I'm trying to show the following. Suppose $(M, g)$ is a $2$-dimensional Riemannian manifold with connection $\nabla$. Suppose also that $\nabla$ is metric compatible, and that length extremizing curves $\gamma$ for $g$ are geodesics in the sense that

$$\nabla_{\gamma '}\gamma ' \propto \gamma '$$

Then $\nabla$ is torsion-free. I've got stuck partway through the proof - does anyone know how to finish it off correctly? My working is below.

Let $p$ be an arbitrary point in $M$ and choose a vector field $T$ around $p$ such that $\{T, \nabla_T T\}$ forms a local frame for $M$. In other words

$$ \nabla_T T \not\propto T$$

We now immediately know that the integral curve $\gamma$ of $T$ through $p$ is not length extremizing. Thus there exists a variation $\tilde{\gamma}(t,s)$ of $\gamma$ and a vector field $N = \tilde{\gamma}_*(\partial/\partial s)$ such that

$$L'(s) = \int_{-\epsilon}^\epsilon N_{\tilde{\gamma}(t,s)}|T_{\tilde{\gamma}(t,s)}|dt$$

is nonzero at $s = 0$. Writing this out more explicitly we get

\begin{align} L'(s) &= \int |T|^{-1} g(\nabla_N T, T)dt\\ &= \int |T|^{-1} g (\nabla_T N,T) + Tors(N,T) dt \end{align}

But here I'm stuck. I have no idea how to use this information to prove that the torsion must be $0$. I think perhaps some kind of contradiction argument might work, but I can't quite see how to implement it. Any ideas?

  • $\begingroup$ I suspect you will be better served using the first variation along a minimizer instead. $\endgroup$ – Ted Shifrin May 13 '13 at 12:55
  • $\begingroup$ I have changed the geodesy tag to geodesic on this and some of your other questions. You might want to read meta.math.stackexchange.com/questions/11167/… $\endgroup$ – Henry Oct 6 '13 at 1:06

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