If the fields $b$ and $c$ have conformal weight $\lambda$ and $1-\lambda$ and action is:

$$S = \frac{1}{2\pi} \int d^2z \, b \bar \partial c,$$

under conformal transformations $z \rightarrow z+\epsilon(z)$, we can get the energy-momentum tensor: $$ T(z) = - \lambda :b \partial c: + (1-\lambda) :(\partial b) c:. $$ But this symmetry $z \rightarrow z+\epsilon(z)$ is local, shouldn't this be the energy-momentum tensor for global translations: $z+ a$? I am getting confused as to why a local symmetry is giving us a stress-energy tensor.

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    $\begingroup$ Well, the Noether current associated with a diffeomorphism $x \to x + \epsilon$ is $j^\mu = \epsilon_\nu T^{\mu\nu}$. A conformal transformation is just one diffeomorphism which maps the metric to a Weyl transform of itself. That's the reason the stress tensor appears. $\endgroup$
    – Gold
    Dec 1, 2022 at 21:21
  • $\begingroup$ But Noether theorem works only for global symmetries: continuous global symmetries lead to conservation laws. My issue is with $z+ \epsilon(z)$, the function of z, which is local. $\endgroup$ Dec 2, 2022 at 0:28
  • $\begingroup$ That is just a warning that you shouldn't try to apply Noether's theorem to "gauge symmetries". It doesn't literally mean that the small parameter is not allowed to depend on position at all. If it did, even rotational symmetry would fail to have a conservation law. $\endgroup$ Dec 13, 2022 at 16:34
  • $\begingroup$ The conformal diffeomorphisms on the worldsheet come from residual gauge invariance though. $\endgroup$ Dec 23, 2022 at 1:51

1 Answer 1


When the background metric is fixed, as in this case, the transformation should be thought of as a physical symmetry, taking the point z to point z'. This is now a global symmetry with the conserved currents and not a diffeomorphism! - From Tong's notes on String theory.


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