If $u \in \Lambda^{p}(V)$ and $v \in \Lambda^{q}(V)$, does $u \wedge v = (-1)^{pq} v \wedge u$?

Question

Consider the exterior algebra $$\Lambda^{\ast}(V)= \bigoplus_{m=0}^\infty \Lambda^m(V),$$ for a finite dimensional $F$-vector space.

If $u \in \Lambda^{p}(V)$ and $v \in \Lambda^{q}(V)$, does $u \wedge v = (-1)^{pq} v \wedge u$?

How do we go about proving this? I was thinking of trying to prove it for the images of the simple tensors first and then going from there (possibly extending this by linearity?). Also note that if $p$ or $q$ are even, then we have $u \wedge v = v \wedge u$.

Answer 1

Let $V$ be a vector space, $\wedge V$ is the quotient of the tensor algebra $T(V)$ by the ideal generated by $u\otimes v+v\otimes u, u,v\in V$. This implies that $u\wedge v=-v\wedge u$.

You have $u_1\wedge ...\wedge u_p\wedge v_1...\wedge v_q =(u_1\wedge ...\wedge u_p\wedge v_1)\wedge v_2...\wedge v_q =(-1)^pv_1\wedge u_1...\wedge u_p\wedge v_2...\wedge v_q$ and recursively you have

$u_1\wedge...\wedge u_p\wedge v_1...\wedge v_q=(-1)^{pq}v_1\wedge...\wedge v_q\wedge u_1...\wedge u_p$.

Answer 2

Yes, you can prove it first for the images of the simple tensors. For these, you just need to count how many times you swap two $dx$'s ($pq$ times). And yes, by linearity, this is enough.