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I'm trying to prove that geodesic flow on the cotangent bundle $T^* M$ is generated by the Hamiltonian vector field $X_H$ where

$$H = \frac{1}{2}g^{ij}p_i p_j$$

but I am stuck. Could somebody show me how to complete the calculation, or where I've made a mistake? Cheers!

I know that vector field for geodesic flow is

$$X = p^i \partial/\partial x^i - \Gamma^i_{jk}p^j p^k \partial /\partial p^i$$

so I must verify that

$$X \ \lrcorner\ \omega = - dH$$

where $\omega = dp_i \wedge dx^i$. It's easy to check that

$$X \ \lrcorner\ \omega = - p^i dp_i - \Gamma^l_{jk}g_{li}p^j p^k dx^i$$

$$-dH = -g^{ij}p_j dp_i -\frac{1}{2}\partial_i g^{jk}p_j p_k dx^i$$

Here I am stuck. Using the explicit formula for the Christoffel symbols in terms of the metric doesn't seem to work! Have I done something wrong, or am I missing something?

N.B. I'm aware of solutions that do not use coordinates, but I'd like to understand this one which does!

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  • $\begingroup$ Could you try writing out the explicit christoffel formulas in terms of metric derivatives? It has to work :). $\endgroup$
    – Alex R.
    Commented May 17, 2013 at 1:56
  • $\begingroup$ @Alex - I really don't see how this would work. The Christoffel formula gives three derivatives of $g$ with respect to different combinations of indices, and I have only one in the expression for $dH$. If you can get it to work, could you let me know how?! $\endgroup$ Commented May 17, 2013 at 9:01

1 Answer 1

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I've worked out my mistake - the vector field $X$ above is incorrect. The subtlety comes in raising and lowering the relevant indices.

Firstly we can all agree that the following vector field on the tangent bundle generates geodesic flow

$$Y = p^i \partial/\partial x^i - \Gamma^i_{jk} p^j p^k \partial/\partial p^i$$

where $(x^i, p^i)$ are coordinates on $TM$.

Now we want to use the tangent-cotangent isomorphism to determine a vector field on $T^*M$ which also gives geodesics after projection. Write out the ODEs for geodesic flow

$$\dot{x}^i = p^i $$ $$\dot{p}^i = -\Gamma^i_{jk}p^j p^k$$

Now define $p_i = g_{ij}p^j$ and calculate

\begin{align*}\dot{p}_i &= g_{ij,k}\dot{x}^k p^j + g_{ij}\dot{p}^j \\&= g_{ij,k}p^j p^k - g_{il}\Gamma^l_{jk}p^jp^k \\ &= p^j p^k(g_{ij,k} - g_{ij,k} + \frac{1}{2}g_{jk,i}) \end{align*}

Thus the correct vector field for geodesic flow on $T^* M$ is in fact

\begin{align*}X &= g^{ij}p_j \frac{\partial}{\partial x^i} + \frac{1}{2}\frac{\partial g_{jk}}{\partial x^i} p^j p^k \frac{\partial }{\partial p_i}\\ &= g^{ij}p_j \frac{\partial}{\partial x^i} - \frac{1}{2}\frac{\partial g^{jk}}{\partial x^i} p_j p_k \frac{\partial }{\partial p_i} \end{align*}

by raising and lowering indices, and using the derivative of the inverse metric. Now this agrees with $-dH$ in the question, after contracting with $\omega$.

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  • $\begingroup$ How have you found the vector field $X$ on $T^*M$ from $Y$ on $TM$? I know there will be the canonical isomorphism, but I don't know how to use it. $\endgroup$ Commented Sep 2, 2020 at 9:47

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