joneall said:
You are very careful to keep the mathematical statements correct. Perhaps you could indicate to me a book for learning that myself. I'd be grateful. Is Schutz''s "Geometrical methods of mathematical physics" good, for instance? Or is it not necessary for what I am trying to understand?
I think the modern abstract formalism is not necessary for a first encounter with GR. I'm myself not an expert in GR (rather in theoretical high-energy hadron/nuclear physics), and I've learned GR out of interest for myself, using Landau/Lifshitz vol. II, which uses the Ricci calculus of general-covariant tensor analysis.
The idea is to start with a definition of a covariant derivative of vector fields ##V^{\mu}(q)##, which leads to a 2nd-rank tensor field ##\nabla_{\nu} V^{\mu}##. We work in holonomous bases and co-bases. Thus one would like to have a transformation behavior under general transformations of the coordinates (diffeomorphism)
$$\nabla_{\nu}' V^{\prime \mu}=\frac{\partial q^{\rho}}{\partial q^{\prime \nu}} \frac{\partial q^{\prime \mu}}{\partial q_{\sigma}} \nabla_{\rho} V^{\sigma}.$$
The partial derivative obviously doesn't fulfill this requirement since
$$\partial_{\nu}' V^{\prime \mu}=\frac{\partial q^{\rho}}{\partial q^{\prime \nu}} \partial_{\rho} \left (\frac{\partial q^{\prime \mu}}{\partial q^{\sigma}} V^{\sigma} \right),$$
i.e., you get an extra term
$$\partial_{\nu}' V^{\prime \mu}=\frac{\partial q^{\rho}}{\partial q^{\prime \nu}} \frac{\partial q^{\prime \mu}}{\partial q^{\sigma}} \partial_{\rho} V^{\sigma} +\frac{\partial q^{\rho}}{\partial q^{\prime \nu}} \frac{\partial^2 q^{\prime \mu}}{\partial q^{\sigma} \partial q^{\rho}} V^{\sigma}.$$
Thus we need a connection, symbolized by Christoffel symbols
$$\nabla_{\nu} V^{\mu}=\partial_{\nu} V^{\mu} + {\Gamma^{\mu}}_{\nu \rho} V^{\rho},$$
where the Christoffel symbol is not a tensor but transforms under general coordinate transformations such as to cancel the "bad term" from the partial derivative:
$${\Gamma^{\prime \alpha}}_{\beta \gamma}=\frac{\partial q^{\prime \alpha}}{\partial q^{\mu}} \frac{\partial q^{\nu}}{\partial q^{\prime \beta}} \frac{\partial q^{\rho}}{\partial q^{\prime \gamma}} {\Gamma^{\mu}}_{\nu \rho} - \frac{\partial q^{\prime \nu}}{\partial q^{\beta}} \frac{\partial q^{\prime \rho}}{\partial q^{\gamma}} \frac{\partial^2 q^{\prime \alpha}}{\partial q^{\prime \nu} q^{\prime \rho}}.$$
Thus the Christoffel symbols are not a tensor but transform such that the "bad term" for the transformation of the partial derivative is cancelled.
The choice of the Christoffel symbols seem to be quite arbitrary, and without further constraints they indeed are. The only thing we can say is that the socalled torsion,
$${\Sigma^{\mu}}_{\nu \rho}={\Gamma^{\mu}}_{\nu \rho}-{\Gamma^{\mu}}_{\rho \nu}$$
is a tensor since then the additional term in the transformation of the Christoffel symbol cancels, because the partial derivatives commute.
You can specify the connection further by assuming that for a scalar field the covariant derivative should be the partial derivative, leading to a vector. Also for the partial derivative the product rule should hold, i.e.,
$$\nabla_{\mu} S=\partial_{\mu} S, \quad \nabla_{\rho}(V^{\mu} W^{\nu})=(\nabla_{\rho} V^{\mu})W^{\nu} + V^{\mu} \nabla_{\rho} W^{\nu}.$$
Applying this to the scalar ##V^{\mu} W_{\mu}## one deduces the rule for covariant vector components
$$\nabla_{\nu} W_{\mu}=\partial_{\nu} W_{\mu} - {\Gamma^{\rho}}_{\nu \mu} W_{\rho}.$$
The parallel transport of a vector along a curve is defined by
$$\mathrm{D}_{\tau} V^{\mu} = \mathrm{d}_{\tau} V^{\mu} + \mathrm{d}_{\tau} q^{\nu} {\Gamma^{\mu}}_{\nu \rho} V^{\rho}=\mathrm{d}_{\tau} q^{\nu} \nabla_{\nu} V^{\mu}.$$
Finally since we have a metric we'd also like to have the scalar products of two vectors invariant under parallel transport, leading to
$$\nabla_{\rho} g_{\mu \nu}=0.$$
This also ensures that indices can be lowered and raised for covariant derivatives as usual, i.e.,
$$\nabla_{\mu} V_{\rho}=g_{\rho \sigma} \nabla_{\mu} V^{\sigma}.$$
If we further assume that the torsion tensor vanishes identically this leads to the unique definition of the then symmetric Christoffel symbols as derivatives of the metric:
$${\Gamma^{\mu}}_{\nu \rho}=\frac{1}{2} g^{\mu \sigma}(\partial_{\nu} g_{\sigma \rho} + \partial_{\rho} g_{\sigma \nu}-\partial_{\sigma} g_{\nu \rho}).$$