Here is the exercise 3.3 in the Simmons-Duffin CFT lecture notes:

Show that in $d \ge 3$,

$$[Q_\epsilon,T^{\mu\nu}] = \epsilon^\rho \partial_\rho T^{\mu\nu} + (\partial^\rho \epsilon_\rho)T^{\mu\nu} - \partial_\rho \epsilon^{\mu} T^{\rho \nu} + \partial^\nu \epsilon_{\rho} T^{\rho \mu}$$

Argue as follows. Assume that only the stress tensor appears on the right-hand side. Using linearity in $\epsilon$, dimensional analysis, and the conformal Killing equation, show that the above equation contains all terms that could possibly appear. Fix the relative coeffiecients using conservation, tracelessness, and symmetry under $\mu \leftrightarrow \nu$. Fix the overall coefficient by matching with the equation $[P_\mu, O(x)] = \partial_\mu O(x)$

In this notation $\epsilon^\mu$ is a conformal Killing vector (in flat space), defined as

$$\partial_\mu \epsilon_\nu + \partial_\nu \epsilon_\mu = \frac{2}{d} (\partial^\rho \epsilon_\rho)\delta_{\mu\nu}$$

And the charge $Q_\epsilon$ is defined with respect to $\epsilon$ and a timelike surface $\Sigma_t$,

$$Q_{\epsilon}(\Sigma_t) = \int_{\Sigma_t} dS_{\mu} \epsilon_\nu T^{\mu\nu} (x)$$

And the momentum operator $P_\mu$ is defined as

$$P_{\mu} = \int_{\Sigma_t} dS_{\nu} T^{\mu\nu} (x)$$

Assuming only $T^{\mu\nu}$ is on the right, I see how to argue using dimensional analysis and linearity in $\epsilon$ that the terms should be restricted to exactly those with exactly one partial derivative, one factor of $\epsilon^\mu$, and one $T^{\mu\nu}$ (because $Q_\epsilon$ is a dimension zero operator, so we need equal numbers of $\partial_\mu$ and $\epsilon^\mu$ factors, and we need exactly one $\epsilon^\mu$ due to linearity).

But, I'm not seeing how to use the conformal Killing equation to reduce the number of factors. I can see how certain terms like $\epsilon^\mu \partial_\rho T^{\rho \nu}$ are zero by conservation, and how others can be changed into each other by the conformal Killing equation. But I don't see for example how we can discount terms like $\epsilon_\rho \partial^{\nu} T^{\rho \mu}$ which aren't in this expansion and can't be turned into other terms by the Killing equation.

Any thoughts?

  • 1
    $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files. $\endgroup$ – Qmechanic Jul 23 '18 at 7:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.