# Connections and Ricci identity

1. Mar 10, 2015

### CAF123

1. The problem statement, all variables and given/known data
Given $\nabla$ a torsionless connection, the Ricci identity for co-vectors is $$\nabla_a \nabla_b \lambda_c - \nabla_b \nabla_a \lambda_c = -R^d_{\,\,cab}\lambda_d.$$
Prove $R^a_{[bcd]} = 0$ by considering the co-vector field $\lambda_c = \nabla_c f$
2. Relevant equations
$$R^a_{[bcd]} = 0 = \frac{1}{3!} \left(R^a_{\,\,bcd} + R^a_{\,\,cdb} + R^a_{\,\,dbc} - R^a_{\,\,bdc} - R^a_{\,\,cbd} - R^a_{\,\,dcb}\right)$$

3. The attempt at a solution
Input the given form for the covector into the Ricci identity in the question. Then since $\nabla_c f = e_c(f),$ we have
$$\nabla_a \nabla_b e_c(f) - \nabla_b \nabla_a e_c(f) = -R^d_{\,\,cab}e_d(f).$$ True for all functions f, so $$\nabla_a \nabla_b e_c - \nabla_b \nabla_a e_c = -R^d_{\,\,cab}e_d.$$ Then since $\nabla_a e_b = \Gamma^d_{ba} e_d$ we can simplify the above to give $$\nabla_a \Gamma^d_{cb}e_d - \nabla_b \Gamma^d_{ca}e_d = -R^d_{\,\,cab}e_d$$ which can then be further rewritten like $$\nabla_a \Gamma^d_{cb} + \Gamma^{\alpha}_{cb}\Gamma^d_{\alpha a} - \nabla_b \Gamma^d_{ca} - \Gamma^{\alpha}_{ca}\Gamma^d_{\alpha b} = -R^d_{\,\,cab}.$$ I was then going to relabel all indices to get terms like that in the equation in 'Relevant Equations' and sum them all up and I hoped to get zero, but it is not. Have I made an error in the above somewhere? Thanks!

Last edited: Mar 10, 2015
2. Mar 12, 2015

### strangerep

I'm a bit rusty on this stuff, but,... since no one else has replied,...

Where did this problem come from? Is it from a textbook? Online notes? (If the latter, please provide a link.)

Since $R^a_{bcd}$ is already skewsymmetric in the last 2 indices, you can simplify the rhs down to 3 terms, which is a sum of cyclically permuted b,c,d, indices. In that form it's called the "first Bianchi identity". Proving that might be less work. Check out the associated formulas on Wikipedia.

Also, I don't understand how you went from your 2nd-last line to your last line.

3. Mar 13, 2015

### strangerep

I just realized... the correct method is sketched in Wald p39.

(I guess I'm rustier than I realized.)

Last edited: Mar 14, 2015