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  1. 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.

  2. 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!

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Did you consider the (co-)tangent bundle of the circle? Hardly a vector space. Operations on bundles are understood here. Exterior powers are constructed analogously to the direct sum and tensor product of bundles: fiberwise. –  t.b. Aug 15 '11 at 13:04
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Thanks! By "hardly a vector space", do you mean it is or is not a vector space? I am still far from understanding differential form. –  Tim Aug 15 '11 at 13:37
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@Tim: The (co)tangent bundle of a manifold is not automatically equipped with the structure of a vector space. I think it is very important to understand precisely what a vector bundle is and isn't before trying to understand what higher differential forms are. –  Zhen Lin Aug 15 '11 at 13:42
    
I meant it isn't. –  t.b. Aug 15 '11 at 15:14
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2 Answers 2

up vote 2 down vote accepted

Does it mean that a differential form of degree k is a mapping $:M \to ∧^k(T^∗M)$?

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 $p$.

If I am correct, "the kth exterior power of" should be followed by a vector space.

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).

I was wondering if a "cotangent bundle" $T^∗M$ of a differentiable manifold M is a vector space?

No, there are no algebraic operations defined on points of a manifold, a priori.

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 $∧^k(T^∗M)$ become an alternating multilinear $T_p M×⋯×T_p M \to R$, as stated in the following?

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.

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The space of alternating linear maps $f:V^k \rightarrow K$ is naturally isomorphic to the dual of the exterior algebra $\wedge^k(V)^*$, which is in turn naturally isomorphic to $\wedge^k(V^*)$ for finite dimensional spaces.

Added later:

We have a differential form $\beta: M \rightarrow \wedge^k(T^*M)$, $p \mapsto \beta_p \in \wedge^k(T^*_pM)$.

$T^*_pM$ is a vector space, so we have the natural isomorphism given above, i.e. we can identify $\beta_p$ with an alternating multilinear map.

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