Confusion regarding wedge products

Question

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?

Answer 1

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.

Answer 2

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.