1
$\begingroup$

Let $M$ be a smooth manifold and for each point $p \in M$, let $T_pM$ denote the tangent space at $p \in M$. We define the set $TM = \bigsqcup_{p \in M}T_pM$ and equip it with the initial topology and canonical projection $\pi$ to construct the tangent bundle $TM \overset{\pi}{\to} M$ (the standard way to construct the tangent bundle, just including for completeness). Similarly, let $T_p^*M$ denote the cotangent space at $p \in M$ and define the set $T^*M = \bigsqcup_{p \in M}T_p^*M$. We construct the cotangent bundle $T^*M \overset{\pi'}{\to}M$ analogous to how we did for the tangent bundle.

Let $\Gamma(TM):=$ { $\sigma: M \to TM| \pi \circ \sigma = \text{id}_M, \ \sigma \ \text{smooth}$} be the $C^{\infty}(M)$-module of smooth vector fields on $M$.

Let $\Gamma(T^*M):=$ { $\omega: M \to T^*M| \pi' \circ \omega = \text{id}_M, \ \omega \ \text{smooth}$} be the $C^{\infty}(M)$-module of smooth covector fields on $M$.

Let $\Gamma(TM)^*:=$ { $\psi: \Gamma(TM) \overset{\sim}{\to}C^{\infty}(M)$} denote the dual of $\Gamma(TM)$.

Does there exist a canonical isomorphism $\Phi: \Gamma(TM)^* \to \Gamma(T^*M)$?

$\endgroup$
2
  • $\begingroup$ Do they even have the same dimension? $\endgroup$
    – pancini
    Aug 4, 2023 at 19:41
  • $\begingroup$ @pancini They're not even guaranteed a basis in ZFC because $C^{\infty}(M)$ isn't a division ring, so a notion of dimension is in general undefined for $\Gamma(TM)$, unless it admits a basis (someone please correct me if there is an alternative definition of dimension that can work here). However, locally once you choose a chart at a point, you can express the vector / covector fields in terms of the induced chart basis. $\endgroup$
    – Druizr
    Aug 4, 2023 at 19:44

1 Answer 1

3
$\begingroup$

Given $\alpha\in \Gamma(T^*M)$ there is a natural map $\alpha': \Gamma(TM)\to C^\infty(M)$. This is given by applying the natural pairing $\langle \cdot,\cdot\rangle:T_p^*M\times T_pM\to \mathbb{R}$ at each point, $\alpha': X\mapsto (p\mapsto \langle \langle \alpha_p, X_p\rangle)$. It is clear that such a map is injective and the fact that it is an isomorphism is a consequence of the Tensor Characterization Lemma. This lemma can be found in John Lee's book Introduction to Smooth manifolds, specifically chapter 12.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .