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Cayley八元组与双切线

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青青子衿 发表于 2024-6-6 00:14 |阅读模式
本帖最后由 青青子衿 于 2024-6-8 01:01 编辑 Plane Quartics, Cayley Octads, and Bitangents.

\begin{gather*}
\left(\left(1-\mu^{2}\nu^{2}\right)x^{2}+\left(1-\lambda^{2}\nu^{2}\right)y^{2}+\left(1-\lambda^{2}\mu^{2}\right)z^{2}\right)^{2}\\
=4\left(1-\lambda^{2}\mu^{2}\nu^{2}\right)\left(\left(1-\nu^{2}\right)x^{2}y^{2}+\left(1-\mu^{2}\right)x^{2}z^{2}+\left(1-\lambda^{2}\right)y^{2}z^{2}\right)\\
\\
\left(\frac{ux^{2}+vy^{2}-wz^{2}}{y^{2}}\right)^{2}+g(\frac{x^{2}}{y^{2}}-\lambda^{2})(\mu^{2}\frac{x^{2}}{y^{2}}-1)=0\\
\\
\left\{\begin{split}
X&=\frac{x}{y}\\
Y&=\frac{ux^{2}+vy^{2}-wz^{2}}{y^{2}}\\
u&=\lambda^{2}\mu^{4}\nu^{2}-2\lambda^{2}\mu^{2}\nu^{2}+\lambda^{2}\mu^{2}+\mu^{2}\nu^{2}-2\mu^{2}+1\\
v&=\lambda^{4}\mu^{2}\nu^{2}-2\lambda^{2}\mu^{2}\nu^{2}+\lambda^{2}\mu^{2}+\lambda^{2}\nu^{2}-2\lambda^{2}+1\\
w&=(\lambda^{2}\mu^{2}-1)^{2}\\
g&=4(\lambda^{2}-1)(\mu^{2}-1)(\nu^{2}-1)(\lambda^{2}\mu^{2}\nu^{2}-1)
\end{split}\right.\\

\end{gather*}


  1. \left(\frac{ux^{2}+vy^{2}-wz^{2}}{y^{2}}\right)^{2}+g(\frac{x^{2}}{y^{2}}-\lambda^{2})(\mu^{2}\frac{x^{2}}{y^{2}}-1)=0
  2. \left(\left(1-\mu^{2}\nu^{2}\right)x^{2}+\left(1-\lambda^{2}\nu^{2}\right)y^{2}+\left(1-\lambda^{2}\mu^{2}\right)z^{2}\right)^{2}-4\left(1-\lambda^{2}\mu^{2}\nu^{2}\right)\left(\left(1-\nu^{2}\right)x^{2}y^{2}+\left(1-\mu^{2}\right)x^{2}z^{2}+\left(1-\lambda^{2}\right)y^{2}z^{2}\right)=0
  3. u=\lambda^{2}\mu^{4}\nu^{2}-2\lambda^{2}\mu^{2}\nu^{2}+\lambda^{2}\mu^{2}+\mu^{2}\nu^{2}-2\mu^{2}+1
  4. v=\lambda^{4}\mu^{2}\nu^{2}-2\lambda^{2}\mu^{2}\nu^{2}+\lambda^{2}\mu^{2}+\lambda^{2}\nu^{2}-2\lambda^{2}+1
  5. w=(\lambda^{2}\mu^{2}-1)^{2}
  6. g=4(\lambda^{2}-1)(\mu^{2}-1)(\nu^{2}-1)(\lambda^{2}\mu^{2}\nu^{2}-1)
  7. (x^{2}-\lambda^{2}y^{2})(\mu^{2}x^{2}-y^{2})=0
复制代码




\begin{gather*}
\\
y^2=x^5+b_4x^4+b_3x^3+b_2x^2+b_1x+b_0\\
\\
\left(\frac{P_1(x)}{Q_1(x)}y\right)^2=
\left(\frac{p_1(x)}{q_1(x)}\right)^4+a_3\left(\frac{p_1(x)}{q_1(x)}\right)^3+a_2\left(\frac{p_1(x)}{q_1(x)}\right)^2+a_1\frac{p_1(x)}{q_1(x)}+a_0\\
\\
\int_{0}^{\frac{p_1(x)}{q_1(x)}}\dfrac{{\mathrm{d}}t}{\sqrt{t^4+a_3t^3+a_2t^2+a_1t+a_0}}=

\int_{0}^{x}\dfrac{\frac{p_1'(t)q_1(t)-p_1(t)q_1'(t)}{q_1^2(t)}\cdot\frac{Q_1(t)}{P_1(t)}}{\sqrt{t^5+b_4t^4+b_3t^3+b_2t^2+b_1t+b_0}}{\mathrm{d}}t\\
\\
\left(\frac{P_2(x)}{Q_2(x)}y\right)^2=
\dfrac{\left(\frac{p_2(x)}{q_2(x)}\right)^4+a_3\left(\frac{p_2(x)}{q_2(x)}\right)^3+a_2\left(\frac{p_2(x)}{q_2(x)}\right)^2+a_1\frac{p_2(x)}{q_2(x)}+a_0}{\left(c_1\frac{p_2(x)}{q_2(x)}+c_0\right)^2}\\
\\
\int_{0}^{\frac{p_2(x)}{q_2(x)}}\dfrac{c_1t+c_0}{\sqrt{t^4+a_3t^3+a_2t^2+a_1t+a_0}}{\mathrm{d}}t=

\int_{0}^{x}\dfrac{\frac{p_2'(t)q_2(t)-p_2(t)q_2'(t)}{q_2^2(t)}\cdot\frac{Q_2(t)}{P_2(t)}}{\sqrt{t^5+b_4t^4+b_3t^3+b_2t^2+b_1t+b_0}}{\mathrm{d}}t
\end{gather*}

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