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.

Advertisements
Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: