Fix a real vector space $V$ of finite dimension. Let's denote by $\Lambda^p(V)$ the vector space of $p$-forms on $V$ (i.e. alternating $p$-tensors). Then we have the product $\wedge : \Lambda^p(V) \times \Lambda^q(V) \to \Lambda^{p + q}(V)$ given by $(\omega \wedge \eta)(X_1, \ldots, X_{p + q}) = \frac{1}{p! q!} \sum_{\sigma \in S_{p + q}} sgn\ \sigma \cdot \omega(X_{\sigma(1)}, \ldots, X_{\sigma(p)}) \eta(X_{\sigma(p + 1)}, \ldots, X_{\sigma(p + q)})$. How can I prove that $\wedge$ is associative? I've tried developing it from the definition but in the end I don't get things that look nice or are obviously equal. Specifically, I get:
$(\omega \wedge \eta) \wedge \theta(X_1, \ldots, X_{p + q + r}) = \frac{1}{(p + q)! r! p! q!} \sum_{\sigma \in S_{p + q + r}} \sum_{\tau \in S_{p + q}} sgn\ \sigma\ sgn \tau \cdot \omega(X_{\sigma \tau(1)}, \ldots, X_{\sigma \tau(p)}) \eta(X_{\sigma \tau(p + 1)}, \ldots, X_{\sigma \tau(p + q)}) \theta(X_{\sigma(p + q + 1)}, \ldots, X_{\sigma(p + q + r)})$
$\omega \wedge (\eta \wedge \theta)(X_1, \ldots, X_{p + q + r}) = \frac{1}{p! (q + r)! q! r!} \sum_{\sigma \in S_{p + q + r}} \sum_{\tau \in S_{q + r}} sgn\ \sigma\ sgn\ \tau \cdot \omega(X_{\sigma(1)}, \ldots, X_{\sigma(p)}) \eta(X_{\sigma(p + \tau(1))}, \ldots, X_{\sigma(p + \tau(q))}) \theta(X_{\sigma(p + \tau(q + 1))}, \ldots, X_{\sigma(p + \tau(q + r))})$
Best Answer
The wedge product is defined as: let $\omega\in \Omega^k(V)$ and $\tau\in \Omega^l(V)$; then
$$\omega \wedge \tau=\frac{1}{k!l!}A(\omega\otimes\tau);$$
where $$A(\omega)(x_1,...,x_k)=\sum_{\sigma\in S_{k}}sgn(\sigma)\omega(x_{\sigma(1)},...x_{\sigma(k)})=\sum_{\sigma\in S_{k}}sgn(\sigma)\sigma\omega.$$ $$ $$
LEMMA: If $\omega\in\Omega^k(V)$, $\tau\in\Omega^l(V)$ then: $A(A(\omega)\otimes\tau)=k!A(\omega\otimes\tau).$
This lemma follows from the definition of $A(\omega)$. Now we can show the associativity of the wedge product.
Let $\omega\in\Omega^k(V)$, $\tau\in\Omega^l(V)$, $\eta\in\Omega^r(V);$ then, by definition,
$$(\omega\wedge\tau)\wedge\eta=\frac{1}{(k+l)!r!}A((\omega\wedge\tau)\otimes\eta)$$
$$=\frac{1}{(k+l)!r!}\frac{1}{k!l!}A(A(\omega\otimes\tau)\otimes\eta) $$ and by the lemma above we have $$=\frac{(k+l)!}{(k+l)!r!k!l!}A((\omega\otimes\tau)\otimes\eta)=\frac{1}{r!k!l!}A((\omega\otimes\tau)\otimes\eta).$$
Similarly
$$\omega\wedge(\tau\wedge\eta)=\frac{1}{k!l!r!}A(\omega\otimes(\tau\otimes\eta)),$$
since the tensor product is associative we can conclude.
The tensor product is associative, in fact:
$$(\omega\otimes\tau)\otimes\eta(v_1,...,v_k,v_{k+1},...,v_{k+l},v_{k+l+1},...,v_{k+l+r})=\omega(v_1,...,v_k)\tau(v_{k+1},...,v_{k+l})\eta(v_{k+l+1},...,v_{k+l+r})=\omega\otimes(\tau\otimes\eta)(v_1,...,v_k,v_{k+1},...,v_{k+l},v_{k+l+1},...,v_{k+l+r}).$$