Wedge/cross products and associativity

[Rasalhague]

The wedge product is said to be associative (e.g. http://mathworld.wolfram.com/WedgeProduct.html ). The cross product is not associative. But the Hodge star is said to give an isomorphism between 3-dimensional (mono?)vectors and their bivectors. How is this possible when "if two objects are isomorphic, then any property which is preserved by an isomorphism and which is true of one of the objects, is also true of the other" ( http://en.wikipedia.org/wiki/Isomorphism ). Is the isomorphism given by the Hodge star with respect to vector addition only, and not also the cross and wedge products?

A component expression for the wedge product can be computed by a determinant in the same way as that of the cross product, suggesting that the wedge should be just as non-associative as the cross product. How is this apparent contradiction resolved: am I misunderstanding something about the cross product, the wedge product, the Hodge star, associativity or isomorphism or...?

 

[adriank]

The cross product of two vectors in a 3-dimensional space satisfies a × b = *(a ∧ b). So if you write out a × (b × c) and (a × b) × c in terms of wedge products and Hodge stars, you'll get different, inequivalent expressions. The main issue is that the wedge product doesn't commute with the isomorphism given by the Hodge star.

 

[Rasalhague]

Thinking further about this, *(a /\ b /\ c) is a scalar, whereas a x (b x c) and (a x b) x c are 1-vectors, so the comparison I was trying to draw doesn't make sense for the same reason as it doesn't make sense to ask whether the dot product is associative.

a x b = *(a /\ b)

a x (b x c) = a x *(b /\ c) = *( a /\ *(b /\ c)

(a x b) x c = *(a /\ b) x c = *(*(a /\ b) /\ c)

Does isomorphic here mean that vectors of the form a x b (I think that's all of the 1-vectors, isn't it?) are isomorphic to 2-vectors of the form a /\ b (2-blades) under vector addition? So that

*(a /\ b + c /\ d) = *(a /\ b) + *(c /\ d).