
#1
Jan2413, 09:23 AM

P: 7

I'm taking multivariate calculus and my teacher just introduced the concept of cross products a week ago. Reading the Wikipedia page, I see that cross products only work in three and seven dimensions, which is puzzling.
One use of the cross product for our class is to find the vector orthogonal to the 2 given vectors. My question is can this be generalized to n dimensions to find the vector orthogonal to the n1 given vectors? Also what is the formal method/operation of doing this? For example, given [itex]u = \left(1,0,0,0\right)[/itex], [itex]v = \left(0,1,0,0\right)[/itex], [itex]w = \left(0,0,1,0\right)[/itex], the vector orthogonal to u, v, and w is given by: [tex] \left\begin{array}{cccc} e_{1} & e_{2} & e_{3} & e_{4} \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}\right = e_{4} [/tex] I read a bit about Hodge duality, exterior products, and kvectors. Much of it was confusing, so could you clarify if you use them as I have little background in linear algebra or tensor theory. 



#2
Jan2413, 12:14 PM

Mentor
P: 10,791

Consider a nxnmatrix where the first n1 columns (or rows) are filled with n1 vectors. Now, for each entry in the remaining column (or row), use the determinant of the (n1)x(n1)matrix you get by removing the last column (or row) and the row (or column) your entry is in.
This might look complicated, but it is easy to show for the conventional crossproduct: $$\begin{pmatrix} a_1 & b_1 && \color{red}{a_2b_3b_2a_3}\\ \color{red}{a_2} & \color{red}{b_2} && a_3b_1b_3a_1 \\ \color{red}{a_3} & \color{red}{b_3} && a_1b_2b_1a_2 \end{pmatrix} $$ It gives a vector which is orthogonal to all other n1 vectors. 



#3
Jan2413, 01:23 PM

Sci Advisor
HW Helper
PF Gold
P: 4,768

Here is an argument based on the fact that the scalar product is a nondegenerate bilinear form, meaning the map R^{n}>(R^{n})^{*}: x><x,  > is surjective. (Indeed, if f:R^{n}>R is any linear map, then as a matrix, it is a 1xn matrix, i.e. a vector; call it x. Then f = <x,  >. Here, (R^{n})^{*} is the set of all linear maps R^{n}>R.)
Consider now the linear map f:R^{n}>R defined as "w > determinant of the matrix whose first n1 columns are the n1 vectors v_{1},...,v_{n1} you want the crossproduct of, and whose n^{th} column is just w". This is linear by properties of the determinant. So, there exists a vector x such that f = <x,  >. This x is the crossproduct of v_{1},...,v_{n1} in the sense that <x,v_{k}> = 0 for all k (by the property of the determinant that says that if the columns of A are linearly dependent, then det(A)=0). 



#4
Jan2413, 04:19 PM

Sci Advisor
HW Helper
P: 9,421

Generalizing Cross Product
the generalized cross product is discussed, as here, on pages 834 of spivak's calculus on manifolds.




#5
Jan2413, 09:23 PM

P: 7

Thank you for the responses. The Spivak reading was especially helpful. So, correct me if my interpretation is incorrect:
The generalized cross product in n dimensions has arity n1, but its operands are all 1vectors and thus the output will also be a 1vector. I guess this might be a different question altogether, but what makes 7 dimensions amenable to a nontrivial cross product with only 2 operands instead of 6? 



#6
Jan2513, 03:35 AM

HW Helper
P: 2,149

^It is a matter of which properties to maintain. As the Wikipedia page states "The sevendimensional cross product is (...) the only other nontrivial bilinear product of two vectors that is vector valued, anticommutative and orthogonal. This follows from the CayleyDickson construction, which yields algebras of order 2^n which in turn yield cross products in 2^n1 space. The 0 and 1 dimensional cases are trivial and the 15,31,2^n1,... cases do not have the desired properties, this leaves only the 3 and 7 dimensional cases.



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