Starting from the Ricci identity

$$V_{a ; b ; c}-V_{a ; c ; b}=V_{e} R_{a b c}^{e}$$

give an expression for the curvature tensor $R_{a b c}^{e}$ of the Levi-Civita connection in terms of the Christoffel symbols and their partial derivatives. Using local inertial coordinates, or otherwise, establish that

$$R_{a b c}^{e}+R_{b c a}^{e}+R_{c a b}^{e}=0 .$$

A vector field with components $V^{a}$ satisfies

$$V_{a ; b}+V_{b ; a}=0$$

Show, using equation $(*)$ that

$$V_{a ; b ; c}=V_{e} R_{c b a}^{e}$$

and hence that

$$V_{a ; b} ; b+R_{a}{ }^{c} V_{c}=0,$$

where $R_{a b}$ is the Ricci tensor. Show that equation $(* *)$ may be written as

$$\left(\partial_{c} g_{a b}\right) V^{c}+g_{c b} \partial_{a} V^{c}+g_{a c} \partial_{b} V^{c}=0$$

If the metric is taken to be the Schwarzschild metric

$$d s^{2}=-\left(1-\frac{2 M}{r}\right) d t^{2}+\left(1-\frac{2 M}{r}\right)^{-1} d r^{2}+r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)$$

show that $V^{a}=\delta^{a}{ }_{0}$ is a solution of $(* * *)$. Calculate $V^{a} ; a$.

Electromagnetism can be described by a vector potential $A_{a}$ and a Maxwell field tensor $F_{a b}$ satisfying

$$F_{a b}=A_{b ; a}-A_{a ; b} \quad \text { and } \quad F_{a b} ; b=0 .$$

The divergence of $A_{a}$ is arbitrary and we may choose $A_{a} ; a=0$. With this choice show that in a general spacetime

$$A_{a ; b}{ }^{; b}-R_{a}{ }^{c} A_{c}=0 .$$

Hence show that in the Schwarzschild spacetime a tensor field whose only non-trivial components are $F_{t r}=-F_{r t}=Q / r^{2}$, where $Q$ is a constant, satisfies the field equations $(* * * *)$.