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What is the meaning of the exponential map of a covariant derivative on manifolds?

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    $\begingroup$ By a "covariant derivative" on a manifold, do you mean covariant derivative operator arising from choice of (pseudo-)Riemannian metric? That operator on "vector fields along parametric curves" can be expressed in terms of the Levi-Civita connection, and any connection on any vector bundle gives a "covariant derivative" operator on sections along parametric curves, which in turn uniquely determines the connection. So it is more natural to speak of an exponential map associated to a connection on the tangent bundle (such as L-C arising from a metric). Now go to Wikipedia for "exponential map". $\endgroup$
    – BCnrd
    Commented May 2, 2010 at 3:42

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I think what the questioner is getting at here is the relation between a connection (or covariant derivative) on the tangent bundle of a manifold and what is called a "geodesic spray" (which is a more convenient way of representing the "exponential map"). This is the subject of a very old paper by Ambrose, Singer, and myself (in 1960). Here is Kuranishi's Math Reviews article for that paper.

MR0126234 (23 #A3530) 53.45 (53.55) Ambrose,W.; Palais, R. S.; Singer, I. M. Sprays. An. Acad. Brasil. Ci. 32 1960 163–178

Let M be a $C^1$ manifold. When an affine connection of M is given, we can associate, for each tangent vector x at a point m in M, the geodesic $\alpha_x$ with tangent x at m. Conversely, we define a spray on M by saying that it is an assignment which gives, for each tangent vector x of M, a $C^1$ curve $\alpha_x$ in M satisfying certain conditions which are satisfied by the geodesics. A spray obtained by an affine connection is called a geodesic spray. The first theorem says that any spray is a geodesic spray and, moreover, we can prescribe the torsion form of the affine connection which gives the spray.

Let $M_m^k$ be the space of kth order tangent vectors of M at m. $M_m^k$ contains $M_m^1$. By an dissectionof M, we mean an assignment which gives, for each point m in M, a linear complementary space $M_m^c$ of $M_m^1$ in $M_m^2$ such that the assignment is of class $C^\infty$. Elements in $M_m^c$ are called pure. Clearly a dissection gives rise to a spray. Namely, we demand that the second-order tangent vectors of $\alpha_x$ are pure. The second theorem says that this correspondence of dissections and sprays is injective as well as surjective.

Reviewed by M. Kuranishi

Copyright American Mathematical Society 1962, 2010

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There are many ways to talk about covariant derivatives and the exponential maps on manifolds. I will discuss the exponential map from the coordinate point of view.

The covariant derivative can be used to give an equation for geodesics, curves with minimal length (on a Riemannian manifold) or curves with maximum length (on a Lorentzian manifold). Given a particular point $p\in M$, where $M$ is the manifold the exponential map takes a vector $v$ to the point which is a unit distance along the geodesic whose derivative at $p$ is $v$.

In other words $\exp_p:U\subset T_pM\to M$ a homeomorphism, where $U$ is some subset, $\exp_p(v)=\gamma_v(1)$ where $\gamma_v$ is the curve given as the solution to the differential equation $\nabla_{\gamma'_v}\gamma'_v=0$ with initial conditions $\gamma'_v(0)=v$ and $\gamma_v(0)=p$ and $\nabla$ is the covariant derivative. Thus, for example, $\exp_p(tv)=\gamma_{tv}(1)=\gamma_v(t)$. We need to use a subset $U$ of $T_pM$ since $\exp$ may fail to be a homeomorphism on all of $T_pM$.

Basically the exponential map of a covariant derivative is a uniquely defined homeomorphism from a subset of the tangent space at $p$ to the manifold $M$. I like to think of it as giving a representation of the covariant derivative on the manifold, but this isn't exactly correct.

As I mentioned there are many ways to understand the exponential map, particular in the context of fibre bundles and Lie groups, may I suggest Kobayashi and Nomizu's excellent book, "Foundations of Differential Geometry" particularly volume 1 for the Riemannian case, where as O'Neill's "Semi-Riemannian Geometry With Applications to Relativity" does adequate justice to the Lorentzian and semi-Riemannian cases.

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  • $\begingroup$ Ben, thank you very much! In your answer, you used `` homeomorphism" several times. I would like to know that '${exp}_{p}: U\in {T}_{p}M \rightarrow M$´ is really a homeomorphism? $\endgroup$
    – user6011
    Commented May 10, 2010 at 19:51
  • $\begingroup$ It is a diffeomorphism from a neighbourhood of $0 \in T_p M$ onto its image. You can see this by noting that the derivative of $exp_p$ at 0 is the indentity map and using the inverse function theorem. $\endgroup$ Commented May 10, 2010 at 21:36
  • $\begingroup$ So ${exp}_{P}(U) \ne M$. $\endgroup$
    – user6011
    Commented May 10, 2010 at 21:52
  • $\begingroup$ Lucas is right, $\exp$ is a diffeomorphism from a subset of the tangent space onto it's image. Yes in general $\exp_p(U)\neq M$. This is the case even when $U=T_pM$. To see this take a look at the Hopf-Rinow Theorem and consider the case of $S^2$, which we know can't be covered by a single chart. For interests sake there is no corresponding Hopf-Rinow theorem for Lorentzian manifolds and in general the three equivalent conditions of the Hopf-Rinow theorem are independent for Lorentzian metrics. $\endgroup$
    – Ben Whale
    Commented May 11, 2010 at 4:12

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