In many String Theory texts (e.g. Polchinski), when discussing the bosonic string in presence of background fields $G_{\mu\nu}$, $B_{\mu\nu}$ and $\phi$, respectively symmetric and antisymmetric tensor and scalar fields, three $beta$ coefficients are introduced ($\beta^G_{\mu\nu}$, $\beta^B_{\mu\nu}$, $\beta^\phi$).

The coefficients multiply terms corrispondent to the three background fields in the Weyl variation of the action. To guarantee Weyl invariance of the theory, they must be put equal to zero.

Then it is also commonly said that it can be shown that $\beta^G_{\mu\nu}=\beta^B_{\mu\nu}=0 \implies \nabla_\nu \beta^\phi = 0$.

I have some difficulties to prove the result, although elementary.

I will use here the same notation of the original paper of Callan et al. (https://doi.org/10.1016/0550-3213(85)90506-1).

The $\beta$ coefficients are

$\beta^G_{\mu\nu} = R_{\mu\nu}-\frac{1}{4}H_\mu^{\ \lambda\sigma}H_{\nu\lambda\sigma}+2\nabla_\mu\nabla_\nu\phi+O(\alpha^\prime)$,

$\beta^B_{\mu\nu}=\nabla_\lambda H^\lambda_{\ \mu\nu}-2(\nabla^\lambda \phi) H_{\lambda\mu\nu} + O(\alpha^\prime)$ and

$\beta^\phi = \frac{D-26}{48\pi^2}+\frac{\alpha^\prime}{16\pi^2}(4(\nabla \phi)^2 -4\nabla^2 \phi -R +\frac{1}{12} H^{\mu\nu\lambda}H_{\mu\nu\lambda})$,

where $R_{\mu\nu}$ is the Ricci tensor, $R=R^\mu_{\ \mu}$ and $H_{\mu\nu\lambda} = 3\nabla_{[\mu} B_{\nu\lambda]}$.

The Bianchi identies applied to $R_{\mu\nu}$ and $H_{\mu\nu\lambda}$ give respectively $\nabla^\mu R_{\mu\nu} = \frac{1}{2}\nabla_\nu R$ and $H_{\mu\nu\lambda}\nabla^\mu H_{\rho}^{\ \nu\lambda}=\frac{1}{6}\nabla_\rho (H_{\mu\nu\lambda}H^{\mu\nu\lambda})$.

In the mentioned paper, the main passage to prove that $\beta^\phi$ is constant, consists in showing that $0 = \nabla^\mu \beta^G_{\mu\nu} = \nabla^\mu \left(R_{\mu\nu} - \frac{1}{4} H_{\mu\rho\lambda}H_{\nu}^{\ \rho\lambda} + 2\nabla_\mu \nabla_\nu \phi \right) = \nabla_\nu \left(-2(\nabla\phi)^2 + 2\nabla^2\phi + \frac{1}{2}R - \frac{1}{24}H_{\mu\nu\lambda}H^{\mu\nu\lambda} \right)$.

Now, if I apply the Bianchi identities to the first two terms of $\nabla^\mu\beta^G_{\mu\nu}$, apparently the last equality holds if $R_{\mu\nu}\nabla^\mu \phi = 0$, which does not seem to me to hold in general. What am I missing?


1 Answer 1


The terms in $\nabla^\mu \beta^G_{\mu\nu} $ which are not related to the Bianchi identities you discussed are $$- \frac{1}{4} \nabla^\mu\left(H_{\mu\rho\lambda}\right)H_{\nu}^{\ \rho\lambda} + 2\nabla^2 \nabla_\nu \phi = -2\nabla^\mu \phi\left(R_{\mu\nu}+2\nabla_\mu\nabla_\nu \phi \right)+2\nabla^2 \nabla_\nu \phi $$ where the second equality follows from using both $\beta^B=\beta^G=0$.

As long as there is no torsion $\nabla_\mu\nabla_\nu\phi=\nabla_\nu\nabla_\mu\phi$, so the second term above can be written as $-2\nabla_\nu \left(\nabla\phi\right)^2$ which is one of the terms you are looking for. Although you didn't show it explicitly, it sounds like you did all this already and the remaining problematic terms are $$-2R_{\mu\nu}\nabla^\mu \phi+2\nabla^2 \nabla_\nu \phi$$

The second term looks like one of the terms you need but covariant derivatives acting on vectors don't commute, and they are related to the curvature tensor (in fact this is often how the curvature tensor is introduced). $$[\nabla_\mu,\nabla_\nu]\nabla^\mu \phi = R_{\mu\nu}\nabla^\mu\phi.$$ So the extra term involving the Ricci tensor is exactly what you need to be able to commute the $\nabla_\nu$ derivative out front so you can get $2\nabla_\nu \nabla^2\phi$.

  • $\begingroup$ Yes, you are right! I missed blindly that term… I must stop late night calculations. Thanks! $\endgroup$
    – Slz2718
    Dec 14, 2021 at 8:01

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