# Can the exponential map be used to define geodesics (and hence, generalisations of geodesics)?

Let $$(M,g)$$ be a (connected, paracompact, $$C^{\infty}$$-smooth) Riemannian manifold with Riemannian metric $$g$$. The exponential map is defined for each point $$p \in M$$ to be the map $$\exp_p : T_p M \to M$$ that sends a tangent vector $$v \in T_p M$$ to the endpoint of the unique geodesic $$\gamma$$ satisfying $$\gamma(0)=p$$ and $$\gamma'(0)=v$$.

In particular, the notion of a geodesic allows us to define an exponential map. My question concerns the reversal of this construction:

Suppose that for each $$p \in M$$, we have a map $$\sigma_p : T_p M \to M$$ with some desirable properties. Can one define a notion of $$\sigma$$-geodesic that coincides with notion of a geodesic when $$\sigma =\exp$$?

If this can be done, I suspect that it is well-known, but I've never come across such a construction.

• One way to generalize geodesics is replacing the Levi-Civita connection by any other (torsion-free if you want) connection on $TM$. There will also be a corresponding notion of exponential map. I suspect you might be able to extract a connection out of your family of maps $\sigma_p$ by pulling vector fields back along $\sigma_p$, differentiating at $0 \in T_ M$ and pushing forward along $\sigma_p$. Dec 2, 2022 at 21:40
• I don't know if this would be sufficient, but you could try something like the following, assuming $M$ is complete: 1) If $\gamma(t) = \exp_p(tv)$, then $\gamma'(0) = v$ 2) If $q = exp_p(v_p)$ and $v_q$ is the parallel transpoart of $v_p$ to $T_qM$ along the curve $t\mapsto\exp_p(tv)$, then $\exp_q(-v_q) = p$. 3) Given $p \in M$ there exists a neighborhood $B$ of $0 \in T_pM$ such that if $v \in B$, then $d(p,\exp_p(v)) = |v|$. Dec 2, 2022 at 22:51
• I think spray to search for in this context Dec 2, 2022 at 23:57

Exponential map in your definition is closely related to the smooth family of smooth curves smoothly depending on the position such that in every point in every direction there exists precisely one curve at this point in this direction. In literature, such a family is sometimes called path structure.

There are two differences between your structure and path structure: the first one is that you curves are parameterised, and the curves of path structure are not. But this an additional structure, so my answer below will has sense also for your setup.

The other difference is that for a path structure there exists precisely one, or, in the case of irreversible path structures, at most two curves from the family. You definition allows that (for three points A,B,C) the curves from A and B to C come to C with the same velocity vector which is not possible for geodesics. Or, even more wild, take 3 points A, B, C and consider the trajectory $$\gamma_{ABC}$$ of your exponential map which starts from A and then passes through first B and then C (assume such trajectory exists, i.e., choose B and C on a trajectory from A). Your definition does not require that the segment of $$\gamma_{ABC}$$ from $$B$$ to $$C$$ is a trajectory of your exponential map starting from $$B$$ and going to $$C$$, though this property may be essential for geodesics.

But, returning to path structures, most path structures do not come from an affine connection (known since at least Cartan for dim 2, essentially the same proof works for all dimensions, see https://arxiv.org/abs/1101.2069). In dimension 2, every reversible path structure is locally Finsler-metrisable https://arxiv.org/abs/1002.0243, the question whether every irreversible path structures is metrisable is still open.
In higher dimensions most path structures are, even microlocally, not Finsler-metrisable, see (the open access paper) https://link.springer.com/article/10.1007/s10714-022-03006-2 .

Let’s define $$\gamma:\mathbb{R}\to M$$ as a $$\sigma$$-geodesic iff for any $$a,b,c\in\mathbb{R}$$, there are vectors $$v,w$$ in the tangent space at $$\gamma(a)$$ with $$\sigma_{\gamma(a)}(v)=\gamma(b)$$ $$\sigma_{\gamma(a)}(w)=\gamma(c)$$ $$(c-a)v=(b-a)w$$

If $$\sigma=\exp$$, and $$\gamma$$ is a geodesic, this property clearly holds.

If $$\sigma=\exp$$, and this property holds for $$\gamma$$, then let $$P(a,k)$$ be the statement that $$d(\gamma(a),\gamma(b))=k|a-b|$$ whenever $$\gamma([a,b])$$ is entirely within the injectivity radius of $$\gamma(a)$$. Then there is some $$k$$ with $$P(0,k)$$, and for that $$k$$, the set $$\{a:P(a,k)\}$$ is both open and closed in $$\mathbb{R}$$. So in fact $$P(a,k)$$ holds for all $$a$$, $$d(\gamma(a),\gamma(b))=k|a-b|$$ for all sufficiently close $$a$$ and $$b$$, and therefore $$\gamma$$ is a geodesic.

• I ask you to comment on the following two questions: Does your definition of the distance satisfy the triangle inequality? Do the curves $\gamma$ are distance-minimising, and if yes, why? Thank you Dec 5, 2022 at 10:27
• I’ve only defined geodesics, I haven’t defined distances. And, yes, since the curves $\gamma$ with this property are geodesics they are locally distance-minimizing.
– user44143
Dec 5, 2022 at 11:31
• There is no way to define the metric from the data given here: $dx^2+dy^2$ and $dx^2+2dy^2$ are different metrics with the same $\exp$ and the same geodesics.
– user44143
Dec 5, 2022 at 13:48
• I agree with Matt F., there are example of metrics such that their geodesics are the same (as parameterized curves) but the metrics are different; in particular one can not use the parameterized geodesics to recover the metric in the general case. Dec 14, 2022 at 11:47