2次元直交行列
得られる式の本数と未知数の関係より、2次元では1軸、3次元では3軸の自由度がある
一般に
$ \pmb{A}\pmb{A}^\top=\pmb{I}\iff \sum_k[\pmb{A}]^\mathsf{E\bar{F}}_{ik}[\pmb{A}]^\mathsf{\bar{E}F}_{jk}=\llbracket i=j\rrbracket
$ \sf E,Fともに正規直交基底とすると、
$ \sum_{k}{[\pmb{A}]^\mathsf{EF}_{ik}}^2=1
$ i< j\implies\sum_{k}[\pmb{A}]^\mathsf{EF}_{ik}[\pmb{A}]^\mathsf{EF}_{jk}=0
$ \pmb{A}\pmb{A}^\topが対称tensorになるため、独立した式は$ \frac12 n(n+1)本得られる
$ \pmb{A}は$ n^2個の未知数があるから、未定になる変数は$ n^2-\frac12 n(n+1)=\frac12 n(n-1)個ある
対角成分の個数と非対角成分の個数/2の和
2次元直交tensorなら1個
3次元直交tensorなら3個
4次元直交tensorなら6個
計算メモ
$ \pmb{A}^\top=\sum_{i,j}[\pmb{A}]^\mathsf{E\bar{F}}_{ij}(\bar{\pmb{e}}_i\otimes\pmb{f}_j)^\top=\sum_{i,j}[\pmb{A}]^\mathsf{E\bar{F}}_{ij}\pmb{f}_j\otimes\bar{\pmb{e}}_i=\sum_{i,j}{[\pmb{A}]^\mathsf{E\bar{F}}}^\top_{ji}\pmb{f}_j\otimes\bar{\pmb{e}}_i
$ \pmb{A}^\top=\sum_{i,j}\left[\pmb{A}^\top\right]^\mathsf{\bar{F}E}_{ji}\pmb{f}_j\otimes\bar{\pmb{e}}_i
$ [\pmb{A}\pmb{A}^\top]^\mathsf{E\bar{E}}=[\pmb{A}]^\mathsf{E\bar{F}}[\pmb{A}^\top]^\mathsf{F\bar{E}}=[\pmb{A}]^\mathsf{E\bar{F}}{[\pmb{A}]^\mathsf{\bar{E}F}}^\top
$ {[\pmb{A}]^\mathsf{\bar{E}F}}^\top=[\pmb{A}^\top]^\mathsf{F\bar{E}}=[\pmb{A}^{-1}]^\mathsf{F\bar{E}}={[\pmb{A}]^\mathsf{E\bar{F}}}^{-1}
2次元直交tensorが取りうる形を求める
https://kakeru.app/d6ea1afc49ab736bd32b8d2e7bd039ef https://i.kakeru.app/d6ea1afc49ab736bd32b8d2e7bd039ef.svg
$ a_{00}^2+a_{01}^2=1\iff\exist\theta\in\R;a_{00}=\cos\theta\land a_{01}=\sin\thetaより、
$ AA^\top=I\iff \exists\theta\in\R;A=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\lor A=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}
$ A:=\begin{pmatrix}a_{00}&a_{01}&a_{02}\\a_{10}&a_{11}&a_{12}\\a_{20}&a_{21}&a_{22}\end{pmatrix}とする
$ AA^\top=I
$ \iff \begin{dcases}{a_{00}}^2+{a_{01}}^2+{a_{02}}^2&=1\\{a_{10}}^2+{a_{11}}^2+{a_{12}}^2&=1\\{a_{20}}^2+{a_{21}}^2+{a_{22}}^2&=1\\a_{00}a_{10}+a_{01}a_{11}+a_{02}a_{12}&=0\\a_{10}a_{20}+a_{11}a_{21}+a_{12}a_{22}&=0\\a_{20}a_{00}+a_{21}a_{01}+a_{22}a_{02}&=0\end{dcases}
$ a_{00}=a_{11}=a_{22}=0のとき
$ \begin{dcases}{a_{01}}^2+{a_{02}}^2&=1\\{a_{10}}^2+{a_{12}}^2&=1\\{a_{20}}^2+{a_{21}}^2&=1\\a_{02}a_{12}&=0\\a_{10}a_{20}&=0\\a_{21}a_{01}&=0\end{dcases}
$ \implies A=\begin{pmatrix}0&\pm1&0\\0&0&\pm1\\\pm1&0&0\end{pmatrix}\lor A=\begin{pmatrix}0&\pm1&0\\0&0&\pm1\\\pm1&0&0\end{pmatrix}^\top
複号任意
$ a_{00}=a_{11}=0\neq a_{22}のとき
$ \begin{dcases}{a_{01}}^2+{a_{02}}^2&=1\\{a_{10}}^2+{a_{12}}^2&=1\\{a_{20}}^2+{a_{21}}^2+{a_{22}}^2&=1\\a_{02}a_{12}&=0\\a_{10}a_{20}+a_{12}a_{22}&=0\\a_{21}a_{01}+a_{22}a_{02}&=0\end{dcases}
$ \iff\begin{dcases}{a_{22}}^2{a_{01}}^2+{a_{22}}^2{a_{02}}^2&={a_{22}}^2\\{a_{10}}^2{a_{22}}^2+{a_{12}}^2{a_{22}}^2&={a_{22}}^2\\{a_{20}}^2+{a_{21}}^2+{a_{22}}^2&=1\\a_{02}a_{12}&=0\\a_{10}a_{20}+a_{12}a_{22}&=0\\a_{21}a_{01}+a_{22}a_{02}&=0\end{dcases}
$ \iff\begin{dcases}({a_{22}}^2+{a_{21}}^2){a_{01}}^2&={a_{22}}^2\\({a_{22}}^2+{a_{20}}^2){a_{10}}^2&={a_{22}}^2\\{a_{20}}^2+{a_{21}}^2+{a_{22}}^2&=1\\a_{02}a_{12}&=0\\a_{10}a_{20}+a_{12}a_{22}&=0\\a_{21}a_{01}+a_{22}a_{02}&=0\end{dcases}
この時点で$ a_{01}\neq 0\neq a_{10}である
$ a_{02}=0\implies a_{21}=0\implies |a_{01}|=1
$ a_{12}=0\implies a_{20}=0\implies |a_{10}|=1
$ a_{02}\neq0\implies
$ \begin{dcases}({a_{22}}^2{a_{02}}^2+{a_{21}}^2{a_{02}}^2){a_{01}}^2&={a_{22}}^2{a_{02}}^2\\{a_{21}}^2+{a_{22}}^2&=1\\a_{20}&=0\\a_{12}&=0\\|a_{10}|&=1\\a_{21}a_{01}+a_{22}a_{02}&=0\end{dcases}
$ \iff\begin{dcases}{a_{02}}^2{a_{01}}^2&={a_{22}}^2{a_{02}}^2\\{a_{21}}^2+{a_{22}}^2&=1\\a_{20}&=0\\a_{12}&=0\\|a_{10}|&=1\\a_{21}a_{01}+a_{22}a_{02}&=0\end{dcases}
$ \iff\begin{dcases}|a_{02}|&=|a_{21}|\\{a_{21}}^2+{a_{22}}^2&=1\\a_{20}&=0\\a_{12}&=0\\|a_{10}|&=1\\a_{21}a_{01}+a_{22}a_{02}&=0\end{dcases}
$ \iff\begin{dcases}|a_{02}|&=|a_{21}|\\|a_{01}|&=|a_{22}|\\{a_{21}}^2+{a_{22}}^2&=1\\a_{20}&=0\\a_{12}&=0\\|a_{10}|&=1\\a_{21}a_{01}+a_{22}a_{02}&=0\end{dcases}
最後の式は、4つのうちのどれかが一つだけ異符号という条件
これは$ A=\begin{pmatrix}0&\cos\theta&-\sin\theta\\\pm 1&0&0\\0&\sin\theta&\cos\theta\end{pmatrix}\lor A=\begin{pmatrix}0&\cos\theta&\sin\theta\\\pm 1&0&0\\0&\sin\theta&-\cos\theta\end{pmatrix}を表す
2022-07-16 05:39:52 図形とvectorで考えればいいことに気づいた
証明
正規直交基底$ \mathsf{E}:=(\pmb{e}_i)_iを使う
$ \pmb{a}_i:=\sum_j[\pmb{A}]^\mathsf{EE}_{ji}\pmb{e}_j とすると、$ \pmb{A}=\sum_{i,j}[\pmb{A}]^\mathsf{EE}_{ij}\pmb{e}_i\otimes\pmb{e}_j=\sum_i\pmb{a}_i\otimes\pmb{e}_i となる
よって
$ \pmb{A}^\top\pmb{A}=\pmb{I}
$ \iff \sum_{i,j}(\pmb{a}_i\otimes\pmb{e}_i)^\top(\pmb{a}_j\otimes\pmb{e}_j)=\pmb{I}
$ \iff \sum_{i,j}(\pmb{e}_i\otimes\pmb{a}_i)(\pmb{a}_j\otimes\pmb{e}_j)=\pmb{I}
$ \iff \sum_{i,j}\pmb{a}_i\cdot\pmb{a}_j(\pmb{e}_i\otimes\pmb{e}_j)=\pmb{I}
$ \underline{\iff \pmb{a}_i\cdot\pmb{a}_j=\llbracket i=j\rrbracket\quad}_\blacksquare
これを用いると、2次元は極座標、3次元は球座標の話に持ち込める
$ a_{ij}:=[A]^\mathsf{EE}_{ij} とする
2次元のとき
$ \pmb{A}^\top\pmb{A}=\pmb{I}
$ \iff\begin{dcases}|\pmb{a}_0|&=1\\|\pmb{a}_1|&=1\\\pmb{a}_0\cdot\pmb{a}_1&=0\end{dcases}
$ \begin{aligned}\iff&\exist\theta,s\in\R;\\&\begin{dcases}\pmb{a}_0&=\pmb{e}_0\cos\theta+\pmb{e}_1\sin\theta\\|\pmb{a}_1|&=1\\\pmb{a}_1&=s(-\pmb{e}_0\otimes\pmb{e}_1+\pmb{e}_1\otimes\pmb{e}_0)\pmb{a}_0=s(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0\end{dcases}\end{aligned}
最後の式は、$ \begin{pmatrix}a_{01}\\a_{11}\end{pmatrix}=s\begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}a_{00}\\a_{10}\end{pmatrix}($ \pmb{a}_0に垂直なvector)を表している
$ \begin{aligned}\iff&\exist\theta,s\in\R;\\&\begin{dcases}\pmb{a}_0&=\pmb{e}_0\cos\theta+\pmb{e}_1\sin\theta\\|s|&=1\\\pmb{a}_1&=s(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0\end{dcases}\end{aligned}
$ \underline{\iff\exist\theta\in\R;[\pmb{A}]^\mathsf{EE}=\begin{pmatrix}\cos\theta&\mp\sin\theta\\\sin\theta&\pm\cos\theta\end{pmatrix}\text{\quad(複号同順)}\quad}_\blacksquare
3次元のとき
$ \pmb{A}^\top\pmb{A}=\pmb{I}
$ \iff\begin{dcases}|\pmb{a}_0|&=1\\|\pmb{a}_1|&=1\\|\pmb{a}_2|&=1\\\pmb{a}_0\cdot\pmb{a}_1&=0\\\pmb{a}_1\cdot\pmb{a}_2&=0\\\pmb{a}_2\cdot\pmb{a}_0&=0\end{dcases}
$ \begin{aligned}\iff&\exist\theta,\phi,s,t\in\R;\\&\begin{dcases}\pmb{a}_0&=\pmb{e}_0\cos\theta\cos\phi+\pmb{e}_1\cos\theta\sin\phi+\pmb{e}_2\sin\theta\\|\pmb{a}_1|&=|\pmb{a}_2|=1\\\pmb{a}_1&=s(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0+t(\pmb{e}_2\wedge\pmb{e}_0)\pmb{a}_0\\\pmb{a}_2\cdot\pmb{a}_0&=0\\\pmb{a}_1\cdot\pmb{a}_2&=0\end{dcases}\end{aligned}
$ \pmb{a}_1の展開方法
$ a_{00}a_{01}+a_{10}a_{11}+a_{20}a_{21}=0
$ \iff \exist s,t\in\R;a_{01}=-a_{10}s-a_{20}t\land a_{11}=a_{00}s\land a_{12}=a_{00}t
$ \iff\exist s,t\in\R;\pmb{a}_1=(-a_{10}s-a_{20}t)\pmb{e}_0+a_{00}s\pmb{e}_1+a_{00}t\pmb{e}_2
$ \iff\exist s,t\in\R;\pmb{a}_1=(-s\pmb{a}_0\cdot\pmb{e}_1-t\pmb{a}_0\cdot\pmb{e}_2)\pmb{e}_0+\pmb{a}_0\cdot\pmb{e}_0s\pmb{e}_1+\pmb{a}_0\cdot\pmb{e}_0t\pmb{e}_2
$ \iff\exist s,t\in\R;\pmb{a}_1=-s(\pmb{e}_0\otimes\pmb{e}_1)\pmb{a}_0-t(\pmb{e}_0\otimes\pmb{e}_2)\pmb{a}_0+s(\pmb{e}_1\otimes\pmb{e}_0)\pmb{a}_0+t(\pmb{e}_2\otimes\pmb{e}_0)\pmb{a}_0
$ \iff\exist s,t\in\R;\pmb{a}_1=s(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0+t(\pmb{e}_2\wedge\pmb{e}_0)\pmb{a}_0
$ ((\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0)\cdot(\pmb{e}_2\wedge\pmb{e}_0)\pmb{a}_0=\pmb{a}_0\cdot(\pmb{e}_1\wedge\pmb{e}_0)^\top(\pmb{e}_2\wedge\pmb{e}_0)\pmb{a}_0
$ =\pmb{a}_0\cdot(\pmb{e}_0\wedge\pmb{e}_1)(\pmb{e}_2\wedge\pmb{e}_0)\pmb{a}_0
$ =\pmb{a}_0\cdot(\pmb{e}_0\otimes\pmb{e}_1-\pmb{e}_1\otimes\pmb{e}_0)(\pmb{e}_2\otimes\pmb{e}_0-\pmb{e}_0\otimes\pmb{e}_2)\pmb{a}_0
$ =\pmb{a}_0\cdot(0+\pmb{e}_1\otimes\pmb{e}_2-0-0)\pmb{a}_0
$ =a_{10}a_{20}
直交しないのか。面倒だな……
直交する2軸で表現したい
あ、$ \pmb{a}_0の全微分から候補を見つければいいんだ
各変数の曲線の接vectorは$ \pmb{a}_0と直行する
$ \mathrm{d}\pmb{a}_0=(-\pmb{e}_0\cos\theta\sin\phi+\pmb{e}_1\cos\theta\cos\phi)\mathrm{d}\phi+(-\pmb{e}_0\sin\theta\cos\phi-\pmb{e}_1\sin\theta\sin\phi+\pmb{e}_2\cos\theta)\mathrm{d}\theta
$ =(-\pmb{e}_0(\pmb{a}_0\cdot\pmb{e}_1)+\pmb{e}_1(\pmb{a}_0\cdot\pmb{e}_0))\mathrm{d}\phi+\frac{1}{\cos\theta}(-\pmb{e}_0\sin\theta\cos\theta\cos\phi-\pmb{e}_1\sin\theta\cos\theta\sin\phi+\pmb{e}_2(\cos\theta)^2)\mathrm{d}\theta
$ =(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0\mathrm{d}\phi+\frac{1}{\cos\theta}(-\pmb{e}_0(a_0\cdot\pmb{e}_0)(a_0\cdot\pmb{e}_2)-\pmb{e}_1(a_0\cdot\pmb{e}_2)(a_0\cdot\pmb{e}_1)+\pmb{e}_2(1-(a_0\cdot\pmb{e}_2)^2))\mathrm{d}\theta
$ =(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0\mathrm{d}\phi+\frac{1}{\cos\theta}(-\pmb{e}_0(a_0\cdot\pmb{e}_0)(a_0\cdot\pmb{e}_2)-\pmb{e}_1(a_0\cdot\pmb{e}_2)(a_0\cdot\pmb{e}_1)+\pmb{e}_2(1-(a_0\cdot\pmb{e}_2)^2))\mathrm{d}\theta
$ \mathrm{d}\thetaのvectorがうまく展開できない
でも$ \mathrm{d}\phiのvectorはわかったし、あとはcross積から計算すればいいだろう
$ \pmb{a}_0\times(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0=
https://kakeru.app/3c485d02040e62c9d2991965b5d10748 https://i.kakeru.app/3c485d02040e62c9d2991965b5d10748.svg
結局、微分係数の式と同じ形になった
ノルムは
$ |\pmb{a}_0\times(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0|^2={a_{20}}^2({a_{00}}^2+{a_{11}}^2)+({a_{00}}^2+{a_{11}}^2)^2={a_{00}}^2+{a_{11}}^2=(\cos\theta)^2
やっぱそうなるか
あとは$ (\pmb{e}_2\wedge\pmb{e}_0)\pmb{a}_0を$ (\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0と$ \pmb{a}_0\times(\pmb{e}_1\wedge\pmb{e}_0)\pmb{e}_1で表せればいい
展開がめんどくさすぎる……
$ \pmb{a}_0と直交する何らかの単位vector$ (\theta,\phi)\mapsto\pmb{a}_{\bot0},(\theta,\phi)\mapsto\pmb{a}_{\bot1}\quad(\pmb{a}_{\bot0}\cdot\pmb{a}_{\bot1}=0)とパラメタ$ \varphiを用いて
$ \pmb{a}_1=\pmb{a}_{\bot0}\cos\varphi+\pmb{a}_{\bot1}\sin\varphi
と表せてほしいのだが……
$ (\pmb{e}_1\wedge\pmb{e}_0)^\top(\pmb{e}_1\wedge\pmb{e}_0)=(\pmb{e}_0\otimes\pmb{e}_1-\pmb{e}_1\otimes\pmb{e}_0)(\pmb{e}_1\otimes\pmb{e}_0-\pmb{e}_0\otimes\pmb{e}_1)
$ =\pmb{e}_0\otimes\pmb{e}_0+\pmb{e}_1\otimes\pmb{e}_1-(0+0)
$ =\pmb{e}_0\otimes\pmb{e}_0+\pmb{e}_1\otimes\pmb{e}_1
対角成分が残るのかtakker.icon
$ |(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0|^2={a_{00}}^2+{a_{10}}^2=|\pmb{a}_0\times(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0|^2
うん？長さ同じなんだtakker.icon
あ、そうか。$ |\pmb{a}_0|=1だから、同じ長さで当然なんだ
てことは$ \exist s,t\in\R;\pmb{a}_1=s(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0+t(\pmb{a}_0\times(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0)から
$ |\pmb{a}_1|=(s^2+t^2)(\cos\theta)^2=1
$ \implies \exist\varphi\in\R;s\cos\theta=\cos\varphi\land t\cos\theta=\sin\varphi
ってできるわけだ
$ \begin{aligned}\iff&\exist\theta,\phi,s,t\in\R;\\&\begin{dcases}\pmb{a}_0&=\pmb{e}_0\cos\theta\cos\phi+\pmb{e}_1\cos\theta\sin\phi+\pmb{e}_2\sin\theta\\|\pmb{a}_1|&=|\pmb{a}_2|=1\\\pmb{a}_1&=s(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0+\frac{t}{(\cos\theta)^2}(a_{10}a_{20}(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0-a_{00}\pmb{a}_0\times(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0)\\\pmb{a}_2\cdot\pmb{a}_0&=0\\\pmb{a}_1\cdot\pmb{a}_2&=0\end{dcases}\end{aligned}
$ \begin{aligned}\iff&\exist\theta,\phi,s,t\in\R;\\&\begin{dcases}\pmb{a}_0&=\pmb{e}_0\cos\theta\cos\phi+\pmb{e}_1\cos\theta\sin\phi+\pmb{e}_2\sin\theta\\|\pmb{a}_1|&=|\pmb{a}_2|=1\\\pmb{a}_1&=s(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0+t\pmb{a}_0\times(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0\\\pmb{a}_2\cdot\pmb{a}_0&=0\\\pmb{a}_1\cdot\pmb{a}_2&=0\end{dcases}\end{aligned}
$ \begin{aligned}\iff&\exist\theta,\phi,\varphi\in\R;\\&\begin{dcases}\pmb{a}_0&=\pmb{e}_0\cos\theta\cos\phi+\pmb{e}_1\cos\theta\sin\phi+\pmb{e}_2\sin\theta\\|\pmb{a}_2|&=1\\\pmb{a}_1\cos\theta&=\cos\varphi(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0+\sin\varphi\pmb{a}_0\times(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0\\\pmb{a}_2\cdot\pmb{a}_0&=0\\\pmb{a}_1\cdot\pmb{a}_2&=0\end{dcases}\end{aligned}
$ \begin{aligned}\iff&\exist\theta,\phi,\varphi\in\R;\\&\begin{dcases}\pmb{a}_0&=\pmb{e}_0\cos\theta\cos\phi+\pmb{e}_1\cos\theta\sin\phi+\pmb{e}_2\sin\theta\\\pmb{a}_1\cos\theta&=\cos\varphi(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0+\sin\varphi\pmb{a}_0\times(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0\\\pmb{a}_2\cos\theta&=\mp\sin\varphi(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0\pm\cos\varphi\pmb{a}_0\times(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0\end{dcases}\end{aligned}
$ \frac{(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0}{\cos\theta}=-\pmb{e}_0\sin\phi+\pmb{e}_1\cos\phiと$ \frac{\pmb{a}_0\times(\pmb{e}_1\wedge\pmb{e}_0)\pmb{a}_0}{\cos\theta}=\pmb{e}_0\tan\theta\cos\phi+\pmb{e}_1\tan\theta\sin\phi-\pmb{e}_2で展開する
2022-07-16 08:51:01 これ以上書くのめんどくさくなってきた……takker.icon
もう回転行列と空間反転行列の積で表せるってわかったし、十分だろう
ただ$ \tan\thetaがでてきたのは気になる
$ \theta=\frac12\pi のときは$ \exist\varphi\in\R;[\pmb{A}]^\mathsf{EE}=\begin{pmatrix}0&\cos\varphi&\mp\sin\varphi\\0&\sin\varphi&\pm\cos\varphi\\1&0&0\end{pmatrix}\text{\quad(複号同順)} になるっぽい