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[几何] 只有$E^3$和$E^7$能定义外积

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hbghlyj 发表于 2022-3-29 09:24 |阅读模式
本帖最后由 hbghlyj 于 2024-12-26 13:29 编辑 令$E^n$为$n$维欧氏空间,若存在运算$E^{n} \times E^{n} \stackrel{\times}{\longrightarrow} E^{n}$满足
(i)(双线性)$\forall \lambda_{1}, \lambda_{2}, \mu_{1}, \mu_{2} \in \mathbb{R}, u_{1}, u_{2}, v_{1}, v_{2} \in E^{n},$
$\left(\lambda_{1} u_{1}+\lambda_{2} u_{2}\right) \times\left(\mu_{1} v_{1}+\mu_{2} v_{2}\right)=\lambda_{1} \mu_{1}\left(u_{1} \times v_{1}\right)+\lambda_{1} \mu_{2}\left(u_{1} \times v_{2}\right)+\lambda_{2} \mu_{1}\left(u_{2} \times v_{1}\right)+\lambda_{2} \mu_{2}\left(u_{2} \times v_{2}\right)$
(ii)(垂直性)$(u \times v) \cdot u=(u \times v) \cdot v=0, \forall u, v \in E^{n}$
(iii)(平行四边形面积公式)$|u \times v|^{2}=|u|^{2} \cdot|v|^{2}-(u \cdot v)^{2}, \forall u, v \in E^{n}$
则$n=3,7$.
When Does a Cross Product on R^n Exist.pdf

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 楼主| hbghlyj 发表于 2022-4-2 07:23
本帖最后由 hbghlyj 于 2024-12-26 13:27 编辑 seven-dimensional cross product
As the cross product is bilinear the operator ${\bf x}×$– can be written as a matrix, which takes the form
$$T_{\mathbf {x} }={\begin{bmatrix}0&-x_{4}&-x_{7}&x_{2}&-x_{6}&x_{5}&x_{3}\\x_{4}&0&-x_{5}&-x_{1}&x_{3}&-x_{7}&x_{6}\\x_{7}&x_{5}&0&-x_{6}&-x_{2}&x_{4}&-x_{1}\\-x_{2}&x_{1}&x_{6}&0&-x_{7}&-x_{3}&x_{5}\\x_{6}&-x_{3}&x_{2}&x_{7}&0&-x_{1}&-x_{4}\\-x_{5}&x_{7}&-x_{4}&x_{3}&x_{1}&0&-x_{2}\\-x_{3}&-x_{6}&x_{1}&-x_{5}&x_{4}&x_{2}&0\end{bmatrix}}$$The cross product is then given by$$\mathbf {x} \times \mathbf {y} =T_{\mathbf {x} }\mathbf {y} .$$
news.ycombinator.com/item?id=19622798
arxiv.org/abs/math/0204357
math.stackexchange.com/questions/706011/why-is-cross-product-only-defined-in-3-and-7-dimensions

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 楼主| hbghlyj 发表于 2024-12-26 21:26

批评

alexkritchevsky.com/2019/01/26/exterior-3.html
Some people will tell you that there’s also a 7-dimensional cross product. They are basically wrong. Well, there is one, sort of, but it’s the wrong generalization, and it’s useless for geometry. \(a \times b = \star (a \wedge b)\) is the definition that extends to other dimensions the way you want – it’s just that it only maps vectors to vectors in \(\Bbb{R}^3\). Alternatively, you could define it to take \(n-1\) vectors to another vector in any \(\Bbb{R}^n\), but at that point why aren’t you just using \(\wedge\)?

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