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While studying smooth manifolds and differential forms I have come across multiple definitions of the wedge product, and I have been having some trouble seeing the equivalence between them.

At the most general level, if we have two alternating tensors $f$ and $g$, of size $k$ and $\ell$ respectively, then $f \wedge g$ is another alternating tensor of size $(k + \ell)$. This suggests the wedge product is a sort of "multiplication operator" amongst tensors. If we apply this definition to two differential forms $\omega$ and $\tau$, then $\omega \wedge \tau$ makes sense because at each point $p$ in a manifold $M$, $\omega_p$ and $\tau_p$ are just alternating tensors so $\omega_p \wedge \tau_p = (\omega \wedge \tau)_p$ is an example of the above definition.

My first question: is there any interpretation of $\omega \wedge \tau$ that is not pointwise?

On the other hand, I have read that the wedge product of $n$ vectors is the same as the determinant of the $n$ vectors, or equivalently, the volume of the parallelepiped which they span. Using $\mathbb{R}^3$ as an example, we use linearity to obtain something of the form $$c_1 (e_1 \wedge e_2) + c_2(e_1\wedge e_3) + c_3 (e_2 \wedge e_3)$$ for some constants $c_i$.

My second question: in the above example the basis vectors $e_i$ are not alternating multilinear functions, so how does the wedge product make sense here? How are we to interpret it in view of the definition I gave?

My third question: my current interpretation of differential forms comes straight from their definition, so they are a function that assigns to each point in the manifold an alternating tensor. Does there exist a geometric interpretation of the wedge product of differential forms (or even just differential forms themselves for that matter) similar to the wedge product of vectors?

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    $\begingroup$ For the second question: you are conflating wedge product on covectors with wedge product on vectors. When you take the wedge product on differential forms (pointwise), you are taking this on covectors / multilinear forms or whatever you'd like to call them, but the latter wedge product is of vectors directly. The wedge product (and in more generality, the tensor product) can take some number of copies of vectors and covectors so you got to make sure you're understanding what is going on. $\endgroup$ Commented Feb 15, 2023 at 1:00
  • $\begingroup$ @OsamaGhani Thank you. I had a feeling something more general was going on and that explains it. The books I am using only define the wedge product in terms of alternating tensors and in the context of smooth manifolds/differential forms. $\endgroup$
    – CBBAM
    Commented Feb 15, 2023 at 1:05

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The global interpretation you are probably looking for makes use of the language of fibre bundles.

First, a differential $k$-form on a manifold $M$ is a section of the $k$-th exterior power of the cotangent bundle over $M$. This just means that if $\omega$ is a $k$-form on $M$ then you can look at it at a point $p$ of $M$, and this will give you a skew-symmetric tensor $\omega_p \colon \Lambda^k T_pM \to \mathbb R$. The map $p \in M \mapsto \omega_p \in \Lambda^k T_p^*M$ is smooth.

Now take a $k$-form $\omega$ and an $l$-form $\eta$ on $M$. Then $\omega \wedge \eta$ is a $(k+l)$-form, and its global interpretation is exactly as above. You are right to consider $\wedge$ as a product in the set of differential forms over $M$.

Regarding your second question, a $k$-vector is a section of the $k$-th exterior power of the tangent bundle over $M$ (e.g. $e_1 \wedge e_2$ as in your question is a $2$-vector). You can define a wedge product for $k$-vectors too, just take the same formal properties of the wedge product for forms and apply them to $k$-vectors.

I comment on your third question. Formally, vectors and covectors are completely different geometric objects. Likewise, $k$-forms and $k$-vectors are totally different. Nonetheless, they share the same properties with respect to $\wedge$, and sometimes they can be identified in a canonical way. For instance, when you have a metric tensor $g$ on your manifold (an inner product at each tangent space), then there is a canonical isomorphism $X_p \in T_pM \mapsto g_p(X_p,{}\cdot{}) \in T_p^*M$ transforming a tangent vector to a covector. You can extend this identification to $k$-vectors and $k$-forms as well. It is then obvious that the geometric interpretation you can give to operations with $k$-forms is the same as if you were using $k$-vectors.

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  • $\begingroup$ Thank you for your response, this is exactly what I was looking for. If I could accept this as an answer as well I would have. $\endgroup$
    – CBBAM
    Commented Feb 15, 2023 at 7:58
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First: The form $\omega\wedge\tau$ will necessarily be pointwise, since it's a tensor, and tensors are pointwise. I'm not sure what kind of answer you're hoping for.

Third: You can use Poincare duality and try to interpret closed forms in terms of that language. However, you should understand the use of forms; a major point of their existence is to be integrated. My advice it to ditch the need for geometric intuition and learn to compute in local coordinates with these objects.

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  • $\begingroup$ Thank you for your answer. I suppose part of my trouble is if someone asks me what is a 1-form, I will say it picks out a particular component of the tangent vector it acts on (or equivalently it assigns a function to each point, if for example the form is a sum of 1-forms). If you ask me what is an $n$-form, I will say that it assigns to each point an $n$-tensor. In general, is there not more that can be said (particularly geometrically)? $\endgroup$
    – CBBAM
    Commented Feb 15, 2023 at 1:00
  • $\begingroup$ I am tempted to say that is is the volume of the parallelepiped spanned by $n$ vectors, but that assumes that the form is exact which is not always true. It also assumes some notion of volume spanned by a set of functions which to my knowledge doesn't make any sense. $\endgroup$
    – CBBAM
    Commented Feb 15, 2023 at 1:01
  • $\begingroup$ I wouldn't bother trying to squeeze any geometry out of general forms. Remember 0-forms are functions; is there any general geometry you can say for arbitrary smooth functions? I think your uneasiness with forms will start dissipating when you start working with them more. $\endgroup$
    – Mr. Brown
    Commented Feb 15, 2023 at 1:12
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    $\begingroup$ Thank you for your help. I will take your advice and put off these questions until I first learn about integrating forms. $\endgroup$
    – CBBAM
    Commented Feb 15, 2023 at 1:17

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