Associativity of wedge product in a vector space 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?
 A: 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. 
