# Higher order variations of Riemannian geodesics

Consider a mapping $\Gamma$ from the Euclidean plane or an open subset to a Riemannian manifold $M$ so that each $\Gamma(s,\cdot)$ is a geodesic. There is a well established theory of the first order variations of this family of geodesics: $J(t)=\partial_s\Gamma(s,t)|_{s=0}$ is a Jacobi field along the geodesic $\Gamma(0,\cdot)$ and it is related to, for example, the index form and the differential of the geodesic flow.

What is known about higher order variations? I am interested in the (covariant) derivatives $\partial_s^k\Gamma(s,t)|_{s=0}$ for $k\geq2$, mainly for $k=2$ at first. Is there a "higher order Jacobi equation" that the resulting vector fields satisfy? Where can I find the known properties of such variation fields?

If these objects have not been studied, is there an obstruction to their definition or study? Is there a good definition with some nice properties out there? Second order variations strike me as a natural geometric object to look at, but I was unable to find any mentions of them.

## Details on the definition

The original question had no definition and the next edit had a bad one. Now I put more thought into it. The question should be meaningful without reading this, but I want to give a proper definition just in case.

Consider a smooth curve $s\mapsto\sigma(s)$ and a vector field $V$ along it. The covariant derivative of $V$ can be defined as $$(D_sV)^m=\dot V^m+\Gamma^m_{ij}\dot\sigma^iV^j,$$ where the dot stands for the coordinate derivative with respect to $s$. This definition makes sense whether $\sigma$ is injective or not. If $\sigma$ stays still ($\dot\sigma$ vanishes), then the Christoffel symbol term vanishes. (This is expected naively: the Christoffel symbol helps compare vectors on different tangent spaces in a local coordinate system, and when $\dot\sigma=0$, the tangent space stays the same.)

The vector field $V$ is a section of the pullback bundle $\sigma*TM$ over the interval on which $\sigma$ is defined. If $\sigma$ is (locally) injective, then it can be realized as a restriction of a (locally defined) vector field on $M$, otherwise not in general.

Let us then look into the covariant derivatives I used in the question. Consider $t$ fixed. Then $s\mapsto\Gamma(s,t)$ is a curve on the manifold. The first order derivative $\partial_s\Gamma(s,t)$ is a vector field along this curve. The second derivative is a covariant derivative along this curve in the usual way as described above, producing another vector field along this curve. You can continue to any order like this. For any $k$, such $k$th order covariant differentiation in $s$ produces a vector in $T_{\Gamma(s,t)}M$. We can then evaluate this vector field at $s=0$. Considering different values of $t$ gives a vector at each $T_{\Gamma(0,t)}M$. This gives rise to a $k$th order variation vector field as a vector field along $\Gamma(0,\cdot)$. This seems to be actually well defined whether the curves $\Gamma(s,\cdot)$ are geodesic or not. In short, it is $s$-covariant differentiation; I see $s$ as a parameter and $t$ as time, and covariant derivatives should not mix these two.

If the first order variation (Jacobi field) vanishes, then the second order covariant variation agrees with the second order coordinate derivative. Similarly, if the first two orders vanish, the third one is the third coordinate derivative. If I calculated correctly, the second order covariant derivative of a vector field with my definition is $$(D^2_sV)^m = \ddot V^m +\Gamma^m_{ij,k}\dot\sigma^k\dot\sigma^iV^j +\Gamma^m_{ij}\ddot\sigma^iV^j +2\Gamma^m_{ij}\dot\sigma^i\dot V^j +\Gamma^m_{ij}\Gamma^j_{kl}\dot\sigma^i\dot\sigma^kV^l.$$ Plugging in $V=\dot\sigma$ gives $$(D^3_s\sigma)^m = \dddot\sigma^m +(\Gamma^m_{ij,k}+\Gamma^m_{ij}\Gamma^j_{kl}) \dot\sigma^i\dot\sigma^j\dot\sigma^k +3\Gamma^m_{ij}\dot\sigma^i\ddot\sigma^j,$$ where the first covariant derivative is understood simply as $D_s\sigma=\dot\sigma$. It seems to work similarly for higher orders. (My first definition lead to $\dot\sigma=0\implies D_s\dot\sigma=0$, which was clearly undesirable.)

• Ok, now with the clarification you have a concrete question. If I am interpreting correctly, what you denote by $\dot{\sigma}$ in your parenthetical is to be replaced by the vector field $\partial_s \Gamma$ (which I'll call $V$ in the following) Let's assume for simplicity that the mapping $\Gamma(s,t) \to M$ is injective so that the vector field $V$ is well-defined. Then the "higher order Jacobi equation" you are looking for can just be obtained by commuting the standard Jacobi equation with $\nabla_V$; this will be a Jacobi equation again except with some inhomogeneities. – Willie Wong Aug 20 '18 at 12:42
• There is, however, a problem with your "definition". If my interpretation in my previous comment is correct, it does not make sense to call what you defined a "second variation". For when the first variation vanishes, the second (and all higher) variation as you defined will vanish also (since what you call $\dot{\sigma}$ would also vanish). On the other hand, when the first variation vanishes, the second derivative $\partial^2_s\Gamma(s,t) |_{s = 0}$ is actually well defined as a (generically non-vanishing) vector field along $\Gamma(0,t)$, and I would expect any definition of higher order – Willie Wong Aug 20 '18 at 12:50
• variation to agree with the naively interpreted $\partial^k_s \Gamma(s,t)|_{s = 0}$ in the case where $\partial_s \Gamma \ldots \partial^{k-1}_s\Gamma$ all vanish at $s = 0$. – Willie Wong Aug 20 '18 at 12:52
• @WillieWong Many thanks! My definition was indeed flawed. The first term of the covariant derivative with respect to the parameter on a curve should be just the coordinate derivative. I wrote a substantial update, and now it seems to satisfy the property you expected: if all derivatives up to an order vanish, the next one will agree with the naive coordinate derivative. I apologize for the badly written definition and thank you again for helping me write it up properly. I agree that one should simply commute $D_s$ with the Jacobi equation, but I have yet to compute what it would look like. – Joonas Ilmavirta Aug 20 '18 at 18:07
• Thinking a bit harder about your question: I think my comments above are not strictly correct. The problem is that if $\sigma:(-1,1)\to M$ is a curve and $\dot{\sigma}(0) =0$, then the pushforward $V = \sigma_* \partial_s$ can in general have a cusp singularity at $s = 0$. (Example, let $M = \mathbb{R}$ and look at $\sigma(s) = s^3$. The vector field $V$ on $M$ is $3x^{2/3} \partial_x$.) This means that even though $\dot{\sigma}(0) = 0$ we can still have $\nabla_{V} V$ be non-vanishing at $s = 0$. – Willie Wong Aug 21 '18 at 19:07