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) }.


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).}


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.