Forgot password?
 Register account
View 288|Reply 2

[几何] 二次曲线的Steiner定义

[Copy link]

3159

Threads

7941

Posts

610K

Credits

Credits
63770
QQ

Show all posts

hbghlyj Posted 2023-4-27 02:55 |Read mode
设 $f$ 是通过点 $X$ 的线束和通过点 $Y$ 的线束之间的一个射影变换。
定理4.1. 集合 $\displaystyle {\cal C} :=\{\ell \cdot f(\ell) : \ell$ 是通过 $ X$的任意直线$\displaystyle \} $ 是二次曲线。
设$L$是通过$X,Y$的直线.
$X=f^{-1}(L)\cdot L$,所以$\mathcal C$通过$X$.
$Y=L\cdot f(L)$,所以$\mathcal C$通过$Y$.
img132[1].gif
如果$T$是把$X$映射到$[0:1:0]$和把$Y$映射到$[1:0:0]$的一个射影变换,则通过$X$和$Y$的线束被映射到平面 $z = 1$ 上的垂直线和水平线。${\cal C}$ 被映射到函数 $g$ 的图象。

习题4.5 利用 $T \circ f \circ T^{-1}$ 是通过$[0:1:0]$的线束(垂直线)到通过 $ [1:0:0]$ 的线束(水平线)的射影变换以证明$g$的形式为$$ g(x) = (ax + b) /(cx+d)$$由此得出 $T({\cal C})$ 是平面 $z = 1$ 上的双曲线,并用它来证明定理 4.1。

3159

Threads

7941

Posts

610K

Credits

Credits
63770
QQ

Show all posts

 Author| hbghlyj Posted 2023-4-27 03:10

试着写写✏️

习题4.5的证明:
设$T \circ f \circ T^{-1}$将垂直线$x+a=0$映射到水平线$y+b=0$. 则$[1:a]\mapsto[1:b]$是一维射影变换,可以写成$$[x_0:x_1]\mapsto[cx_1+dx_0:ax_1+bx_0]$$即$$\left[1:\frac{x_1}{x_0}\right]\mapsto\left[1:{a\frac{x_1}{x_0}+b\over c\frac{x_1}{x_0}+d}\right]$$所以$$g(x)={ax + b\over cx+d}$$$T(\mathcal C)$即$g$的图象是双曲线, 所以$\mathcal C=T^{-1}(T(\mathcal C))$也是二次曲线.

由定理1推出习题4.5:
假设$g$的图象是二次曲线. 它通过$(1:0:0)$和$(0:1:0)$说明$g$的$x^2$项和$y^2$项系数为0, 于是可以整理为$y=(ax + b) /(cx+d)$的形式.

3159

Threads

7941

Posts

610K

Credits

Credits
63770
QQ

Show all posts

 Author| hbghlyj Posted 2023-4-27 03:26
Steiner conic
MathWorld
Another theorem due to Steiner lets lines $x$ and $y$ join a variable point on a conic section to two fixed points on the same conic section. Then $x$ and $y$ are projectively related.
134134g87wi5g5kzjpzk7i.png Harold Scott Macdonald Coxeter, Projective geometry, Springer (2003)
8.32

Construct tangents $p$ and $q$, at the fixed points $P$ and $Q$, meet in $D$, the pole of $PQ$ as in Figure 8.3 A.
Let $c$ be a fixed line through $D$ (but not through $P$ or $Q$), meeting $x$ in $B$, and $y$ in $A$. By 8.31 (SEYDEWITZ'S THEOREM), $BA$ is a pair of the involution of conjugate points on $c$. Hence, when the point $R=x\cdot y$ varies on the conic, we have $x$ projective to $B$, $B$ projective to $A$, and $A$ projective $y$, as desired.
135456mpdzoxsnd9sn7mks.gif

Mobile version|Discuz Math Forum

2025-5-31 10:53 GMT+8

Powered by Discuz!

× Quick Reply To Top Edit