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Reference:

Chapter 11.3.1 of Freedman and Van Proeyen's Supergravity textbook.

\begin{eqnarray} \notag \delta(a,\lambda) \phi(x) &=& \left(a^\mu(x) P_\mu -\frac{1}{2}\lambda^{\mu\nu}(x)M_{\mu\nu}\right) \phi(x) \\\notag &=& \left(a^\mu(x) \partial_\mu +\lambda^{\mu\nu}(x) x_\nu \partial_\mu\right) \phi(x)\\\notag &=& \left(a^\mu(x) +\lambda^{\mu\nu}(x) x_\nu\right) \partial_\mu \phi(x) \\\notag &=:& \xi^\mu(x) \partial_\mu \phi(x) \\\notag &=& L_\xi \phi(x) \\ &=:& \delta_{\text{gct}} \phi(x) \end{eqnarray}

where we generalized the old spacetime translation vector $a^\mu(x)$ to curved spacetime with $\xi^\mu(x)= a^\mu(x)+ \lambda^{\mu\nu}(x) x_\nu$. So we will have general coordinate transformations (GCTs) parametrized by $\xi^\mu(x)$ and local Lorentz transformations (LLTs) parametrized by $\lambda^{ab}(x)$.

I am trying to understand the introduction of "covariant GCTs" (CGCTs) in the context of gauged spacetime translations. CGCTs are defined by equation 11.61 in the reference above

\begin{equation} \delta_{\text{cgct}} (\xi) = \delta_{\text{gct}}(\xi) - \delta(\xi^\mu B_\mu) \end{equation}

This is motivated by the following:

Consider the standard transformation of scalar fields given by equations 11.1 and 11.2 in the reference above

\begin{equation} \delta(\epsilon) \phi^i(x) = - \epsilon^A(x) t_A{}^i{}_j \phi^j \end{equation}

Now, we showed above what the transformation of the scalar field under GCTs is, so let's say that the symmetry ($T_A = - (t_A)^i{}_j$) is GCT, i.e. $\partial_\mu$, and the parameter ($\epsilon$) is $\xi$.

Then we have, as before,

\begin{equation} \delta(\xi) \phi^i(x) = \xi^\mu(x) \partial_\mu \phi^i(x) \end{equation}

The authors then state on page 228,

"This is correct, but it has the undesirable property that it does not transform covariantly under internal symmetry. We fix this by adding a field-dependent gauge transformation and thus define

\begin{equation} \delta_{\text{cgct}}(\xi) \phi^i = \xi^\mu \partial_\mu \phi^i(x)+(\xi^\mu A_\mu{}^A)t_{A}{}^i{}_j \phi^j" \end{equation}

where $\phi^i$ and $\xi$ are still functions of spacetime, the $(x)$ has just been neglected for brevity.

My confusion lies in "but it has the undesirable property that it does not transform covariantly under internal symmetry."

Can anyone expound on this?

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1 Answer 1

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By successively acting on the field with a general coordinate transformation and an internal symmetry transformation, it can be seen that: $$ \delta_\epsilon\delta_\xi \phi = \xi^\mu \partial_\mu \delta_\epsilon \phi = \xi^\mu \partial_\mu (\epsilon^A t_A(\phi)) $$ Since $\epsilon$ is also dependent on $x$, it can be seen that $\xi^\mu \partial_\mu \phi$ doesn't transform properly under internal symmetry. Therefore, the general coordinate transformation is redefined to accommodate the additional contribution from gauge and internal symmetries.

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  • $\begingroup$ I am sorry I was not able to include math symbols within the sentence $\endgroup$ Aug 14, 2020 at 4:18

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