Orthogonality by a quadric

Let $Q 𝑛$ be a non-singular quadric in $P G (𝑛, 𝕂)$ belonging to the quadratic form $𝑄 𝑛$ and let

𝐺 (𝐱, 𝐲) = 𝑄 𝑛 (𝐱 + 𝐲) - 𝑄 𝑛 (𝐱) - 𝑄 𝑛 (𝐲)

be the corresponding bilinear form.¹ Then for $𝑋 \in Q 𝑛$ #m/thm/geo

Let $𝑌 \neq 𝑋$ be an arbitrary point. Then $𝐺 (𝐱, 𝐲) \neq 0$ iff the line $𝑋 𝑌$ is a secant of $Q 𝑛$ , i.e. $| 𝑋 𝑌 \cap Q 𝑛 | = 2$ .
Let $𝑌 \notin Q 𝑛$ . Then $𝐺 (𝐱, 𝐲) = 0$ iff the line $𝑋 𝑌$ is a tangent of $Q 𝑛$ at $𝑋$ , i.e. $𝑋 𝑌 \cap Q 𝑛 = {𝑋}$ .
Let $𝑋 \neq 𝑌 \in Q 𝑛$ . Then $𝐺 (𝐱, 𝐲) = 0$ iff the line $𝑋 𝑌$ is a line of $Q 𝑛$ , i.e. completely contained in $Q 𝑛$ .

Proof

Any point other than $𝑋$ on the line $𝑋 𝑌$ has homogenous coördinates $𝐲 + 𝜆 𝐱$ .

𝑄 𝑛 (𝐲 + 𝜆 𝐱) = 𝜆 𝐺 (𝐲, 𝐱) + 𝑄 𝑛 (𝐲) + 𝜆 2 𝑄 𝑛 (𝐱) = 𝜆 𝐺 (𝐲, 𝐱) + 𝑄 𝑛 (𝐲)

So if $𝑌 \in Q 𝑛$ then $𝑄 𝑛 (𝐲 + 𝜆 𝐱) = 𝜆 𝐺 (𝐲, 𝐱)$ , giving cases ^2 and ^3. If $𝑌$ is arbitrary and $𝐺 (𝐲, 𝐱) \neq 0$ , then $𝑄 𝑛 (𝐲 + 𝜆 𝐱) = 0$ iff $𝜆 = - 𝑄 𝑛 (𝐲) / 𝐺 (𝐱, 𝐲)$ , giving case ^1.

#state/tidy | #lang/en | #SemBr

2020. Finite geometries, ¶4.50, pp. 104–105 ↩

Orthogonality by a quadric

Footnotes