有理曲线$(\frac{t+t^2}{1+t^5+t^6+t^9+t^{10}},\frac{1+t^3+t^4+t^5+t^6}{t+t^8})$

hbghlyj

ParametricPlot[{(t + t^2)/(1 + t^5 + t^6 + t^9 + t^10), (1 + t^3 + t^4 + t^5 + t^6)/(t + t^8)}, {t, -2, 2}]

参数曲线$(x,y)=\left(\frac{t+t^2}{1+t^5+t^6+t^9+t^{10}},\frac{1+t^3+t^4+t^5+t^6}{t+t^8}\right)$，如何重新有理参数化为$(g_1(u),g_2(u))$，使$u$可表为$t$的有理函数且可表为坐标$x,y$的有理函数（因此曲线上一点$x,y$对应于唯一的参数$u$）

hbghlyj

arxiv.org/pdf/2101.01059 说这个问题是 rational minimal basis theorem，对于1个或2个参数是可行的。

具体的问题如何解决呢

hbghlyj

按照神奇的Lüroth定理求出$\Psi(t,t_1)$

1. PolynomialGCD[(t+t^2)(1+t1^5+t1^6+t1^9+t1^10)-(1+t^5+t^6+t^9+t^10)(t1+t1^2),(1+t^3+t^4+t^5+t^6)(t1+t1^8)-(1+t1^3+t1^4+t1^5+t1^6)(t+t^8)]

复制代码

$\Psi(t,t_1)=t-t_1$是一次的。可见$t$已经是$x,y$的有理函数

把$\psi(t)x-f(t), \, \psi(t)y-\varphi(t)$关于$t$辗转相除直到1次，可将$t$表示为$x,y$的有理函数
$t=\frac{16 x^9 y^9+51 x^8 y^9-76 x^8 y^8+62 x^7 y^9-174 x^7 y^8+112 x^7 y^7+40 x^6 y^9-137 x^6 y^8+221 x^6 y^7-116 x^6 y^6+19 x^5 y^9-46 x^5 y^8+141 x^5 y^7-242 x^5 y^6+13 x^5 y^5+7 x^4 y^9-8 x^4 y^8+23 x^4 y^7-174 x^4 y^6+83 x^4 y^5+24 x^4 y^4+x^3 y^9-2 x^3 y^8-x^3 y^7-46 x^3 y^6+53 x^3 y^5-2 x^3 y^4+11 x^3 y^3-x^2 y^8+6 x^2 y^6+18 x^2 y^5-46 x^2 y^4-59 x^2 y^3-32 x^2 y^2+2 x y^4+16 x y^3+14 x y^2+5 x y-1}{2 x y \left(16 x^8 y^8+60 x^7 y^8-84 x^7 y^7+89 x^6 y^8-254 x^6 y^7+141 x^6 y^6+66 x^5 y^8-266 x^5 y^7+335 x^5 y^6-62 x^5 y^5+29 x^4 y^8-110 x^4 y^7+268 x^4 y^6-238 x^4 y^5-64 x^4 y^4+11 x^3 y^8-14 x^3 y^7+50 x^3 y^6-214 x^3 y^5+46 x^3 y^4+70 x^3 y^3+3 x^2 y^8-18 x^2 y^6-60 x^2 y^5+80 x^2 y^4+46 x^2 y^3-8 x^2 y^2-6 x y^4-48 x y^3-42 x y^2-14 x y+3\right)}$

即$K(t)=K(x,y), x=\frac{f(t)}{\psi(t)},y=\frac{\varphi(t)}{\psi(t)}$，对任何域$K$.

hbghlyj

再举一个例子，Ampersand curve的有理参数化:
\begin{cases}x=\frac{f(t)}{\psi(t)}={3645 - 864\sqrt3 t + 63t^2 + 12\sqrt3 t^3\over2754 - 864\sqrt3 t + 306t^2 - 24\sqrt3 t^3+4t^4}\\
y=\frac{\varphi(t)}{\psi(t)}={-1620\sqrt3 + 1647t - 162\sqrt3 t^2 - 3t^3 + 2\sqrt3 t^4\over2754 - 864\sqrt3 t + 306t^2 - 24\sqrt3 t^3+4t^4}\end{cases}

1. PolynomialGCD[(3645-(864 Sqrt[3]) t+63 t^2+12 Sqrt[3] t^3)(2754-864 Sqrt[3] t1+306 t1^2-24 Sqrt[3] t1^3+4 t1^4)-(2754-(864 Sqrt[3]) t+306 t^2-(24 Sqrt[3]) t^3+4 t^4)(3645-864 Sqrt[3] t1+63 t1^2+12 Sqrt[3] t1^3),(-1620 Sqrt[3]+1647 t-(162 Sqrt[3]) t^2-3 t^3+2 Sqrt[3] t^4)(2754-864 Sqrt[3] t1+306 t1^2-24 Sqrt[3] t1^3+4 t1^4)-(2754-(864 Sqrt[3]) t+306 t^2-(24 Sqrt[3]) t^3+4 t^4)(-1620 Sqrt[3]+1647 t1-162 Sqrt[3] t1^2-3 t1^3+2 Sqrt[3] t1^4)]

复制代码

$\Psi(t,t_1)=6(t-t_1)$是一次的。可见$t$已经是$x,y$的有理函数

把$\psi(t)x-f(t), \, \psi(t)y-\varphi(t)$关于$t$辗转相除直到1次，可将$t$表示为$x,y$的有理函数
$t=\frac{3 \left(648076 \sqrt{3} x^5-883278 x^4 y-1503194 \sqrt{3} x^4+234120 \sqrt{3} x^3 y^2+3058001 x^3 y-275762 \sqrt{3} x^3-877196 x^2 y^3-291512 \sqrt{3} x^2 y^2-1827939 x^2 y+1461552 \sqrt{3} x^2+359072 \sqrt{3} x y^4+243840 x y^3-281160 \sqrt{3} x y^2-1320984 x y+577944 \sqrt{3} x-117312 y^5-177600 \sqrt{3} y^4+1327104 y^3-751536 \sqrt{3} y^2+397764 y\right)}{815298 x^5-334488 \sqrt{3} x^4 y-1910887 x^4+265300 x^3 y^2+1264756 \sqrt{3} x^3 y-341851 x^3-360176 \sqrt{3} x^2 y^3-264416 x^2 y^2-838924 \sqrt{3} x^2 y+1861896 x^2+407616 x y^4+127040 \sqrt{3} x y^3-318240 x y^2-579024 \sqrt{3} x y+735012 x-37632 \sqrt{3} y^5-280320 y^4+589824 \sqrt{3} y^3-969408 y^2+168624 \sqrt{3} y}$

即$K(t)=K(x,y),x=\frac{f(t)}{\psi(t)},y=\frac{\varphi(t)}{\psi(t)}$，对任何域$K$.