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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$\begingroup$ Try Winitzki: sites.google.com/site/winitzki/linalg $\endgroup$– Qiaochu YuanNov 13, 2017 at 7:18
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$\begingroup$ What do you know about the Levi-Civita tensor? $\endgroup$– J. M. ain't a mathematicianNov 13, 2017 at 7:19
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1$\begingroup$ Just extend the determinant version of the cross product en.wikipedia.org/wiki/Cross_product#Matrix_notation to 4-by-4 matrices. $\endgroup$– Angina SengNov 13, 2017 at 7:23
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$\begingroup$ That's funny: someone just wrote an answer about this generalization yesterday. $\endgroup$– Viktor VaughnNov 13, 2017 at 7:55
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$\begingroup$ @Quasicoherent thank you for the link. I missed that when looking for previous questions $\endgroup$– 14159Nov 13, 2017 at 13:05
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1 Answer
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Yes, you can generalize using the matrix determinant, e.g. $$\det\left(\begin{array}{llll}\mathbf{i}&x_1&y_1&z_1\\\mathbf{j}&x_2&y_2&z_2\\\mathbf{k}&x_3&y_3&z_3\\\mathbf{l}&x_4&y_4&z_4\\\end{array}\right).$$
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$\begingroup$ does it work for n-1 vectors as well? I know it is computationally expensive, I am asking theoretically $\endgroup$– 14159Nov 13, 2017 at 13:04
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$\begingroup$ 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. $\endgroup$– pshmath0Nov 13, 2017 at 14:46
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$\begingroup$ 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? $\endgroup$– 14159Dec 11, 2017 at 8:34
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1$\begingroup$ @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. $\endgroup$– pshmath0Dec 11, 2017 at 8:54
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$\begingroup$ 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. $\endgroup$– pshmath0Dec 11, 2017 at 9:02