# Determining the rate of spread of geodesics when the sectional curvature is zero

I have posted this question in mathSE a few weeks ago (and proposed a bounty) but so far got no response.

In the book Riemanian geometry (by do-Carmo), the following result is proved (Corollary 2.9 page 115):

$\text{(1) } \|J(t) \|^2=t^2-\frac 13kt^4+R(t)$ where $(\frac{R(t)}{t^4}) \stackrel{t\rightarrow 0}{\longrightarrow} 0$.

($J(t)$ is a Jacobi field, $k$ is the sectional curvature of the relevant plane)

An immediate corollary (2.10 in the book) is:

$\text{(2) } \|J(t) \|=t-\frac 16kt^3+\tilde R(t)$ where $(\frac{\tilde R(t)}{t^3}) \stackrel{t\rightarrow 0}{\longrightarrow} 0$

Question:

In the case $k=0$ (I only assume $K_p(\sigma)=0$, i.e the sectional curvature at a given point w.r.t a given plane is zero), how can we determine if the geodesics spread faster or slower compared to the Euclidean case? (Then $\|J(t) \|=t$).

In other words, what governs the next term of the taylor expansion?

(Perhaps the values of the sectional curvature in nearby points or its derivatives?)

The proof of $\text{(1)}$ was via calculation of the derivatives of $f(t)=\|J(t)\|^2$ (using the metricity of the connection). It turns out that: $$f(0)=f'(0)=f^{(3)}(0)=0, f''(0)=2 ,f^{(4)}(0)= -8k.$$

I am asking what is $f^{(5)}(0)$?

If $J(t)=\exp_p(tv(s))$, $v(0)=v,v'(0)=w, \sigma = \operatorname{span} \{v,w\}\subseteq T_pM$, then: $J(0)=0,J'(0)=w$.

By Jacobi's equation $J''(0)=J^{(2)}(0)=0$ as well. An easy calculation shows that:

$$f^{(5)}(0)= 20 \langle J^{(2)}(0),J^{(3)}(0) \rangle + 10 \langle J^{'}(0),J^{(4)}(0) \rangle + 2 \langle J(0),J^{(5)}(0) \rangle = 10 \langle w,J^{(4)}(0) \rangle$$

So, the question amounts to calculating $J^{(4)}(0)$.

I have tried to differentiate the Jacobi equation twice, but couldn't obtain anything useful. (do-Carmo differentiated it once, in order to calculate $J^{(3)}(0)=-R(v,w)v$).

• Finding the power series solution to the Jacobi equation at $t = 0$ doesn't work? I would expect the covariant derivative of the curvature tensor to appear. – Deane Yang Jan 18 '17 at 18:18
• Here's a blog with the computations you're looking for: cuhkmath.wordpress.com/2012/03/16/taylor-expansion-of-metric (Disclaimer: This is a link to someone else's work; I am not claiming any credit for it, and can make no guarantees regarding its accuracy.) – Graham Cox Feb 6 '17 at 14:59
• @GrahamCox Thanks, this indeed seems helpful. – Asaf Shachar Feb 7 '17 at 15:02

## 1 Answer

I'm not 100% sure this works, since I haven't carried it out carefully. Here's what I would do:

1. Fix an orthonormal frame $e_1, \dots, e_n$ that is parallel along the geodesic and $e_n = T$ is tangent to the geodesic.

2. Let $J_1, \dots, J_n$ be the Jacobi fields along the geodesic such that $J(0) = 0$ and $\nabla_TJ_k(0) = e_k$

3. Then $J_i = A_i^ke_k$, where the matrix $A$ satisfies the second order ODE $$A'' + KA = 0,\ A(0) = 0,\ A'(0) = I$$ where $K$ is a symmetric-matrix-valued function given in terms of the curvature tensor.

4. It is now straightforward to compute a power series solution to the matrix ODE in terms of the power series of $K$, which can in turn be written as a power series whose coefficients are given by covariant derivatives of the curvature tensor.