# Two linear connections have the same geodesics if their difference tensor is antisymmetric

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.

• 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. Dec 21, 2021 at 10:35
• 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 $D_p$. Nov 14, 2023 at 18:46

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 $$\gamma\vert_J$$.