Equation with Riemann curvature tensor

[paweld]

Can anyone prove the following formula:
[tex]
R_{abf}^{\phantom{abf}e} \Gamma_{cd}^f = R_{abc}^{\phantom{abc}f} \Gamma_{fd}^e + R_{abd}^{\phantom{abd}f} \Gamma_{cf}^e
[/tex]
I found it in "General Relativity" by Wald (in slightly different notation).

 

[javierR]

If you write it out in terms of the derivatives of metric components, you should be able to manipulate the expression until you get the indices arranged as you want.

 

[Entropia]

Yes, this formula can be proven using the definition of the Riemann curvature tensor and the Christoffel symbols. The Riemann curvature tensor is defined as:

R_{abcd} = \frac{\partial \Gamma_{bd}^{\phantom{bd}e}}{\partial x^a} - \frac{\partial \Gamma_{ad}^{\phantom{ad}e}}{\partial x^b} + \Gamma_{ac}^{\phantom{ac}f} \Gamma_{fb}^{\phantom{fb}e} - \Gamma_{bc}^{\phantom{bc}f} \Gamma_{fa}^{\phantom{fa}e}

Using this definition, we can rewrite the given equation as:

R_{abf}^{\phantom{abf}e} \Gamma_{cd}^f = \frac{\partial \Gamma_{df}^{\phantom{df}e}}{\partial x^a} - \frac{\partial \Gamma_{af}^{\phantom{af}e}}{\partial x^b} + \Gamma_{ac}^{\phantom{ac}f} \Gamma_{fb}^{\phantom{fb}e} - \Gamma_{bc}^{\phantom{bc}f} \Gamma_{fa}^{\phantom{fa}e} \cdot \Gamma_{cd}^f

Using the product rule and rearranging terms, we get:

R_{abf}^{\phantom{abf}e} \Gamma_{cd}^f = \frac{\partial \Gamma_{df}^{\phantom{df}e}}{\partial x^a} \Gamma_{cd}^f - \frac{\partial \Gamma_{af}^{\phantom{af}e}}{\partial x^b} \Gamma_{cd}^f + \Gamma_{ac}^{\phantom{ac}f} \Gamma_{fb}^{\phantom{fb}e} \Gamma_{cd}^f - \Gamma_{bc}^{\phantom{bc}f} \Gamma_{fa}^{\phantom{fa}e} \Gamma_{cd}^f

Now, using the definition of the Christoffel symbols, we can rewrite the first two terms as:

\frac{\partial \Gamma_{df}