau_i\tau_j$$
If we know the normalization $\operatorname{tr}(\tau_i\tau_j)$, then we can complete the expression.

This post imported from StackExchange Physics at 2015-01-22 11:33 (UTC), posted by SE-user 0celo7