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Let $M$ be a Riemannian manifold. Let us look at the Riemannian exponential function $\exp_x: T_x M \supset \mathcal{D} \longrightarrow M$.

The derivative of the exponential map can be expressed in Terms of Jacobi Fields. Is there any slick way to express the second derivative $$ \nabla d \exp_x|_X: T_x M \times T_x M \longrightarrow T_{\exp_x(X)}M$$ in terms of integrals over curvature or Jacobi Fields? What about its Taylor expansion in $X \in T_x M$ about $0$?

(Just to clarify what I mean with the above expression: We can pullback the tangent bundle $TM$ and the Levi-Civita connection to $T_x M$. Look at the vector bundle $T^*T_x M \otimes \exp_x^* TM \rightarrow T_x M$. The first factor is naturally isomorphic to $T^*_x M$ and is just flat; for the second one, we use the pullback of the Levi-Civita Connection. Hence we have a vector bundle with connection, and $d \exp_x$ is a section of it. Hence we can form $\nabla$ of it, which will be a section of $T^*_x M \times T^*_x M \otimes \exp_x^* TM$; that is the quantity I am interested in.)


Edit: Let $c(t, \varepsilon) := \exp_x(t(X+\varepsilon Y))$. Then $$ (d \exp_x|_{t(X+\varepsilon Y)}) \cdot t Z = J(t, \varepsilon),$$ where $J(t, \varepsilon)$ is the Jacobi field along $c(t, \varepsilon)$ with $J(0, \varepsilon) = 0$ and $\frac{\nabla}{\partial t} J(0, \varepsilon) = Z$. Now $$(\nabla d \exp_x|_{tX})[tZ, tZ] = \frac{\nabla}{\partial \varepsilon} J(t, 0) =: K(t)$$

Now Deane Zang claims in the comments that $K(t)$ fulfills a second order equation obtained by differentiating the equation for $J(t, \varepsilon)$, $$\left(\frac{\nabla}{\partial t}\right)^2 J(t, \varepsilon) = R\left(\frac{\partial}{\partial t} c(t, \varepsilon), J(t, \varepsilon)\right)\frac{\partial}{\partial t} c(t, \varepsilon).$$ However, when I differentiate this (with respect to $\varepsilon$, I suppose?), I get all kinds of terms, $J(t, 0)$ and its derivative, derivatives of the curvature tensor, and especially second derivatives of $\exp$ in direction $X$ and $Z$. So this seems far from obtaining a good equation for $K$.

Edit 2: So after working things out, I get that the above mentioned $K(t)$ along $\exp_x(tX)$ fulfills the differential equation $$\left(\frac{\nabla}{d t}\right)^2 K = R(\dot{c}, K)\dot{c} + 2R(\dot{c}, J^Y)\frac{\nabla}{d t}J^Z + 2R(\dot{c}, J^Z)\frac{\nabla}{d t}J^Y + \left(\frac{\nabla}{d t} R\right) (J^Z, \dot{c})J^Y + (\nabla_{J^Z} R)(\dot{c}, J^Y)\dot{c},$$ where $J^Y$ and $J^Z$ are the Jacobi fields with initial conditions zero and $Y$ or $Z$ respectively. The initial conditions on $K$ should be zero and zero. This equation is indeed invariant under exchanging $Y$ and $Z$ although this is not obvious (one has to use the second Bianchi identity).

Thanks to Dean for putting me onto the right track. I was a bit surprised though that derivatives of the curvature tensor show up, but probably this is not surprising after all.

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  • $\begingroup$ Since the first derivative of the exponential map are Jacobi fields, it makes sense that the second derivative are derivatives of the Jacobi fields. And just as Jacobi fields satisfy a linear second order ODE along a geodesic ray, their derivatives satisfy the second order ODE that you get by differentiating the first ODE. $\endgroup$ – Deane Yang Feb 22 '14 at 20:33
  • $\begingroup$ I do not really understand by what I should differentiate the Jacobi Equation. Could you elaborate? $\endgroup$ – Matthias Ludewig Feb 23 '14 at 3:37
  • $\begingroup$ If derivative of the exponential map can be described in terms of Jacobi fields, then the second derivative should have a corresponding description in terms of the covariant derivatives of the Jacobi fields. And since a Jacobi field satisfies the Jacobi equation, its covariant derivative satisfies an equation obtained by taking the covariant derivative of both sides of the Jacobi equation. $\endgroup$ – Deane Yang Feb 23 '14 at 6:24
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    $\begingroup$ I don't know about this question but I did learn recently that the 4th jet of the function (s,t) -> d(exp(sv), exp(tw)) where the `exp' is taken at a fixed pt. p yields the Riem curv tensor R_p (v,w,v,w) $\endgroup$ – Richard Montgomery Mar 13 '14 at 6:01
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    $\begingroup$ @CSA, Matthias Ludewig. Apologies for not remembering the ref. I would start looking probably in Berger's Geometry in the 2nd half of the 20th century, or Cheeger-Ebin. (I'm not near access to books right now). $\endgroup$ – Richard Montgomery May 26 '17 at 18:55

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