How to prove $\mathrm{Tr}[(\partial_\mu U)U^\dagger]=0$? I am studying ChPT by referring to "A Primer for Chiral Perturbation Theory" by  Stefan Scherer.
I'm having a problem with the consideration of terms that appear in the Lagrangian.
The textbook says only $\mathrm{Tr}(\partial^\mu U\partial_\mu U^\dagger)$ is important and other terms such as $\mathrm{Tr}[(\partial_\mu \partial^\mu U) U^\dagger]$ are irrelevant because $\mathrm{Tr}[(\partial_\mu U)U^\dagger]=0$.
Proving $\mathrm{Tr}[(\partial_\mu U)U^\dagger]=0$ is also an exercise and my question.
Below is the corresponding part from the textbook.
Do you have any ideas?

 A: I think the following derivation works: start with the fact that $U$ is unitary such that $U^{-1} = U^\dagger$. Therefore we can rewrite
\begin{equation}
\mathrm{Tr}(\partial_\mu U U^\dagger) = \partial_\mu \mathrm{Tr}(\ln U)
\end{equation}
where I used the cyclicity and linearity of the trace.
Now, using your notation we can write
\begin{equation}
\partial_\mu \mathrm{Tr}(\ln U) = \frac{i}{F_0} \partial_\mu \phi_a(x) \mathrm{Tr}(\Lambda^a)
\end{equation}
Since the trace only works on the generators forming the Lie algebra. But we know that the generators of $\mathrm{SU}(N)$ are traceless, hence your desired result follows.
A: *

*Even in the absence of an additional relation, a term of the form ${\rm Tr} (\partial_\mu A \,\,  \partial^\mu B)$ in the Lagrangian  would be (physically) equivalent to $- {\rm Tr} [ (\partial_\mu \partial^\mu A) B]$ as they differ only by a 4-gradient $\partial_\mu(\ldots)$, which does not effect the equations of motion.


*There is an easy way to see that ${\rm Tr} [(\partial_\mu U) U^\dagger]=0$ holds. Instead of using Stefan Scherer's hint, take advantage of Julian Schwinger's slick formula $\delta \det A= \det A \, \,{\rm Tr} (A^{-1} \delta A)$, holding for any invertible square matrix $A$. Inserting $A= U \in {\rm SU}(N)$ and $\delta A = \delta a^\mu \, \partial_\mu U$ (with an arbitrary infinitesimal 4-vector $\delta a^\mu$) shows the desired relation.


*The proof of the general relation $\delta \det A = \det A \, \,                                 {\rm Tr}(A^{-1} \delta A) $ is quite simple:
$\ \delta \det A = \det(A+\delta A)- \det A= \det[A(\mathbf{1}+A^{-1} \delta A)]-\det A = \det A[\det(\mathbf{1} +A^{-1} \delta A)-1]$,
$\qquad$ and finally keeping only terms linear in $\delta A$.
A: I should have checked the answer is attached at the end of the book...
The important things are the hint $[\phi, U^\dagger]=0$ and the cyclic property of trace.
$U$ and $U^\dagger$ can be rearranged so that they are next to each other like a c-number in the trace even though $\phi$ and $\partial_\mu \phi$ do not commute.
\begin{align*}
\mathrm{Tr}[(\partial_\mu U)U^\dagger]&=\cdots
= \mathrm{Tr}[i\frac{\partial_\mu \phi}{F_0}U U^\dagger]
=\mathrm{Tr}[i\frac{\partial_\mu \phi}{F_0}]
=i\frac{\partial_\mu \phi_a}{F_0}\mathrm{Tr}[\Lambda_a]
=0
\end{align*}
A: Following the detailed hint,
$$
\begin{align}
(\partial_\mu U) U^\dagger
&= \left( \frac{i \partial_\mu \phi}{F_0} + \frac{(i \partial_\mu \phi) i \phi + i \phi (i \partial_\mu \phi)}{2! F_0^2} + \cdots \right) U^\dagger \\
&= \frac{i \partial_\mu \phi \ U^\dagger}{F_0} + \frac{(i \partial_\mu \phi \ U^\dagger) i \phi + i \phi (i \partial_\mu \phi \ U^\dagger)}{2! F_0^2} + \cdots 
\end{align} 
$$
I used $[\phi, U^\dagger] = 0$ in the last line above to get $i \partial_\mu \phi$ and $U^\dagger$ together in each term. Now taking the trace will allow to move each $(i \partial_\mu \phi \ U^\dagger)$ factor to the front of every term,
$$
\begin{align}
\text{Tr}((\partial_\mu U) U^\dagger)
&= \text{Tr}\left(
\frac{i \partial_\mu \phi}{F_0} U^\dagger + 2 \frac{i \partial_\mu \phi}{F_0} U^\dagger \frac{i \phi}{2! F_0} + \cdots
\right) \\
&= \text{Tr}\left(
\frac{i \partial_\mu \phi}{F_0} U^\dagger \left(1 + \frac{i \phi}{1! F_0} + \cdots \right)
\right) \\
&= \text{Tr}\left( \frac{i \partial_\mu \phi}{F_0} U^\dagger U \right) \\
&= \frac{i \partial_\mu \phi_a}{F_0} \text{Tr} (\Lambda_a) \\
&= 0
\end{align}
$$
