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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!
Best Answer
Yes, if you understand this to hold for every point $p \in M$: $$ d: p \to T^*_p M ∧ ...∧ T^*_p M $$ i.e. a k-form $d$ maps every point in M to an element of the k-exterior product of the cotangent space at $p$.
Strictly speaking, yes, but in differential geometry it is often understood that one talks about the construction on all vector spaces at all points p of M (e.g. tangent or cotangent spaces).
No, there are no algebraic operations defined on points of a manifold, a priori.
That's the part that John M already addressed in his answer. Here is an elementary example:
A 1-form eats a vector field, for example, in cartesian coordinates in $\mathbb{R}^n$, we have a global vector field $$ \partial_x $$ and a global 1-form $$ d x $$ , and for every point $p \in \mathbb{R}^n$ we have the relation $$ d x_p (\partial_x)_p = 1 $$ (If you are not sure about this, you should try to plug in the definitions of the gadgets on the left side and see if you can compute the result.)
A two form would be, for example, $$ d x \wedge d y $$ which we can feed two vector fields $\partial_u, \partial_v$, but we know of the relation $$ (d x \wedge d y)_p (\partial_u, \partial_v)_p = - (d x \wedge d y)_p (\partial_v, \partial_u )_p $$ by definition of the wedge-product. Note that you first choose a base point p, then your two-form gives you an element in $T^*_p M \wedge T^*_p M$, which you can then feed two tangent vectors at $p$ to get a real number.
HTH.