# Derive the Jacobi equation from a specific geodesic variation

Construction of a specific geodesic variation: Let $$\tau = \{\exp_{x_0}(tX):t\in[0,1]\}$$ be a geodesic on a complete Riemannian manifold $$(M,g)$$, starting from $$x_0\in M$$ with velocity $$X\in T_{x_0} M$$. Let $$V$$ be a vector attached to $$x_0$$ that is not parallel to $$X$$. Suppose that both $$X$$ and $$V$$ have unit length, i.e., $$g_{x_0}(X,X) = g_{x_0}(V,V) = 1$$. Now we introduce the geodesic starting from $$x_0$$ with velocity $$V$$ by $$y_s := \exp_{x_0}(sV).$$ Denote the parallel transport of $$X$$ along $$\gamma = \{y_s\}$$ by $$X(s) := \Gamma(\gamma)_0^s (X).$$ Then we get a family of geodesics starting from $$y_s$$ with velocity $$X(s)$$, $$\tau^s := \{ \exp_{y_s} (tX(s)):t\in[0,1]\},$$ which forms a variation of the geodesic $$\tau$$.

So the following vector field along $$\tau$$ should be the Jacobi field, $$J(t) := \frac{\partial}{\partial s}\bigg|_{s=0} \exp_{y_s} (tX(s)),$$ which means $$$$\tag{1} \frac{D^2}{dt^2} J(t) + R(J(t), \dot \tau(t))\dot \tau(t) =0,$$$$ where $$D$$ denotes the covariant derivative with respect to the Levi-Civita connection, $$R$$ the Riemann curvature tensor.

The question is, how to derive the Jacobi equation (1) from this contruction? I tried a lot but failed. This question acturally arises from this paper at its equation (26). Could anyone figure this out? TIA...

We have $$\exp_{y_s} (tX(s))$$ as a coordinate-field, not a vector field. However, when we do operations like $$\frac{D}{ds}$$, they correspond to covariant derivatives along the directions of the field given by $$\frac{D}{ds}\exp_{y_s} (tX(s))$$. $$\frac{D^2}{dt} J(t) = \frac{D}{dt} \frac{D}{dt} (\frac{D}{ds} \exp_{y_s} (tX(s))) = \\= \frac{D}{dt} \frac{D}{ds} \frac{D}{dt}\exp_{y_s} (tX(s)) = \frac{D}{dt} \frac{D}{ds} \dot \tau(t) = \\=\frac{D}{ds} \frac{D}{dt} \dot \tau(t) \; - R(ds,dt)\dot \tau(t) =\\= -R(J(t), \dot \tau(t))\dot \tau(t)$$ We can commute the two innermost covariant derivatives because we start off with a coordinate field. (it works roughly for the same reason that $$s_{;ab} = s_{;ba}$$ for a scalar field). For commuting the two outermost covariant derivatives we apply the Ricci identity. Also $$\frac{D}{dt} \dot \tau(t)$$ vanishes because they are geodesics.
The Jacobi equation is a second order differential equation whose solutions are completely determined by the values of $$J(p)$$ and $$\dot J(p)$$ for a given parameter $$p$$ along the geodesic. At $$\tau(0) = x_0$$ we have $$J(0) = V$$ and $$\dot J(0) = \frac{D}{dt} \frac{D}{ds} \exp_{y_s} (tX(s))) = \\= \frac{D}{ds} \frac{D}{dt} \exp_{y_s} (tX(s))) = \frac{D}{ds} \dot \tau(t)$$ at values $$s= 0,t=0$$. Given that the initial geodesic family is formed by parallel-transporting $$\dot \tau(0)$$ along the geodesic spanned by $$ds$$, we get that $$\frac{D}{ds} \dot \tau(t) = 0$$ at $$s=0,t=0$$ . Thus our initial conditions are $$J(0) = V$$, $$\dot J(0) = 0$$, which, along with the Jacobi equation, uniquely determine the field along the geodesic.
• Oh, I was just trying to differentiate the exponential map to get $J$ explicitly, it looks really complicated... Your appraoch is handy. Thank you so much! Aug 26 at 13:41
• By the way, I am still curious about how $J$ looks like... The problem is the base point of the exponential map in $J$ depends on the parameter $s$. Aug 26 at 13:44
• I've checked my calculations and added a derivation for $J$ and $\dot J$ along $\tau$ at $x_0$ Aug 26 at 18:39