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