coordinates for $T^*M$, $T^*T^*M$, $T^*T^*... T^*M$

84 Views Asked by At

This relates to my previous question on the tautological one form. I'm reading John Lee's Smooth Manifold book. Anyway, I wanted to clear up on some notation, which is confusing at first, but I think I have made sense of it now.

Lets say $M$ is a smooth manifold with cotangent bundle $T^*M$ and $dim(T^*M)=n$. Then for the case of the tangent bundle, we'll just replace $T^*M$ with $TM$ below.

On $T^*M$ we have:

  • $\omega_i dx^i$ giving coordinates $(x^1, \dots, x^n, \omega_1, \dots, \omega_n)$

On $T^*T^*M$ we have:

  • $\omega_i dx^i+ \sum \xi_i d\omega_i$ giving coordinates $(x^1, \dots, x^n, \omega_1, \dots, \omega_n, \xi_1, \dots, \xi_n)$

Then we can just continue to generalize this, so for $T^*T^*T^*M$ :

  • $\omega_i dx^i+\sum \xi_i d\omega_i+\sum \gamma_i d\xi_i$ giving coordinates $(x^1, \dots, x^n, \omega_1, \dots, \omega_n, \xi_1, \dots, \xi_n, \gamma_1, \dots, \gamma_n)$

Is this right?

But the index in $\xi_i d\omega_i$ no longer follows Einstein's summation notation? So an explicit $\sum$ is always needed?

Then, for the pullback mapping $\pi^* : T^*M \to T^*T^*M$, its role is to add $n$ 0's to the last $n$ coordinates:

  • $\omega_i dx^i \to \omega_i dx^i + \sum 0 d\xi_i$

And we can generalize this as $\pi^{**} : T^*T^*M \to T^*T^*T^*M$:

  • $\omega_i dx^i + \sum \gamma_i d\xi_i \to \omega_i dx^i + \sum \gamma_i d\xi_i + \sum 0 d\gamma_i $

and so on....