# Why is the wedge product skew commutative? [closed]

Suppose we have a ring $$R$$ and some $$R-$$module $$N.$$ Let $$a \in \bigwedge^{k} N,$$ and $$b \in \bigwedge^{l} N.$$

Why is it that $$a \wedge b = (-1)^{kl} \; b \wedge a?$$

## closed as off-topic by Shaun, Dietrich Burde, Xander Henderson, egreg, LeucippusApr 7 at 1:24

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If $$a = e_1 \wedge e_2 \wedge \ldots \wedge e_k$$, and $$b = f_1 \wedge f_2 \wedge \ldots \wedge f_l$$, then in order to rearrange $$a \wedge b$$ into $$b \wedge a$$, I have to move each of the $$l$$ terms $$f_1, \ldots, f_l$$ "across" the $$k$$ terms $$e_1, \ldots, e_k$$. Each time I move an $$f$$ "across" an $$e$$, I introduce a minus sign. So in all there are $$kl$$ minus signs introduced.
By moving "across", I mean using the identity $$e_i \wedge f_j$$ = $$-f_j \wedge e_i$$.
• I was thinking the same, but I don't know how to justify the identity $e_{i} \wedge f_{j} = - f_{j} \wedge e_{i}$ in the first place. – Confused Student Apr 6 at 18:02
• (Or when it's not an axiom then it is an axiom that $v\wedge v=0$, and therefore $$0=(v+w)\wedge(v+w) = (v\wedge v)+(w\wedge w)+(v\wedge w)+(w\wedge v) = (v\wedge w)+(w\wedge v)$$ so $v\wedge w$ and $w\wedge v$ are always additive inverses.) – Henning Makholm Apr 6 at 18:15
• @AddledStudent are you also wondering why $v\wedge v = 0$ (or equivalently $v\wedge w = -w\wedge v$) is a useful axiom, i.e. why the wedge algebra is a useful thing? – Ben Blum-Smith Apr 6 at 20:20
• It should be mentioned that $a$ and $b$ may not have this form, but can instead be sums of expressions of this form. – Eric Wofsey Apr 6 at 21:51