Thread: cross product question View Single Post

 Originally posted by matt grime In generality, given a finite dimenisional vector space V there is the n-fold tensor product, which can saefly be thought of as: let v_i be a basis of V, then a basis of $\bigotimes^nV$ is given by the set of all symbols $$v_{i_1}\otimes\ldots \otimes v_{i_n}$$ ( I am, slightly contradictarily, being a gentleman here by not defining that in a basis free way, if you want I can but it is more confusing for the non-algebraist, trust me)
while i am not myself an algebraist, i do have a fair amount of algebra. i am, for example, familiar with the universal construction of a tensor product. as well as the coordinate definition of a tensor product of vector spaces. so whatever.

mostly, i am just looking to establish what is the correct terminology, and why.
 now the permutation group on n elements acts on the n-fold tesnor product by swapping factors in the expresion. the exterior algebra is that sub-space where swapping the elements is the same as multiplying by -1
sure.

 eg in the two fold tensor, it is the subspace spanned by expressions like $u\otimes v - v\otimes u$ It is not hard to see that in an r dimensional vector space this is only non-zero if n<=r, and it has dimension r choose n. The element $u\otimes v - v\otimes u$ is labelled $u\wedge v$ (can't recall tex for it off hand; it's probably \wedge) The reason why its so damn easy and confusing at the same time for R^3 is because 3 choose 2 is 3 and is therefore a 3 dimensional vector space. Different groups label these different things all over the place, but the wedge is usually called the exterior product.
yes, as i understand it, wedge product and exterior product are synonymous.

 Let's stick in the 3-d case: where does the rule $\imath\times\jmath=k$ come from? Note i'm using x for ease of type setting, it is the wedge above. well, the answer is: there is an exact pairing from between the first and second exterior algebra
sure, because $\binom{3}{1}$ and $\binom{3}{2}$ are equal, there is an isomorphism between the two given by the Hodge dual (which depends on the orientation and the inner product we give to the vector space)

 basically, there is a unique element in the third exterior algebra's base: ixjxk corresponding to the invariant element $\hat{\imath}\otimes\hat{\jmath} \otimes \hat{k} -\hat{\jmath}\otimes \hat{\imath}\otimes\hat{k} + \hat{\jmath}\otimes\hat{k}\otimes\hat{\imath} -\hat{\imath}\otimes \hat{k}\otimes\hat{\jmath} + \hat{k}\otimes\hat{\imath}\otimes\hat{\jmath} - \hat{k}\otimes\hat{\jmath} \otimes \hat{\imath}$ in the tensor algebra.
but this exactly illustrates my point of why the exterior algebra is not the cross product algebra. first of all, the exterior algebra is associative, and the cross product algebra is not. it follows that in the cross product algebra,

$$(\hat{\imath}\times\hat{\jmath})\times\hat{k}=\hat{\imath}\times(\hat{\ jmath}\times\hat{k})=0$$

whereas in the exterior algebra, you get the invariant element $e_1\wedge e_2\wedge e_3\neq 0$.

if we assign a product on the exterior algebra $e_i\times e_j=\star(e_i\wedge e_j)$, then we get an explicit correspondence with the vector cross product algebra in R3

 so the paring matches ixj with k and jxk with i and so on. Is that clear in the slightest? I think that might answer the question but I'm not sure.
well, this isn't really my question. i know what the exterior algebra is and how it relates to the cross product.

i guess all i really want to know is: what does the term "outer product" mean to you? some people are saying "outer product"="exterior product", i think saski might have even said "outer product = Clifford product (geometric product)", but i guess we ruled that out.

on the other hand, i think "outer product"="tensor product"

i really just wanted you to weigh in with your opinion of what the term outer product should mean.