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.