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Does the covariant derivative of the metric metric always vanish? I.e. $$\nabla_a g_{bc}=0$$ Are there situations where this can be assumed to not hold? For instance in case of an asymmetric metric?

Danu
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eeqesri
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1 Answers1

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Asymmetric metrics don't particularily make sense, because a metric exists for distance calculation as $ds^2=g_{\mu\nu}dx^\mu dx^\nu$, which is a symmetric expression, thus if we had instead of $g$ a tensor $h_{\mu\nu}=g_{\mu\nu}+a_{\mu\nu}$ with $g$ being symmetric and $a$ being antisymmetric, then the antisymmetric part would just cancel.

Are there situations where this can be assumed to not hold?

Yes. A linear connection ($\nabla$) is technically a distinct, unrelated object to a metric tensor $g$. If a linear connection satisfies $\nabla_\sigma g_{\mu\nu}=0$, then we say that $\nabla$ is metric compatible.

The fundamental theorem of Riemannian geometry says that a metric compatible symmetric connection is unique, and is given by $$ \Gamma^\sigma_{\mu\nu}=\frac{1}{2}g^{\sigma\lambda}(\partial_\mu g_{\nu\lambda}+\partial_\nu g_{\mu\lambda}-\partial_\lambda g_{\mu\nu}). $$

So that we use this unique connection, it is a choice on our part. Why do we make this choice?

  • Flat space/spacetime: In flat spacetime, we have a natural connection, one that in a cartesian frame is given by the partial derivatives $\partial_\mu$. Since the metric in these coordinates is given by $\eta_{\mu\nu}$ (or $\delta_{\mu\nu}$), and these objects have constant coefficients, we have $\partial_\sigma\eta_{\mu\nu}=0$, and the connection is manifestly symmetric. These are coordinate-independent properties, so they also hold in whacky curvy coordinates, where the $\Gamma$ coefficients are not identically zero. So our "God given" connection is precisely the unique compatible, symmetric connection.

  • General relativity: We would be execused if we used this connection from the get-go, because it matches the properties of the flat-space connection, and because $\nabla g=0$ is a very natural assumption (it means parallel transport does not change the length of vectors). We assume symmetricity on top of that, because it can be shown that the equivalence principle is valid only then. $$ $$ However, if we keep the Einstein-Hilbert action (from which the equations of GR come) as it is, but assume that the connection is symmetric, but not metric compatible, we get the same equations of motion, along with the compatibility condition, so it seems mathematics enforces the compatibility on us. This is called Palatini-formalism. If we also allows for torsion (nonsymmetric connection), we get further equations, but as it turns out, this will only affect fermionic fields (Einstein-Cartan theory). So it turns out, letting the connection be more general doesn't change much.

Bence Racskó
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