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 $\Phi: \Gamma(TM)^* \to \Gamma(T^*M)$?

  • $\begingroup$ 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


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.


You must log in to answer this question.

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