I'm given the exercise to show that the exterior product is the same as the cross product if we identify $\bigwedge^2(\mathbb{R}^3)$ with $\mathbb{R}^3$.
So let $x,y \in \mathbb{R}^3$ with $x=a_1 e_1+a_2e_2+a_3e_3$ and $y=b_1 e_1+b_2e_2+b_3e_3$
Then we have that (by using the distributivity and the alternating property of the wedge product
$x \wedge y = (a_1 b_2-a_2 b_1)(e_1 \wedge e_2) + (a_3 b_1-a_1 b_3)(e_3 \wedge e_1)+(a_2b_3-a_3b_2)(e_3 \wedge e_2$ ).
If I now identify $e_1 \wedge e_2$ with $e_3$, $e_2 \wedge e_3$ with $e_1$ and $e_3 \wedge e_1$ with $e_2$ the sum above is obviously equal to the cross product.
Is this what the exercise asks for? I'm just pretty confused as I don't completely understand what is meant by "identifying". I mean if I understand the wedge product correctly $e_1 \wedge e_2$ is not the same as $e_3$ as $e_1 \wedge e_2$ is a 2-vector but $e_3$ a 1-vector. Also $x \wedge y$ is a 2-vector, while $x \times y$ is just a vector. So does this exercise even make any sense or am I misunderstanding something?
Best Answer
The idea is that the wedge product gives you a canonical application $\mathbb{R}^3\times \mathbb{R}^3\to \bigwedge^2(\mathbb{R}^3)$, and that there is then a more-or-less canonical isomorphism $\bigwedge^2(\mathbb{R}^3)\to \mathbb{R}^3$, and composing the two yields a map $\mathbb{R}^3\times \mathbb{R}^3\to \mathbb{R}^3$ which is precisely the cross product.
I think one gets a clearer idea of what's happening by looking at a more general case. Let's say that $E$ is a $3$-dimensional vector space (let's say over the real numbers, but it doesn't matter). Then there is a canonical map $E\times E\to \bigwedge^2(E)$, namely the wedge product. But there is no canonical equivalent of the cross product $E\times E\to E$. That is because although $E$ and $\bigwedge^2(E)$ both have dimension $3$, and are therefore isomorphic, they are not canonically isomorphic.
So what do we need to add to get a natural map $E\times E\to E$, or equivalently a natural isomorphism (an identification, as your question puts it) $\bigwedge^2(E)\simeq E$? What happens is that although there is no natural identification between $E$ and $\bigwedge^2(E)$, there is a natural duality. Namely, we have the wedge product $E\times \bigwedge^2(E)\to \bigwedge^3(E)$, where $\bigwedge^3(E)$ has dimension $1$. If we choose a non-zero element $u\in \bigwedge^3(E)$, we get an identification $\bigwedge^3(E)\simeq \mathbb{R}$, and therefore a (non-degenerate) pairing $E\times \bigwedge^2(E)\to \mathbb{R}$ which defines an isomorphism $\bigwedge^2(E)\simeq E^*$ (where $E^*$ is the dual of $E$). Using this, we get close to our cross product, since the wedge product $E\times E\to \bigwedge^2(E)$ now becomes $E\times E\to E^*$.
To finish, we now need an identification $E\simeq E^*$, which is equivalent to a non-degenerate bilinear form $E\times E\to \mathbb{R}$. You may think of a scalar product (there are other possibilities, but let's say scalar products are more palatable to beginners).
In the end, to transform our initial wedge product $E\times E\to \bigwedge^2(E)$ into a "cross product" $E\times E\to E$, we needed a scalar product on $E$, and a choice of a "volume form" $u\in \bigwedge^3(E)$. In particular, if $E$ is a euclidean space (that is, it has a choice of scalar product), then there are two natural choices for $u$, corresponding to "volume $1$" forms, the two choices corresponding to the two possible orientations of $E$. That means that there is a canonical cross product on any oriented eucliean space of dimension $3$, which is the familiar statement (explicitly, one defines $u\times v$ by the formula $$ \langle u\times v, w\rangle = \det(u,v,w) $$ for all $w\in E$, where $\det$ is the determinant corresponding to any direct orthonormal basis).
Of course, $\mathbb{R}^3$ is naturally an oriented euclidean space, so we can skip this whole theory and just use the classical formulas. But then we miss what's happening behind the curtain, and it looks like we are just doing random identification, as your question suggests.