Wedge Product and Cross Product I am trying to understand the wedge product. I know that:

*

*It acts only on p-forms (and hence will act on the dual vector space for my question ahead),

*It is different from cross-product in general (henceforth, I will restrict my question to $\textbf{R}^3$ only).

I am following the notes by David Tong. On page 77, he says:

As a more specific example, consider $M = \textbf{R}^3$ and $ω = ω_µdx^µ$ and $η = η_µdx^µ$. We then have
$$ω ∧ η = (ω_1dx^1 + ω_2dx^2 + ω_3dx^3
) ∧ (η_1dx^1 + η_2dx^2 + η_3dx^3)
= (ω_1η_2 − ω_2η_1)dx^1 ∧ dx^2 + (ω_2η_3 − ω_3η_2)dx^2 ∧ dx^3 + (ω_3η_1 − ω_1η_3)dx^3 ∧ dx^1$$
Notice that the components that arise are precisely those of the cross-product acting
on vectors in $\textbf{R}^3$. This is no coincidence: what we usually think of as the cross-product between vectors is really a wedge product between forms

I don't understand the last line. Is it because the dual space and normal vector space are the same? So, working with duals will fetch us the result. Because clearly, vector and dual vector have different components and bases in general.
 A: The relationship between the wedge and cross products is given by taking the Hodge star. Generally speaking, if $V$ is a vector space, with no extra structure we can always talk about the wedge product as an operation
$$\Lambda^k(V) \otimes \Lambda^{\ell}(V) \ni \alpha \otimes \beta \mapsto \alpha \wedge \beta \in \Lambda^{k+\ell}(V).$$
(If you want to think in terms of forms then $V$ is the cotangent space, the dual of the tangent space. I don't want to write things in terms of the tangent space to reduce the number of duals floating around.) This makes sense for any vector space, not necessarily finite-dimensional, over any field (and in fact more generally). Now some special things happen if $V$ has extra structure:
First, if $V$ is $n$-dimensional and has an orientation, by which I mean we have chosen a nonzero vector $\omega \in \Lambda^n(V)$, then the wedge product gives a map $\Lambda^k(V) \otimes \Lambda^{n-k}(V) \to \Lambda^n(V) \cong F$ (where $F$ is the underlying field) which you can check is nondegenerate. Hence it induces an ($SL(V)$-equivariant) isomorphism
$$\Lambda^k(V) \cong \Lambda^{n-k}(V)^{\ast}.$$
(Without an orientation we instead get a $GL(V)$-equivariant isomorphism $\Lambda^k(V) \cong \Lambda^{n-k}(V)^{\ast} \otimes \Lambda^n(V)$.)
Next, if $V$ is real and has an inner product, then each of the exterior powers $\Lambda^k(V)$ inherits an inner product. It is slightly annoying to write down this inner product in general (we need to use a Gram determinant) but it is determined by the condition that if $\{ e_i \}$ is an orthonormal basis of $V$ then the wedge products of all distinct $k$-tuples of the $e_i$ are an orthonormal basis of $\Lambda^k(V)$. This inner product induces an ($O(V)$-equivariant) isomorphism
$$\Lambda^k(V) \cong \Lambda^k(V)^{\ast}.$$
Combining, if $V$ has both an orientation and an inner product, we get an isomorphism
$$\star: \Lambda^k(V) \cong \Lambda^{n-k}(V)^{\ast} \cong \Lambda^{n-k}(V)$$
called the Hodge star. It is determined by the condition that
$$\alpha \wedge (\star \beta) = \langle \alpha, \beta \rangle \omega$$
where $\omega$ is the orientation.
Now suppose that $V$ has an inner product, an orientation (determined, say, by a wedge product of the elements of an orthonormal basis), and furthermore is exactly $3$-dimensional. Then we can apply the Hodge star to the wedge product $V \otimes V \to \Lambda^2(V)$ to produce a map
$$V \otimes V \ni v \otimes w \mapsto \star(v \wedge w) \in V.$$
This is the cross product! This makes for a nice if slightly tedious exercise.
If $V$ is $3$-dimensional and has an inner product but we don't choose an orientation then the cross product does not take values in $V$ but actually in $V \otimes \Lambda^3(V)$; this is why people sometimes say the cross product is not really a vector but a "pseudovector".
