过3点与1直线相切的二次曲线

hbghlyj

Exercise 3.6
7. Let $P^5(\mathbf{R})=P\left(\mathbf{R}^6\right)$ be the space of all conics in $P^2(\mathbf{R})$.
The conics which pass through three non-collinear points form a projective plane $P(V) \subset P^5(\mathbf{R})$.
Show that the conics parametrized by this plane which are tangent to a given line form a conic in $P(V)$.

hbghlyj

经过$A_1,A_2,A_3$的二次曲线构成射影平面$P(V)$,其中$V=\Set{λ_1C_1+λ_2C_2+λ_3C_3|λ_i∈ℝ}\congℝ^3$.

有了1#的结果后, 就可以证明: MSE 最多有4条二次曲线过3点与2直线相切. 即 最多4个$C∈P(V)$与$L_1,L_2$相切. 这是因为$$\Set{C∈P(V)|C与L_1相切}$$与$$\Set{C∈P(V)|C与L_2相切}$$是$P(V)$中的2条二次曲线,最多有4个交点.

青青子衿

hbghlyj 发表于 2023-4-27 07:55
最多有4条二次曲线过3点与2直线相切.

二次曲线条数的计数问题属于泛代数几何的枚举几何(enumerative geometry)问题.

hbghlyj

7. 重述如下:
在$ℝ^2$中, 直线$L$,
3个不共线的点$A_1,A_2,A_3\notin L$.
$C_1,C_2,C_3$为经过$A_1,A_2,A_3$的二次曲线, 且$C_1,C_2,C_3$线性无关.
求证: 存在$A∈ℝ^{3×3}$使得

$\{λ_1C_1+λ_2C_2+λ_3C_3\;λ=(λ_1,λ_2,λ_3)∈\mathbb P(ℝ^3)\}$与$L$相切当且仅当$λ^TAλ=0$

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Suppose the bilinear forms $C_1(r,s),C_2(r,s),C_3(r,s)$ are linearly independent.
Then $C_1(r,s)=0,C_2(r,s)=0,C_3(r,s)=0$ are conics in $\mathbb{RP}^2$ passing through 3 non-collinear points $A_1,A_2,A_3$.
Then all the bilinear forms of conics pass through $A_1,A_2,A_3$ form a projective space$$\mathbb P(V)=\{λ_1C_1+λ_2C_2+λ_3C_3:λ=[λ_1:λ_2:λ_3]∈\mathbb P(ℝ^3)\}$$
Let the bilinear form $B(r,s)=x_1C_1(r,s)+x_2C_2(r,s)+x_3C_3(r,s)$ be an element of $\mathbb P(V)$.
The conic $B(r,s)=0$ is tangent to the line $\mathbb P(\langle u,v\rangle)$ in $\mathbb P(\mathbb R^3)$.
if and only if the equation $B(\alpha u+\beta v,\alpha u+\beta v)=0$ in $[\alpha:\beta]\in\mathbb P(\mathbb R^2)$ has 1 solution.
if and only if the determinant $B(u,v)^2-B(u,u)B(v,v)=0$
Rewrite by the definition of $B(r,s)$:
$$\small(x_1C_1(u,v)+x_2C_2(u,v)+x_3C_3(u,v))^2-(x_1C_1(u,u)+x_2C_2(u,u)+x_3C_3(u,u))(x_1C_1(v,v)+x_2C_2(v,v)+x_3C_3(v,v))$$This is a bilinear form in $[x_1:x_2:x_3]$.