# Conserved currents for Lagrangian given by a trace

Let the Lagrange density $\mathcal{L}$ be given by $\mathcal{L}=\mathrm{Tr}\left(\partial_\mu U^\dagger \partial^\mu U\right)$, where $U=U(x)\in U(N)$. Suppose there are two matrices $A,B\in SU(N)$ and consider transformation $U(x)\rightarrow A^\dagger U(x)B$. One can easily check that $\mathcal{L}$ is invariant under such transformation, hence by Nother's theorem there exist conserved currents $j_{A,B}$ depended of choice of matrices $A,B$. My question is how to derive these currents ? I know the general formula given in the proof of Noether's theorem but I don't how to treat derivatives $\frac{\partial \mathcal{L}}{\partial (\partial_\mu U(x))}$ in this case. Moreover, we need to obtain $\delta U(x)$. I think the last one can be obtained using representation of $A,B$, which are near to unity, by the exponent of generators of appropriate Lie algebra.

• This is evidently a trivial homework problem your instructor assigned to ensure you understand the symbols involved. Would it help you to write the matrices as sums of the respective two indices of the respective two matrices U and its h.c.? If the two matrices are in U(N), they are expandible around the identity, A ~ $1\!\!\!1 +iaT$, so then $\delta U= ia T U$, etc... exploiting the cyclicity of the trace, don't you have your answer? You only need appreciate what Gursey's elegant chiral notation actually means. – Cosmas Zachos Aug 18 '16 at 14:58
• This is an exercise which I found reading a book about field theory. I tried to expand $A\sim 1+i\sum_{j}^{N^2-1}a_j T^j$, but now I see it is enough to take $A\sim 1+iaT$ as you wrote. The question is how to treat the derivative ? $U(N)$ is not a Banach space - we cannot calculate it. – mikis Aug 18 '16 at 15:04
• Banach? Aren't you overthinking it? I encouraged you to take derivatives w.r.t. matrix elements. In any case, in any way you like, see how you obtain the right-invariant, so left current $j_A\propto i U\partial_\mu U^\dagger$ and the right, so left-invariant one, $j_B \propto i U^\dagger \partial_\mu U$. – Cosmas Zachos Aug 18 '16 at 15:30
• Ok. My main problem was what it the meaning of $\frac{\partial}{\partial (\partial_\mu U)}$, where $U\in U(N)$ or more generally, if $G$ is a Lie group and $g\in G$ what is $\frac{\partial}{\partial (\partial_\mu g)}$. The problem of thinking of it directly is that we do not have a linear structure on $U(N)$, so formally we don't know how to understand these derivatives. To sum up, these derivatives means $\left( \frac{\partial}{\partial (\partial_\mu U_{ab})}\right)_{a,b}$, right ? – mikis Aug 18 '16 at 15:41
• I'm still baffled as to what is bothering you and why you imagine there is no (bi)linear structure. Do you not read this Tr$\partial_\mu U^\dagger \partial_\mu U=\sum_{a,b}\partial_\mu U^\dagger_{ab} \partial_\mu U_{ba}$? So you consider each $U_{ab}$ as a separate variable, properly interpreting the h.c. partner. – Cosmas Zachos Aug 18 '16 at 15:49

Seeing that it is tagged as a homework exercise, I'm probably not supposed to solve the problem, but I can give some hints as to how I would tackle the problem. Assuming one wants to use the standard approach to derive the Noether current, I would suggest that one does not overthink the problem and simply treat the $U$'s as (matrix-valued) functions. One can then apply the derivatives as functional derivatives in the usual way (but I'll use the normal notation for derivatives). One would need a rule for applying the functional derivatives. Making the indices of the matrices explicit, one would have $$\frac{\partial [U(x)]_{ab} }{\partial[U(y)]_{cd}} = \delta(x-y)\delta_{a}^{c}\delta_{b}^{d} .$$ You can now generalize this to the case where $U$ is replaced by $\partial_{\mu} U$.
Another important thing to remember is that the derivative of the adjoint is not zero. $$\frac{\partial U^{\dagger} }{\partial U}\neq 0 .$$
For this one can use $$U^{\dagger} = U^{\dagger}UU^{\dagger} .$$