22
$\begingroup$

I'm trying to understand a little bit about symplectic geometry, in particular the tautological 1-form on the cotangent bundle. I'm following Ana Canas Da Silva's notes.

On page 10 she describes the coordinate free definitions and gives an exercise to find the expression in the local coordinates $\sum_{i=1}^n \xi_i dx_i$. I've tried to do this exercise but can't seem to be able to do it which is really annoying since everywhere I look it is said to be trivial and as a consequence never formally proved.

$\endgroup$
24
$\begingroup$

Let $(x^i)$ be local coordinates on our base manifold $M$ and let $(x^i, \xi_j)$ be the induced coordinates on the cotangent bundle $T^* M$. Let $\pi : T^*M \to M$ be the projection $(x^i, \xi_j) \mapsto (x^i)$. It induces a $C^\infty (M)$-linear map on $1$-forms, which I will write as $\pi^* : \Omega^1 (M) \to \Omega^1 (T^* M)$. In coordinates, this sends a $1$-form $\phi = \phi_i \, \mathrm{d} x^i$ (summation convention) to $(\phi_i \circ \pi) \, \mathrm{d} x^i$. As usual this induces a $\mathbb{R}$-linear map on the fibres, namely $\pi^*_{(x, \xi)} : T^*_x M \to T^*_{(x, \xi)} (T^* M)$, sending the covector $p$ to the covector $(p, 0)$. (We must be careful and distinguish between covectors and $1$-forms here, to avoid confusion.)

The tautological $1$-form on $T^* M$ is defined to be $\pi^*_{(x, \xi)} \xi$ at each point $(x, \xi)$ in $T^* M$. Why does this formula even make sense? Well, $\xi$ by definition is an element of $T_x^* M$, so it typechecks. Thus the point $(x, \xi)$ is mapped to the covector $(\xi, 0)$ in $T^*_{(x, \xi)} (T^* M)$, and so the tautological $1$-form in coordinates is given by $$\xi_i \, \mathrm{d} x^i$$ as claimed. (The coefficient of $\mathrm{d} \xi_j$ is $0$, of course.)

(Perhaps the reason no-one likes writing this out in full is because the tautological nature of the construction makes it quite confusing, unless one keeps track of the types of all the expressions involved.)

$\endgroup$
  • $\begingroup$ Could someone explain the last part in a little more detail please? I'm not sure how to go from one goes from $(x,\xi )$ mapping to $(\xi,0)$ to writting the one from in coordinates. $\endgroup$ – T. Stark Sep 12 at 17:43
1
$\begingroup$

EASY METHOD:

Suppose that $(p,\omega_p)\in Q \equiv T^*(M)$, where $p \in M, \omega_p \in T_p^*(M)$, represents a point in the cotangent bundle $Q$ whose fibres are the cotangent linear spaces at each point in the base manifold. We have the canonical fibre bundle projection,

\begin{align*} \pi: \quad Q &\to M \\ (p,\omega_p) &\mapsto p \ \end{align*}

which induces the pullback of 1-forms,

\begin{align*} \pi^*: \quad \Omega (M) &\to \Omega (Q) \\ \omega &\mapsto \pi^*(\omega)\equiv \theta. \ \end{align*}

In local coordinates, if we write $ \omega = \sum_i \alpha_i \ \text{d}x^i $, then it is straightforward to compute the local representation of the pullback:

$$ \theta = \pi^*(\sum_i \alpha_i \ \text{d}x^i) = \sum_i (\alpha_i\circ\pi) \ \pi^*(\text{d}x^i)\ ``=" \sum_i \alpha_i \ \text{d}x^i$$

Note the quotation marks I used to emphasize that one should be careful about abuse of notation and use them smartly for convenience knowing that it should not be taken too literally.

ALTERNATIVE METHOD:

For a vector field $X \in \mathfrak{X}(Q)$, one could locally decompose it into a component parallel to the fibre $T_p^*(M)$ and another parallel to the base $M$ for each $p \in M$ :

$$ X = \sum_j (X_{\parallel}^j\circ\pi)\frac{\partial}{\partial x^j} + \sum_i X_{\perp}^i \frac{\partial}{\partial \alpha^i} $$

where we have again considered $ \omega = \sum_i \alpha_i \ \text{d}x^i $. By this decomposition, we mean that

$$ \pi_*(X)= \sum_j X_{\parallel}^j \ \frac{\partial}{\partial x^j}$$

And therefore, we have that

\begin{align*} \theta_{(p,\omega_p)}(X) = \pi_{(p,\omega_p)}^*(\omega_p)(X) &= \omega_p (\pi_{*_{(p,\omega_p)}}(X)) \\ &= \sum_i \alpha_i(p)\ \text{d}x^i (\sum_j X_{\parallel}^j(p) \ \frac{\partial}{\partial x^j}) \\ &= \sum_i (\alpha_i\circ\pi)(p,\omega_p)(X_{\parallel}^i\circ\pi)(p,\omega_p)\ \end{align*}

Thus: $$ \theta (X) = \sum_i (\alpha_i\circ\pi) (X_{\parallel}^i\circ\pi) = \sum_i (\alpha_i\circ\pi) \ \pi^*(\text{d}x^i)(X) $$

which is the same result that we convinced ourselves of before.

$\endgroup$
  • $\begingroup$ I know that's an old answer...anyway, could I ask you why $\pi^*(dx^i) = dx^i$ in your third equation? $\endgroup$ – Lo Scrondo Jan 26 at 19:11
  • 1
    $\begingroup$ @LoScrondo It's not really. Note that the equality is within quotes emphasising that it is a notational abuse. $\endgroup$ – Nanashi No Gombe Jan 27 at 19:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.