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I am confused about the affine connection in General Relativitiy.

I've heard it said that there is a degree of arbitrariness in the connection coefficients but that they can be uniquely specified by choosing $\nabla_\mu g_{\alpha\beta}=0$ and by additionally requiring that the metric is torsionless. I don't understand why texts state that the co-efficients are arbitrary without these two conditions.

For example, in polar co-ordinates, I can solve for the connection co-efficients without coming across any arbitrariness whatsoever. I don't knowlingly assert either of those two conditions - I just mechanically solve for the co-efficients.

\begin{align} \partial_a \mathbf{e_b}=\Gamma^c_{ab}\mathbf{e_c}\newline \partial_\theta \mathbf{e_r} = \partial_\theta(\cos\theta\mathbf{e_x} +\sin\theta\mathbf{e_y})=-\sin\theta\mathbf{e_x} + \cos\theta\mathbf{e_y}= \frac{1}{r}\mathbf{e_\theta} \newline \implies\Gamma^\theta_{\theta r}=1/r \end{align}

What gives? Was there any arbitrariness here? Did I implicitly make any assumptions? Would different (curved) geometries have introduced arbitrariness that's not present in flat geometries?

More confusing still, under Cartan's vector formalisation, where the coordinate induced basis vectors {$\mathbf{e_\mu}$} are identified with the set of partial derivatives {$\partial_\mu$}, I cannot see how any metric would ever give rise to torsion. Shouldn't $\Gamma^c_{ab}$ always be equal to $\Gamma^c_{ba}$ due to the commutativity laws of partial derivatives? I am not seeing any freedom to choose anything with these coefficients.

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  • $\begingroup$ I'm not sure what you're trying to say. The Christoffel symbols for a connection $\nabla$ are given by $\nabla_a e_b = \Gamma^c_{ab} e_c$ and you chose to compute them for the flat connection $\nabla = \partial$ on $\mathbb{R}^n$ in polar coordinates. $\endgroup$ – ACuriousMind Jan 16 '17 at 21:37
  • $\begingroup$ Numerous texts state that the connection coefficients have a degree of arbitrariness about them but that a unique connection can always be found by enforcing the two conditions I mentioned above. I'm trying to understand why the texts state there is any arbitrariness whatsoever in determining connection coefficients. I didn't see any arbitrariness when I differentiated the basis vectors in my example; I didn't knowingly make any assumptions. Is this because I used a flat space example or did I implicitly assume something? Would the connection coefficients be arbitrary in some other geometry? $\endgroup$ – Big AL Jan 16 '17 at 22:07
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Given a vector bundle $E\to M$, we just define a connection as a rule $$\nabla:\Gamma(E)\otimes\Gamma(TM)\to \Gamma(E), (\sigma,V)\mapsto \nabla_V\sigma$$satisfying three conditions: tensoriality in $V$, $\Bbb R$-linearity in $\sigma$, and a product rule. So if we just say "let $\nabla$ be a connection on $TM$," there is a lot of ambiguity! In fact, if $\nabla$ is a connection, then so is $\alpha \nabla$ for any $\alpha\in\Bbb R-\{0\}$. The fact that the requirements $T=0$ and $\nabla g=0$ force $\nabla$ to be the Levi-Civita connection uniquely is quite remarkable.

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