27.02.2018

# Geodesic flows: notes 6

## Geodesic flows, notes 6: geometry of the sphere bundle

Here we use the definitions and notations of notes 4.

### Computations in local coordinates

Given a local coordinate chart (U,x)(U,x) on MM, there are corresponding coordinates (x,y)(x,y) for TMTM and (x,y,X,Y)(x,y,X,Y) for TTMTTM, so that if ξT(x,y)TMξ∈T(x,y)TM, then

ξ=Xjxj+Ykyk.ξ=Xj∂xj+Yk∂yk.

Definition.The vector field δxjT(x,y)TMδxj∈T(x,y)TM is defined as

δxj:=xjΓljkykyl,δxj:=∂xj−Γjklyk∂yl,

where the ΓljkΓjkl are the Christoffel symbols.

Lemma 1. The vectors {δxj}nj=1{δxj}j=1n are a basis for the subspace H(θ)H(θ), where θ=(x0,y0)θ=(x0,y0).

Proof. We will show that

Kθ(δxj)=0, and dθπ(δxj)=xj.Kθ(δxj)=0, and dθπ(δxj)=∂xj.

The claim follows from this, since now δxjker(Kθ)=H(θ)δxj∈ker⁡(Kθ)=H(θ), and since dπθ:H(θ)TxMdπθ:H(θ)→TxM is an isomorphism, so that a basis vector maps to a basis vector.

To compute Kθ we pick a curve σ:(ϵ,ϵ)TMσ:(−ϵ,ϵ)→TMσ(t)=(x(t),y(t))σ(t)=(x(t),y(t)), s.t.

σ(0)=(x0,y0),σ˙(0)=(x˙(0),y˙(0))=δxj.σ(0)=(x0,y0),σ˙(0)=(x˙(0),y˙(0))=δxj.

Notice that x˙(0)=ej=(0,..,1,0,..,0)x˙(0)=ej=(0,..,1,0,..,0) and y˙(0)=(Γ1jkyk,..,Γnjkyk)y˙(0)=(−Γjk1yk,..,−Γjknyk).

Recall that the covariant derivative of the vector field Z(t)=Zj(t)xjZ(t)=Zj(t)∂xj along a curve α(t)α(t) is defined as

α˙Z=(Z˙k(t)+α˙iZaΓkia)k.∇α˙Z=(Z˙k(t)+α˙iZaΓiak)∂k.

And that by definition

Kθδxj=(x˙y)|t=0.Kθδxj=(∇x˙y)|t=0.

Since x˙(0)=ejx˙(0)=ej, we have that

Kθδxj=(y˙k+yaΓkja)xk|t=0.Kθδxj=(y˙k+yaΓjak)∂xk|t=0.

On the other hand we know that y˙(0)=(Γ1jkyk,..,Γnjkyk)y˙(0)=(−Γjk1yk,..,−Γjknyk). Substituting this in the expression for KθδxjKθδxj, gives then that

Kθδxj=(Γkjaya+yaΓkja)xk|t=0=0.Kθδxj=(−Γjakya+yaΓjak)∂xk|t=0=0.

We thus see that δxjH(θ)δxj∈H(θ).

It remains to show that dθπ(δxj)=xjdθπ(δxj)=∂xj. Since dθπdθπ is a linear map, we need only show that

dθπ(xj)=xj,dθπ(yk)=0.dθπ(∂xj)=∂xj,dθπ(∂yk)=0.

This holds because firstly we have that

(dθπ(xj))f=xj(fπ)=xjf,(dθπ(∂xj))f=∂xj(f∘π)=∂xjf,

And secondly that

dθπ(yk)=yk(fπ)=0.dθπ(∂yk)=∂yk(f∘π)=0.◻

Lemma 2. The vectors {yk}nk=1{∂yk}k=1n are a basis for the subspace V(θ)V(θ)θ=(x0,y0)θ=(x0,y0).

Proof. By similar but simpler computations as in Lemma 1 one sees that

dθπ(yk)=0 and Kθ(yk)=xk.dθπ(∂yk)=0 and Kθ(∂yk)=∂xk.

It follows that xkker(dθπ)=V(θ)∂xk∈ker⁡(dθπ)=V(θ). Moreover, we had that Kθ:V(θ)TxMKθ:V(θ)→TxM is an isomorphism, so that basis vectors are mapped to basis vectors.

Recall that the Sasaki metric was defined by the formula

η,ξSasaki:=(dθπ)η,(dθπ)ξ+Kθη,Kθξ,⟨η,ξ⟩Sasaki:=⟨(dθπ)η,(dθπ)ξ⟩+⟨Kθη,Kθξ⟩,

for ξ,ηTθTMξ,η∈TθTM.

Given vectors ξ,ηTθTMξ,η∈TθTM and writing them in the basis given by Lemma 1 and Lemma 2, i.e.

ξ=Xiδxi+Ykyk,η=X~iδxi+Y~kyk,ξ=Xiδxi+Yk∂yk,η=X~iδxi+Y~k∂yk,

we get that

ξ,ηSasaki=Xixi,X~ixi+Ykxk,Y~kxk=gjkXjX~k+gjkYjY~k.

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