Proving wedge product is associative

Question

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:

$(\omega \wedge \eta) \wedge \theta(X_1, \ldots, X_{p + q + r}) = \frac{1}{(p + q)! r! p! q!} \sum_{\sigma \in S_{p + q + r}} \sum_{\tau \in S_{p + q}} sgn\ \sigma\ sgn \tau \cdot \omega(X_{\sigma \tau(1)}, \ldots, X_{\sigma \tau(p)}) \eta(X_{\sigma \tau(p + 1)}, \ldots, X_{\sigma \tau(p + q)}) \theta(X_{\sigma(p + q + 1)}, \ldots, X_{\sigma(p + q + r)})$

$\omega \wedge (\eta \wedge \theta)(X_1, \ldots, X_{p + q + r}) = \frac{1}{p! (q + r)! q! r!} \sum_{\sigma \in S_{p + q + r}} \sum_{\tau \in S_{q + r}} sgn\ \sigma\ sgn\ \tau \cdot \omega(X_{\sigma(1)}, \ldots, X_{\sigma(p)}) \eta(X_{\sigma(p + \tau(1))}, \ldots, X_{\sigma(p + \tau(q))}) \theta(X_{\sigma(p + \tau(q + 1))}, \ldots, X_{\sigma(p + \tau(q + r))})$

Answer 1

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:

$$(\omega\otimes\tau)\otimes\eta(v_1,...,v_k,v_{k+1},...,v_{k+l},v_{k+l+1},...,v_{k+l+r})=\omega(v_1,...,v_k)\tau(v_{k+1},...,v_{k+l})\eta(v_{k+l+1},...,v_{k+l+r})=\omega\otimes(\tau\otimes\eta)(v_1,...,v_k,v_{k+1},...,v_{k+l},v_{k+l+1},...,v_{k+l+r}).$$

Answer 2

Lemma. Let $\omega$ be a $k$-linear form and $\tau$ an $l$-linear form. Then $$ A(A(\omega)\otimes\tau) = k!A(\omega\otimes\tau) $$

Proof. Let's start by introducing some notation.

For any given index set $I\subset\{1,\dots,n\}$ let $\pi_I$ denote the projection from $\mathbb V\times\cdots\times\mathbb V$ ($n$ times) onto the coordinates with indexes in $I$ in $\mathbb V\times\cdots\times\mathbb V$ ($|I|$ times).

If $\sigma$ is a permutation of indexes and $\omega$ an $n$-form, let's denote with $\omega_\sigma$ the $n$-form given by $$ \omega_\sigma(v_1,\dots,v_n) = \omega(v_{\sigma(1)},\dots,v_{\sigma(n)}). $$ Let $K=\{1,\dots,k\}$ and $L=\{k+1,\dots,k+l\}$. By definition \begin{align*} A(\omega)\otimes\tau &= (A(\omega)\circ\pi_K)(\tau\circ\pi_L)\\ &= \sum_{\varsigma\in S_k}\textrm{sg}(\varsigma) (\omega_\varsigma\circ\pi_K)(\tau\circ\pi_L). \end{align*} Therefore, \begin{align*} A(A(\omega)\otimes\tau) &= \sum_{\sigma\in S_{k+l}}\textrm{sg}(\sigma) \sum_{\varsigma\in S_k} \textrm{sg}(\varsigma)(\omega_\varsigma\circ\pi_K)_\sigma (\tau\circ\pi_L)_\sigma\\ &= \sum_{\sigma\in S_{k+l}}\sum_{\varsigma\in S_k} \textrm{sg}(\varsigma\sigma) (\omega\circ\pi_K)_{\varsigma\sigma}(\tau\circ\pi_L)_\sigma\\ &= \sum_{\varsigma\in S_k}\sum_{\sigma\in S_{k+l}} \textrm{sg}(\varsigma\sigma)(\omega\circ\pi_K)_{\varsigma\sigma} (\tau\circ\pi_L)_{\varsigma\sigma}\\ &= \sum_{\varsigma\in S_k}\sum_{\sigma\in S_{k+l}} \textrm{sg}(\sigma) (\omega\circ\pi_K)_\sigma(\tau\circ\pi_L)_\sigma\\ &= \sum_{\varsigma\in S_k}A(\omega\otimes\tau)\\ &= k!A(\omega\otimes\tau). \end{align*}