I have a problem with the underlined step of the demonstration (picture below) . Why can $f$ and $g$ be switched ? I can't seem to grasp this part, since I know tensor product is not, in general, commutative. It has to do with the permutation somehow. Someone care to explain?
Anticommutativity wedge product
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differential-geometry
exterior-algebra
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0You are not commuting $f\wedge g$, but rather $f$ evaluated at $(v_{\sigma(1)},...,v_{\sigma(k)})$ with $g$ evaluated at $(v_{\sigma(k+1)},...,v_{\sigma(k+l)})$. These are elements of the field $\mathbb K$ and thus commute. – 2017-01-22
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0As @s.harp said, they are just numbers. – 2017-01-22