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It's my first question! I hope I'm correctly formatting it.

I'm trying to prove that two connections $\nabla, \widetilde{\nabla}$ on a manifold determine the same geodesics iff their difference tensor is alternating. The difference tensor of two connections is the tensor $D(X,Y) = \widetilde{\nabla}_XY - \nabla_XY$, so this is saying that the geodesics are the same for the two connections iff $D(X,Y) = -D(Y,X)$ for all vector fields $X, Y$ on $M$.

Since $D$ is a tensor it can be written as a sum of its symmetric and alternating parts, $D = S + A$. Since $A(X,X) \equiv 0$ we have $D(X,X) = S(X,X)$. So if $S(X,X) = 0$ then so does $D(X,X)$, and if $D(X,X)$ is always $0$, then $$ D(X+Y,X+Y) =S(X+Y, X+Y) = S(X,X) + S(Y,Y) + 2S(X,Y) = 2S(X,Y)$$

Then $S(X,Y) \equiv 0$. So it's enough to check that having the same geodesics is the same as $D(X,X) \equiv 0$. This observation was made in Spivak Vol. II Ch. 6, Addendum 1, but his proof that this latter condition is equivalent to the connections having the same geodesics doesn't make sense to me.

Does anyone know how to show that if the connections have the same geodesics then the difference tensor is $0$ on the diagonal? That's the direction I'm stuck on.

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Assume that the tensor $D$ vanishes. Let $\alpha$ be a geodesic with respect to one of the connections, say $\nabla$. Then $$\nabla_TT\equiv 0$$where $T=\dot \alpha$. Using the identity $D\equiv 0$, we get $$\bar\nabla_TT=0$$ i.e. the connection $\bar\nabla$ has the same geodesic.
Conversely, we assume that $\nabla$ and $\bar\nabla$ have the same geodesics. Let $X$ be any vector field. We will prove that $D$ vanishes point-wise. Let $p\in M$ with nbd $U$ and let $\gamma$ be the geodesic with $\gamma(0)=p,\dot\gamma(0)=X_p$(ODE guarantees that there is a unique such geodesic), then $$D(X_p,X_p)=(\nabla_XX-\bar\nabla_XX)_p=0$$i.e. $D$ vanishes at $p$ and consequently at every point.

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  • $\begingroup$ I think your proof is assuming that $X$ agrees with $\gamma'$ on the image of $\gamma$ in a nbhd of $p$, but then we're not dealing with all $X$, only very special ones. $\endgroup$ – John Samples Dec 3 '16 at 0:04
  • $\begingroup$ Nevermind, I understand. It is using the tensoriality of $D$, i.e. the $C^\infty$-linearity in the second variable, to allow us to pick $X$ such that it's equal to $\gamma'$ on the image of $\gamma$. $\endgroup$ – John Samples Dec 3 '16 at 6:42

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