Index form and Jacobi field

Assume ${\gamma(t):[0,a]\rightarrow M}$ is a normal geodesic, ${\gamma(t,s):[0,a]\times[-\epsilon,\epsilon]\rightarrow M}$ is one parameter smooth variation of curve ${\gamma(t)}$. Denote ${\gamma_s(t)=\gamma(t,s)}$.

$\displaystyle V(t)=\frac{\partial \gamma_s}{\partial s}\bigg|_{s=0}$

is called the variation field. As we all know if the variation is geodesic one, then ${V}$ is a Jacobi field. If ${L(s)=L(\gamma_s(t))}$, then

$\displaystyle L''(0)=\langle\nabla_VV,T\rangle|_0^a+\int_0^a |\dot{V}(t)|^2-R(T,V,T,V)dt$

Let us define $\mathcal{V}=\{V|V\perp \dot{\gamma}\text{ and } V \text{ is piecewise smooth along }\gamma\}$ and ${\mathcal{V}_0=\{V\in\mathcal{V}|V(0)=V(a)=0\}}$ is a subspace of ${\mathcal{V}}$. Index form

$\displaystyle I(X,Y)=\int_{0}^a\langle\dot{X},\dot{Y}\rangle-R(T,X,T,Y)dt$

for ${X,Y\in\mathcal{V}}$.

Thm1: If ${\gamma}$ does not have conjugate point, then ${I(X,X)>0}$ for any ${X\in \mathcal{V}_0}$ and ${X\neq 0}$. This is equivalent to say index form is positive definite.

Thm2(Minimality of Jacobi field): Assume ${\gamma}$ does not have conjugate point. Suppose ${V,X\in \mathcal{V}}$ and ${V(0)=X(0)}$ and ${V(a)=X(a)}$. If ${V}$ is a Jacobi field on ${\gamma}$, then

$\displaystyle I(V,V)\leq I(X,X)$

Thm1 and Thm2 are equivalent.
Thm1 ${\Rightarrow}$ Thm2:

${V-X\in \mathcal{V}}$, then

$\displaystyle 0\leq I(V-X,V-X)=I(V,V)-2I(V,X)+I(X,X)$

$\displaystyle =\langle\dot{V},V\rangle|_0^a-2\langle\dot{V},X\rangle|_0^b+I(X,X)=I(V,V)-I(X,X)$

Thm2 ${\Rightarrow}$ Thm1: Obviously

One way to prove Thm1 is to prove Thm2. However, we will prove Thm1 directly.

Proof: Suppose ${\gamma(t)=\exp_ptv}$, ${t\in [0,a]}$. Since ${\gamma}$ does not have conjugate point, then ${d\exp}$ is non-degenerate in ${tv}$, ${t\in [0,a]}$. There exists a neighborhood ${U}$ of ${tv}$ such that ${d\exp:U\rightarrow M}$ is an immersion. From the fact that

${\sigma:[0,a]\rightarrow \exp_pU}$ is any piecewise smooth curve connecting ${\gamma(0)}$ and ${\gamma(a)}$, then ${L(\gamma)\leq L(\sigma)}$ when the equality holds, ${\sigma}$ must be monotone reparametirzation of ${\gamma}$

After proving this, we can construct ${\gamma_s\subset \exp_pU}$ a variation of ${\gamma}$ such that the variation field is ${X}$, then

$\displaystyle 0\leq L''(0)=I(X,X)$

Suppose ${I(X,X)=0}$ for some ${X\neq 0}$, then for any ${Y\in \mathcal{V}_0}$

$\displaystyle 0\leq I(X+\epsilon Y,X+\epsilon Y)$

$\displaystyle 0\leq I(X-\epsilon Y,X-\epsilon Y)$

this means ${I(X,Y)=0}$, ${\forall Y\in \mathcal{V}_0}$. This means ${X}$ must be a Jacobi field. But ${X\in \mathcal{V}_0}$ and ${\gamma}$ has no conjugate point, ${X}$ must be 0.