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hbghlyj Posted 2023-3-11 20:55 |Read mode

二次曲线的坐标系中一个点等同于一个映射.具体如下定义: $p$为平面上一点, $k$为二次曲线上一点,过两点的直线和$C$交于另外一点$l$,定义函数\(f_{p}(k) = l\)

过点$p$的直线束给出了二次曲线上点的一个映射,记为\(f_{p}\).例如图中\(f_{p}(k) = l\)

①当$p$在曲线外部时,\(f_{p}\)是一个双射,其不动点为$p$对曲线的两个切点\(t_1\),\(t_1\)

②当$p$点在曲线内部时,\(f_{p}\)是一个双射且没有不动点

③当$p$点在曲线上时,\(f_{p}\)不是双射,因为此时\(f_{p}\)将曲线上除$p$外所有点都映射为$p$自身

此时当$p$与$k$重合时,\(f_{p}\)的定义失效,但是,我们将会看到,经过参数化,可补充定义\(f_{p}(p) = 0\).

④当$p$不在曲线上时\(f_{p}^{- 1} = f_{p}\),也就是说, $p$是对合

II. Projective Geometry

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 Author| hbghlyj Posted 2023-3-11 21:18
椭圆标准参考系size(8cm); pair P = (-1, 2.5); label("$p=(x,y)$", P, N); transform T=rotate(60)*shift(-.3,1.5); path ellipse=T*scale(1.2,0.6)*unitcircle; pair O=rotate(60)*(-.3,1.5); dot((0,0),blue+linewidth(1.5mm)); dot(O,red+linewidth(1.5mm)); draw(O--P,red,Arrow); draw(ellipse); draw(T*((-2,0)--(2,0)),red,Arrow); draw(T*((0,-2.5)--(0,2)),red,Arrow); // Draw the blue coordinate axes draw((0,-1)--(0,3), blue, Arrow); draw((-3,0)--(1,0), blue, Arrow); draw((0,0)--P,blue,Arrow);

曲线的参数化是\(\mathbb{R}\)到曲线上的点的一个双射:设有参数化\(\xi\),那么\(F_{p} = \xi^{- 1} \circ f_{p} \circ \xi\)是一个\(\mathbb{R \rightarrow R}\)的函数.称\(F_{p}\)是\(f_{p}\)在参数化\(\xi\)下的一个表示

\(f_{p}\)描述了点之间的映射,要想把它表达成具体的函数,先要给二次曲线参数化:

以椭圆为例,对于标准椭圆方程\(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\)有参数化\(\xi(\theta) = \left( a\cos\theta,b\sin\theta \right),\theta \in \lbrack 0,2\pi)\)

利用三角代换公式\(\cos\theta = \frac{1 - \omega^{2}}{1 + \omega^{2}},\sin\theta = \frac{2\omega}{1 + \omega^{2}},\)其中\(\omega = \tan\frac{\theta}{2} \in ( - \infty, + \infty\rbrack,\)得到参数化\(\psi:\)

\(\psi(\omega) = e^{i\Omega}\left( \frac{a\left( 1 - \omega^{2} \right)}{1 + \omega^{2}} + \frac{2b\omega i}{1 + \omega^{2}} \right) + u + vi,\)其中\(\omega\mathbb{\in R,}\)

\(e^{i\Omega}\)是旋转因子,\(u + vi\)是平移常数,其实和我们所要讨论的几何无关.写出它们只是为了强调\((x,y)\)是\(p\)相对于椭圆标准参考系下的笛卡尔坐标,下文将直接采用\(p = x + iy\).

设点\(p = x + yi,\)根据直线方程:\(\left| \begin{matrix} \operatorname{Re}{\phi(\omega')} & \operatorname{Im}{\phi(\omega'}) & 1 \\ \operatorname{Re}{\phi(\omega)} & \operatorname{Im}{\phi(\omega)} & 1 \\ x & y & 1 \\ \end{matrix} \right| = 0\)

即,\(\left| \begin{matrix} a\left( 1 - {\omega'}^{2} \right) & 2b\omega' & 1 + \omega'^2 \\ a\left( 1 - \omega^{2} \right) & 2b\omega & 1 + \omega^{2} \\ x & y & 1 \\ \end{matrix} \right| = 0\)

\(c_{1} + a \times c_{3}\),得\(\left| \begin{matrix} 2a & 2b\omega' & 1 + \omega'^2 \\ 2a & 2b\omega & 1 + \omega^{2} \\ x + a & y & 1 \\ \end{matrix} \right| = 0\)

\(r_{1} - r_{2}\),再约去\(\left( \omega - \omega' \right)\)得\(\left| \begin{matrix} 0 & 2b & \omega' + \omega \\ 2a & 2b\omega & 1 + \omega^{2} \\ x + a & y & 1 \\ \end{matrix} \right| = 0\)

按第一行展开,解得\(\omega' = \frac{ay\omega + bx - ab}{(bx + ab)\omega - ay}\),所以在参数化\(\psi\)下,

\begin{equation}\label1 F_{p}(\omega) = \frac{ay\omega + bx - ab}{(bx + ab)\omega - ay} \end{equation}

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 Author| hbghlyj Posted 2023-3-11 21:39

无迹矩阵和伽马矩阵

一个矩阵$A$的迹\(\text{Tr}A = 0\)则称为无迹矩阵,容易验证,

\[\text{Tr}(A \cdot B) = \text{Tr}(B \cdot A),\text{Tr}(A \pm B) = \text{Tr}A \pm \text{Tr}B\]

\(\begin{pmatrix} a & b \\ c & - a \\ \end{pmatrix}^{2} = \begin{pmatrix} a^{2} + bc & 0 \\ 0 & a^{2} + bc \\ \end{pmatrix}\),所以,若$A$是二阶无迹矩阵,则\(A^{2} = - |A|E\).

引入三个特殊的无迹矩阵,称之为伽马矩阵\(\gamma^{1} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix},\gamma^{2} = \begin{pmatrix} 0 & - i \\ i & 0 \\ \end{pmatrix},\gamma^{3} = \begin{pmatrix} 1 & 0 \\ 0 & - 1 \\ \end{pmatrix}\)

容易验证下面的等式:

①\(\gamma^{1} \cdot \gamma^{2} = \gamma^{3}{,\gamma}^{2} \cdot \gamma^{3} = \gamma^{1},\gamma^{3} \cdot \gamma^{1} = \gamma^{2},\gamma^{2} \cdot \gamma^{1} = - \gamma^{3}{,\gamma}^{3} \cdot \gamma^{2} = - \gamma^{1},\gamma^{1} \cdot \gamma^{3} = - \gamma^{2}\)

②\(\gamma^{1} \cdot \gamma^{1} = \gamma^{2} \cdot \gamma^{2} = \gamma^{3} \cdot \gamma^{3} = 1\)

(上面的两组等式可以合并为\(\gamma^{i} \cdot \gamma^{j} = \delta^{{ij}}E + i\varepsilon^{{ijk}}\gamma^{k}\)

此处采用哑指标求和规则,\(\delta^{{ij}}\)为Dirac记号,定义为\(i = j,\delta^{{ij}} = 1;i \neq j,\delta^{{ij}} = 0,\)

另一个符号\(\delta^{{ijk}}\)为Levi-Civita记号,定义为\(\varepsilon^{123} = 1,\varepsilon^{{ijk}} = - \varepsilon^{{jik}} = \varepsilon^{{jki}}\))

显然任何二阶无迹矩阵可以表示为伽马矩阵的线性组合\begin{equation} M = m^{1}\gamma^{1} + m^{2}\gamma^{2} + m^{3}\gamma^{3}\label2\end{equation}

反过来,任何一个二阶无迹矩阵$M$都对应着一个三维矢量\(\overrightarrow{T}\)使得$[\overrightarrow{T}]=M$

式\eqref{2}和三维矢量的点积\(\overrightarrow{A} \cdot \overrightarrow{B} = A^{1}B^{1} + A^{2}B^{2} + A^{3}B^{3}\)是类似的,这启发我们对任一个三维矢量\(\overrightarrow{T} = \left( T^{1},T^{2},T^{3} \right)\)定义它对应的二阶无迹矩阵\(\lbrack T\rbrack = T^{1}\gamma^{1} + T^{2}\gamma^{2} + T^{3}\gamma^{3} = \begin{pmatrix} T^{3} & T^{1} - iT^{2} \\ T^{1} + iT^{2} & - T^{3} \\ \end{pmatrix}\).

于是形式上可以写出\(\overrightarrow{T} \cdot \overrightarrow{\gamma} = \lbrack T\rbrack\),其中\(\overrightarrow{\gamma} = (\gamma^{1},\gamma^{2},\gamma^{3})\)

对于平面上一个点$p=x+iy$,定义它对应的三维矢量\(\overrightarrow{p} = (bx, - iab,ay)\).

那么点$p$对应的矩阵为\((p) = \begin{pmatrix} ay& bx - ab \\ bx + ab & - ay \\ \end{pmatrix}\)

由于分式线性函数在复合运算下和二阶矩阵的一维线性空间在矩阵乘法运算下是同态的,即\(F_{p}(\omega) = \lbrack p\rbrack(\omega)\),因此映射\(F_{p}\)在复合运算下和一维矩阵空间\(\left\{ \lambda\lbrack p\rbrack \mid \lambda \in \mathbb{R}^{*} \right\}\)在矩阵乘法运算下是同态的.所以可以用\(\lbrack p\rbrack\)来描述\(F_{p}\),进而描述点$p$本身.

无穷远点的引入:\(\overrightarrow{p}\)和三维矢量并非一一对应,因为\(p\)的第二个分量\(-iab≠0\),称\(\overrightarrow{k}=(bu,0,av)\)所描述的点\(k\)为斜率为\(\frac{v}{u}\)的平行直线的无穷远点.

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 Author| hbghlyj Posted 2023-3-11 22:00
符号约定:用大写或小写英文字母表示点,点\(p\)对应的参数\(\sigma(p)\)也记为\(p\),两点\(A,B\)间距离用\({AB}\)表示,用\(L_{{AB}}\)代表过两点\(A,B\)的直线,用希腊字母\(\xi,\phi,\psi,\cdots\)代表参数化
到目前为止我们的讨论都建立在一个特定的参数化(1)上,但后文将证明上述结果对于一大类参数化都是成立的,更关键地是这些参数化的集合包含了射影坐标系!(读者应该将参数化理解为建立一个一维坐标系)

\(\lbrack p\rbrack\)的重要等式

  1. \(\lbrack p\rbrack^{2} = {\overrightarrow{p}}^{2}E\)
  2. \(\lbrack p\rbrack \cdot \lbrack q\rbrack \cdot \lbrack p\rbrack = 2\overset{\rightarrow}{p} \cdot \overset{\rightarrow}{q}\lbrack p\rbrack - {\overset{\rightarrow}{p}}^{2}\lbrack q\rbrack\)
  3. \(p\)在曲线上\(\Leftrightarrow \lbrack p\rbrack^{2} = 0\)
  4. \(\lbrack p\rbrack \cdot \lbrack q\rbrack + \lbrack q\rbrack \cdot \lbrack p\rbrack = 2\overrightarrow{p} \cdot \overrightarrow{q}E\)
  5. \(k \in L_{{pq}} \Leftrightarrow \lbrack k\rbrack = \lambda\lbrack p\rbrack + (1 - \lambda)q\)(和定比分点公式形式上一致)
  6. \(\text{tr}\left( \left\lbrack p_{n} \right\rbrack\cdots\left\lbrack p_{1} \right\rbrack \cdot \lbrack q\rbrack \cdot \left\lbrack p_{1} \right\rbrack\cdots\left\lbrack p_{n} \right\rbrack \right) = 0\)
  7. \(\lbrack n\rbrack \cdot \lbrack p\rbrack \cdot \lbrack m\rbrack - \lbrack m\rbrack \cdot \lbrack p\rbrack \cdot \lbrack n\rbrack = 2i\left\lbrack \overrightarrow{n},\overrightarrow{p},\overrightarrow{m} \right\rbrack E\)
    于是存在非零实数\(\kappa\)和点\(h\)满足\(\lbrack p_{n}\rbrack\cdots\lbrack p_{1}\rbrack \cdot \lbrack q\rbrack \cdot \lbrack p_{1}\rbrack\cdots\lbrack p_{n}\rbrack = \kappa\lbrack h\rbrack\).
证明
  1. \(\lbrack p\rbrack^{2} = \begin{pmatrix} {ay} & bx - ab \\ bx + ab & - ay \\ \end{pmatrix}^{2} = \begin{pmatrix} a^{2}y^{2} + b^{2}x^{2} - a^{2}b^{2} & 0 \\ 0 & a^{2}y^{2} + b^{2}x^{2} - a^{2}b^{2} \\ \end{pmatrix}\)
    或者这样证明:$[p] \cdot [ p ]=\sum_{i}^{3} p^{i} \gamma^{i}\left(\sum_{j}^{3} p^{j} \gamma^{j}\right)=\sum_{i, j}^{3} p^{i} p^{j}\left(\gamma^{i} \cdot \gamma^{j}\right) $
    $=\sum_{i, j}^{3} p^{i} p^{j} \delta^{i j}+{\color{red}{i \sum_{i, j, k}^{3} p^{i} p^{j} \varepsilon^{i j k} \gamma^{k} }}=\vec{p}^{2} $
    因为\(\varepsilon^{{ijk}}\)关于$i,j$反对称而\(p^{i}p^{j}\)关于$i,j$对称,所以红色部分为0
  2. \begin{array}{l} [p] \cdot[q] \cdot[p] = \sum\limits_{i}^{3} p^{i} \gamma^{i}\left(\sum\limits_{j}^{3} q^{j} \gamma^{j}\right)\left(\sum\limits_{t}^{3} p^{t} \gamma^{t}\right) = \sum\limits_{i, j}^{3} p^{i} q^{j} \delta^{i j}\left(\sum\limits_{t}^{3} p^{t} \gamma^{t}\right) \\
    +i \sum\limits_{i, j, k}^{3} p^{i} q^{j} \varepsilon^{i j k} \gamma^{k}\left(\sum\limits_{t}^{3} p^{t} \gamma^{t}\right) = \vec{p} \cdot \vec{q}[p]+i \sum\limits_{i, j, k, t}^{3} p^{i} q^{j} p^{t} \varepsilon^{i j k} \gamma^{k} \cdot \gamma^{t} = \vec{p} \cdot \vec{q}[p]+ \\
    {\color{red}{i \sum\limits_{i, j, k, t}^{3} p^{i} q^{j} p^{t} \varepsilon^{i j k} \delta^{k t} E}}-\sum\limits_{i, j, k, t, g}^{3} p^{i} q^{j} p^{t} \varepsilon^{i j k} \varepsilon^{k t g} \gamma^{g} = \vec{p} \cdot \vec{q}[p]-\sum\limits_{i, j, k, t, g}^{3} p^{i} q^{j} p^{t} \delta^{i t} \delta^{j g} \gamma^{g}+ \\
    \sum\limits_{i, j, k, t, g}^{3} p^{i} q^{j} p^{t} \delta^{i g} \delta^{j t} \gamma^{g} = \vec{p} \cdot \vec{q}[p]-\vec{p}^{2}[q]+\vec{p} \cdot \vec{q}[p] = 2 \vec{p} \cdot \vec{q}[p]-\vec{p}^{2}[q]
    \end{array} 因为\(\varepsilon^{{ijk}}\)关于\(i,j\)反对称而\(p^{i}p^{j}\)关于\(i,j\)对称,所以红色部分为\(0\)
    或者利用(4),\(\lbrack p\rbrack \cdot \lbrack q\rbrack \cdot \lbrack p\rbrack = \lbrack p\rbrack \cdot \left( \lbrack q\rbrack \cdot \lbrack p\rbrack \right) = \lbrack p\rbrack\left( 2\overrightarrow{p} \cdot \overrightarrow{q}E - \lbrack p\rbrack \cdot \lbrack q\rbrack \right) = 2\overrightarrow{p} \cdot \overrightarrow{q}\lbrack p\rbrack - {\overrightarrow{p}}^{2}\lbrack q\rbrack\)
  3. \(p\)在曲线上\(\Leftrightarrow {\overrightarrow{p}}^{2} = a^{2}y^{2} + b^{2}x^{2} - a^{2}b^{2} = 0\),利用(1)即得证
    由此可补充定义\(\lbrack p\rbrack p = 0,\)即\(\lbrack p\rbrack(x) = \left\{ \begin{matrix} 0 & x = p \\ p & x \neq p \\ \end{matrix} \right.\ \)
    \begin{aligned} \lbrack p] \cdot[q]+[q] \cdot[p]=2 \sum_{i, j}^{3} p^{i} q^{j} \delta^{i j}E+i \sum_{i, j, k}^{3} p^{i} q^{j}\left(\varepsilon^{i j k}+\varepsilon^{j i k}\right) \gamma^{k}=2 \vec{p} \cdot \vec{q} E\\
    \end{aligned} 当\(\lbrack p\rbrack \cdot \lbrack q\rbrack + \lbrack q\rbrack \cdot \lbrack p\rbrack = O\)时,称点$p$和点$q$关于曲线对偶.显然,若$p,q$对偶,则$q,p$对偶.
    利用(2)(4),\(p,q\)对偶$⇔[p]·[q]+[q]·[p]=O⇔\overrightarrow{p} \cdot \overrightarrow{q} = 0 \Leftrightarrow \frac{x_{P}x_{Q}}{a^{2}} + \frac{y_{P}y_{Q}}{b^{2}} = 1$
    \[\Leftrightarrow {\overrightarrow{p}}^{2}{\overrightarrow{q}}^{2} = 1 \Leftrightarrow \lbrack p\rbrack \cdot \lbrack q\rbrack \cdot \lbrack p\rbrack = - {\overrightarrow{p}}^{2}\lbrack q\rbrack \Leftrightarrow \lbrack q\rbrack \cdot \lbrack p\rbrack \cdot \lbrack q\rbrack = - {\overrightarrow{q}}^{2}\lbrack p\rbrack\]
  4. \begin{align*} &\lambda[p]+(1-\lambda)[q]=\lambda\left(\begin{array}{cc}
    a y & b x-a b \\
    b x+a b & -a y
    \end{array}\right)+(1-\lambda)\left(\begin{array}{cc}
    a v & b u-a b \\
    b u+a b & -a v
    \end{array}\right)\\
    &=\left(\begin{array}{cc}
    a( \lambda y+(1-\lambda) v) & b( \lambda x+(1-\lambda) u)-a b \\
    b( \lambda x+(1-\lambda) u)+a b & -a( \lambda y+(1-\lambda) v)
    \end{array}\right)=[k]\end{align*} 其中$p=x+iy,q=u+iv,k=λp+(1-λ)q$.
  5. \(\text{tr}\left( \left\lbrack p_{n} \right\rbrack\cdots\left\lbrack p_{1} \right\rbrack \cdot \lbrack q\rbrack \cdot \left\lbrack p_{1} \right\rbrack\cdots\left\lbrack p_{n} \right\rbrack \right) = \text{tr}\left( \left\lbrack p_{1} \right\rbrack^{2}\cdots\left\lbrack p_{n} \right\rbrack^{2} \cdot \lbrack q\rbrack \right) = \text{tr}\left( {\overrightarrow{p_{1}}}^{2}\cdots{\overrightarrow{p_{n}}}^{2} \cdot \lbrack q\rbrack \right) = 0\).

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 Author| hbghlyj Posted 2023-3-11 22:09

帕斯卡定理

在二次曲线上取六点\(a,b,c,d,e,f\), \(L_{{ab}} \cap L_{{de}} = p,L_{{bc}} \cap L_{{ef}} = q,L_{{cd}} \cap L_{{fa}} = r,\) 则\(p,q,r\)共线

证明:设\(L_{{de}} \cap L_{{qr}} = p'\),命题等价于证明\(a,p',b\)共线,等价于证明\(\lbrack p'\rbrack(b) = a\),

而\(a = \lbrack r\rbrack \cdot \lbrack q\rbrack \cdot \left\lbrack p' \right\rbrack \cdot \lbrack r\rbrack \cdot \lbrack q\rbrack(b),\)只需证明\(\exists k\mathbb{\in R}\)使\(\lbrack r\rbrack \cdot \lbrack q\rbrack \cdot \left\lbrack p' \right\rbrack \cdot \lbrack r\rbrack \cdot \lbrack q\rbrack = k\left\lbrack p' \right\rbrack.\)

因为\(p' \in L_{{qr}},\)所以存在\(\alpha,\beta\mathbb{\in R}\)使\(\lbrack p'\rbrack = \alpha\lbrack q\rbrack + \beta\lbrack r\rbrack,\)代入左边得

\(\lbrack r\rbrack \cdot \lbrack q\rbrack \cdot \left( \alpha\lbrack q\rbrack + \beta\lbrack r\rbrack \right) \cdot \lbrack r\rbrack \cdot \lbrack q\rbrack = \lbrack q\rbrack^{2}\lbrack r\rbrack^{2}\left( \alpha\lbrack q\rbrack + \beta\lbrack r\rbrack \right) = k\left\lbrack p' \right\rbrack\),其中\(k = \lbrack q\rbrack^{2}\lbrack r\rbrack^{2}\)


也可以利用三次曲线的加法性质证明:将直线\({pq}\)与二次曲线并成一个三次曲线

\[(A + B) + C = D + C = Q = A + F = A + (B + C)\]

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