Projective correlation

Orthogonal complement polarity

Let 𝑈 ≤𝕂𝑛+1 be a (𝑘 +1)-dimensional vector subspace. The orthogonal complement 𝜏(𝑈) defined as

𝜏(𝑈)={𝑣∈𝕂𝑛:𝑈𝖳𝑣={0}}

is a (𝑛 −𝑘)-dimensional vector subspace, and 𝜏 :PG(𝑛,𝕂) →PG(𝑛,𝕂)𝐨𝐩 defines a projective polarity. geo Moreover, if 𝑈 =colsp⁡𝑀 for some matrix 𝑀, then 𝜏(𝑈) =ker⁡𝑀𝖳.

Proof

First we will show the column/kernel characterization always exists. Let (𝑢𝑖)𝑘𝑖=0 be a basis for 𝑈, and let 𝑀 =[𝑢1,…,𝑢𝑘] so colsp⁡𝑀 =𝑈. Then 𝑣 ∈ker⁡𝑀 iff 𝑢𝖳𝑖𝑣 =0 for all 𝑖 =1,…,𝑘. Since any 𝑢 ∈𝑈 is a finite linear combination of 𝑢𝑖, it follows 𝑢𝖳𝑣 =0 for any 𝑢 ∈𝑈. Thus 𝜏(𝑈) =ker⁡𝑀.

Since rank⁡𝑀 =rank⁡𝑀𝖳 =dim⁡𝑈 it follows from the Rank-nullity theorem that dim⁡𝑈 +dim⁡𝜏(𝑈) =𝑘 +1.

Note that 𝜏 clearly reverses inclusion of vector subspaces: If 𝑈 ≤𝑉 then certainly all vectors orthogonal to 𝑉 are orthogonal to 𝑈, i.e. 𝜏(𝑉) ≤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 PG(𝑛,𝕂) can be written as a collineätion followed by 𝜏.

Properties

1. The orthogonal complement commutes with any field automorphism.
2. Let 𝐴 ∈PGL𝑛+1(𝕂). Then 𝜏𝐴𝜏 =(𝐴−1)𝖳 =(𝐴𝖳)−1.

Proof of 1–2

Let 𝜎 ∈Aut⁡(𝕂) and 𝑈 =colsp⁡𝑀. Then

𝜏(𝜎𝑈)=𝜏(𝜎colsp⁡𝑀)=𝜏colsp⁡(𝜎𝑀)=ker⁡((𝜎𝑀)𝖳)=ker⁡(𝜎𝑀𝖳)

so

𝑣∈𝜏(𝜎𝑈)⟺(𝜎𝑀𝖳)𝑣=0⟺𝜎−1((𝜎𝑀𝖳)𝑣)=0⟺𝑀𝖳(𝜎−1𝑣)=0

Similarly

𝜎(𝜏𝑈)=𝜎(𝜏colsp⁡𝑀)=𝜎ker⁡𝑀𝖳

so

𝑣∈𝜎(𝜏𝑈)⟺𝜎−1𝑣∈ker⁡𝑀𝖳⟺𝑀𝖳(𝜎−1𝑣)=0

Therefore 𝜏(𝜎𝑈) =𝜏(𝜎𝑈), proving ^P1. check

Let 𝑈 =colsp⁡𝑀 and 𝐴 ∈PGL𝑛+1(𝕂) Then

𝑣∈𝐴𝜏𝑈⟺𝐴−1𝑣∈ker⁡𝑀𝖳⟺𝑀𝖳𝐴−1𝑣=0

thus

𝐴𝜏𝑈=ker⁡𝑀𝖳𝐴−1=ker⁡((𝐴−1)𝖳𝑀)𝖳=𝜏colsp⁡((𝐴−1)𝖳𝑀)=𝜏(𝐴−1)𝖳𝑈

But since 𝜏2 =1, it follows 𝜏𝐴𝜏𝑈 =(𝐴−1)𝖳𝑈, proving ^P2.

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