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This is a follow up question on an answer to my previous question.

Let $M$ be a smooth $n$ manifold and let $U\subseteq M$ be a domain. Let $T_xU$ denote the tangent space to $U$ at point $x$. Let $\omega$ denote a differential $p$ form.

Previously, I understood that a differential $1$-form was a linear map $\omega : T_x U \to \mathbb R$. Now, in the answer the author writes that on all of $U$ the $1$ form is a (linear) map that takes a vector field and maps it to a scalar.

This made me wonder: Is the tangent space of a manifold a vector field?

(The problem I have with this question is that a vector field is a map taking a point and returning a vector but picking a point on a tangent space yields many vectors but I can't quite figure out what I don't understand here)

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self-learner
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2 Answers2

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A $1$-form is a map $\omega:U\rightarrow T^*U$. Now, if $X:U\rightarrow TU$ is a vector field then you can consider $\omega_p(X_p)$, in that sense, you can apply the $1$-form to a vector field. $TU$ is a vector $bundle$ not a vector field.

azarel
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For each $x \in U$, the tangent space $T_x U$ is a vector space. A "differential $1-$form at $x$" is a linear map $T_x U \to \mathbb R$; that is, it sends a vector to a scalar.

Now when we say differential $1-$form we normally mean a field of such maps - i.e. a function that sends each $x$ to a differential $1-$form $\omega_x : T_x U \to \mathbb R$. Such a form can act on any tangent vector, and can therefore act on a whole field of them at once: for a form $\omega$ and a vector field $X$ we get a scalar function $\omega(X)(p) = \omega_p (X_p).$

Anthony Carapetis
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