Torsion tensor symmetrisation Given that the affine connection can be written as: $$\Gamma ^a{}_{bc}= \bigg\{ {a \atop bc} \bigg\} - \frac{1}{2}(T^a{}_{bc}+T_c{}^a{}_b-T_{bc}{}^a) $$
Where $\big\{ {a \atop bc} \big\}$ denotes the metric connection, given by: $$\bigg\{ {a \atop bc} \bigg\}=\frac{1}{2}g^{ad}(\partial_bg_{dc}+\partial_cg_{bd}-\partial_dg_{bc})$$
And the $T^\alpha{}_{\beta \gamma}$ are the torsion tensors, defined as $T^\alpha{}_{\beta \gamma}=\Gamma^\alpha{}_{\beta \gamma}-\Gamma^\alpha{}_{\gamma \beta}$.
Now, defining an index symmetrisation operation as $\Gamma^a{}_{(bc)}\equiv\frac{1}{2}(\Gamma^a{}_{bc}+\Gamma^a{}_{cb})$, I need to prove that: $$\Gamma ^a{}_{(bc)}= \bigg\{ {a \atop bc} \bigg\} + T_{(bc)}{}^a$$
I've tried proving it by just plugging in the symmetrisation operation the expressions for both the affine connections, and it's clear that the metric connection terms are by definition equal (assuming symmetry in the metric functions), so they just sum together. Instead, the part involving the torsion tensors really puzzle me, cause it seems by definition that they are antisymmetric, so they should just cancel out. I'm guessing it has something to do with the raised and lowered indexes that only allows the one with the rightmost index raised to not cancel... But I'm completely stuck. Any help will be much appreciated!
 A: So here's one way of doing it, probably a little clumsy but trying to be as explicit as possible. Also note that I use both symmetry and antisymmetry brackets at the same time, so it's just about being careful with both. Stating with
$$
\Gamma^{a}{}_{bc} = \big\{{}^a_{bc}\big\} - \frac{1}{2} \big( T^{a}{}_{bc} + T_{c}{}^{a}{}_b - T_{bc}{}^{a} \big) \ ,
$$
we can write this in terms of the connection (using antisymmetry brackets) as
$$
\Gamma^{a}{}_{bc} = \big\{{}^a_{bc}\big\} - \frac{1}{2} \big(2\Gamma^{a}{}_{[bc]} +2 g_{cd} g^{ae} \Gamma^{d}{}_{[eb]} - 2 g_{bd}g^{ae} \Gamma^{d}{}_{[ce]} \big) \ .
$$
Then symmetrising one finds
$$ \begin{align}
\Gamma^{a}{}_{(bc)} &=  \big\{{}^a_{bc}\big\} - \frac{1}{2} \Big(0+2g_{(c|d} g^{ae}\Gamma^{d}{}_{[e|b)]}+2g_{(b|d} g^{ae}\Gamma^{d}{}_{[e|c)]}  \Big) \\
&= \big\{{}^a_{bc}\big\} - \frac{1}{2} g^{ae} \big( g_{(c|d}\Gamma^{d}{}_{e|b)}-g_{(c|d}\Gamma^{d}{}_{b)e}+g_{(b|d}\Gamma^{d}{}_{e|c)}+g_{(b|d}\Gamma^{d}{}_{c)e}\big) \\
&= \big\{{}^a_{bc}\big\} - \frac{1}{2} g^{ae} \big(2g_{(c|d}\Gamma^{d}{}_{e|b)} -2g_{(c|d}\Gamma^{d}{}_{b)e}\big) \\
&=  \big\{{}^a_{bc}\big\}  + g^{ae} g_{(c|d}\Gamma^{d} {}_{b)e} - g^{ae}g_{(c|d}\Gamma^{d}{}_{e|b)} \\
&= \big\{{}^a_{bc}\big\} +g^{ae} g_{d(c} T^{d}{}_{b)e} \\
&=  \big\{{}^a_{bc}\big\} + T_{(bc)}{}^{a} \ \ ,
\end{align} 
$$
which gives the desired outcome. The first term on the first line vanishes from $[(ab)]=0$, and the Christoffel symbol of course stays the same. The rest of the lines are just manipulating indices whilst trying to stay clear and explicit. Hope this helps.
