I’m going to keep the same notation as before.  In particular, we’re studying how the energy integral behaves with respect to variations of curves.  Now I want to prove the second variation formula when ${c}$ is a geodesic.Now to compute ${\frac{d^2}{d^2 u} E(u)|_{u=0}}$, for further usage. We already showed$\displaystyle E'(u) = \int_I g\left( \frac{D}{dt} \frac{\partial}{\partial u} H(t,u), \frac{\partial}{\partial t} H(t,u) \right)$ Differentiating again yields the messy formula for ${E''(u)}$:

$\displaystyle \int_I g\left( \frac{D}{du} \frac{D}{dt} \frac{\partial}{\partial u} H, \frac{\partial}{\partial t} H \right) + \int_I g\left( \frac{D}{dt} \frac{\partial}{\partial u} H, \frac{D}{du}\frac{\partial}{\partial t} H(t,u) \right).$

Call these ${I_1(u), I_2(u) }$.

${I_2}$

Now ${I_2(0)}$ is the easiest, since by symmetry of the Levi-Civita connection we get$\displaystyle I_2(0) = \int g\left( \frac{D}{dt} \frac{\partial}{\partial u} H(t,u), \frac{D}{dt}\frac{\partial}{\partial u} H(t,u) \right) = \int g\left( \frac{D}{dt} V, \frac{D}{dt} V \right).$ For vector fields along ${c}$ ${E,F}$ with ${E(a)=F(a)=E(b)=F(b)=0}$, we have$\displaystyle \int g\left( \frac{D}{dt} E, \frac{D}{dt} F \right) = - \int g\left( \frac{D^2}{dt^2} E , F \right).$ This is essentially a forum of integration by parts. Indeed, the difference between the two terms is

$\displaystyle \frac{d}{dt} g\left( \frac{D}{dt} E, F \right).$

So if we plug this in we get

$\displaystyle \boxed{ I_2(0) = -\int g\left( \frac{D^2}{dt^2} V , V \right).}$

${I_1}$

Next, we can write

$\displaystyle I_1(0) = \int_I g\left( \frac{D}{du} \frac{D}{dt} \frac{\partial}{\partial u} H(t,u) |_{u=0}, \dot{c}(t) \right)$ Now ${R}$ measures the difference from commutation of ${\frac{D}{dt}, \frac{D}{du}}$. In particular this equals

$\displaystyle \int_I g\left( \frac{D}{dt} \frac{D}{du} \frac{\partial}{\partial u} H(t,u) |_{u=0}, \dot{c}(t) \right) + \int_I g\left( R(V(t), \dot{c}(t)) V(t), \dot{c}(t)) \right).$

By antisymmetry of the curvature tensor (twice!) the second term becomes

$\displaystyle \int_I g\left( R( \dot{c}(t), V(t)) \dot{c}(t), V(t), \right).$

Now we look at the first term, which we can write as

$\displaystyle \int_I \frac{d}{dt} g\left( \frac{D}{du} \frac{\partial}{\partial u} H(t,u), \dot{c}(t)\right)$

since ${\ddot{c} \equiv 0}$. But this is clearly zero because ${H}$ is constant on the vertical lines ${t=a,t=b}$. If we put everything together we obtain the following “second variation formula:”

Theorem 1 If ${c}$ is a geodesic, then$\displaystyle \boxed{\frac{d^2}{du^2}|_{u=0} E(u) = \int_I g\left( R( \dot{c}(t), V(t)) \dot{c}(t) - \frac{D^2 V}{Dt^2}, V(t) \right).}$

Evidently that was some tedious work, and the question arises: Why does all this matter? The next goal is to use this to show when a geodesic cannot minimize the energy integral—which means, in particular, that it doesn’t minimize length. Then we will obtain global comparison-theoretic results.

As I hinted a couple of posts back, I am interested in discussing the application of the calculus of variations to differential geometry. So fix a Riemannian manifold with metric denoted either by ${g}$ or ${\left \langle \cdot, \cdot \right \rangle}$, and let ${c: I \rightarrow M}$ be a smooth path in ${M}$. Define the energy as

$\displaystyle E(c) := \frac{1}{2} \int g(c',c').$

The energy integral is closely related to the length function, though it is easier to deal with. Now we are interested in studying a variation ${H}$ of the curve ${c}$ and how the energy integral behaves with respect to ${H}$. Recall that ${H: I \times(-\epsilon,\epsilon) \rightarrow M}$ is a smooth map with ${H(t,0)=c(t), H(a,u)=c(a), H(b,u)=c(b)}$ for all ${t,u}$. The last two conditions mean that ${H}$ is a family of curves that keep the endpoints fixed.

Define ${H_u}$ as the curve ${J(\cdot, u)}$, and consider the function of ${u}$,

$\displaystyle E(u) := E(H_u) = \frac{1}{2} \int_I g\left( \frac{\partial}{\partial t} H(t,u) , \frac{\partial}{\partial t} H(t,u) \right).$

Ultimately, we are interested in curves that minimize the energy integral, at least locally. This means that for any variation ${H}$ as above, ${E(u)}$ should have a local minimum at ${u=0}$. So we will compute ${\frac{d}{du} H(u)|_{u=0}}$, and, eventually, the second derivative too. The evaluations involve nothing more than a rehash of many standard tricks we have repeated already.

If we differentiate under the integral sign, legal because of all the smoothness, we get

$\displaystyle \frac{d}{du} E(u) = \frac{1}{2} \int_I \frac{\partial}{\partial u} g\left( \frac{\partial}{\partial t} H(t,u) , \frac{\partial}{\partial t} H(t,u) \right)$

By the formula for differentiating inner products and the symmetry of the Levi-Civita connection, this becomes

$\displaystyle \int_I g\left( \frac{D}{du} \frac{\partial}{\partial t} H(t,u), \frac{\partial}{\partial t} H(t,u) \right) = \int_I g\left( \frac{D}{dt} \frac{\partial}{\partial u} H(t,u), \frac{\partial}{\partial t} H(t,u) \right)$

If we let ${V(t)}$ be the variation vector field and set ${u=0}$, we find that

$\displaystyle \frac{d}{du} E(u) |_{u=0} = \int_I g\left( \frac{D}{dt} V(t), \dot{c}(t) \right),$

which using similar identities becomes

$\displaystyle \int_I \frac{d}{d t} g\left( V(t), \dot{c}(t) \right) - \int_I g\left( V(t), \ddot{c}(t) \right)$

(Note that ${\ddot{c}(t) := \frac{D}{dt} \dot{c}(t)}$ by definition.) The left integral is ${g(V(b), \dot{c}(b)) - g(V(a), \dot{c}(a)) = 0}$, since ${V(b)=V(a)=0}$ for a variation keeping endpoints fixed. Hence we have the following:

Theorem 1 (First Variation Formula)

For variations fixing the endpoints

$\displaystyle \boxed{ \frac{d}{du}E(u)|_{u=0} = -\int_I g\left( V(t), \ddot{c}(t) \right) .}$

In particular, since ${V(t)}$ can be really chosen arbitrary with ${V(a)=V(b)=0}$, we see that a curve locally minimizes the energy ${E}$ only if it is a geodesic. There is a reason for this. First, by Cauchy-Schwarz, we have

$\displaystyle l(c) \leq \sqrt{E(c)} \sqrt{length(I)}.$

There is equality precisely when ${c}$ moves at constant speed, i.e. ${|c'|}$ is constant. In particular,

$\displaystyle \left(\frac{l(c)}{\sqrt{length(I)}}\right)^2 \leq E(c)$

with equality holding for a geodesic. If ${c}$ is not a geodesic, we can always find a shorter geodesic path between ${c(a),c(b)}$ that necessarily makes ${E}$ smaller by the above inequality (and equality).

It is of interest to consider functions on the space of curves ${I \rightarrow M}$, where ${I}$ is an interval and ${M}$ is a smooth manifold. To study maxima and minima, it is of interest to consider variations of curves, holding the endpoints fixed. Let ${c:[a,b] \rightarrow M}$ be smooth. A variation of ${c}$ is a smooth map$\displaystyle H: [a,b] \times (-\epsilon, \epsilon) \rightarrow M$

with ${H(t,0) = c(t)}$, and ${H(a,u)=c(a), H(b,u)=c(b)}$ for all ${t,u}$. For a variation ${H}$ of ${c}$, define the variation vector field (which is an analog of a “tangent vector”)

$\displaystyle V(t) = \frac{\partial}{\partial u} H ;$ this is a vector field along ${c}$. Similarly we can define the “velocity vector field” ${\dot{c}}$ along ${c}$. If ${M}$ is provided with a connection, we can define the “acceleration vector field” ${A(t) = \frac{D}{dt} \dot{c}}$, where ${\frac{D}{dt}}$ denotes covariant differentiation.

Given a vector field ${V}$ along ${c}$, we can construct a variation of ${c}$ with ${V}$ as the variation vector field: take ${(t,u) \rightarrow \exp_{c(t)}(u V(t))}$.

Variations of geodesics and Jacobi fields Let ${M}$ now be a manifold with a symmetric connection ${\nabla}$.Let ${H}$ be a variation of a geodesic ${c}$ such that for any ${u \in (-\epsilon,\epsilon)}$, ${t \rightarrow H(t,u)}$ is a geodesic as well. Then the variation vector field satisfies a certain differential equation. Now$\displaystyle \frac{D^2}{dt^2 } V(t) = \frac{D}{dt} \frac{D}{dt} \frac{\partial}{\partial u } H |_{u=0} = \frac{D}{dt} \frac{D}{du} \frac{\partial}{\partial t} H |_{u=0}.$

We have used the symmetry of ${\nabla}$. Now we can write this as

$\displaystyle \frac{D}{du} \frac{D}{d t} \frac{\partial}{\partial t} H |_{u=0} + R\left( \frac{\partial H}{\partial t}, \frac{\partial H}{\partial u}\right) \frac{\partial H}{\partial t} |_{u=0}.$ By geodesy, the first part vanishes, and the second is

$\displaystyle R( \dot{c}(t), V(t)) \dot{c}(t)$ We have shown that ${V}$ satisfies the Jacobi equation

$\displaystyle \boxed{ \frac{D^2}{dt^2 } V(t) = R( \dot{c}(t), V(t)) \dot{c}(t).}$ Any vector field along ${c}$ satisfying this is called a Jacobi field.

The differential of the exponential map Let ${p \in M}$, and consider the exponential map ${\exp_p: U \rightarrow M}$ where ${U}$ is a neighborhood of the origin in ${T_p(M)}$. Let ${X,Y \in T_p(M)}$. Now the map$\displaystyle H: (t,u) \rightarrow \exp_p( tX + ut Y)$ takes horizontal lines to geodesics in ${M}$ when ${t,u}$ are small enough. This can be viewed as a vector field along the geodesic ${t \rightarrow \exp(tX)}$. The variation vector field ${J}$ is thus a Jacobi field, and also at ${t=1}$ is ${(\exp_p)_{*X}(Y) \in T_{\exp_p(X)}(M).}$ Note that ${J}$ satisfies ${J(0)=0}$ and

$\displaystyle \frac{D}{dt} J(t)|_{t=0} = \frac{D}{du} \frac{\partial}{\partial t} H(t,u) |_{t,u=0} = \frac{D}{du} (X + uY)|_{u=0} = Y.$

Proposition 1 Suppose ${X,Y}$ are sufficiently small. Let ${J}$ be the Jacobi field along the geodesic ${\gamma(t) := \exp_p(tX)}$ with ${J(0) = 0, \frac{D}{dt} J(t)|_{t=0} = Y}$ (i.e. using the ODE theorems). Then$\displaystyle J(1)= (\exp_p)_{*X}(Y) \in T_{\exp_p(X)}(M).$

I will next explain how to use this fact to prove the Cartan-Hadamard theorem on manifolds of negative curvature.

Source: These notes, from an introductory course on differential geometry.