# Question about gauge transformations of a Non-linear sigma model

Consider a set of scalar fields $\phi^i$ ($i = 1, 2, \ldots, N$) which we now would to couple to a set of gauge vector fields $A_\mu^A$ where $A = 1, 2, \ldots \text{dim}(G)$ ($G$ is generically a non-Abelian gauge group). The (minimal coupling) prescription to do so is the usual replacement of the standard derivative by the gauge covariant derivative:

$$\partial_\mu \phi^i \longrightarrow D_\mu \phi^i \equiv \partial_\mu\phi^i - A_\mu^A K_A{}^i$$

where $K_A^i$ are Killing vectors that satisfy the Killing condition $\mathcal{L}_{K_A}g_{ij} = 0$ and the algebra $\mathcal{L}_{K_A}K_{B} = [K_A, K_B] = f_{AB}{}^C K_C$ (here $f_{AB}{}^C$ are the structure constants of the Lie algebra of $G$, and $K_A \stackrel{\Delta}{=} K_A{}^i\partial_i$).

An infinitesimal gauge transformation is given by

$$\delta\phi^i = \Lambda^A K_A{}^i$$ $$\delta A_\mu^A = \partial_\mu \Lambda^A + f^A{}_{BC}A_\mu^B\Lambda^C$$

The parameters $\Lambda^A$ are arbitrary functions of spacetime. This is all from the book by Cecotti entitled "Supersymmetric Field Theories". For an excerpt click here.

Now, the author states

$$\label{eq:A}\delta D_\mu\phi^i = \Lambda^A(\partial_j K_A^i)D_\mu\phi^j + A_\mu^B\Lambda^C[K_B{}^j\partial_j K_C^i-K_C{}^i\partial_j K_B{}^i]-f_{BC}{}^A A_\mu^B \Lambda^C K_A^i = \Lambda^A(\partial_j K_A^i)D_\mu\phi^j \qquad \text{(#)}$$

I do not understand how this equation (the first equality) has been arrived at.

If spacetime is denoted by $\Sigma$ and labeled by coordinates $x^\mu$, and target space (the space where $\phi^i$'s live) is denoted by $\mathcal{M}$ (so the index $i$ labels coordinates of $\mathcal{M}$, i.e. the fields $\phi^i$), then my understanding is that $\delta\phi^i = \Lambda^A K_A^i$ is a diffeomorphism on target space, and I would expect

$$\delta K_A^i(\phi) \stackrel{?}{=} \Lambda^B \partial_B K^j \partial_j K_A^i - \partial_j(\Lambda^B K_B^i)K_A^j = \Lambda^B (f_{BA}{}^C K_C{}^i) - (\partial_j \Lambda^B)K_B^i K_A^j$$

where we can take $\partial_j \Lambda^B = 0$ because $\Lambda^B$ (as a Lie-algebra valued parameter) is not a function of $\phi^i$'s.

In the book by Cecotti, the transformation law for $K_A^i$ is not given. However, in a book by Tomás Ortín entitled "Gravity and Strings" (for an excerpt, click here), the transformation law for $K_A^i$ is (see Appendix J, equation J.8 if you wish to refer to the book)

$$\delta K_A{}^i = \Lambda^B K_B{}^j \partial_j K_A{}^i$$

which is just the first term (the transport term'') of what I wrote above.

My first question is: why doesn't the transformation law for $K_A{}^i$ contain a derivative of the transformation parameter, given that the transformation parameter is local (in target space) since $K_A{}^i$ depends in general on the fields $\phi^1, \ldots, \phi^N$?

My second question is is related to the actual derivation of equation (#): I did the naive thing of writing

$$\delta D_\mu\phi^i = \delta(\partial_\mu \phi^i - A_\mu^A K_A{}^i)$$

and then taking the $\delta$ in, using the transformation laws of $\phi^i$, $A_\mu^A$ and $K_A{}^i$. I expected to get (#) but I don't even get the first term this way. What am I doing wrong?

The key is to look at the transformation of $\partial_\mu\phi^i$ first:

$$\delta(\partial_\mu\phi^i) = (\partial_\mu\Lambda^A)K_A{}^i + \Lambda^A(\partial_\mu K_A{}^i) = (\partial_\mu\Lambda^A)K_A{}^i + \Lambda^A(\partial_j K_A{}^i)(\partial_\mu\phi^j)$$

where, in going from the first equality to the second, we have made use of the fact that $\phi^i: \Sigma \rightarrow \mathcal{M}$ is a map from $x^\mu$ to $\phi^i$ so one can write

$$\partial_\mu K_A{}^i(\phi) = \frac{\partial \phi^j}{\partial x^\mu}\partial_j K_A{}^i$$

since this is just like a "coordinate" transformation and this is just the chain rule to write spacetime derivatives in terms of target space derivatives.

The first term in $\delta(\partial_\mu\phi^i)$ has a $\partial_\mu \Lambda^A$ which is generically nonvanishing since the gauge parameter $\Lambda^A$ is local (i.e. a function of $x^\mu$). But if this term were not present, one would have the expected transformation of a target space vector (as this is just a target space diffeomorphism). So, we want a covariant derivative $D_\mu\phi^i$ with the property that

$$\delta(D_\mu\phi^i) = \Lambda^A(\partial_j K_A{}^i)D_\mu\phi^j$$

Now we can either accept this (as the definition of how a covariant derivative must transform) to figure out the transformation law that $K_A{}^i$ must obey. OR we can note that $K_A{}^i$ has a target space vector index $i$ so it must transform as a target space vector under a target space diffeomorphism, i.e. as

$$\delta K_A{}^i = \Lambda^B K_B{}^j\partial_j K_A{}^i$$

which has the form of derivative of the parameter times the vector. Note that the vector index in this setting is $j$.

• Note that one just has the transport term in the transformation law for $K_A{}^i$. Now, one might wonder that given a vector $V^i$, since the Lie derivative of $V^i$ along another field $\xi^j$ is $$\mathcal{L}_{\xi}V^{i} = \xi^j\partial_{j} V^{i} - (\partial_{j} \xi^{i})V^{j}$$ why doesn't one also get the second term, which has the form of a derivative $\partial_j(\Lambda^B K_B{}^i)K_A{}^j$. I don't quite understand this yet. – leastaction Oct 5 '17 at 2:36