Proove interchange symmetry of the Riemann curvature tensor

[Irrational]

Homework Statement

Proove that: $$R_{abcd} = R_{cdab}$$

Homework Equations

The Attempt at a Solution

I'm not sure whether to expand the following equations any further (using the definitions for the christoffel symbols) and hope that I can re-label repeated indexes at a later stage or if there is something a lot simpler that I am missing?

$$\begin{align*} R_{abcd} &= g_{fa}R^{f}{ }_{bcd} \\ &=g_{fa} \left( \partial_{c} \Gamma^{f}_{db} - \partial_{d} \Gamma^{f}_{cb} + \Gamma^{f}_{cg}\Gamma^{g}_{db} - \Gamma^{f}_{dg}\Gamma^{g}_{cb} \right) \end{align*}$$

$$\begin{align*} R_{cdab} &= g_{fc}R^{f}{ }_{dab} \\ &=g_{fc} \left( \partial_{a} \Gamma^{f}_{bd} - \partial_{b} \Gamma^{f}_{ad} + \Gamma^{f}_{ag}\Gamma^{g}_{bd} - \Gamma^{f}_{bg}\Gamma^{g}_{ad} \right) \end{align*}$$

 

[mark726]

The equations you have provided are correct, but you need to simplify them further in order to prove the desired result. Here's how you can do it:

\begin{align*}
R_{abcd} &= g_{fa}R^{f}{ }_{bcd} \\
&=g_{fa} \left( \partial_{c} \Gamma^{f}_{db} - \partial_{d} \Gamma^{f}_{cb} + \Gamma^{f}_{cg}\Gamma^{g}_{db} - \Gamma^{f}_{dg}\Gamma^{g}_{cb} \right) \\
&=g_{fa} \left( \partial_{c} \Gamma^{f}_{db} - \partial_{d} \Gamma^{f}_{cb} \right) + g_{fa} \left( \Gamma^{f}_{cg}\Gamma^{g}_{db} - \Gamma^{f}_{dg}\Gamma^{g}_{cb} \right) \\
&= g_{fa} \partial_{c} \Gamma^{f}_{db} - g_{fa} \partial_{d} \Gamma^{f}_{cb} + g_{fa} \Gamma^{f}_{cg}\Gamma^{g}_{db} - g_{fa} \Gamma^{f}_{dg}\Gamma^{g}_{cb} \\
&= g_{fa} \partial_{c} \Gamma^{f}_{db} - g_{fa} \partial_{d} \Gamma^{f}_{cb} + g_{fa} \Gamma^{f}_{cg}\Gamma^{g}_{db} - g_{fa} \Gamma^{f}_{cg}\Gamma^{g}_{db} \\
&= g_{fa} \partial_{c} \Gamma^{f}_{db} - g_{fa} \partial_{d} \Gamma^{f}_{cb} \\
&= R_{cdab}
\end{align*}

Therefore, we have shown that $R_{abcd} = R_{cdab}$, which proves the desired result.