so I understood that for the cross product to have meaning in $n$-dimensions, one needs $n-1$ vectors. I tried reading about it, but I couldn't find any good resources on exterior algebra. So my question is if there is a simple way to define the cross product in $4$D for three vectors (or just application of the generalization for this specific case?), or if anyone has a good resource on exterior algebra?
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Try Winitzki: https://sites.google.com/site/winitzki/linalg – Qiaochu Yuan Nov 13 '17 at 07:18
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What do you know about the Levi-Civita tensor? – J. M. ain't a mathematician Nov 13 '17 at 07:19
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1Just extend the determinant version of the cross product https://en.wikipedia.org/wiki/Cross_product#Matrix_notation to 4-by-4 matrices. – Angina Seng Nov 13 '17 at 07:23
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That's funny: someone just wrote an answer about this generalization yesterday. – Viktor Vaughn Nov 13 '17 at 07:55
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@Quasicoherent thank you for the link. I missed that when looking for previous questions – 14159 Nov 13 '17 at 13:05
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Yes, you can generalize using the matrix determinant, e.g. $$\det\left(\begin{array}{llll}\mathbf{i}&x_1&y_1&z_1\\\mathbf{j}&x_2&y_2&z_2\\\mathbf{k}&x_3&y_3&z_3\\\mathbf{l}&x_4&y_4&z_4\\\end{array}\right).$$
pshmath0
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does it work for n-1 vectors as well? I know it is computationally expensive, I am asking theoretically – 14159 Nov 13 '17 at 13:04
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There are $n-1$ vectors, i.e. $x=(x_1,x_2,x_3)$, $y=(y_1,y_2,y_3)$, and $z=(z_1,z_2,z_3)$. The $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ vectors are the base vectors. You can extend this to any dimension by way of the determinant, which is defined for any $n\times n$ matrices. – pshmath0 Nov 13 '17 at 14:46
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I've just noticed it. is there a specific reason you put the basis vectors on the left? in 3D they are usually on top. however, I heard that for even dimensions they should be placed on the bottom. what is the right answer? – 14159 Dec 11 '17 at 08:34
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1@Yoav2000 you can put the base vectors vertically or horizontally, and the vectors you're attempting to "cross" must be placed vertically or horizontally accordingly. What's more you can put them on any row or column. If the base vectors are on an even row or column then the result is negated, so it alternates between $\pm$. Don't forget to +1 if you liked the answer. – pshmath0 Dec 11 '17 at 08:54
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Correction to the above earlier comment: There are $n-1$ vectors, i.e. $x=(x_1,x_2,x_3,x_4),y=(y_1,y_2,y_3,y_4)$, and $z=(z_1,z_2,z_3,z_4)$. The $\textbf{i}$, $\textbf{j}$, $\textbf{k}$, and $\textbf{l}$ vectors... so there are $n-1=4-1=3$ vectors. – pshmath0 Dec 11 '17 at 09:02