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?


1 Answer 1


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.