Given a smooth manifold $M$, a smooth vector field $V$, and some $x_0 \in M$, we can define the exponential map through the map $\phi$, where $\phi$ is defined on some open interval of $\mathbb{R}$, $\phi(t_0) = x_0$, and $x = \phi(t)$ satisfies the differential equation $$\frac{dx}{dt} = V(x),$$ and we let $\phi(t) = \exp(t\cdot V, x_0)$, so $\exp(v, x) = \phi(1)$.

This is taken from Theorem 3.1 of J-M Souriau's Structure of Dynamical Systems. At this point of the book, he hasn't introduced any concept of an affine connection, only assuming a Hausdorff manifold, but here we have somehow defined (and shown the existence of a unique) exponential map. Is there some natural affine connection he has assumed, or can we find the connection corresponding to this given definition of an exponential map? I saw this related question: Exponential maps depends on Riemannian metric?, but I'd like to pinpoint where exactly the metric or associated connection would show up.

I know that the exponential map can be defined in terms of geodesics, and I guess the ODE $dx/dt = V(x)$ is the geodesic equation, which can be solved componentwise with respect to some a suitable coordinate basis on the tangent bundle (e.g., Prop 6.1.2 here). But what metric or connection does this geodesic equation correspond to?

  • $\begingroup$ He should not have called this map exponential. What you need for exponential map is an affine connection, not a Riemannian metric. $\endgroup$ Commented Nov 27, 2020 at 3:26
  • $\begingroup$ The use of $\exp$ of here is more akin to the Lie exponential map then the Riemannian exponential map. Despite having the same name, these are two different objects, and the former doesn't require a metric. $\endgroup$
    – Kajelad
    Commented Nov 27, 2020 at 5:52
  • $\begingroup$ I have seen people denote the flow $\Phi_X$ of a vector field $X \in \mathfrak{X}(M)$ just by ${\rm e}^{tX}p = \Phi_X(t,p)$, just because the group property of the flow then reads ${\rm e}^{(t+s)X} = {\rm e}^{tX}{\rm e}^{sX}$, and one has ${\rm e}^0 = 1$, where $1$ is to be interpreted as the identity $M \to M$. I think this is a big stretch. Go figure. $\endgroup$
    – Ivo Terek
    Commented Nov 27, 2020 at 7:39

2 Answers 2


The author seems to refer to exponential map for any "time 1 of a flow". This is a sort of generalization of:

  • the exponential map of a Riemannian manifold, which is in fact the time 1 map of the geodesic flow (defined on the unit tangent bundle)
  • the exponential map of a Lie group, which is the time 1 map of a natural flow of left invariant vector fields

These notions are the same in some particular cases - when a Lie group is equipped with a bi-invariant metric. In general, they differ, but they share some common properties, for example, $\psi_t\psi_s = \psi_{t+s}$ and $\psi_0=\mathrm{id}$, which is a very-well known property of the complex exponential function $\exp : \mathbb{C}\to \mathbb{C}^*$.

There is no reason for a time 1 map of a flow to be induced by some geodesic flow of a particular riemannian metric, so there may be no connextion inducing this flow and giving sense to the term "exponential map" as understood as a riemannian notion.


He should not have called this map exponential. The most common name for it is the "time 1 flow of the vector field $V$."

Here is how this relates to the exponential map in differential geometry. What you need for exponential map is an affine connection $\nabla$ on the tangent bundle $TM$, not a Riemannian metric. (Although, a Riemannian metric defines the Levi-Civita connection that you can then use.) The connection $\nabla$ defines a certain vector field $W$ on $TM$. The time one flow $F$ of $W$ is closely related to the exponential map of $M$: Given a point $p\in M$ and an tangent vector $v\in T_pM$, consider the vector $W(v)\in T_vTM$. Then $F(W(v))=(p',v')$, where $p'\in M$ and $v'\in T_{p'}M$. Lastly, $\exp_p(v)=p'$.

My favorite reference for this is "Riemannian Geometry" by do Carmo.

  • $\begingroup$ Ah, so you can recover the exponential map from the time 1 flow of an appropriate vector field. In what way does $\nabla$ define the vector field $W$? $\endgroup$
    – gwtw14
    Commented Nov 27, 2020 at 22:58
  • $\begingroup$ @gwtw14 As I said, check do Carmo's book (discussion of the geodesic flow). When I have more time I might add an explanation. $\endgroup$ Commented Nov 27, 2020 at 23:24
  • $\begingroup$ Just took a look at Do Carmo. Just to confirm, is the vector field you're talking about the geodesic field? So the flow of any vector field is just defined by some ODE, and the exponential map is the flow of this geodesic field, which depends on the connection. $\endgroup$
    – gwtw14
    Commented Nov 28, 2020 at 0:06
  • 1
    $\begingroup$ @gwtw14: Right. A more conceptual construction of the geodesic vector field on $TM$ is as follows. Each affine connection determines a direct sum decomposition of each $T_vTM$ into "horizontal" and "vertical" components, where the horizontal component projects isomorphically to $T_pM$ ($v\in T_pM$). Thus, for each $v\in T_pM$ we obtain a unique horizontal vector $W(v)\in T_vTM$ which projects to $v$. This yields the geodesic vector field $W$ on $TM$. $\endgroup$ Commented Nov 28, 2020 at 3:01

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