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Given two linear connections $\triangledown^a$ and $\triangledown^b$ on a manifold M, their assosciated difference tensor is $D(X,Y)=\triangledown^a_XY-\triangledown^b_XY$. I have confidently verified that this is a $(2,1)$-type tensor and that if it is antisymmetric, then $\triangledown^a$ and $\triangledown^b$ have the same geodesics. I am now trying to show the converse.


Proof Attempt:

Assume $\triangledown^a$ and $\triangledown^b$ have the same geodesics and let $D(-,-)$ be their difference tensor. We want to show that $D(X_p,Y_p)=-D(Y_p,X_p)$ for all vectors in $T_pM$; Equivalently, I will show that $D(X_p,X_p)=0$ for all $X_p \in T_pM$

Let $X_p$ be a vector in $T_pM$, then by hypothesis and the existense and uniqueness of geodesics, there exists a unique geodesic $\gamma$ for $\triangledown^a$ and $\triangledown^b$such that $\gamma(0)=p$ and $\gamma'(0)=X_p$, where $\gamma$ is defined on some neighborhood of $0$ in $\mathbb{R}$. Extend $X_p$ to a vector field on the image of $\gamma$ by $X_{\gamma(t)}=\gamma'(t)$.

Then we have that:

$D(X_p,X_p)=\triangledown^a_XX-\triangledown^b_YY=\triangledown^a_{\gamma'(0)}\gamma'(0)-\triangledown^b_{\gamma'(0)}\gamma'(0)=0+0=0$

Thus $D$ is antisymmetric.


Is this proof is okay? The part about extending $X_p$ to a vector field feels a bit sketchy for some reason.

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    $\begingroup$ Yes, it seems that to use the last equation, we have to extend $X_p$ in a neighbourhood of the image of the curve. But, here we have only managed to extend it to the image of the curve. $\endgroup$ Dec 21, 2021 at 10:35
  • $\begingroup$ You can use Problem 4-8 in Lee's Introduction to Riemannian Manifolds. If $X_p \neq 0$ then there is a connected neighborhood $J$ of $0$ in $I$ such that every smooth vector field along $\gamma\vert_J$ is extendible. So there is a smooth vector field $\tilde{X}$ defined on a neighborhood of $\gamma(J)$ such that $\tilde{X} \circ \gamma\vert_J = \gamma'\vert_J$. If $X_p = 0$, then $D(X, X)_p = D_p(X_p, X_p) = D_p(0, 0) = 0$ by linearity of $D_p$. $\endgroup$
    – Tob Ernack
    Nov 14, 2023 at 18:46

1 Answer 1

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You can use Problem 4-8 (a) in Lee's Introduction to Riemannian Manifolds.

Suppose $\gamma: I \to M$ is your geodesic, and $\gamma'(0) = X_p$.

If $X_p \neq 0$ then there is a connected neighborhood $J$ of $0$ in $I$ such that every smooth vector field along $\gamma\vert_J$ is extendible (this is Problem 4-8 (a) in Lee). So there is a smooth vector field $\tilde{X}$ defined on a neighborhood of $\gamma(J)$ such that $\tilde{X} \circ \gamma\vert_J = \gamma'\vert_J$. Then we have $$D(X, X)\vert_p = D_p(X_p, X_p) = D_p(\tilde{X}_p, \tilde{X}_p) = \nabla^a _{\gamma\vert_J'(0)} \tilde{X} - \nabla^b _{\gamma\vert_J'(0)} \tilde{X} = D_t^a \gamma'\vert_J(0) - D_t^b \gamma'\vert_J(0) = 0$$

If $X_p = 0$, then $D(X, X)\vert_p = D_p(X_p, X_p) = D_p(0, 0) = 0$ by the fact that $D$ acts pointwise and linearly.

So in either case, $D(X, X)\vert_p = 0$ and this holds for any $p \in M$.


Now let's prove Problem 4-8 (a) in Lee.

Let $\gamma: I \to M$ be a smooth curve and suppose $\gamma'(t_0) \neq 0$ for some $t_0 \in I$ (I'll assume that $t_0$ is not an endpoint, otherwise you have to extend $\gamma$ to a smooth curve defined on an open interval containing $I$).

By continuity, $\gamma'(t) \neq 0$ for all $t$ in some neighborhood $I' \subseteq I$ of $t_0$. We can also assume that $\gamma$ is injective on $I'$ by shrinking further if needed. Then $\gamma\vert_{I'}$ is an injective smooth immersion, so its image $\gamma(I')$ is an immersed submanifold in $M$. This means we can find a neighborhood $U$ of $\gamma(t_0)$ in $M$ such that $U \cap \gamma(I')$ is an embedded submanifold in $M$.

Now let $J$ be a connected neighborhood of $t_0$ in $I'$ such that $\gamma(J) \subseteq U \cap \gamma(I')$. Then $S = \gamma(J)$ is also an embedded submanifold of $M$. Note that $\gamma\vert_J: J \to S$ is a diffeomorphism.

Given any smooth vector field $V: J \to TM$ along $\gamma\vert_J$, we can define a smooth section of the restricted bundle $TM\vert_S$ by $\sigma = V \circ \gamma\vert_J^{-1}: S \to TM$. Since $S$ is embedded, the extension lemma for sections of restricted bundles (Lee, Introduction to Smooth Manifolds, Problem 10-9) shows that $\sigma$ admits an extension $\tilde{\sigma}$ on a neighborhood of $S$ in $M$. This means that $\tilde{\sigma}$ is a vector field defined on a neighborhood of $S$ in $M$ that agrees with the values of $V$ on $\gamma\vert_J$.

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