# polarities and forms

Through out this article we assume $\dim V\neq 2$. This is not a true constraint as there are only trivial dualities for $\dim V\leq 2$.

###### Proof.

To see this, let $d:PG(V)\rightarrow PG(V)$ be a duality. We may express this as an order preserving map $d:PG(V)\rightarrow PG(V^{*})$. Then by the fundamental theorem of projective geometry  it follows $d$ is induced by a bijective   semi-linear transformation $\hat{d}:V\rightarrow V^{*}$.

An semi-linear isomorphism        of $V$ to $V^{*}$ is equivalent     to specifying a non-degenerate sesquilinear form. In particular, define the form $b:V\times V\rightarrow k$ by $b(v,w)=(v)(w\hat{d})$ (notice $w\hat{d}\in V^{*}$ so $w\hat{d}:V\rightarrow k$).

Now, if $b:V\times V\rightarrow k$ is a non-degenerate sesquilinear form. Then define

 $\hat{b}:V\rightarrow V^{*}:v\mapsto b(-,v):V\rightarrow k$

which is semi-linear, as $b$ is sesquilinear, and bijective, since $b$ is non-degenerate. Therefore $\hat{b}$ induces an order preserving bijection $PG(V)\rightarrow PG(V^{*})$, that is, a duality. ∎

We write $W^{\perp}$ for the image of the induced duality of a non-degenerate sesquilinear form $b$. Notice that $W^{\perp}=\{w\in V:b(v,W)=0\}$. (Although the form may not be reflexive     , we still use the $\perp$ notation, but we now demonstrate that we can indeed specialize to the reflexive case.) Notice then that

 $\dim W^{\perp}=\dim V-\dim W.$
###### Proof.

Let $b$ be the sesquilinear form induced by the polarity $p$. Then suppose we have $v,w\in V$ such that $0=b(v,w)=(v)(w\hat{p})$. So $\langle v\rangle\leq\langle w\rangle^{\perp}=\langle w\rangle p$. But $p$ has order 2 so $\langle v\rangle^{\perp}=\langle v\rangle p\geq\langle w\rangle$. But this implies $b(w,v)=0$ so $b$ is reflexive.

Likewise, given a reflexive non-degenerate sesquilinear form $b$ it gives rise do a duality $d$ induced by $\hat{b}$. By the reflexivity, $b(W,W^{\perp})=0$ implies $b(W^{\perp},W)=0$ also. As $(W^{\perp})^{\perp}=\{v\in V:b(v,(W^{\perp})^{\perp})=0\}$ it follows $W\leq(W^{\perp})^{\perp}$. But by dimension  arguments:

 $\dim(W^{\perp})^{\perp}=\dim V-\dim W^{\perp}=\dim V-(\dim V-\dim W)=\dim W$

we conclude $W=(W^{\perp})^{\perp}$. Thus $d$ is a polarity. ∎

From the fundamental theorem of projective geometry it follows if $\dim V\neq 2$ then every order preserving map is induced by a semi-linear transformation of $V$. In similar   fashion we have

###### Proposition 3.

$P\Gamma L^{*}(V)=P\Gamma L(V)\rtimes\mathbb{Z}_{2}$, meaning that every order reversing map $f:PG(V)\rightarrow PG(V)$ can be decomposed as a $f=sr$ where $s$ is induced from a semi-linear transformation and $r$ is a polarity.

###### Proof.

Let $d$ be any duality of $PG(V)$. Then $d^{2}$ is order preserving. Thus $d^{2}$ is a projectivity  so by the fundamental theorem of projective geometry $d^{2}$ is induced by a semi-linear transformation $s$. Therefore $P\Gamma L(V)$ has index 2 in $P\Gamma L^{*}(V)$. Finally it suffices to provide any polarity of $PG(V)$ to prove $P\Gamma L^{*}(V)=P\Gamma L(V)\rtimes\mathbb{Z}_{2}$. For this use any reflexive non-degenerate sesquilinear form. ∎

###### Remark 4.

The group $P\Gamma L^{*}(V)$ is the automorphism group  of $PSL(V)$. In particular, the polarities account for the graph automorphisms  of the Dynkin diagram of $A_{d-1}$, $d=\dim V$. When $\dim V=2$ there is no graph automorphism, just as there are no dualities (points are hyperplanes   when $\dim V=2$.)

## References

 Title polarities and forms Canonical name PolaritiesAndForms Date of creation 2013-03-22 15:58:13 Last modified on 2013-03-22 15:58:13 Owner Algeboy (12884) Last modified by Algeboy (12884) Numerical id 7 Author Algeboy (12884) Entry type Topic Classification msc 51A05 Related topic polarity Related topic Projectivity Related topic ProjectiveGeometry Related topic Isometry2 Related topic ProjectiveGeometry3 Related topic ClassicalGroups Related topic Polarity2 Related topic DualityWithRespectToANonDegenerateBilinearForm