Let $M$ be a smooth manifold. A differential form of degree $k$ is a smooth section of the $k$th exterior power of the cotangent bundle of $M$.
Does it mean that a differential form of degree $k$ is a mapping $: M \rightarrow \wedge^k(T^*M) $?
If I am correct, "the $k$th exterior power of" should be followed by a vector space. I was wondering if a "cotangent bundle" $T^*M$ of a differentiable manifold $M$ is a vector space? The fiber is a vector space, but I am not sure if the total space, as the union of a family of vector spaces indexed by points in $M$, also is.
At a point of $M$, how does the definition of a $k$-form above lead to an alternating multilinear map, i.e., how does an element of$\wedge^k(T^*M)$ become an alternating multilinear $T_p M\times \cdots \times T_p M \to \mathbb{R}$, as stated in the following?
At any point $p∈M$, a $k$-form $β$ defines an alternating multilinear map
$\beta_p\colon T_p M\times \cdots \times T_p M \to \mathbb{R}$
(with $k$ factors of $T_pM$ in the product), where $T_pM$ is the tangent space to $M$ at $p$. Equivalently, $β$ is a totally antisymmetric covariant tensor field of rank $k$.
Quotes are from Wikipedia. Thanks and regards!