pmb_phy
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Hurkyl said:Parallel transport respecting a metric is supposed to preserve local geometry -- i.e. it's supposed to preserve the metric. I.e. if \tau denotes parallel transport along some curve, we're supposed to have \langle \tau(v), \tau(w) \rangle = \langle v, w \rangle.
For the circle, any metric can be described by an everywhere nonzero periodic scalar function g, and the inner product by
\langle X, Y \rangle = g \cdot X \cdot Y
(again, ordinary function multiplication)
The metric compatability condition can be expressed locally by
\nabla_X g = 0
for any vector field X, which translates here to
X g' + X g = 0.
This implies g has the form g(t) = K e^{-t} -- but this cannot be both periodic and nonzero. Therefore, there does not exist any metric compatable with this connection.
If you're uncomfortable with my assertion that the derivative of the metric is expressed by the derivative of g, we have the alternative differential formulation of metric compatability:
X \langle Y, Z \rangle = \langle \nabla_X Y, Z\rangle + \langle Y, \nabla_X Z \rangle
which you can check again leads to a contradiction.
Please be patient with me since I'm not well versed in the finer details of differential geometry and tensor calculus. Especially the notation that you're using. Then again I'm here to learn! :)
Pete
ps - Why is my smiley face function, as well as all the format functions, not working for me?