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Given a Riemannian manifold $M$, let $\gamma: (a,b) \to M$ be a geodesic and $E$ a parallel vector field along $\gamma$. Define $\varphi: (a,b) \to M$ by $t \mapsto \exp_{\gamma(t)}(E(t))$. Is there a "nice" expression for $\varphi'(t)$?

This question originates in an attempt to understand the proof of corollary 1.36 in Cheeger and Ebin's "Comparison Theorems in Riemannian Geometry."

Yakov Shlapentokh-Rothman
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    Yes, use the chain rule together with the expression of the differential of exp in terms of Jacobi fields – Sebastian Dec 30 '10 at 10:19
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    @Sebastian, I am confused how to compute the differential of exp. I know the result that $(exp_p)*:TT_pM \to TM$ is $(exp_p)|V(W) = J_W(1)$ where $J$ is the Jacobi field with $J(0) = 0$ and $D_tJ = W$. However, this is only half of the full differential of exp that I think one would need, i.e. $(exp): TTM \to TM$. How do you compute the other half, coming from the fact that $p$ is now varying? – Otis Chodosh Dec 30 '10 at 15:11
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    If $p$ varies, then you just get a Jacobi field $J$ where $J(0) \ne 0$. The point is that the first variation of any family of geodesics is called a Jacobi field and satisfies the Jacobi equation along the original geodesic. Any Jacobi field can be decomposed into three different components, one tangent to the geodesic, two orthogonal. Of the two orthogonal, one vanishes at the origin but has nonzero derivative there and the second is nonzero at the origin but has zero derivative there. I am pretty sure this is all explained in Cheeger-Ebin (since I probably learned it all from there). – Deane Yang Dec 30 '10 at 15:16
  • Thanks. For some reason I was hoping for an answer that was more concrete than Jacobi fields, in hindsight that was a little silly... – Yakov Shlapentokh-Rothman Dec 30 '10 at 17:42
  • What condition on the manifold will imply that $\phi$ is a geodesic too? It's definitely true when the manifold is $\mathbb{R^n}$, but otherwise seems to be not true. Is there a sufficient condition? Thanks! – Learning math Feb 12 '15 at 13:42
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    I am also interested in this question when $\gamma$ is not necessarily a curve. Can anyone provide a general answer? Namely, if $(M,g)$ is a Riemannian manifold and $X \in \mathbb{T}(M)$ a smooth vector field (which may be taken as a vector field along the image of a path or a more general map), we can exponentiate $X$ to give a smooth map $$\mathrm{exp}\colon M\to M$$ $$p\mapsto \mathrm{exp}_p(X(p))$$ How can one express the derivative $d_p \mathrm{exp}\colon \mathbb{T}_pM \to \mathbb{T}_pM$ in terms of X? – Aerinmund Fagelson Apr 06 '16 at 13:59

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Let $x(u,t) = \exp_{\gamma(t)}(u E(t))$. For fixed $t$, as $u$ ranges from 0 to 1, the curve $x(\cdot, t)$ is a geodesic segment from $\gamma(t)$ to $\exp_{\gamma(t)}(E(t))$. Then $\phi'(t) = J(1)$ where $J$ is the Jacobi field along this geodesic segment, with the initial conditions $J(0) = \gamma'(t)$ and $(\nabla_{d x/d u} J)(0) = 0$.

Why? As $t$ varies, $x$ is a variation through geodesics, so for fixed $t$, $\frac{\partial x}{\partial t}(u,t)$ is a Jacobi field (call it $J$) along the geodesic $x(\cdot, t)$. Then $\phi'(t) = \frac{\partial x}{\partial t}(1,t) = J(1)$ while $\gamma'(t) = \frac{\partial x}{\partial t}(0,t) = J(0)$. And using the fact that $\nabla$ is torsion-free and $E$ is parallel we have $(\nabla_{\partial x/\partial u} \frac{\partial x}{\partial t})(0,t) = (\nabla_{\partial x/\partial t} \frac{\partial x}{\partial u})(0,t)= \nabla_{\partial x/\partial t}E(t) = 0.$

(I guess this does not need $\gamma$ to be a geodesic....?)

Michael Pauley
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    This does need $\gamma$ to be a geodesic, otherwise the derivative of the Jacobi field at 0 does not vanish. – zzz May 07 '16 at 21:20
  • @zzz Could you explain why? – Dreamer Sep 06 '21 at 22:05
  • There is no need for $\gamma(t)$ to be a geodesic as long as the $\exp$ mapping can be defined to start from a point on $\gamma(t)$. But some smoothness conditions are needed for $\nabla_{\partial / \partial{t}} \nabla_{\partial / \partial{u}} = \nabla_{\partial / \partial{u}} \nabla_{\partial / \partial{t}} $ to hold though; otherwise, things break down – Chee Feb 23 '23 at 05:15