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Let $M$ denote a smooth manifold. Then a covector at $p \in M$ is an element of the dual space of $T_p M$. We can organize covectors into a bundle over $M$, and then define a $1$-form on $M$ to be a section of this bundle.

Question. Is there a more direct approach to defining $1$-forms, like so:

A $1$-form on $M$ is a linear way of turning sections of $TM \rightarrow M$ into smooth functions $M \rightarrow \mathbb{R}$ satisfying some smoothness or "locality" conditions.

(I'm also interested in defining arbitrary $k$-forms in this way.)

goblin GONE
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  • I don't understand, you gave the exact answer to your question in your definition : a section of $T^M$ is exactly the data of a function $f \in \mathfrak X(M)^$. If you want this to be smooth, well take local trivialization and ask it to be smooth. What more do you want ? –  Jul 02 '17 at 15:05
  • Maybe you are annoyed with checking that all of these different constructions form bundles? If so, you can construct all them from the associated principle $GL_n(V)$ bundle $P$, and the appropriate $GL(V)$ representation $W$, as a balanced product $P \times_G W$. $W = V^$ would give you the cotangent bundle, $\wedge^k (V^)$ the bundle of $k$ forms. – Elle Najt Jul 02 '17 at 15:17

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$\newcommand{\O}{\mathcal{O}}$ $\newcommand{\D}{\mathcal{D}}$ Let $\O_M=C^\infty(M)$ be the ring of smooth functions on $M$. Let $\D_M$ be the set of $\Bbb R$-linear derivations of the ring $\O_M$. We can identify the elements of $\D_M$ with the smooth tangent fields on $M$. Then $\D_M$ is an $\O_M$-module, and we can identify the smooth $1$-forms with the module $\text{Hom}_{\O_M}(\D_M,\O_M)$.

This point of view is developed systematically in the book Smooth Manifolds and Observables by "Jet Nestruev" (Springer GTM 220).

Angina Seng
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You essentially said the same thing twice. A covector is a way of turning a vector into a scalar. Therefore a covector field (i.e., a differential 1-form on $M$) is a way of turning a vector field into a scalar field, i.e., a function on $M$.

Mikhail Katz
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  • But surely it's not an arbitrary (linear) way of turning vector fields into scalar fields? – goblin GONE Jul 02 '17 at 14:52
  • Requiring smoothness is a separate issue! You wanted a global definition but the definition you gave is already a global definition. Perhaps you can clarify your question. Are you looking for a global way of defining smoothness of functions or something? @goblin – Mikhail Katz Jul 02 '17 at 14:54
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As Mikhail Katz says, a $1$-form $\alpha$ is a machine that eats vector fields and spits out functions. Nothing more than that. As many others before me commented, you already wrote this in your post. If smoothness is the issue that bothers you, just add the requirement $$X\;\mathrm{is\;smooth}\Rightarrow\alpha(X)\mathrm{\;is\;smooth.}$$

Amitai Yuval
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