While I was reading Operators, Geometry and Quanta: Methods of Spectral Geometry in Quantum Field Theory I encountered a statement which seems to me not-so-correct. First of all they define the differential operator on a manifold $\mathcal{M}$ this way:

$$ L^\mathrm{non-cov}=-(g^{\mu\nu}(x)\partial_\mu\partial_\nu + a^\mu(x)\partial_\mu + b(x)), $$

where $a^\mu$ and $b$ are matrix valued. Then they want to rewrite $L$ in a covariant form and he claim that

$$ L^\mathrm{comv} =-(g^{\mu\nu}(x)\nabla_\mu\nabla_\nu+E) $$


$$\begin{align} \nabla_\mu &= \nabla^\mathrm{R}_\mu+\omega_\mu\\ \omega_\mu &=\frac{1}{2}g_{\mu\nu}(a^\nu + g^{\alpha\beta}\Gamma_{\alpha\beta}^\nu) \\ E &= b -g^{\mu\nu}(\partial_\mu\omega_\nu+\omega_\mu\omega_\nu-\omega_\alpha\Gamma_{\mu\nu}^\alpha) \end{align}$$

where $\nabla^\mathrm{R}$ is theusual covariant derivative defined in terms of the Christoffel symbols $\Gamma$. They also call $\omega$ a "gauge" connection (with quote) and they explain after why they do this. But here is irrelevant. The point is that $\omega$ doesn't seem to me to be a vector since it contains a Christoffel symbol which is not antisymmetrized. However it seems that in order to prove that $L^\mathrm{non-cov}=L^\mathrm{com}$ one needs to use

$$ \nabla^\mathrm{R}_\mu\omega_\nu= \partial_\mu\omega_\nu-\Gamma_{\mu\nu}^\lambda\omega_\lambda. $$

but the latter equation doesn't make sense since $\omega$ is not a vector. In fact let's consider a scalar function $\phi$ on $\mathcal{M}$, we have:

$$ g^{\mu\nu}\nabla_\mu\nabla_\nu\phi=g^{\mu\nu}(\nabla^\mathrm{R}_\mu+\omega_\mu)(\partial_\nu+\omega_\nu)\phi= \nabla^\mathrm{R}_\mu\nabla_\nu\phi+(\nabla^\mathrm{R}_\mu\omega_\nu)\phi+\omega_\nu\partial_\nu\phi+\omega_\nu\partial_\mu\phi+\omega_\mu\omega_\nu\phi $$

then if one uses the incriminated formula gets the result. My question is: why am I allowed to use

$$ \nabla^\mathrm{R}_\mu\omega_\nu= \partial_\mu\omega_\nu-\Gamma_{\mu\nu}^\lambda\omega_\lambda. $$

if $\omega$ is not a vector?

  • 1
    $\begingroup$ I don't know how much we can say without having the book, though a possible answer would be that $a^\nu$ is not a vector. It could have a transformation law cancelling that of the Christoffel symbols. $\endgroup$ – Javier Sep 27 '16 at 0:51

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