Uniqueness of affine connections 
This is a problem from Carmelli book on general relativity. 
the conceptual problem is, given a spacetime, and hence a metric, can there exist more than one affine connection for which one can take the difference and show that the difference is a tensor? As the affine connection is defined using the metric, given the metric, this should be unique, what I think.
 A: The standard way to obtain the coefficients of a metric compatible connection goes as follows.  First, we demand that 
$$\nabla_\mu g_{ab} = 0$$
for all $\mu,a,b$.  Plugging in the connection coefficients gives us that
$$\nabla_\mu g_{ab} = \partial_\mu g_{ab} - \Gamma^i_{a\mu} g_{ib} - \Gamma^i_{b\mu} g_{ai}=0$$
Next we permute the indices to get two additional equations:
$$\nabla_b g_{\mu a} = \partial_b g_{\mu a} - \Gamma^i_{\mu b} g_{ia} - \Gamma^i_{ab} g_{\mu i}=0$$
$$\nabla_a g_{b\mu} = \partial_a g_{b\mu} - \Gamma^i_{ba} g_{i\mu} - \Gamma^i_{\mu a} g_{bi}=0$$
For a given choice of $(\mu,a,b)$, we have three equations here with six unknowns (the connection coefficients).  Nevertheless, we can forge ahead.  Adding the second and the third equation together and subtracting the first yields
$$\big(\partial_a g_{b\mu} + \partial_b g_{\mu a} - \partial_\mu g_{ab}\big) + 2\Gamma^i_{[a\mu]}g_{bi} + 2\Gamma^i_{[b\mu]}g_{ai} - 2\Gamma^i_{(ab)}g_{\mu i} = 0$$
where $\Gamma^i_{[ab]} \equiv \frac{1}{2}\left(\Gamma^i_{ab} - \Gamma^i_{ba}\right)$ and $\Gamma^i_{(ab)} \equiv \frac{1}{2}\left(\Gamma^i_{ab} + \Gamma^i_{ba}\right)$.
This can be rearranged to yield (utilizing the symmetry of $g$)
$$\Gamma^i_{(ab)} = \frac{1}{2}g^{i\mu}\big(\partial_a g_{b\mu} + \partial_b g_{\mu a} - \partial_\mu g_{ab}\big) + \Gamma^i_{[a\mu]} + \Gamma^i_{[b\mu]}$$
This is all that can be obtained by demanding that the connection be metric compatible.  We need more input to uniquely obtain the connection.  If we demand that the connection be torsion-free (i.e. symmetric in its lower two indices), then the two connection terms on the right vanish and the left hand side simply becomes $\Gamma^\mu_{ab}$, and so
$$\Gamma^i_{ab} = \frac{1}{2}g^{i\mu}\big(\partial_a g_{b\mu} + \partial_b g_{\mu a} - \partial_\mu g_{ab}\big)$$
uniquely determines the connection coefficients from the metric.  This is the Levi-Civita connection.  Any other metric-compatible connection can have torsion in the sense that a connection $\Gamma$ can be written  
$$\Gamma^i_{jk} = \overline{\Gamma}^i_{jk} + K^i_{jk}$$
where $\overline{\Gamma}$ is the Levi-Civita connection and $K$ is the so-called contorsion tensor.
