0
$\begingroup$

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.

Improve this question
$\endgroup$
2
  • $\begingroup$ 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 ♦
    Commented Jan 16, 2017 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
    Commented Jan 16, 2017 at 22:07

1 Answer 1

Reset to default
0
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.