Editing Polyakov action (section)

=== Weyl transformation ===
Assume the [[Weyl transformation]]:
: <math> h_{ab} \to \tilde{h}_{ab} = \Lambda(\sigma) h_{ab}, </math>
then
: <math>\begin{align}
                                    \tilde{h}^{ab} &= \Lambda^{-1}(\sigma) h^{ab}, \\
  \operatorname{det} \left( \tilde{h}_{ab} \right) &= \Lambda^2(\sigma) \operatorname{det} (h_{ab}).
\end{align}</math>
And finally:
: {|
| <math> \mathcal{S}', </math>
| <math> = {T \over 2}\int \mathrm{d}^2 \sigma  \sqrt{-\tilde{h}} \tilde{h}^{ab} g_{\mu\nu} (X) \partial_a X^\mu (\sigma) \partial_b X^\nu(\sigma), </math>
|-
|
|<math> = {T \over 2}\int \mathrm{d}^2 \sigma  \sqrt{-h} \left( \Lambda \Lambda^{-1} \right) h^{ab} g_{\mu \nu} (X) \partial_a X^\mu (\sigma) \partial_b X^\nu(\sigma) = \mathcal{S}. </math>
|}
And one can see that the action is invariant under [[Weyl transformation]]. If we consider ''n''-dimensional (spatially) extended objects whose action is proportional to their worldsheet area/hyperarea, unless ''n'' = 1, the corresponding Polyakov action would contain another term breaking Weyl symmetry.

One can define the [[stress–energy tensor]]:
: <math> T^{ab} = \frac{-2}{\sqrt{-h}} \frac{\delta S}{\delta h_{ab}}. </math>
Let's define:
: <math> \hat{h}_{ab} = \exp\left(\phi(\sigma)\right) h_{ab}. </math>
Because of [[Weyl symmetry]], the action does not depend on <math> \phi </math>:
: <math> \frac{\delta S}{\delta \phi} =  \frac{\delta S}{\delta \hat{h}_{ab}} \frac{\delta \hat{h}_{ab}}{\delta \phi} = -\frac12 \sqrt{-h} \,T_{ab}\, e^{\phi}\, h^{ab} = -\frac12 \sqrt{-h} \,T^a_{\ a} \,e^{\phi} = 0 \Rightarrow T^{a}_{\ a} = 0, </math>
where we've used the [[functional derivative]] chain rule.