存在唯一的二次曲面 包含三条相互不相交的线

hbghlyj

AG Lent 2020 Supplementary Examples.pdf

1. Given three mutually disjoint lines in P³, prove that there is a unique quadric surface containing these three lines.

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考虑与3直线相交的第四直线，然后第四条直线形成曲面，这是二次曲面

math.stackexchange.com/questions/1204689/there-is-a-unique-quadric-through-three-disjoint-lines

Theodore Shifrin - Abstract Algebra: A Geometric Approach (page 337)

Proposition 4.10. Let $\ell_1, \ell_2, \ell_3$ be lines in $\mathbb{P}^3$ in general position. Then there is a unique quadric surface containing them all.

Proof. Let's start with a specific case. Suppose
\[\tag∗
\begin{aligned}
& \ell_1=\left\{x_0=x_1=0\right\} \\
& \ell_2=\left\{x_2=x_3=0\right\} \\
& \ell_3=\left\{x_0=x_2, x_1=x_3\right\} .
\end{aligned}
\]
As in the proof of Lemma 4.8, for each point $P_3 \in \ell_3$, there is a unique line through $P_3$ intersecting both $\ell_1$ and $\ell_2$; we now calculate this line explicitly. Let $P_3=[\alpha, \beta, \alpha, \beta]$. The plane spanned by $P_3$ and $\ell_2$ consists of all points of the form $[*, *, \alpha, \beta]$ (where $*$ denotes an arbitrary real number), and so it intersects $\ell_1$ in the point $P_1=$ $[0,0, \alpha, \beta]$. Now the line $\ell$ through $P_1$ and $P_3$ is given by
\[
\overleftrightarrow{P_1 P_3}=\{[s(0,0, \alpha, \beta)+t(\alpha, \beta, \alpha, \beta)]: s, t \in \mathbb{R} \text { not both zero }\} .
\]
We can rewrite this explicitly as follows:
\[
\begin{aligned}
& x_0=t \alpha \\
& x_1=t \beta \\
& x_2=(s+t) \alpha \\
& x_3=(s+t) \beta .
\end{aligned}
\]
Note that every point on this line satisfies the equation
\[\tag{∗∗}\label{segre}
x_0 x_3=x_1 x_2 .
\]
And, conversely, every point on the surface $(* *)$ can be expressed in the form $(*)$ for some values of $s, t, \alpha$, and $\beta$. (For example, if $x_0 \neq 0$, then we may take $x_0=1=\alpha t, x_1=\beta t, x_2=(s+t) \alpha=1+s \alpha$, and solve for $x_3: x_3=x_1 x_2 / x_0=t \beta(1+s \alpha)=t \beta+(\alpha t)(s \beta)=(s+t) \beta$, as required. If $x_0=0$, then either $x_1=0$ or $x_2=0$ : if $x_0=x_1=0$, then set $t=0$, and solve $[\alpha, \beta]=\left[x_2, x_3\right]$; if $x_0=x_2=0$, then set $\alpha=0, \beta=1$, and solve $[t, s+t]=\left[x_1, x_3\right]$.)

hbghlyj

hbghlyj 发表于 2024-3-13 09:51
也许在论坛以前的一些帖子中出现 @青青子衿

kuing.cjhb.site/forum.php?mod=viewthread&tid=2897

hbghlyj

2. How many lines in P³ intersect four given disjoint lines L₁, L₂, L₃, L₄ in P³?

相当于两个二次曲面相交？

hbghlyj

谢谢

hejoseph

如果方向为 $\{X,Y,Z\}$，且过点 $(x_0,y_0,z_0)$ 的直线是二次曲线 $a_{11}x^2+a_{22}y^2+a_{33}z^2+2a_{12}xy+2a_{13}xz+2a_{23}yz+2a_{14}x+2a_{24}y+2a_{34}z+a_{33}=0$ 的直母线，设
\begin{align*}
f(x,y,z)&=a_{11}x^2+a_{22}y^2+a_{33}z^2+2a_{12}xy+2a_{13}xz+2a_{23}yz+2a_{14}x+2a_{24}y+2a_{34}z+a_{44}，\\
f_1(x,y,z)&=a_{11}x+a_{12}y+a_{13}z+a_{14}，\\
f_2(x,y,z)&=a_{12}x+a_{22}y+a_{23}z+a_{24}，\\
f_3(x,y,z)&=a_{13}x+a_{23}y+a_{33}z+a_{34}，\\
\Phi(x,y,z)&=a_{11}x^2+a_{22}y^2+a_{33}z^2+2a_{12}xy+2a_{13}xz+2a_{23}yz，\\
\end{align*}
则有
\[
\left\{
\begin{aligned}
&f(x_0,y_0,z_0)=0，\\
&\Phi(X,Y,Z)=0，\\
&f_1(x_0,y_0,z_0)X+f_2(x_0,y_0,z_0)Y+f_3(x_0,y_0,z_0)Z=0。
\end{aligned}
\right.
\]
这里已知三条直母线，可根据上面结果列出九个关于 $f(x,y,z)$ 系数的线性方程，只要方程不是线性相关的就能求得其系数比，二次曲面也就确定了。

hejoseph

设三条两两互相互为异面直线的直线方程是 $a_i:\dfrac{x - x_i}{l_i} = \dfrac{y - y_i}{m_i} = \dfrac{z - z_i}{n_i}$（$i = 1,2,3$），直线 $\dfrac{x - t}{l} = \dfrac{y - u}{m} = \dfrac{z - v}{n}$ 与上述三直线均共面，则
\[
\begin{vmatrix}
t - x_i & u - y_i & v - z_i \\
l_i & m_i & n_i \\
l & m & n
\end{vmatrix}
= 0 ，
\]
即
\[
(n_iu - m_iv - n_iy_i + m_iz_i)l - (n_it - l_iv - n_ix_i + l_iz_i)m + (m_it - l_iu - m_ix_i + l_iy_i)n = 0 ，
\]
这三个关于 $l$、$m$、$n$ 组成的线性方程组有非零解，得
\[
\begin{vmatrix}
n_1u - m_1v - n_1y_1 + m_1z_1 & n_1t - l_1v - n_1x_1 + l_1z_1 & m_1t - l_1u - m_1x_1 + l_1y_1 \\
n_2u - m_2v - n_2y_2 + m_2z_2 & n_2t - l_2v - n_2x_2 + l_2z_2 & m_2t - l_2u - m_2x_2 + l_2y_2 \\
n_3u - m_3v - n_3y_3 + m_3z_3 & n_3t - l_3v - n_3x_3 + l_3z_3 & m_3t - l_3u - m_3x_3 + l_3y_3
\end{vmatrix}
= 0 ，
\]
展开上面的行列式，得到的是关 $t$、$u$、$v$ 次数不超过 $2$ 的多项式，因此曲面方程是唯一确定的。

另外，因为存在直母线的二次曲面只有二次柱面、二次锥面、单叶双曲面、双曲抛物面，而有互为异面直线的直母线的二次曲面只有单叶双曲面或双曲抛物面，若给定单叶双曲面或双曲抛物面的三条互为异面直线的直母线，则这个单叶双曲面或双曲抛物面内有无数条直母线与给定的直母线都相交，由上面的计算可知曲面是唯一确定的。

hbghlyj

hejoseph 发表于 2024-3-15 06:45
另外，因为存在直母线的二次曲面只有二次柱面、二次锥面、单叶双曲面、双曲抛物面，而有互为异面直线的直母线的二次曲面只有单叶双曲面或双曲抛物面，若给定单叶双曲面或双曲抛物面的三条互为异面直线的直母线，则这个单叶双曲面或双曲抛物面内有无数条直母线与给定的直母线都相交，由上面的计算可知曲面是唯一确定的。

@青青子衿还说，三条互为异面直线一般决定一单叶双曲面，除非它们都平行于一平面（但不互相平行）此时决定一双曲抛物面。

这个怎么从上面的方程得出呢

hbghlyj

hbghlyj 发表于 2024-3-27 00:32
Three skew lines always define a one-sheeted hyperboloid, except in the case where they are all parallel to a single plane but not to each other. In this case, they determine a hyperbolic paraboloid (Hilbert and Cohn-Vossen 1999, p. 15).

上面@hejoseph得出的二次曲线方程为：

1. f=Det[{{n1*u-m1*v-n1*y1+m1*z1,n1*t-l1*v-n1*x1+l1*z1,m1*t-l1*u-m1*x1+l1*y1},{n2*u-m2*v-n2*y2+m2*z2,n2*t-l2*v-n2*x2+l2*z2,m2*t-l2*u-m2*x2+l2*y2},{n3*u-m3*v-n3*y3+m3*z3,n3*t-l3*v-n3*x3+l3*z3,m3*t-l3*u-m3*x3+l3*y3}}]

复制代码

如果是双曲抛物面，二次项部分能分解成两个一次式，则二次项系数构成的矩阵是奇异的。计算行列式：

1. Det[{{Coefficient[f, t^2], Coefficient[f, t u]/2, Coefficient[f, t v]/2},
2. {Coefficient[f, t u]/2, Coefficient[f, u^2], Coefficient[f, u v]/2},
3. {Coefficient[f, t v]/2, Coefficient[f, u v]/2, Coefficient[f, v^2]}}] // Factor

复制代码

得到
-(1/4) (-l3 m2 n1+l2 m3 n1+l3 m1 n2-l1 m3 n2-l2 m1 n3+l1 m2 n3) (-m2 n1 x1+m1 n2 x1+m2 n1 x2-m1 n2 x2+l2 n1 y1-l1 n2 y1-l2 n1 y2+l1 n2 y2-l2 m1 z1+l1 m2 z1+l2 m1 z2-l1 m2 z2) (m3 n1 x1-m1 n3 x1-m3 n1 x3+m1 n3 x3-l3 n1 y1+l1 n3 y1+l3 n1 y3-l1 n3 y3+l3 m1 z1-l1 m3 z1-l3 m1 z3+l1 m3 z3) (-m3 n2 x2+m2 n3 x2+m3 n2 x3-m2 n3 x3+l3 n2 y2-l2 n3 y2-l3 n2 y3+l2 n3 y3-l3 m2 z2+l2 m3 z2+l3 m2 z3-l2 m3 z3)
可见，它有因式(-l3 m2 n1+l2 m3 n1+l3 m1 n2-l1 m3 n2-l2 m1 n3+l1 m2 n3)
这个正好等于Det[{{l1,m1,n1},{l2,m2,n2},{l3,m3,n3}}]
所以当这三条异面直线都平行于一平面时决定一双曲抛物面

可以想象，双曲抛物面是一组单叶双曲面的极限

hbghlyj

hbghlyj 发表于 2024-3-27 01:49
计算行列式得到

话说，后面还有3个因式（对称的）其中一个是：
-m2 n1 x1 + m1 n2 x1 + m2 n1 x2 - m1 n2 x2 + l2 n1 y1 - l1 n2 y1 - l2 n1 y2 + l1 n2 y2 - l2 m1 z1 + l1 m2 z1 + l2 m1 z2 - l1 m2 z2
怎么证明它不为零呢？

hbghlyj

hbghlyj 发表于 2024-3-27 01:56
其中一个是：
-m2 n1 x1 + m1 n2 x1 + m2 n1 x2 - m1 n2 x2 + l2 n1 y1 - l1 n2 y1 - l2 n1 y2 + l1 n2 y2 - l2 m1 z1 + l1 m2 z1 + l2 m1 z2 - l1 m2 z2
怎么证明它不为零呢？

这个正好等于Det[{{l1, m1, n1}, {l2, m2, n2}, {x1 - x2, y1 - y2, z1 - z2}}]
它不为0等价于$l_1,l_2$异面

hbghlyj

hbghlyj 发表于 2024-3-13 10:20
2. How many lines in P³ intersect four given disjoint lines L₁, L₂, L₃, L₄ in P³?

(Hilbert and Cohn-Vossen §25)
三条异面的直綫 $a, b, c$ 决定一个双曲面 $H$。
设 $H$ 上的直线族为 $S_1,S_2$，$a, b, c$ 所在的族为 $S_1$。
一条任意的第四条直线 $d$ 与 $H$ 一般交于两点 $p_1,p_2$，也可能它与 $H$ 相切 或者 在 $H$ 上。
通过 $p_1$ 可找到一条直綫 $d_1\in S_2$，
通过 $p_2$ 可找到一条直綫 $d_2\in S_2$，因此 $d_1,d_2$ 必与 $a, b, c$ 相交。
反之，每条跟 $a, b, c$ 和 $d$ 相交的直綫必在 $H$ 上，因而必通过 $d$ 和 $H$ 的一个交点。
这样看来，与已知四条直綫相交的直綫一般有两条。

在 $d$ 和 $H$ 相切的情形，則（上面的两条直线$d_1,d_2$重合）只有一条直线与 $a,b,c,d$ 相交。

在 $d$ 在 $H$ 上的情形：
如果 $d\in S_1$，则$S_2$中的任何直綫与 $a,b,c, d$ 相交，因而有无穷多条直綫与 $a,b,c, d$ 相交。
如果 $d\in S_2$，没有直綫与 $a,b,c, d$ 相交。

hbghlyj

hbghlyj 发表于 2024-3-13 09:49
AG Lent 2020 Supplementary Examples.pdf

研究了第1、2题之后，我们继续看下第3题，仿射锥 的题：

3. Given a homogeneous ideal $I$, its vanishing locus $X=\mathbb{V}(I) \subset \mathbb{P}_k^n$ is a projective variety. The affine cone over $X$ is the vanishing locus of the same equations, but in $\mathbb{A}_k^n$. Prove that the affine cone over $\mathbb{P}^1 \times \mathbb{P}^1$ in its Segre embedding inside $\mathbb{P}^3$ is not smooth. Classify all projective varieties $X$ whose affine cone is smooth.

这个题让我们找到所有的射影簇，满足“仿射锥”光滑的条件，怎么找呢

比如说，抛物线$$y-x^2=0$$的“仿射锥”大概是这样的：

hbghlyj

hbghlyj 发表于 2024-3-27 02:08
Classify all projective varieties $X$ whose affine cone is smooth.

例如直线。直线的“仿射锥”是平面，平面是光滑的。

hbghlyj

3. $\ldots$Prove that the affine cone over $\mathbb{P}^1 \times \mathbb{P}^1$ in its Segre embedding inside $\mathbb{P}^3$ is not smooth.

\eqref{segre} 的affine cone $\{(x,y,z,w):wx-yz=0\}$是不光滑的，怎么证明？

$w=0$截面: $yz=0$在原点不光滑

$w=1$截面:

hbghlyj

6. Consider the nodal cubic $C \subset \mathbb{A}^2$ given by the vanishing of $f=y^2-x^2(x+1)$. Prove that it has a singular point at the origin. Draw this curve over the real numbers. Describe the Euclidean topological space associated to $C$ over the complex numbers.

这里“曲线 $C$ 的Euclidean topological space”我猜是一个pinched torus？

nodal cubic是pinched torus吧？

对cuspidal cubic是平面吧？

乱猜的。不确定

hbghlyj

hbghlyj 发表于 2024-6-12 16:08
nodal cubic是pinched torus吧？

math.purdue.edu/~arapura/graph/nodal.html
实曲线以红色标记

Notice that the graph looks similar to the graph of the elliptic curve above. This is because the nodal cubic can be viewed as limit of elliptic curves
$$y^2 = x(x + ε)(x +1)$$
as $ε → 0$. In the process, the yellow curve in the previous graph -- called a vanishing cycle in this context -- shrinks to a point. So the global topology is different; it's no longer a torus. In fact, it's not difficult to see from the above parameterization that after adding a point at infinity, we get a sphere with the two points, $t = 1$ and $t = -1$, pinched together.

topospaces.subwiki.org/wiki/Pinched_torus