homework and exercises - Contracting Indices

Does anyone know how to get from (1) to (2) in the system

$$ \begin{align} \mathrm{g}^{\mu\nu}_{,\rho}+ \mathrm{g}^{\sigma\nu}{{\Gamma}}^{\mu}_{\sigma\rho}+ \mathrm{g}^{\mu\sigma}{{\Gamma}}^{\nu}_{\rho\sigma} -\frac{1}{2}( {{\Gamma}}^{\sigma}_{\rho\sigma}+{{\Gamma}}^{\sigma}_{\sigma\rho} ) \mathrm{g}^{\mu\nu} &=0, \tag1 \\ \mathrm{g}^{[\mu\nu]}_{,\nu} +\frac{1}{2}( {{\Gamma}}^{\rho}_{\rho\nu}-{{\Gamma}}^{\rho}_{\nu\rho} ) \mathrm{g}^{(\mu\nu)} &=0, \tag2 \end{align} $$

by contracting equation (1) once with respect to ($\mu,\rho$), then with respect to ($\nu,\rho$)?

Where $\Gamma$ is not symmetric with respect to the lower Indices.

My attempt so far to solve this problem is: Well, when contracting with respect to μ and ρ I get: $$ -\frac{1}{2} g^{\rho \nu } {\Gamma} _{a\rho}^{a}-\frac{1}{2} g^{\rho \nu} \Gamma _{\rho a}^{a}+g^{a\nu} \Gamma _{a\rho}^{\rho}+g^{\rho a} \Gamma _{\rho a}^{\nu}+g_{,\rho}^{\rho\nu}=0 $$ and when contracting with respect to nu and ρ I get: $$ -\frac{1}{2} g^{\mu \rho } \Gamma _{a\rho}^a-\frac{1}{2} g^{\mu \rho} \Gamma _{\rho a}^a+g_{}^{a\rho } \Gamma _{a\rho}^{\mu }+g^{\mu a} \Gamma _{\rho a}^{\rho }+g_{,\rho }^{\mu \rho }=0 $$ when subtracting these two equations I get: $$ \frac{1}{2} g_{}^{\mu \rho } \Gamma _{a\rho }^a-\frac{1}{2} g_{}^{\rho \nu } \Gamma _{a\rho }^a+\frac{1}{2} g_{}^{\mu \rho } \Gamma _{\rho a}^a-\frac{1}{2} g_{}^{\rho \nu } \Gamma _{\rho a}^a+g_{}^{a\nu } \Gamma _{a\rho }^{\rho }-g_{}^{a\rho } \Gamma _{a\rho }^{\mu }-g_{}^{\mu a} \Gamma _{\rho a}^{\rho }+g_{}^{\rho a} \Gamma _{\rho a}^{\nu }-g_{,\rho }^{\mu \rho }+g_{,\rho }^{\rho \nu }=0 $$ I cant see how this is equal to equation (2)

Answer

The attempt of obtaining

$$ g^{[\mu\nu]}_{,\nu} +\frac{1}{2}( {{\Gamma}}^{\rho}_{\rho\nu}-{{\Gamma}}^{\rho}_{\nu\rho} ) g^{(\mu\nu)} =0, $$ was almost right! The only thing missing was a little care with relabeling indices. We will proceed in three main steps.

1.) So when contracting Eq. (1) with respect to $\mu$ and $\rho$, we get the identity:

$$ -\frac{1}{2} g^{\rho \nu } {\Gamma} _{a\rho}^{a}-\frac{1}{2} g^{\rho \nu} \Gamma _{\rho a}^{a}+g^{a\nu} \Gamma _{a\rho}^{\rho}+g^{\rho a} \Gamma _{\rho a}^{\nu}+g_{,\rho}^{\rho\nu}=0. $$

Now by relabeling the dummy indices in the 3rd term as $ a \leftrightarrow \rho$, we get that the 3rd term can be written as $g^{\rho \nu} \Gamma _{\rho a}^{a}$. Moreover, we can see that now the 2nd and 3rd term can be simplified: adding them together gives $+ \tfrac{1}{2}g^{\rho \nu} \Gamma _{\rho a}^{a}$. By a final index label change $\nu \to \mu$, we get that:

$$ -\frac{1}{2} g^{\rho \mu } {\Gamma} _{a\rho}^{a}+\frac{1}{2} g^{\rho \mu} \Gamma _{\rho a}^{a}+g^{\rho a} \Gamma _{\rho a}^{\mu}+g_{,\rho}^{\rho\mu}=0. \; \; \; \; (A) $$

2.) When contracting Eq. (1) with respect to $\nu$ and $\rho$, we get the identity:

$$ -\frac{1}{2} g^{\mu \rho } \Gamma _{a\rho}^a-\frac{1}{2} g^{\mu \rho} \Gamma _{\rho a}^a+g_{}^{a\rho } \Gamma _{a\rho}^{\mu }+g^{\mu a} \Gamma _{\rho a}^{\rho }+g_{,\rho }^{\mu \rho }=0. $$

Let us also rename the dummy indices in the 4th term as $ a \leftrightarrow \rho$. We can see now that the 4th term is simply $g^{\mu \rho } \Gamma _{a\rho}^a$, and the 1st and 4th term hence together give $\frac{1}{2} g^{\mu \rho } \Gamma _{a\rho}^a$. Moreover, let us perform also the $ a \leftrightarrow \rho$ "dummy index relabeling", yielding $g_{}^{\rho a} \Gamma _{\rho a}^{\mu }$ for the 3rd term. After these manipulations our identity reads as

$$ \frac{1}{2} g^{\mu \rho } \Gamma _{a\rho}^a-\frac{1}{2} g^{\mu \rho} \Gamma _{\rho a}^a+g_{}^{\rho a} \Gamma _{\rho a}^{\mu }+g_{,\rho }^{\mu \rho }=0. \; \; \; \; (B) $$

3.) Now taking (B)-(A), we obtain:

$$ g_{,\rho }^{\mu \rho } -g_{,\rho}^{\rho\mu} + \frac{1}{2}\left( g^{\mu \rho } + g^{\mu \rho }\right) \left( \Gamma _{a\rho}^a - \Gamma _{\rho a}^a \right)=0, $$ which after the $a \to \rho$ and $\rho \to \nu$ relabeling is exactly the same as the desired Eq. (2).