It looks like the Laplace operator in the nonlinear sigma model (say the Polyakov action) is different from the Laplace-Beltrami operator, how can one get the Ricci flow as a low order approximation to the perturbative renormalization group flow?

The Lagrangian in the Polyakov action is the trace of the pullback metric $\phi^*g$, or induced metric $ \phi^*g_{\mu \nu}(x) = g_{ij}(\phi(x)) \partial_\mu \phi^i(x) \partial_\nu \phi^j(x) $ for $\phi: (\Sigma, \gamma) \rightarrow (M, g) $,

$S( g; \phi) = \frac{T}{2} \int_\Sigma dvol \,\gamma^{\mu \nu} \phi^*g_{\mu \nu} = - \frac{T}{2} \int_\Sigma dvol \, g_{ij} \phi^i \partial^\mu\partial_\mu \phi^j $

The usual notations are $\Delta = \partial^\mu \partial_\mu$ and $\Delta_g = Tr_g(- \Delta)$. But this $\Delta_g$ is obviously different from the
Laplace-Beltrami operator $\Delta_g = g^{ij}\partial_i \partial_j$ in the Ricci flow.


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.