Contraction of the second Bianchi identity

Question

The second Bianchi identity is $${R^a}_{b[cd;e]}=0$$

And contracting it with respect to $a$ and $e$ we get $${R^a}_{b[cd;a]}=0 \Leftrightarrow $$ $${R^a}_{bcd;a}+R_{bc;d}-R_{bd;c}=0$$

What I don't understand is why the third term has the minus sign.

Answer 1

This is just a matter of recalling the (skew) symmetries of the Riemann curvature tensor. We have that \begin{align} 3! \cdot R^a_{\phantom{a}b[cd;a]} & = R^a_{\phantom{a}bcd;a} - R^a_{\phantom{a}bdc;a} + R^a_{\phantom{a}bda;c} - R^a_{\phantom{a}bca;d} + R^a_{\phantom{a}bac;d} - R^a_{\phantom{a}bad;c} \\ & = 2R^a_{\phantom{a}bcd;a} - 2R^a_{\phantom{a}bad;c} + 2R^a_{\phantom{a}bac;d} \tag{$\ast$} \\ & = 2R^a_{\phantom{a}bcd;a} - 2R_{bd;c} + 2R_{bc;d}. \tag{$\ast\ast$} \end{align} To get $(\ast)$ I used the skew symmetry of the Riemann curvature tensor in its last two indices three times: $$R^a_{\phantom{a}bcd} = - R^a_{\phantom{a}bdc}.$$ To get $(\ast\ast)$ I used the definition of the Ricci curvature: $$R_{bd} = R^a_{\phantom{a}bad}.$$ Note that we contract with the second index on the bottom to get the Ricci curvature.