Orthogonal complement polarity

Let $𝑈 \leq 𝕂 𝑛 + 1$ be a $(𝑘 + 1)$ -dimensional vector subspace. The orthogonal complement $𝜏 (𝑈)$ defined as

𝜏 (𝑈) = {𝑣 \in 𝕂 𝑛 : 𝑈 𝖳 𝑣 = {0}}

is a $(𝑛 - 𝑘)$ -dimensional vector subspace, and $𝜏 : P G (𝑛, 𝕂) \to P G (𝑛, 𝕂) 𝐨 𝐩$ defines a projective polarity. #m/thm/geo Moreover, if $𝑈 = c o l s p 𝑀$ for some matrix $𝑀$ , then $𝜏 (𝑈) = k e r 𝑀 𝖳$ .

Proof

First we will show the column/kernel characterization always exists. Let $(𝑢 𝑖) 𝑘 𝑖 = 0$ be a basis for $𝑈$ , and let $𝑀 = [𝑢 1, \dots, 𝑢 𝑘]$ so $c o l s p 𝑀 = 𝑈$ . Then $𝑣 \in k e r 𝑀$ iff $𝑢 𝖳 𝑖 𝑣 = 0$ for all $𝑖 = 1, \dots, 𝑘$ . Since any $𝑢 \in 𝑈$ is a finite linear combination of $𝑢 𝑖$ , it follows $𝑢 𝖳 𝑣 = 0$ for any $𝑢 \in 𝑈$ . Thus $𝜏 (𝑈) = k e r 𝑀$ .

Since $r a n k 𝑀 = r a n k 𝑀 𝖳 = d i m 𝑈$ it follows from the Rank-nullity theorem that $d i m 𝑈 + d i m 𝜏 (𝑈) = 𝑘 + 1$ .

Note that $𝜏$ clearly reverses inclusion of vector subspaces: If $𝑈 \leq 𝑉$ then certainly all vectors orthogonal to $𝑉$ are orthogonal to $𝑈$ , i.e. $𝜏 (𝑉) \leq 𝜏 (𝑈)$ . With the observation above, this shows that $𝜏$ is incidence-preserving and is therefore a projective polarity.

It follows that every projective correlation of $P G (𝑛, 𝕂)$ can be written as a collineätion followed by $𝜏$ .

Properties

The orthogonal complement commutes with any field automorphism.
Let $𝐴 \in P G L 𝑛 + 1 (𝕂)$ . Then $𝜏 𝐴 𝜏 = (𝐴 - 1) 𝖳 = (𝐴 𝖳) - 1$ .

Proof of 1–2

Let $𝜎 \in A u t (𝕂)$ and $𝑈 = c o l s p 𝑀$ . Then

𝜏 (𝜎 𝑈) = 𝜏 (𝜎 c o l s p 𝑀) = 𝜏 c o l s p (𝜎 𝑀) = k e r ((𝜎 𝑀) 𝖳) = k e r (𝜎 𝑀 𝖳)

𝑣 \in 𝜏 (𝜎 𝑈) ⟺ (𝜎 𝑀 𝖳) 𝑣 = 0 ⟺ 𝜎 - 1 ((𝜎 𝑀 𝖳) 𝑣) = 0 ⟺ 𝑀 𝖳 (𝜎 - 1 𝑣) = 0

Similarly

𝜎 (𝜏 𝑈) = 𝜎 (𝜏 c o l s p 𝑀) = 𝜎 k e r 𝑀 𝖳

𝑣 \in 𝜎 (𝜏 𝑈) ⟺ 𝜎 - 1 𝑣 \in k e r 𝑀 𝖳 ⟺ 𝑀 𝖳 (𝜎 - 1 𝑣) = 0

Therefore $𝜏 (𝜎 𝑈) = 𝜏 (𝜎 𝑈)$ , proving ^P1. #to/check

Let $𝑈 = c o l s p 𝑀$ and $𝐴 \in P G L 𝑛 + 1 (𝕂)$ Then

𝑣 \in 𝐴 𝜏 𝑈 ⟺ 𝐴 - 1 𝑣 \in k e r 𝑀 𝖳 ⟺ 𝑀 𝖳 𝐴 - 1 𝑣 = 0

thus

𝐴 𝜏 𝑈 = k e r 𝑀 𝖳 𝐴 - 1 = k e r ((𝐴 - 1) 𝖳 𝑀) 𝖳 = 𝜏 c o l s p ((𝐴 - 1) 𝖳 𝑀) = 𝜏 (𝐴 - 1) 𝖳 𝑈

But since $𝜏 2 = 1$ , it follows $𝜏 𝐴 𝜏 𝑈 = (𝐴 - 1) 𝖳 𝑈$ , proving ^P2.

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