每条直线恰好包含三个点

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Melchior不等式得出，在$\mathbb{R}$上任意直线的有限集有$t_2\geqslant3$，但在$\mathbb{C}$上却可以有$t_2=0$请看下面的例子：

Example 5 (Fermat configuration).
复射影曲线$X^k + Y^k + Z^k = 0$的拐点的集合为
$$\mathcal{V}=\bigcup_{i=1}^k\left\{\left[1: \omega^i: 0\right]\right\} \cup\left\{\left[\omega^i: 0: 1\right]\right\} \cup\left\{\left[0: 1: \omega^i\right]\right\}$$
其中$ω^k=−1$.
那么 $\cal V$ 有 $n = 3k$ 个点，很容易证明 $V$ 确定了三条直线，每条直线包含 $k$ 个点，而其他每条直线恰好包含三个点。特别地，$V$ 确定的直线都不是普通直线（恰好包含 2 个点的直线称为普通直线）即$t_2=0$.

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minipages.impan.pl/images/BHarbourne__CracowL … v3NoSolns.pdf#page=3 第3页
arxiv.org/pdf/1705.09946
The Fermat arrangement of $3n$ complex lines and their $n^2+3$ points of intersection for $n=3$.
(The 12 points for $n=3$ are indicated by the three open circles, the three dotted circles and the six black circular dots.
The coordinate axes are represented by dotted lines.
At each coordinate vertex there occur $n$ of the $3n$ lines, defined by the forms shown; the $n^2+3$ points consist of
a complete intersection of $n^2$ points plus the 3 coordinate vertices. This arrangement
does not exist over the reals: one must regard the open circles as representing collinear points, and
likewise the dotted circles as representing collinear points.)

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en.wikipedia.org/wiki/Hesse_configuration#Realizability
kuing.cjhb.site/forum.php?mod=redirect&go … =12053&pid=58348
例如$f=\alpha\left(z_1^3+z_2^3+z_3^3\right)+6 \beta z_1 z_2 z_3$的9个拐点为
\begin{array}{lll}
(0,1,-1) & (0,1,-\omega) & \left(0,1,-\omega^2\right) \\
(-1,0,1) & (-\omega, 0,1) & \left(-\omega^2, 0,1\right) \\
(1,-1,0) & (1,-\omega, 0) & \left(1,-\omega^2, 0\right)
\end{array}其中$ω=\exp(2\pi i/3)$.
这些点确定的12条直线为\begin{array}{lll}
z_1 = 0&z_2=0&z_3=0\\
z_1+z_2+z_3=0&\omega^2 z_1+\omega z_2+z_3=0&\omega z_1+\omega^2 z_2+z_3=0\\
\omega z_1+z_2+z_3=0&z_1+\omega z_2+z_3=0&\omega^2 z_1+\omega^2 z_2+z_3=0\\
\omega^2z_1+z_2+z_3=0&\omega z_1+\omega z_2+z_3=0&z_1+\omega^2 z_2+z_3=0
\end{array}每条直线恰好包含三个点因此Sylvester–Gallai theorem在$\mathbb{C}$上不成立。

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有关的帖子：math.stackexchange.com/questions/3102843/sylv … lex-projective-plane
楼主问，为什么Sylvester–Gallai theorem在$\mathbb{C}$上不成立？
获采纳的答案大概是，因为Sylvester–Gallai theorem的证明使用了一条直线的位于一侧或另一侧的点，这在$\mathbb{C}$上没有意义。复直线上的点没有序关系。