I'm reading a book on differential forms to try to understand wedge products. Given two vectors $u$ and $v$, the wedge product is defined $u\wedge v = u^Tv - v^Tu$.
It is stated that the wedge product is associative, but I'm having trouble proving it. For example, $$\begin{split} u\wedge (v\wedge w) &= u\wedge(v^Tw - w^Tv)\\ &= u\wedge(v^Tw) - u\wedge(w^Tv)\\ &= u^Tv^Tw - w^Tvu - u^Tw^Tv + v^Twu\\ &= u^Tv^Tw - u^Tw^Tv + v^Twu - w^Tvu, \end{split}$$ but $$\begin{split} (u\wedge v)\wedge w &= (u^Tv - v^Tu)\wedge w\\ &= (u^Tv)\wedge w - (v^Tu)\wedge w\\ &= v^Tuw - wu^Tv - u^Tvw + w^Tv^Tu, \end{split}$$ and they're clearly not the same. Is it because associativity does not apply for this definition?
The wedge product of two vectors is a multilinear function. The sign is in front of each term depends if the indices are an odd or even permutation.
$ \vec{u} \wedge \vec{v} = \sum_{\sigma \in S_n} (sgn \sigma) a^{\sigma(1)...\sigma(n)} (\vec{e_{\sigma(1)}} \otimes ... \otimes \vec{e_{\sigma(n)}}) $ n is the sum of the dimensions of the vectors spaces of $\vec{v}$ and $\vec{u}$.
In this case, the wedge product of two vectors, each in $\mathbb{R}^2$.
If $\vec{u}, \vec{v} \in V$
$ \vec{u} \wedge \vec{v} : V^* \times V^* \rightarrow \mathbb{R} \\ \vec{u} \wedge \vec{v} = a^{1,2} (\vec{e_1} \otimes \vec{e_2}) - a^{2,1} (\vec{e_2} \otimes \vec{e_1}) $
It takes, for example, this argument and spits out a number
$ \vec{x}^* \otimes \vec{y}^* \in V^* \times V^* \\ \vec{x}^* \otimes \vec{y}^* = b_{i,j} (\vec{e^i} \otimes \vec{e^j}) $
So, $ (\vec{u} \wedge \vec{v})(\vec{x}^* \otimes \vec{y}^*) = a^{1,2} b_{1,2} - a^{2,1} b_{2,1} $
The wedge product of 3 vectors, each in $V$ acts on an element in $V^* \times V^* \times V^*$. So to show associativity, you have to act on a (0,3)-type tensor and show the components are the same.