Wedge product of vectors and of linear functionals

Let's take the following definition of the wedge product of two vectors (see definition 14 in this lecture notes:

Definition 13 The second exterior power $\Lambda^2V$ of a finite-dimensional vector space is the dual space of the vector space of alternating bilinear forms on $V$ . Elements of $\Lambda^2V$ are called 2-vectors.

This definition is a convenience – there are other ways of defining $\Lambda^2V$ , and for most purposes it is only its characteristic properties which one needs rather than what its objects are. A lot of mathematics is like that – just think of the real numbers. Given this space we can now define our generalization of the cross-product, called the exterior product or wedge product of two vectors.

Definition 14 Given $u,v\in V$ the exterior product $u\wedge v\in\Lambda^2V$ is the linear map to $F$ which, on an alternating bilinear form $B$, takes the value

$$(u\wedge v)(B) = B(u, v)$$

Of course, this nice definition works also in the case when $V$ is a dual space of some other vector space $W$, i.e. when $V=W^*$. But for this special case there is a commonly used other definition for the wedge product:

Second kind definition. $u\wedge v$ is the alternating bilinear form (i.e an element of $\mathcal A^2(W)$, that maps $(x,y)\in W^2$ to $u(x)v(y)-u(y)v(x)$: $$(u\wedge v)(x,y)=u(x)v(y)-u(y)v(x)$$

My question: if I use Definition 14 for the wedge product, then how can I identify it with the bilinear form appearing in the second kind definition?

Edit: Trying to understand md2perpe's comment.

To avoid the confusion, I will denote the second kind wedge product by $\barwedge$. Our inventory is the following.

• $x,y\in W=V^*$,
• $u,v\in V=W^*$,
• $(u,v)\in V^2=(W^*)^2$,
• $B\in \mathcal A^2(V)$
• $u\wedge v: \mathcal A^2(V)\to\mathbb R$
• $u\barwedge v: W^2\to \mathbb R$

So, as md2perpe ponted out, and the last two items show that an identification of the elements of $\mathcal A^2(V)$ (i.e. of the alternating bilinear forms on $V$) with the elements of $W^2=(V^*)^2$ (i.e of pairs of linear functionals on $V$) would solve my problem. He says that the $(x,y)\in W^2$ should be identified with the bilinear form $V^2\to \mathbb R: (u,v)\mapsto x(u)y(v)-x(v)y(u)$. As far as I see, this bilinear form is none other than $x\barwedge y$. A bit circular, but interesting.

• Given Definition 14 take $B = B_{(x,y)}$ defined by $B_{(x,y)}(u,v) = u(x)v(y) - u(y)v(x)$ for $(x,y) \in W^2$ and $(u,v) \in W^*{}^2$. – md2perpe Oct 27 '17 at 19:37