MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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."

share|cite|improve this question
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
@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
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! – Let's talk math Feb 12 '15 at 13:42
up vote 6 down vote accepted

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....?)

share|cite|improve this answer
This does need $\gamma$ to be a geodesic, otherwise the derivative of the Jacobi field at 0 does not vanish. – bianchira May 7 at 21:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.