12.09.2018

# Geodesic flows: notes 6

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

Here we use the definitons and notations of notes 4.

## Computations in local coordinates

Given a local coordinate chart $$(U,x)$$ on $$M$$, there are corresponding coordinates $$(x,y)$$ for $$TM$$ and $$(x,y,X,Y)$$ for $$TTM$$, so that if $$\xi \in T_{(x,y)}TM$$, then $$\xi = X^j \partial_{x^j} + Y^k \partial_{y^k}.$$

Definition.The vector field $$\delta_{x^j} \in T_{(x,y)}TM$$ is defined as $$\delta_{x^j} := \partial_{x^j} - \Gamma^l_{jk} y^k \partial_{y^l},$$ where the $$\Gamma^l_{jk}$$ are the Christoffel symbols.

Lemma 1. The vectors $$\{\delta_{x^j}\}_{j=1}^n$$ are a basis for the subspace $$H(\theta)$$, where $$\theta =(x_0,y_0)$$.

Proof. We will show that $$K_\theta(\delta_{x^j}) = 0, \quad \text{ and } \quad d_\theta\pi(\delta_{x^j}) = \partial_{x^j}.$$ The claim follows from this, since now $$\delta_{x^j} \in \ker(K_\theta)=H(\theta)$$, and since $$d\pi_\theta \colon H(\theta) \to T_x M$$ is an isomorphism, so that a basis vector maps to a basis vector.

To compute $$K_\theta$$ we pick a curve $$\sigma \colon (-\epsilon,\epsilon) \to TM$$, $$\sigma(t)=(x(t),y(t))$$, s.t. $$\sigma(0) = (x_0,y_0), \quad \quad \dot \sigma(0) = (\dot x(0), \dot y(0)) = \delta_{x^j}.$$ Notice that $$\dot x(0)= e_j = (0,..,1,0,..,0)$$ and $$\dot y(0) = (- \Gamma^1_{jk} y^k,..,-\Gamma^n_{jk} y^k)$$.

Recall that the covariant derivative of the vector field $$Z(t) = Z^j(t) \partial_{x^j}$$ along a curve $$\alpha(t)$$ is defined as $$\nabla_{\dot{\alpha}} Z = (\dot{Z}^k(t) + \dot{\alpha}^i Z^a \Gamma^k_{ia}) \partial_k.$$

And that by definition $$K_\theta \delta_{x_j} = (\nabla_{\dot x} y) |_{t=0}.$$ Since $$\dot x(0)= e_j$$, we have that $K_\theta \delta_{x_j} = ( \dot{y}^k + y^a \Gamma^k_{ja} ) \partial_{x^k} |_{t=0}.$ On the other hand we know that $$\dot y(0)= (- \Gamma^1_{jk} y^k,..,-\Gamma^n_{jk} y^k)$$. Substituting this in the expression for $$K_\theta \delta_{x^j}$$, gives then that $$K_\theta \delta_{x_j} = ( - \Gamma^k_{ja} y^a + y^a \Gamma^k_{ja} ) \partial_{x^k} |_{t=0} = 0.$$ We thus see that $$\delta_{x^j} \in H(\theta)$$.

It remains to show that $$d_\theta \pi(\delta_{x^j}) = \partial_{x^j}$$. Since $$d_\theta \pi$$ is a linear map, we need only show that $$d_\theta \pi(\partial_{x^j}) = \partial_{x^j}, \quad d_\theta \pi(\partial_{y^k}) = 0.$$ This holds because firstly we have that $$(d_\theta \pi(\partial_{x^j})) f = \partial_{x^j}(f\circ \pi) = \partial_{x^j} f,$$ And secondly that $$d_\theta \pi(\partial_{y^k}) = \partial_{y^k}(f\circ \pi) = 0. \quad\square$$

Lemma 2. The vectors $$\{\partial_{y^k}\}_{k=1}^n$$ are a basis for the subspace $$V(\theta)$$, $$\theta =(x_0,y_0)$$.

Proof. By similar but simpler computations as in Lemma 1 one sees that $$d_\theta\pi(\partial_{y^k}) = 0 \quad \text{ and } \quad K_\theta(\partial_{y^k}) = \partial_{x^k}.$$ It follows that $$\partial_{x^k} \in \ker(d_\theta\pi)=V(\theta)$$. Moreover, we had that $$K_\theta \colon V(\theta) \to T_x M$$ is an isomorphism, so that basis vectors are mapped to basis vectors.

Recall that the Sasaki metric was defined by the formula $$\langle \eta, \xi \rangle_{\text{Sasaki}} := \langle (d_\theta\pi)\eta, (d_\theta\pi)\xi \rangle + \langle K_\theta\eta, K_\theta\xi \rangle,$$ for $$\xi,\eta \in T_\theta TM$$.

Given vectors $$\xi,\eta \in T_\theta TM$$ and writing them in the basis given by Lemma 1 and Lemma 2, i.e. $$\xi = X^i \delta_{x^i} + Y^k \partial_{y^k}, \quad \eta = \tilde X^i \delta_{x^i} + \tilde Y^k \partial_{y^k},$$

we get that $$\langle \xi, \eta \rangle_{\text{Sasaki}} = \langle X^i \partial_{x^i}, \tilde X^i \partial_{x^i} \rangle + \langle Y^k \partial_{x^k}, \tilde Y^k \partial_{x^k} \rangle = g_{jk} X^j \tilde X^k + g_jk Y^j \tilde Y^k.$$