Reading Ana Cannas da Silva's book, I found the following step defining the tautological form (the "$p_i\wedge dq^i$" form) in a coordinate-free manner.

Let $X$ be a given manifold, its cotangent bundle being $M\equiv T^\ast X$. Let's consider a point $x\in X$, its cotangent space being $T^\ast_pX$, and an element $\xi\in T_p^\ast X$; we have the natural projection: $$ \pi : M \equiv T^\ast X \longrightarrow X \\ \ \ \ \ \ p=(x,\xi)\longmapsto x. $$ We choose to define the tautological form $\alpha_p$ as $$ \alpha_p = \left(\mathrm{d}\pi_p\right)^\ast_p\xi, $$ meaning by this that for each vector $u\in T_xX$ $\alpha_p$ acts as: $$ \alpha_p(u) = \xi \left(\mathrm{d}\pi_p(u)\right) $$ where $\mathrm{d}\pi = \pi_\ast$ is the usual differential (the application, induced by the map $\pi$, between tangent spaces at start and arrival point).

My point is: how is this different to the "pullback" $\pi^\ast$: $$ \left(\pi^\ast\xi\right)_p (u) = \xi_{\pi(p)}\left(\mathrm{d}\pi_p(u)\right)? $$ Is this just a more specific notation, or am I missing something more subtle?


They're basically not different-you're just conflating the pullback of forms with the pullback of cotangent vectors. $(d\pi_p)$ is just a linear map $TT^*X_p\to X_{\pi p}$ so that $(d\pi_p)^*$ is the dual linear map. To be picky, the $\xi$ in Silva's notation is a covector at $p$, not a 1-form, so the notation $(\pi^*\xi)_p$ is not meaningful. Probably this is no big deal, but in case it's still not clear, just remember that a 1-form is a map $X\to T^*X$ whereas a cotangent vector is an element of $T^*X$; and the pullback you're wanting to use is defined on the former, not the latter.

  • $\begingroup$ Ok so essentially my problem is a notational one as I suspected: I was taught that one-forms are linear forms on vectors, therefore I thought that one-forms were the elements of $T^\ast_xX$. $\endgroup$ – Brightsun Mar 7 '14 at 23:48

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