The conformal dilatation and Beltrami forms over quadratic field extensions

TL;DR

This paper develops an algebraic framework for complex dilatation and Beltrami forms over quadratic field extensions, generalizing classical geometric notions to a broader algebraic context with invariant properties.

Contribution

It introduces an algebraic analogue of complex dilatation and Beltrami forms over quadratic field extensions, providing an invariant and generalized approach.

Findings

Establishes a relationship between conformal dilatation and Beltrami form over quadratic extensions.

Shows that working with general field extensions clarifies algebraic aspects without added difficulty.

Provides an invariant version of the classical geometric approach to complex dilatation.

Abstract

The paper is devoted to an algebraic analogue of a geometric approach to the classical notion of complex dilatation suggested in the paper arXiv:1701.06259 [math.CV] by the author. At the same time it provides an invariant version of this geometric approach. From the algebraic point of view it is only natural to work with a general field extension K/k of degree 2 instead of the fields of real and complex number (under the assumption that the characteristic is not equal to 2). Given a k-linear map between two K-vector spaces of dimension 1 over K, there are two natural measures of deviation of this map from being K-linear: its conformal dilatation, defined in terms of quadratic forms over k, and its Beltrami form, directly generalizing the classical complex dilatation. It turns out that these two measures are related in the same way as in the classical case. Working with a general field extension does not lead to any new difficulties compared to the classical case, but only clarifies the algebraic aspects of the theory.

Equations

$$\mathbb{M}(V)=\mathbb{P}Q(V)\setminus \mathbb{A}(V)$$

$$\mathcal{D}:{\mathrm{Hom}}_a(V,V)⟶\text{mathbold}k.$$

$${\mu }_f=(Lf{)}^{-1}\circ (Af)\in {\mathrm{Hom}}_a(V,V).$$

$$\text{mathbold}c(f)=\frac{2{\mu }_f}{1+\mathcal{D}({\mu }_f)}$$

$$M_{ij}(\phi )=\phi (v_i,v_j),$$

$$\det \phi =\det M(\phi )$$

$$v_i^{\prime }=a_{i1}{v}_1+a_{i2}{v}_2+\dots +a_{in}{v}_n$$

$$M^{\prime }(\phi )=AM(\phi )A^t,$$

$$\det {}^{\prime }\phi =\det \phi \cdot (\det A{)}^2,$$

$$f(v_i)=f_{i1}{v}_1+f_{i2}{v}_2+\dots +f_{in}{v}_n,$$

$$f\cdot \phi :(v,w)⟼\phi (f(v),w),$$

$$\det (f\cdot \phi )=\det f\cdot \det \phi ,$$

$$\hat{\phi }:V⟶\mathrm{hom}(V,\text{mathbold}k)$$

$$w⟼\phi (f(v),w)$$

$$\phi (v,w)=\phi (w,v)$$

$$\quad(\hskip 0.50003ptv\hskip 0.50003pt,\hskip 1.99997ptw\hskip 0.50003pt)_{q}\hskip 3.99994pt=\hskip 3.99994pt\bigl{(}q\hskip 0.50003pt(v\hskip 1.99997pt+\hskip 1.99997ptw)\hskip 1.99997pt-\hskip 1.99997ptq\hskip 0.50003pt(v)\hskip 1.99997pt-\hskip 1.99997ptq\hskip 0.50003pt(w)\hskip 1.00006pt\bigr{)}\bigl{/}2\hskip 1.99997pt.$$

$$\det q=\det G(q).$$

$$\det :Q(V)⟶\text{mathbold}k.$$

$$\det {}^{\prime }q=\det q\cdot (\det A{)}^2,$$

$$\phi (f(v),w)=\phi (v,f(w))$$

$$(v,w)⟼(f(v),w{)}_q$$

$$f\cdot q:v⟼(f(v),v{)}_q.$$

$$\det (f\cdot q)=\det f\cdot \det q.$$

$$(v,w{)}_p=(f(v),w{)}_n$$

$$p=(p/n)\cdot n\text{mbox}andf=f\cdot n/n$$

$$G(q)=\left[\begin{array}{cc}(v,v{)}_q& (v,w{)}_q\text{vspace}\ast 1.5\text{medskipamount}\\ (w,v{)}_q& (w,w{)}_q\end{array}\right]$$

$$\det q=(v,v{)}_q(w,w{)}_q-(v,w{)}_q^2=q(v)q(w)-(v,w{)}_q^2.$$

$$q(v+w)=q(v)+q(w).$$

$$\det (p+q)=\det p+\det q.$$

$$p(v)q(w)+q(v)p(w)=0.$$

Full text

††footnotetext: © Nikolai V. Ivanov, 2017.  Neither the work reported in this paper,  nor its preparation were supported by any governmental or non-governmental agency,  foundation,  or institution.

The  conformal  dilatation  and  Beltrami  forms  

over  quadratic  field  extensions

Nikolai V. Ivanov

Contents

Introduction   

1****1.   Bilinear and quadratic forms   

1****2.   The norm and the trace   

1****3.   Quadratic forms in dimension $1$ over $\mathbold{K}$   

1****4.   The structure of a $\mathbold{K}$-vector space on $\mathbold{A}\hskip 0.50003pt(V)$   

1****5.   Conformal structures   

1****6.   Anti-linear maps   

1****7.   The structure of $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$   

1****8.   The conformal dilatation and the Beltrami forms   

1****9.   Identification of $\mathbb{M}\hskip 0.50003pt(V)$ with $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$   

10.   Comparing $\mathbold{c}\hskip 0.50003pt(f)$ and $\mu_{\hskip 0.70004ptf}$   

References

Introduction

The motivation.   The present paper grew out of the desire to present an invariant treatment of the geometric interpretation of the complex dilatation suggested in  [I] .   The main part of  [I]  is devoted to the real linear maps  $\mathbold{C}\longrightarrow\mathbold{C}$  and heavily uses the standard basis  $1\hskip 0.50003pt,\hskip 3.00003pti$  of  $\mathbold{C}$  as a real vector space.   Only the last section of  [I]  is devoted to the general case of real linear maps  $V\longrightarrow W$,   where  $V\hskip 0.50003pt,\hskip 3.00003ptW$  are two complex vector spaces of dimension $1$,   and this case is treated by a reduction to the case  $\mathbold{C}\longrightarrow\mathbold{C}$.   This approach served well to the goals of the paper  [I] ;   see  [I] ,   the end of the Introduction.

A recent paper of M.  Atiyah  [A]  reminded the author that a truly invariant approach requires more than just dealing with maps  $V\longrightarrow W$  from the very beginning .   Namely,   to quote M.  Atiyah,   “one should not distinguish between  $i$  and  $-\hskip 1.99997pti$  whereas one can distinguish between  $1$  and  $-\hskip 1.99997pt1$”.   An attempt to follow this maxim quickly lead to the realization that  one should distinguish  between  $-\hskip 1.99997pt1$  as a purely additive notion and  $i^{\hskip 0.35002pt2}$.

Since  $i^{\hskip 0.35002pt2}\hskip 1.99997pt=\hskip 1.99997pt-\hskip 1.99997pt1$  anyhow,   the only sensible way to distinguish  $-\hskip 1.99997pt1$  and  $i^{\hskip 0.35002pt2}$  is to abandon the axiom  $i^{\hskip 0.35002pt2}\hskip 1.99997pt+\hskip 1.99997pt1\hskip 1.99997pt=\hskip 1.99997pt0$  and replace  $\mathbold{C}\hskip 0.50003pt,\hskip 3.00003pt\mathbold{R}$  by two fields  $\mathbold{K}\hskip 0.50003pt,\hskip 3.00003pt\mathbold{k}$  such that  $\mathbold{K}$  is an extension of  $\mathbold{k}$  of degree  $[\hskip 1.00006pt\mathbold{K}:\mathbold{k}\hskip 1.00006pt]\hskip 1.99997pt=\hskip 1.99997pt2$.   Since quadratic forms behave differently in characteristic $2$,   it is reasonable to assume that the characteristic of  $\mathbold{k}$  is  $\neq\hskip 1.99997pt2$.

This paper is devoted to an analogue of the theory of  [I]  in this situation,   independent on any choices of bases.   Working with a general field extension  $\mathbold{K}/\hskip 0.50003pt\mathbold{k}$  does not lead to any new difficulties compared to the classical case  $\mathbold{K}/\hskip 0.50003pt\mathbold{k}\hskip 1.99997pt=\hskip 1.99997pt\mathbold{C}/\hskip 0.50003pt\mathbold{R}$,   but only clarifies the theory.   In the rest of the introduction is devoted to an outline of this analogue.

Conformal structures and conformal dilatation.   For a vector space $V$ over $\mathbold{k}$ we denote by  $Q\hskip 0.50003pt(V)$  the vector space of all quadratic forms on $V$.   A  conformal structure  on $V$ is a non-zero quadratic form on $V$ considered up to multiplication by non-zero elements of $\mathbold{k}$,   i.e.  a point in the projective space  $\mathbb{P}Q\hskip 0.50003pt(V)$  associated with  $Q\hskip 0.50003pt(V)$.   If  $\dim V\hskip 1.99997pt=\hskip 1.99997pt2$,   then  $\mathbb{P}Q\hskip 0.50003pt(V)$  is a projective plane and the set  $\mathbb{C}\hskip 0.50003pt(V)\hskip 1.00006pt\subset\hskip 1.00006pt\mathbb{P}Q\hskip 0.50003pt(V)$  of conformal structures corresponding to degenerate quadratic forms is a  conic  in  $\mathbb{P}Q\hskip 0.50003pt(V)$.   In other terms,  $\mathbb{C}\hskip 0.50003pt(V)$  is defined by a homogeneous equation of degree $2$.

This paper is concerned mostly with vector spaces $V$ of dimension $1$ over $\mathbold{K}$,   considered as a vector spaces over $\mathbold{k}$.   There is a  canonical conformal structure  on any such  $V$,   denoted by  $c_{\hskip 0.35002ptV}$.   It is invariant under the multiplication by non-zero elements of  $\mathbold{K}$.

Let  $\mathbb{A}\hskip 0.50003pt(V)$  be the projective line polar ***In the main part of the paper the notion of polarity is not used and  $\mathbb{A}\hskip 0.50003pt(V)$  is defined differently .   Theorem  3. Quadratic  forms  in  dimension  $1$  over  $\mathbold{K}$  implies that the two definitions are equivalent .  to  $c_{\hskip 0.35002ptV}$  with respect to the conic  $\mathbb{C}\hskip 0.50003pt(V)$.   Then

[TABLE]

is an affine plane containing  $c_{\hskip 0.35002ptV}$.   One can turn  $\mathbb{M}\hskip 0.50003pt(V)$  into a vector space over $\mathbold{k}$ by designating  $c_{\hskip 0.35002ptV}$  as its zero vector .   The structure of a vector space of dimension  $1$  over  $\mathbold{K}$  on  $V$  allows to extend this structure of a vector space over  $\mathbold{k}$  on  $\mathbb{M}\hskip 0.50003pt(V)$  to a structure of a vector space over  $\mathbold{K}$  on  $\mathbb{M}\hskip 0.50003pt(V)$  in a canonical manner .   There is a canonical quadratic form   $\displaystyle\mathbb{D}\hskip 1.00006pt\colon\hskip 1.00006pt\mathbb{M}\hskip 0.50003pt(V)\longrightarrow\mathbold{k}$   equal to  $1$  on the conic  $\mathbb{C}\hskip 0.50003pt(V)$.

Let  $f\hskip 1.00006pt\colon\hskip 1.00006ptV\longrightarrow W$  be a map linear over  $\mathbold{k}$  between vector spaces  $V\hskip 0.50003pt,\hskip 3.00003ptW$  of dimension  $1$  over  $\mathbold{K}$.   A natural measure of the distortion of the canonical conformal structure by  $f$  is the pull-back   $\displaystyle\mathbold{c}\hskip 0.50003pt(f)\hskip 3.99994pt=\hskip 3.99994ptf^{\hskip 0.70004pt*}\hskip 0.50003ptc_{\hskip 0.35002ptW}\hskip 1.99997pt\in\hskip 1.99997pt\mathbb{P}Q\hskip 0.50003pt(V)$,    called  the  conformal dilatation  of  $f$.

Anti-linear maps and Beltrami forms.   Since    $[\hskip 1.00006pt\mathbold{K}:\mathbold{k}\hskip 1.00006pt]\hskip 1.99997pt=\hskip 1.99997pt2$,   there is only one non-trivial automorphism of  $\mathbold{K}$ fixed on $\mathbold{k}$,   which is denoted by  $z\hskip 1.99997pt\longmapsto\hskip 1.99997pt\overline{z}$.

Let  $V\hskip 0.50003pt,\hskip 3.00003ptW$  be vector spaces of dimension $1$ over $\mathbold{K}$.   A map  $g\hskip 1.00006pt\colon\hskip 1.00006ptV\longrightarrow W$  is said to be  anti-linear  if it is linear over $\mathbold{k}$ and  $\displaystyle g\hskip 0.50003pt(z\hskip 1.00006ptv)\hskip 3.00003pt=\hskip 3.00003pt\overline{z}\hskip 1.99997ptg\hskip 0.50003pt(v)$  for all  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$,  $v\hskip 1.00006pt\in\hskip 1.00006ptV$.   Such maps form a vector space  $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V\hskip 0.50003pt,\hskip 1.49994ptW)$  over $\mathbold{K}$.   There is a canonical quadratic form

[TABLE]

In view of Lemma  7. The  structure  of  $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$  it can be defined by   $\mathcal{D}\hskip 0.50003pt(g)\hskip 1.99997pt=\hskip 1.99997pt-\hskip 1.99997pt\det g$,   where the determinant is taken over $\mathbold{k}$.

If a map  $f\hskip 1.00006pt\colon\hskip 1.00006ptV\longrightarrow W$  is linear over $\mathbold{k}$,   then   $\displaystyle f\hskip 3.99994pt=\hskip 3.99994ptLf\hskip 1.99997pt+\hskip 1.99997ptAf$,   where  $Lf$  is linear over  $\mathbold{K}$  and  $Af$  is anti-linear .   Moreover ,   the maps  $Lf\hskip 0.50003pt,\hskip 3.00003ptAf$  are uniquely determined by $f$.   If  $Lf\hskip 1.99997pt\neq\hskip 1.99997pt0$,   then  $Lf$  is invertible and the composition

[TABLE]

is a natural measure of failure of $f$ to be linear over $\mathbold{K}$,   called the  Beltrami form  of  $f$.

Comparing the conformal dilatation and Beltrami forms.   The results of  [I]  suggest that  $\mathbold{c}\hskip 0.50003pt(f)$  and  $\mu_{\hskip 0.70004ptf}$  should be related by an analogue of the map relating the Klein and the Poincaré models of the hyperbolic plane.   As it turns out ,   this is indeed the case.

There is a canonical isomorphism of  $\mathbb{M}\hskip 0.50003pt(V)$  and  $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$  as vector spaces over $\mathbold{K}$.   Moreover ,   this isomorphism turns the quadratic forms  $\mathbb{D}$,  $\mathcal{D}$  one into the other .   One can use this isomorphism to identify  $\mathbb{M}\hskip 0.50003pt(V)$  and  $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$.

Let  $f\hskip 1.00006pt\colon\hskip 1.00006ptV\longrightarrow W$  be a map linear over  $\mathbold{k}$.   The map  $f$  is called  regular  if  $\mathbold{c}\hskip 0.50003pt(f)\hskip 1.00006pt\in\hskip 1.00006pt\mathbb{M}\hskip 0.50003pt(V)$,   and  exceptional  otherwise.   The exceptional maps are a new feature of the general case.   If the extension  $\mathbold{K}/\hskip 0.50003pt\hskip 0.24994pt\mathbold{k}$  resembles enough the classical case of  $\mathbold{C}/\hskip 0.50003pt\mathbold{R}$,   then all non-zero maps  $f$  are regular .   This is the case if the field $\mathbold{k}$ is ordered and $\mathbold{K}$ is obtained by adjoining a root of a  negative  element to $\mathbold{k}$,   as it easily follows from Theorem  10. Comparing   $\mathbold{c}\hskip 0.50003pt(f)$   and   $\mu_{\hskip 0.70004ptf}$.

Suppose that  $Lf\hskip 1.99997pt\neq\hskip 1.99997pt0$.    If   $\displaystyle 1\hskip 1.99997pt+\hskip 1.99997pt\mathcal{D}\hskip 0.50003pt(\mu_{\hskip 0.70004ptf})\hskip 3.99994pt\neq\hskip 3.99994pt0$,   then  $f$  is regular and

[TABLE]

after the identification of  $\mathbb{M}\hskip 0.50003pt(V)$  with  $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$.    See Theorem  10. Comparing   $\mathbold{c}\hskip 0.50003pt(f)$   and   $\mu_{\hskip 0.70004ptf}$.   A simple calculation  (see   [I] ,   Corollary  3.3)  shows that for  $\mathbold{K}\hskip 1.99997pt=\hskip 1.99997pt\mathbold{C}$,  $\mathbold{k}\hskip 1.99997pt=\hskip 1.99997pt\mathbold{R}$,   and  $V\hskip 1.99997pt=\hskip 1.99997pt\mathbold{C}$  this relation between  $\mathbold{c}\hskip 0.50003pt(f)$  and  $\mu_{\hskip 0.70004ptf}$  is the same as in   [I].

If   $\displaystyle 1\hskip 1.99997pt+\hskip 1.99997pt\mathcal{D}\hskip 0.50003pt(\mu_{\hskip 0.70004ptf})\hskip 3.99994pt=\hskip 3.99994pt0$,    then  $f$  is exceptional.   In this case the relation between  $\mathbold{c}\hskip 0.50003pt(f)$  and  $\mu_{\hskip 0.70004ptf}$  is even simpler .   See Theorem  10. Comparing   $\mathbold{c}\hskip 0.50003pt(f)$   and   $\mu_{\hskip 0.70004ptf}$.

1. Bilinear  and  quadratic  forms

Bilinear forms,  their matrices and determinants.   Let us fix a vector space $V$ over $\mathbold{k}$ of finite dimension $n$ and a basis  $v_{1}\hskip 0.50003pt,\hskip 3.00003ptv_{2}\hskip 0.50003pt,\hskip 3.00003pt\ldots\hskip 0.50003pt,\hskip 3.00003ptv_{\hskip 0.35002ptn}$  of $V$.   A bilinear form  $\varphi\colon V\times V\longrightarrow\mathbold{k}$  gives rise to an  $n\times n$  matrix  $M\hskip 0.50003pt(\varphi)\hskip 1.99997pt=\hskip 1.99997pt(\hskip 0.50003ptM_{\hskip 0.70004pti\hskip 0.70004ptj}\hskip 0.50003pt(\varphi)\hskip 0.50003pt)$  with the entries

[TABLE]

called  the matrix of  $\varphi$  with respect to the basis  $v_{1}\hskip 0.50003pt,\hskip 3.00003ptv_{2}\hskip 0.50003pt,\hskip 3.00003pt\ldots\hskip 0.50003pt,\hskip 3.00003ptv_{n}$.   The determinant

[TABLE]

is called the  determinant  of  $\varphi$  (with respect to the basis  $v_{1}\hskip 0.50003pt,\hskip 3.00003ptv_{2}\hskip 0.50003pt,\hskip 3.00003pt\ldots\hskip 0.50003pt,\hskip 3.00003ptv_{\hskip 0.35002ptn}$).

Change of basis.   If  $v^{\prime}_{1}\hskip 0.50003pt,\hskip 3.00003ptv^{\prime}_{2}\hskip 0.50003pt,\hskip 3.00003pt\ldots\hskip 0.50003pt,\hskip 3.00003ptv^{\prime}_{\hskip 0.35002ptn}$  is some other basis of $V$,   then

[TABLE]

for some invertible matrix  $A\hskip 1.99997pt=\hskip 1.99997pt(a_{\hskip 0.70004pti\hskip 0.70004ptj}\hskip 0.50003pt)$  and

[TABLE]

where  $M^{\prime}\hskip 0.50003pt(\varphi)$  is the matrix of  $\varphi$  with respect to the basis  $v^{\prime}_{1}\hskip 0.50003pt,\hskip 3.00003ptv^{\prime}_{2}\hskip 0.50003pt,\hskip 3.00003pt\ldots\hskip 0.50003pt,\hskip 3.00003ptv^{\prime}_{\hskip 0.35002ptn}$,   and  $A^{t}$  is the transposed matrix of  $A$.   Therefore

[TABLE]

where  $\det^{\prime}\varphi$  is the determinant of  $\varphi$  with respect to the basis  $v^{\prime}_{1}\hskip 0.50003pt,\hskip 3.00003ptv^{\prime}_{2}\hskip 0.50003pt,\hskip 3.00003pt\ldots\hskip 0.50003pt,\hskip 3.00003ptv^{\prime}_{\hskip 0.35002ptn}$.

Bilinear forms and linear maps.   If  $f\colon V\longrightarrow V$  is a linear map,   then

[TABLE]

where  $F\hskip 1.99997pt=\hskip 1.99997pt(f_{\hskip 0.70004pti\hskip 0.70004ptj}\hskip 0.50003pt)$  is the matrix of  $f$  in the basis  $v_{1}\hskip 0.50003pt,\hskip 3.00003ptv_{2}\hskip 0.50003pt,\hskip 3.00003pt\ldots\hskip 0.50003pt,\hskip 3.00003ptv_{\hskip 0.35002ptn}$.   Together with the form  $\varphi$  the map  $f$  gives rise to a new bilinear form,   the  left  product

[TABLE]

of  $f$  and  $\varphi$.   The matrix  $M\hskip 0.50003pt(f\cdot\varphi)$  of  $f\cdot\varphi$  is equal to  $F\hskip 1.00006ptM\hskip 0.50003pt(\varphi)$,   and hence

[TABLE]

where  $\det f\hskip 1.99997pt=\hskip 1.99997pt\det F$  is independent on the choice of the basis.

Non-degenerate bilinear forms.   A bilinear form  $\varphi$  gives rise to a linear map

[TABLE]

taking  $v\hskip 1.00006pt\in\hskip 1.00006ptV$  to the map  $w\hskip 1.99997pt\longmapsto\hskip 1.99997pt\varphi\hskip 0.50003pt(v\hskip 0.50003pt,\hskip 3.00003ptw)$.   The form  $\varphi$  is said to be  non-degenerate  if  $\widehat{\varphi}$  is an isomorphism.   As is well known,  this condition is equivalent to  $\det\varphi\hskip 1.99997pt\neq\hskip 1.99997pt0$.

Suppose that  $\varphi\hskip 0.50003pt,\hskip 3.00003pt\beta$  are bilinear forms on $V$ and  $\varphi$  is non-degenerate.   By the previous paragraph,   for every  $v\hskip 1.00006pt\in\hskip 1.00006ptV$  the homomorphism  $w\hskip 1.99997pt\longmapsto\hskip 1.99997pt\beta\hskip 0.50003pt(v\hskip 0.50003pt,\hskip 3.00003ptw)$  is equal to

[TABLE]

for a unique  $f\hskip 0.50003pt(v)\hskip 1.00006pt\in\hskip 1.00006ptV$.   A trivial verification shows that  $f$  is linear and hence  $\beta\hskip 1.99997pt=\hskip 1.99997ptf\cdot\varphi$.

Quadratic forms.   A bilinear form  $\varphi\colon V\times V\longrightarrow\mathbold{k}$  is called  symmetric  if

[TABLE]

for all  $v\hskip 0.50003pt,\hskip 3.00003ptw\hskip 1.00006pt\in\hskip 1.00006ptV$.   A bilinear form is symmetric if and only if its matrix is symmetric.   A  quadratic form  on $V$ is a map  $q\colon V\longrightarrow\mathbold{k}$  of the form  $v\hskip 1.99997pt\longmapsto\hskip 1.99997pt\varphi\hskip 0.50003pt(v\hskip 0.50003pt,\hskip 3.00003ptv)$,   where  $\varphi$  is a symmetric bilinear form.   The form  $\varphi$  is uniquely determined by $q$ as its  polarization

[TABLE]

A quadratic form $q$ is said to be  non-degenerate  if its polarization is non-degenerate.

The  Gram matrix  $G\hskip 0.50003pt(q)$  of a quadratic form  $q$  is defined as the matrix of its polarization,   and the determinant  $\det q$  of  $q$  is defined as the determinant of its polarization,   i.e.

[TABLE]

A quadratic form  $q$  is non-degenerate if and only if  $\det q\hskip 1.99997pt\neq\hskip 1.99997pt0$.

The determinant map.   The set  $Q\hskip 0.50003pt(V)$  of quadratic forms on $V$ is a vector space over $\mathbold{k}$.   The map  $q\hskip 1.99997pt\longmapsto\hskip 1.99997ptG\hskip 0.50003pt(q)$  is an isomorphism between  $Q\hskip 0.50003pt(V)$  and the space of symmetric  $n\times n$  matrices with entries in  $\mathbold{k}$.   The determinants  $\det q$  lead to the  determinant map

[TABLE]

If  $\det\hskip 0.24994pt^{\prime}$  is the determinant map corresponding to some other basis of  $V$,   then

[TABLE]

where  $A$  is the matrix defined as above.   It follows that up to a multiplication by a non-zero constant the determinant map is independent from the choice of basis of  $V$.

Symmetric maps.   A linear map  $f\hskip 1.00006pt\colon\hskip 1.00006ptV\longrightarrow V$  is said to be  symmetric  with respect to a bilinear form  $\varphi$  if

[TABLE]

for all  $v\hskip 0.50003pt,\hskip 3.00003ptw\hskip 1.00006pt\in\hskip 1.00006ptV$.    If  $\varphi$  is symmetric,   then   $\displaystyle\varphi\hskip 0.50003pt(\hskip 0.50003ptv\hskip 0.50003pt,\hskip 3.00003ptf\hskip 0.50003pt(w)\hskip 0.50003pt)\hskip 3.00003pt=\hskip 3.00003pt\varphi\hskip 0.50003pt(\hskip 0.50003ptf\hskip 0.50003pt(w)\hskip 0.50003pt,\hskip 3.00003ptv\hskip 0.50003pt)$   for all  $v\hskip 0.50003pt,\hskip 3.00003ptw\hskip 1.00006pt\in\hskip 1.00006ptV$.   It follows that  $f$  is symmetric with respect to a symmetric bilinear form  $\varphi$  if and only if the left product  $f\cdot\varphi$  is symmetric.

Quadratic forms and symmetric maps.   A linear map  $f\hskip 1.00006pt\colon\hskip 1.00006ptV\longrightarrow V$  is said to be  symmetric  with respect to a quadratic form  $q$  on  $V$  if  $f$  is symmetric with respect to the polarization of  $q$,   i.e.  if  the bilinear form

[TABLE]

is symmetric.   If  $q$  is a quadratic form with the polarization  $\varphi$  and  $f$  is a linear map symmetric with respect to  $q$,   then the  product  $f\cdot q$  is defined as the quadratic form with the polarization  $f\cdot\varphi$.   Equivalently ,   the  product of  $f$  and  $q$  is the map

[TABLE]

By applying  (1.1)  to  $f$  and the polarization  $\varphi$  of  $q$,   we see that

[TABLE]

Suppose that  $n$  is a non-degenerate quadratic form on  $V$.   Let  $p$  be an arbitrary quadratic form on  $V$.   Then there exist a unique linear map  $f\hskip 1.00006pt\colon\hskip 1.00006ptV\longrightarrow V$  such that

[TABLE]

for all  $v\hskip 0.50003pt,\hskip 3.00003ptw\hskip 1.00006pt\in\hskip 1.00006ptV$.   Since the polarization  $(\hskip 0.50003ptv\hskip 0.50003pt,\hskip 3.00003ptw\hskip 0.50003pt)_{\hskip 0.35002ptp}$  is symmetric,  $f$  is symmetric with respect to  $n$.   By taking  $v\hskip 1.99997pt=\hskip 1.99997ptw$  in the last displayed formula,   we see that  $p\hskip 1.99997pt=\hskip 1.99997ptf\cdot n$.   The map  $f$  is uniquely determined by the forms  $p\hskip 0.50003pt,\hskip 3.00003ptn$.   We will denote it by  $p/n$.

The operations  $p\hskip 1.00006pt\longmapsto p/n$  and $f\hskip 1.00006pt\longmapsto f\cdot n$ are mutually inverse in the sense that

[TABLE]

for all  $p\hskip 1.00006pt\in\hskip 1.00006ptQ\hskip 0.50003pt(V)$  and all maps  $f\hskip 0.50003pt\colon\hskip 0.50003ptV\longrightarrow V$  symmetric with respect to  $n$.

Anisotropic forms.   A quadratic form  $q$  is called  anisotropic  if  $q\hskip 0.50003pt(v)\hskip 1.99997pt\neq\hskip 1.99997pt0$  for  $v\hskip 1.99997pt\neq\hskip 1.99997pt0$.   Suppose that  $q$  is an anisotropic form with the polarization  $\varphi$.   If  $v\hskip 1.99997pt\neq\hskip 1.99997pt0$,   then the map  $w\hskip 1.99997pt\longmapsto\hskip 1.99997pt\varphi\hskip 0.50003pt(v\hskip 0.50003pt,\hskip 3.00003ptw)$  is non-zero because  $\varphi\hskip 0.50003pt(v\hskip 0.50003pt,\hskip 3.00003ptv)\hskip 1.99997pt=\hskip 1.99997ptq\hskip 0.50003pt(v)\hskip 1.99997pt\neq\hskip 1.99997pt0$.   It follows that  $\widehat{\varphi}$  is injective and hence is an isomorphism.   Therefore  $\varphi$  and  $q$  are non-degenerate.

The case of vector spaces of dimension $2$.   Suppose now that the dimension of  $V$  is $2$.   Let  $v\hskip 0.50003pt,\hskip 3.00003ptw$  be a basis of $V$.    If  $q\hskip 1.00006pt\in\hskip 1.00006ptQ\hskip 0.50003pt(V)$,    then

[TABLE]

is the Gram matrix of $q$ with respect to the basis  $v\hskip 0.50003pt,\hskip 3.00003ptw$.   It follows that the determinant map   $\displaystyle\det\colon Q\hskip 0.50003pt(V)\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\mathbold{k}$   with respect to the basis  $v\hskip 0.50003pt,\hskip 3.00003ptw$  has the form

[TABLE]

Therefore the determinant map itself is a non-degenerate quadratic form.

Orthogonality .   Two vectors  $v\hskip 0.50003pt,\hskip 3.00003ptw\hskip 1.00006pt\in\hskip 1.00006ptV$  are said to be  orthogonal  with respect to a form  $q\hskip 1.00006pt\in\hskip 1.00006ptQ\hskip 0.50003pt(V)$  if  $(\hskip 0.50003ptv\hskip 0.50003pt,\hskip 1.99997ptw\hskip 0.50003pt)_{q}\hskip 1.99997pt=\hskip 1.99997pt0$,   i.e.  if

[TABLE]

Suppose now that the dimension of  $V$  is  $2$.   Let  $p\hskip 0.50003pt,\hskip 3.00003ptq$  be two quadratic forms on $V$.   They are said to be  orthogonal  if  $p\hskip 0.50003pt,\hskip 3.00003ptq$  are orthogonal with respect to a determinant map $\det$ considered as a quadratic form on $Q\hskip 0.50003pt(V)$,   i.e.  if

[TABLE]

Since different determinant maps differ by the multiplication by a non-zero element of $\mathbold{k}$,   the property  (1.3)  does not depend on the choice of the determinant map.

1.1. Theorem.   Let  $V$  be a vector space over  $\mathbold{k}$  of dimension  $2$  and  let  $p\hskip 0.50003pt,\hskip 3.00003ptq\hskip 1.00006pt\in\hskip 1.00006ptQ\hskip 0.50003pt(V)$. Suppose that  $v\hskip 0.50003pt,\hskip 3.00003ptw$  is a basis of  $V$  orthogonal with respect to  $p$.   Then the forms  $p\hskip 0.50003pt,\hskip 3.00003ptq$  are orthogonal with respect to any determinant map if and only if

[TABLE]

Proof**.**     By the remark preceding the theorem,   it is sufficient to consider the orthogonality with respect to determinant map defined by the basis  $v\hskip 0.50003pt,\hskip 3.00003ptw$.

The forms  $p\hskip 0.50003pt,\hskip 3.00003ptq$  are orthogonal if and only if  (1.3)  holds,   i.e.  if  and only if

[TABLE]

[TABLE]

The basis  $v\hskip 0.50003pt,\hskip 3.00003ptw$  is orthogonal with respect to $p$ and hence  $(\hskip 0.50003ptv\hskip 0.50003pt,\hskip 1.99997ptw\hskip 0.50003pt)_{p}\hskip 1.99997pt=\hskip 1.99997pt0$  and

[TABLE]

It follows that the  (1.4)  is equivalent to

[TABLE]

or ,   what is the same,   to

[TABLE]

Since  $(p\hskip 1.99997pt+\hskip 1.99997ptq)\hskip 0.50003pt(u)\hskip 1.99997pt=\hskip 1.99997ptp\hskip 0.50003pt(u)\hskip 1.99997pt+\hskip 1.99997ptq\hskip 0.50003pt(u)$  for all  $u\hskip 1.00006pt\in\hskip 1.00006ptV$,  the left hand side of  (1.5)  is equal to

[TABLE]

and hence  (1.5)  is equivalent to  $\displaystyle p\hskip 0.50003pt(v)\hskip 1.00006ptq\hskip 0.50003pt(w)\hskip 1.99997pt+\hskip 1.99997ptq\hskip 0.50003pt(v)\hskip 1.49994ptp\hskip 0.50003pt(w)\hskip 1.99997pt=\hskip 1.99997pt0$.  The theorem follows.   $\blacksquare$

2. The  norm  and  the  trace

The norm and the trace.   Recall that  $z\hskip 1.99997pt\longmapsto\hskip 1.99997pt\overline{z}$  is the only non-trivial automorphism of the field $\mathbold{K}$ fixed on $\mathbold{k}$.   Let  $N\hskip 0.50003pt,\hskip 3.00003pt\operatorname{t{\hskip 0.24994pt}r}\hskip 0.50003pt\hskip 1.00006pt\colon\hskip 1.00006pt\mathbold{K}\longrightarrow\mathbold{k}$  be the maps

[TABLE]

Then  $N$  is the norm of the extension  $\mathbold{K}/\hskip 0.50003pt\mathbold{k}$  and  $\operatorname{t{\hskip 0.24994pt}r}\hskip 0.50003pt$  is the half of the trace of  $\mathbold{K}/\hskip 0.50003pt\mathbold{k}$.   The map  $\operatorname{t{\hskip 0.24994pt}r}\hskip 0.50003pt$  has the advantage of being equal to the identity on  $\mathbold{k}$.   If   $\mathbold{K}/\hskip 0.50003pt\mathbold{k}\hskip 3.99994pt=\hskip 3.99994pt\mathbold{C}/\hskip 0.50003pt\mathbold{R}$,   then  $\overline{z}$  is the complex conjugate of  $z$,   $\operatorname{t{\hskip 0.24994pt}r}\hskip 0.50003pt\hskip 0.50003pt(z)\hskip 3.99994pt=\hskip 3.99994pt\mbox{Re}\hskip 1.99997ptz$   and   $N\hskip 0.50003pt(z)\hskip 3.99994pt=\hskip 1.00006pt\hskip 3.99994pt|\hskip 1.00006ptz\hskip 1.00006pt|^{\hskip 0.35002pt2}$.

The norm as a quadratic form on  $\mathbold{K}$.     Let

[TABLE]

Then  $\langle\hskip 0.50003pt\hskip 0.24994ptz\hskip 0.50003pt,\hskip 1.99997ptu\hskip 0.50003pt\rangle$  is a bilinear form on  $\mathbold{K}$  considered as a vector space over  $\mathbold{k}$.   Since

[TABLE]

for all  $z\hskip 0.50003pt,\hskip 3.00003ptu\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$,   the bilinear form  $\langle\hskip 0.50003pt\hskip 0.24994ptz\hskip 0.50003pt,\hskip 1.99997ptu\hskip 0.50003pt\rangle$  is symmetric.   Since   $z\hskip 1.00006pt\overline{z}\hskip 3.99994pt=\hskip 3.99994pt\operatorname{t{\hskip 0.24994pt}r}\hskip 0.50003pt(z\hskip 1.00006pt\overline{z})$   for all  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$,   it follows that  $N$  is a quadratic form on  $\mathbold{K}$  with polarization  $\langle\hskip 0.50003pt\hskip 0.24994ptz\hskip 0.50003pt,\hskip 1.99997ptu\hskip 0.50003pt\rangle$.   Obviously,   the quadratic form  $N$  is anisotropic and hence is non-degenerate.

The orthogonal complement of  $\mathbold{k}$.   Let  $\mathbold{k}^{\hskip 0.70004pt\perp}$  be the orthogonal complement of  $\mathbold{k}$  in  $\mathbold{K}$  with respect to  $N$,   i.e.  the set of all  $\rho\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$  such that  $\langle\hskip 0.50003pt\rho\hskip 0.50003pt,\hskip 1.99997pta\hskip 0.50003pt\rangle\hskip 1.99997pt=\hskip 1.99997pt0$  for all  $a\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{k}$.   Since  $N$  is anisotropic,  $\mathbold{k}\hskip 1.00006pt\cap\hskip 1.00006pt\mathbold{k}^{\hskip 0.70004pt\perp}\hskip 1.99997pt=\hskip 1.99997pt\varnothing$  and hence  $\mathbold{K}\hskip 1.99997pt=\hskip 1.99997pt\mathbold{k}\oplus\mathbold{k}^{\hskip 0.70004pt\perp}$.    Since the dimension of  $\mathbold{K}$  over  $\mathbold{k}$  is equal to  $2$,   the dimension of  $\mathbold{k}^{\hskip 0.70004pt\perp}$  over  $\mathbold{k}$  is equal to $1$.

The orthogonal complement  $\mathbold{k}^{\hskip 0.70004pt\perp}$  is equal to the kernel of  $\operatorname{t{\hskip 0.24994pt}r}\hskip 0.50003pt\hskip 1.00006pt\colon\hskip 1.00006pt\mathbold{K}\longrightarrow\mathbold{k}$,   as it follows from the fact that  $\langle\hskip 0.50003pt\rho\hskip 0.50003pt,\hskip 1.99997pta\hskip 0.50003pt\rangle\hskip 1.99997pt=\hskip 1.99997pta\hskip 1.00006pt\operatorname{t{\hskip 0.24994pt}r}\hskip 0.50003pt\hskip 0.50003pt(\rho)$  for  $a\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{k}$.   Therefore  $\rho\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{k}^{\hskip 0.70004pt\perp}$  implies that

[TABLE]

3. Quadratic  forms  in  dimension  $1$  over  $\mathbold{K}$

The framework.   For the remaining part of the paper we will assume that  $V$  is a vector space over  $\mathbold{K}$  of dimension  $1$  over  $\mathbold{K}$.   By the restriction of scalars we may consider $V$ also as a vector space over $\mathbold{k}$.   By a quadratic form on $V$ we will understand a quadratic form on $V$ as a vector space over $\mathbold{k}$.   As a vector space over $\mathbold{k}$ the vector space $Q\hskip 0.50003pt(V)$ of quadratic forms on $V$ depends only on the structure of $V$ as a vector space over $\mathbold{k}$.

The norm-like quadratic forms.   A form  $n\hskip 1.00006pt\in\hskip 1.00006ptQ\hskip 0.50003pt(V)$  is said to be  norm-like  if

[TABLE]

for all  $v\hskip 1.00006pt\in\hskip 1.00006ptV$  and  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$.   Let  $\mathbold{N}\hskip 0.50003pt(V)$  be the space of all norm-like forms on  $V$.

The dimension of  $\mathbold{N}(V)$  over  $\mathbold{k}$  is equal to  $1$.   Indeed,   let  $v\hskip 1.00006pt\in\hskip 1.00006ptV$  and  $v\hskip 1.99997pt\neq\hskip 1.99997pt0$.   Then every element of  $V$  is equal to  $z\hskip 1.00006ptv$  for some  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$  and hence every norm-like form is determined by its value at  $v$.   On the other hand,   for every  $a\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{k}$  the map   $\displaystyle z\hskip 1.00006ptv\hskip 1.00006pt\longmapsto\hskip 1.00006ptN\hskip 0.50003pt(z)\hskip 1.00006pta$   is a norm-like form and takes the value $a$ at $v$.

Every non-zero  $n\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{N}(V)$  is anisotropic and hence is non-degenerate.   Indeed,   if  $n\hskip 0.50003pt(v)\hskip 1.99997pt\neq\hskip 1.99997pt0$,   then every non-zero element  $w$  of  $V$  is equal to  $z\hskip 1.00006ptv$  for some non-zero  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$.   Since  $N$  is anisotropic,   together with  (3.1)  this implies that  $n\hskip 0.50003pt(w)\hskip 1.99997pt\neq\hskip 1.99997pt0$.

The polarization of any norm-like form  $n$  is closely related to the polarization  $\langle\hskip 0.50003pt\hskip 0.24994ptz\hskip 0.50003pt,\hskip 1.99997ptu\hskip 0.50003pt\rangle$  of  $N$.   Let  $v\hskip 1.00006pt\in\hskip 1.00006ptV$  and  $z\hskip 0.50003pt,\hskip 3.00003ptu\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$.    Then

[TABLE]

Indeed,   the maps  $z\hskip 1.00006pt\longmapsto\hskip 1.00006ptn\hskip 0.50003pt(z\hskip 1.00006ptv)$  and  $z\hskip 1.00006pt\longmapsto\hskip 1.00006ptN\hskip 0.50003pt(z)\hskip 1.49994ptn\hskip 0.50003pt(v)$  are quadratic forms on $\mathbold{K}$.   By  (3.1)  they are equal.   Hence their polarizations are equal also.   But the left and the right hand sides of  (3.2)  are nothing else but these polarizations.

3.1. Lemma.   A quadratic form  $n$  on  $V$  is norm-like if and only if

[TABLE]

*for every  $v\hskip 0.50003pt,\hskip 3.00003ptw\hskip 1.00006pt\in\hskip 1.00006ptV$  and every  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$.   *

Proof**.**     Suppose that  $n$  is norm-like.   We may assume that  $v\hskip 1.99997pt\neq\hskip 1.99997pt0$.   Then $v$ is a basis of $V$ over $\mathbold{K}$ and hence  $w\hskip 1.99997pt=\hskip 1.99997ptu\hskip 1.00006ptv$  for some  $u\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$.   The identity  (3.2)  implies that

[TABLE]

[TABLE]

On the other hand,   $\displaystyle\operatorname{t{\hskip 0.24994pt}r}\hskip 0.50003pt\hskip 1.00006pt(z\hskip 1.99997pt\overline{u}\hskip 0.50003pt)\hskip 3.99994pt=\hskip 3.99994pt\operatorname{t{\hskip 0.24994pt}r}\hskip 0.50003pt\hskip 1.00006pt(\hskip 0.50003pt\overline{z}\hskip 1.99997ptu)\hskip 3.99994pt=\hskip 3.99994pt\operatorname{t{\hskip 0.24994pt}r}\hskip 0.50003pt\hskip 1.00006pt(1\cdot\overline{z}\hskip 1.99997ptu)$   and hence

[TABLE]

By combining these equalities we see that  (3.3)  holds.   Conversely ,  if  (3.3)  holds,   then

[TABLE]

and hence   $\displaystyle n\hskip 0.50003pt(z\hskip 1.00006ptv)\hskip 3.99994pt=\hskip 3.99994ptN\hskip 0.50003pt(z)\hskip 1.00006ptn\hskip 0.50003pt(v)$   for all  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$,  $v\hskip 1.00006pt\in\hskip 1.00006ptV$.    $\blacksquare$

The anti-norm-like quadratic forms.   A form  $q\hskip 1.00006pt\in\hskip 1.00006ptQ\hskip 0.50003pt(V)$  is said to be  anti-norm-like  if

[TABLE]

for all  $v\hskip 1.00006pt\in\hskip 1.00006ptV$  and  $\rho\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{k}^{\hskip 0.70004pt\perp}$.   Let  $\mathbold{A}\hskip 0.50003pt(V)$  be the space of all anti-norm-like forms on  $V$.

As in the case of norm-like forms,   the identity  (3.4)  can be extended to the polarizations.   Let  $q$  is an anti-norm-like form.   Let  $v\hskip 0.50003pt,\hskip 3.00003ptw\hskip 1.00006pt\in\hskip 1.00006ptV$  and let  $\rho\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{k}^{\hskip 0.70004pt\perp}$.   Then

[TABLE]

Indeed,   the maps  $v\hskip 1.00006pt\longmapsto\hskip 1.00006ptq\hskip 0.50003pt(\rho\hskip 1.00006ptv)$  and  $v\hskip 1.00006pt\longmapsto\hskip 1.00006pt-\hskip 1.99997ptN\hskip 0.50003pt(\rho)\hskip 1.00006ptq\hskip 0.50003pt(v)$  are quadratic forms on $V$.   By  (3.4)  they are equal.   Hence their polarizations are equal also.   But the left and the right hand sides of  (3.5)  are nothing else but these polarizations.

3.2. Lemma.   A quadratic form  $q$  on  $V$  is anti-norm-like if and only if

[TABLE]

*for every  $v\hskip 0.50003pt,\hskip 3.00003ptw\hskip 1.00006pt\in\hskip 1.00006ptV$  and every  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$.   *

Proof**.**     Suppose that  $q$  is anti-norm-like.   Every  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$  has the form  $z\hskip 1.99997pt=\hskip 1.99997pta\hskip 1.99997pt+\hskip 1.99997pt\rho$,   where  $a\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{k}$  and  $\rho\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{k}^{\hskip 0.70004pt\perp}$.     If   $\rho\hskip 1.99997pt=\hskip 1.99997pt0$,     then   $z\hskip 1.99997pt=\hskip 1.99997pta\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{k}$   and hence  (3.6)  holds.   If  $\rho\hskip 1.99997pt\neq\hskip 1.99997pt0$,   then  $\rho^{\hskip 0.35002pt-\hskip 0.70004pt1}$  is defined and  (3.5)  implies that

[TABLE]

On the other hand ,   $\displaystyle(\hskip 0.50003pta\hskip 1.00006ptv\hskip 0.50003pt,\hskip 3.00003ptw\hskip 0.50003pt)_{q}\hskip 3.99994pt=\hskip 3.99994pt(\hskip 0.50003ptv\hskip 0.50003pt,\hskip 3.00003pta\hskip 1.00006ptw\hskip 0.50003pt)_{q}$   because  $a\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{k}$.   Therefore

[TABLE]

It follows that  $(\hskip 0.50003ptz\hskip 1.00006ptv\hskip 0.50003pt,\hskip 3.00003ptw\hskip 0.50003pt)_{q}\hskip 1.99997pt=\hskip 1.99997pt(\hskip 0.50003ptv\hskip 0.50003pt,\hskip 3.00003ptz\hskip 1.00006ptw\hskip 0.50003pt)_{q}$,    i.e.  the identity  (3.6)  holds.

Suppose now that  (3.6)   holds.   If  $v\hskip 1.00006pt\in\hskip 1.00006ptV$  and  $\rho\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{k}^{\hskip 0.70004pt\perp}$,   then

[TABLE]

and hence  $q\hskip 0.50003pt(\rho\hskip 1.00006ptv)\hskip 1.99997pt=\hskip 1.99997pt\rho^{\hskip 0.35002pt2}\hskip 1.00006ptq\hskip 0.50003pt(v)$.    It follows that  $q$  is anti-norm-like.    $\blacksquare$

3.3. Theorem.   *$\mathbold{A}\hskip 0.50003pt(V)$  is equal to the orthogonal complement of  $\mathbold{N}\hskip 0.50003pt(V)$  in  $Q\hskip 0.50003pt(V)$.   *

Proof**.**     Suppose that  $n\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{N}\hskip 0.50003pt(V)$,   $n\hskip 1.99997pt\neq\hskip 1.99997pt0$.   Let  $v\hskip 1.00006pt\in\hskip 1.00006ptV$,  $v\hskip 1.99997pt\ \neq\hskip 1.99997pt\ 0$.   Since $n$ is non-zero,   the property  (3.1)  implies that  $n\hskip 0.50003pt(v)\hskip 1.99997pt\neq\hskip 1.99997pt0$.   Let  $\rho\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{k}^{\hskip 0.70004pt\perp}$.   Then  (3.2)  implies that

[TABLE]

and hence  $\rho\hskip 1.00006ptv$  is orthogonal to  $v$  with respect to $n$.

By Theorem  1. Bilinear  and  quadratic  forms  the forms  $n\hskip 0.50003pt,\hskip 3.00003ptq$  are orthogonal if and only if

[TABLE]

By the property  (3.1)  this equality is equivalent to

[TABLE]

Since  $n\hskip 0.50003pt(v)\hskip 1.99997pt\neq\hskip 1.99997pt0$,   the last equality is equivalent to

[TABLE]

and hence to   $\displaystyle q\hskip 0.50003pt(\rho\hskip 1.00006ptv)\hskip 3.99994pt=\hskip 3.99994pt-\hskip 1.99997ptN\hskip 0.50003pt(\rho)\hskip 1.00006ptq\hskip 0.50003pt(v)$.

It follows that  $n\hskip 0.50003pt,\hskip 3.00003ptq$  are orthogonal if and only if  $q$  is anti-norm-like.    $\blacksquare$

3.4. Theorem.   $\displaystyle Q\hskip 0.50003pt(V)\hskip 3.99994pt=\hskip 3.99994pt\mathbold{N}(V)\hskip 1.99997pt\oplus\hskip 1.99997pt\mathbold{A}\hskip 0.50003pt(V)$.

Proof**.**     Suppose that   $\displaystyle q\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{N}\hskip 0.50003pt(V)\hskip 1.99997pt\cap\hskip 1.99997pt\mathbold{A}\hskip 0.50003pt(V)$.   Let  $\rho\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{k}^{\hskip 0.70004pt\perp}$,  $\rho\hskip 1.99997pt\neq\hskip 1.99997pt0$.   Then

[TABLE]

and hence  $\displaystyle N\hskip 0.50003pt(\rho)\hskip 1.00006ptq\hskip 0.50003pt(v)\hskip 1.99997pt=\hskip 1.99997pt0$  for all  $v\hskip 1.00006pt\in\hskip 1.00006ptV$.   But if  $\rho\hskip 1.99997pt\neq\hskip 1.99997pt0$,   then  $N\hskip 0.50003pt(\rho)\hskip 1.99997pt\neq\hskip 1.99997pt0$.   It follows that  $q\hskip 0.50003pt(v)\hskip 1.99997pt=\hskip 1.99997pt0$  for all  $v$,   i.e.  $q\hskip 1.99997pt=\hskip 1.99997pt0$.   Therefore

[TABLE]

In view of Theorem  3. Quadratic  forms  in  dimension  $1$  over  $\mathbold{K}$,   this implies   $\displaystyle Q\hskip 0.50003pt(V)\hskip 3.99994pt=\hskip 3.99994ptN\hskip 0.50003pt(V)\hskip 1.99997pt\oplus\hskip 1.99997pt\mathbold{A}\hskip 0.50003pt(V)$.    $\blacksquare$

4. The  structure  of  a  $\mathbold{K}$-vector  space  on  $\mathbold{A}\hskip 0.50003pt(V)$

The multiplication maps.   For  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$  let   $\displaystyle m_{\hskip 0.70004ptz}\hskip 1.00006pt\colon\hskip 1.00006ptV\longrightarrow V$   be the map

[TABLE]

We consider  $m_{\hskip 0.70004ptz}$  as a map linear over  $\mathbold{k}$  (although it is linear over  $\mathbold{K}$  also).   Then

[TABLE]

If  $V\hskip 1.99997pt=\hskip 1.99997pt\mathbold{K}$,   this is a well known result of the Galois theory .   Since $V$ is isomorphic to $\mathbold{K}$ as a vector space over $\mathbold{K}$,   this special case implies the general one.

The anti-norm-like quadratic forms and the multiplication maps.   Suppose that  $q$  is an anti-norm-like quadratic form on  $V$  and let  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$.   Lemma  3. Quadratic  forms  in  dimension  $1$  over  $\mathbold{K}$  implies that  $m_{\hskip 0.70004ptz}$  is symmetric with respect to  $q$,   and hence the product  $m_{\hskip 0.70004ptz}\cdot q$  is defined.

4.1. Lemma.   If  $q$  is an anti-norm-like quadratic form on  $V$  and  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$,   then the product  $m_{\hskip 0.70004ptz}\cdot q$  is also anti-norm-like.

Proof**.**     Let  $\rho\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{k}^{\hskip 0.70004pt\perp}$.   By applying  (3.5)  to  $z\hskip 1.00006ptv\hskip 0.50003pt,\hskip 3.00003ptv$  in the roles of  $v\hskip 0.50003pt,\hskip 3.00003ptw$,   we see that

[TABLE]

[TABLE]

It follows that  $m_{\hskip 0.70004ptz}\cdot q$  is anti-norm-like.    $\blacksquare$

4.2. Lemma.   If  $q$  is an anti-norm-like quadratic form on  $V$  and  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$,   then

[TABLE]

*for every determinant map  $\det\hskip 1.00006pt\colon\hskip 1.00006ptQ\hskip 0.50003pt(V)\longrightarrow\mathbold{k}$.   *

Proof**.**     It is sufficient to combine  (1.2)  with  (4.1).    $\blacksquare$

The structure of a vector space over  $\mathbold{K}$  on  $\mathbold{A}\hskip 0.50003pt(V)$.   Let us define the multiplication  $(z\hskip 0.50003pt,\hskip 3.00003ptq)\hskip 1.00006pt\longmapsto\hskip 1.00006pt\ z\hskip 0.50003ptq$  of forms  $q\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{A}\hskip 0.50003pt(V)$  by elements  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$  by the formula

[TABLE]

This turns $\mathbold{A}\hskip 0.50003pt(V)$ into a vector space over  $\mathbold{K}$,   as a trivial verification shows.   By Lemma  4. The  structure  of  a  $\mathbold{K}$-vector  space  on  $\mathbold{A}\hskip 0.50003pt(V)$  any determinant map defines a norm-like quadratic form on  $\mathbold{A}\hskip 0.50003pt(V)$.

The multiplication  $m_{\hskip 1.39998pt\overline{z}}$  is used instead of  the more natural  $m_{\hskip 0.70004ptz}$  in order to avoid  $\overline{z}$  in Lemma  6. Anti-linear  maps.   This makes the identification maps from Section  10. Comparing   $\mathbold{c}\hskip 0.50003pt(f)$   and   $\mu_{\hskip 0.70004ptf}$  linear over  $\mathbold{K}$.

5. Conformal  structures

Conformal structures.   A  conformal structure  on a vector space $U$ over $\mathbold{k}$ is defined as a non-zero quadratic form on $U$ considered up to multiplication by a non-zero element of $\mathbold{k}$.   The conformal structure determined by  $q\hskip 1.00006pt\in\hskip 1.00006ptQ\hskip 0.50003pt(U)$  is called the  conformal class  of $q$ and is denoted by $[\hskip 0.50003ptq\hskip 1.00006pt]$.   It is called  non-degenerate  if the form $q$ is non-degenerate.   The set of all conformal structures on $U$ is nothing else but the  projective space  $\mathbb{P}Q\hskip 0.50003pt(U)$ associated with the  vector space  $Q\hskip 0.50003pt(U)$  of quadratic forms on $V$.

Conformal structures in dimension $1$ over $\mathbold{K}$.   The conformal class  $c_{\hskip 0.35002ptV}\hskip 1.99997pt=\hskip 1.99997pt[\hskip 0.50003ptn\hskip 0.50003pt]$  of a non-zero  $n\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{N}(V)$  does not depend on the choice of  $n$  and is called the  canonical conformal structure  on $V$.   A conformal structure on $V$ is called  exceptional  if it is equal to the conformal class  $[\hskip 0.50003ptq\hskip 0.50003pt]$  of a non-zero  $q\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{A}\hskip 0.50003pt(V)$,   and  regular  regular otherwise.

Let  $\displaystyle\mathbb{P}\mathbold{A}\hskip 0.50003pt(V)\hskip 1.99997pt\subset\hskip 1.99997pt\mathbb{P}Q\hskip 0.50003pt(V)$  be the set of the exceptional conformal structures,   i.e.  be the set of conformal classes  $[\hskip 0.50003ptq\hskip 1.00006pt]$  of quadratic forms  $q\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{A}\hskip 0.50003pt(V)$.   Its complement

[TABLE]

is the set of all regular conformal structures on $V$.   The set  $\mathbb{P}\mathbold{A}\hskip 0.50003pt(V)$  is a line in the projective plane  $\mathbb{P}Q\hskip 0.50003pt(V)$,   and its complement  $\mathbb{M}\hskip 0.50003pt(V)$  is an affine plane.

The structure of a vector space over  $\mathbold{K}$  on  $\mathbb{M}\hskip 0.50003pt(V)$.   Let  $n\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{N}(V)$,  $n\hskip 1.99997pt\neq\hskip 1.99997pt0$.   The map  $\displaystyle q\hskip 1.99997pt\longmapsto\hskip 1.99997pt[\hskip 0.50003ptn\hskip 1.99997pt+\hskip 1.99997ptq\hskip 0.50003pt]$  is a bijection  $\displaystyle\mathbold{A}\hskip 0.50003pt(V)\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\mathbb{M}\hskip 0.50003pt(V)$.   One can use it to transfer the structure of a vector space over  $\mathbold{K}$  from  $\mathbold{A}\hskip 0.50003pt(V)$  to  $\mathbb{M}\hskip 0.50003pt(V)$.   The resulting multiplication by elements  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$  is given by the formula

[TABLE]

While the bijection  $\displaystyle\mathbold{A}\hskip 0.50003pt(V)\hskip 1.99997pt\longrightarrow\hskip 1.99997pt\mathbb{M}\hskip 0.50003pt(V)$  depends on the choice of  $n$,   an immediate verification shows that the resulting multiplication does not .   Therefore,   the above construction defines a canonical structure of a vector space over  $\mathbold{K}$  on  $\mathbb{M}\hskip 0.50003pt(V)$.   The canonical conformal structure  $c_{\hskip 0.35002ptV}$  on $V$ serves as the zero of this vector space.

A canonical quadratic form on  $\mathbb{M}\hskip 0.50003pt(V)$.   Every point of  $\mathbb{M}\hskip 0.50003pt(V)$  has the form  $[\hskip 0.50003ptn\hskip 1.99997pt+\hskip 1.99997ptq\hskip 0.50003pt]$,   where  $n\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{N}(V)$,  $n\hskip 1.99997pt\neq\hskip 1.99997pt0$,   and  $q\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{A}\hskip 0.50003pt(V)$.    Let us define a map  $\mathbb{D}\hskip 1.00006pt\colon\hskip 1.00006pt\mathbb{M}\hskip 0.50003pt(V)\longrightarrow\mathbold{k}$  by

[TABLE]

If   $[\hskip 0.50003ptn^{\prime}\hskip 1.99997pt+\hskip 1.99997ptq^{\prime}\hskip 0.50003pt]\hskip 3.99994pt=\hskip 3.99994pt[\hskip 0.50003ptn\hskip 1.99997pt+\hskip 1.99997ptq\hskip 0.50003pt]$,   then  $n^{\prime}\hskip 1.99997pt=\hskip 1.99997pta\hskip 0.50003ptn$,  $q^{\prime}\hskip 1.99997pt=\hskip 1.99997pta\hskip 0.50003ptq$  for some  $a\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{k}$,   and hence

[TABLE]

It follows that  $\mathbb{D}$  is correctly defined.   If we temporarily fix $n$ and identify  $\mathbb{M}\hskip 0.50003pt(V)$  with  $\mathbold{A}\hskip 0.50003pt(V)$  by the map  $q\hskip 1.00006pt\longmapsto\hskip 1.00006pt[\hskip 0.50003ptn\hskip 1.99997pt+\hskip 1.99997ptq\hskip 0.50003pt]$,   then  $\mathbb{D}$  turns into the map  $-\hskip 1.99997pt\det/\det\hskip 1.00006ptn$  restricted to  $\mathbold{A}\hskip 0.50003pt(V)$.   Since  $\det$  is a quadratic form,   this implies that  $\mathbb{D}$  is a quadratic form on  $\mathbb{M}\hskip 0.50003pt(V)$.   Lemma  4. The  structure  of  a  $\mathbold{K}$-vector  space  on  $\mathbold{A}\hskip 0.50003pt(V)$  implies that  $\mathbb{D}$  is norm-like.

6. Anti-linear  maps

Anti-linear maps.   Let  $U\hskip 0.50003pt,\hskip 3.00003ptW$  be vector spaces over $\mathbold{K}$,   and let  $f$  be a map  $U\longrightarrow W$  linear over $\mathbold{k}$.   The map  $f$  said to be  anti-linear  over  $\mathbold{K}$,   or simply  anti-linear  if

[TABLE]

for all  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$,  $u\hskip 1.00006pt\in\hskip 1.00006ptU$.   Let  $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(U\hskip 0.50003pt,\hskip 1.49994ptW)$  be the space of all anti-linear maps  $U\longrightarrow W$.

6.1. Lemma.   If  $f\hskip 1.00006pt\in\hskip 1.00006pt\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$  and  $n\hskip 0.50003pt\in\hskip 0.50003pt\mathbold{N}(V)$,   then  $f$  is symmetric with respect to $n$.

Proof**.**     If  $v\hskip 0.50003pt,\hskip 3.00003ptw\hskip 1.00006pt\in\hskip 1.00006ptV$  and  $v\hskip 1.99997pt\neq\hskip 1.99997pt0$,   then  $w\hskip 1.99997pt=\hskip 1.99997ptz\hskip 1.00006ptv$  for some  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$,   and

[TABLE]

by Lemma  3. Quadratic  forms  in  dimension  $1$  over  $\mathbold{K}$  and the anti-linearity of  $f$.    It follows that

[TABLE]

and hence $f$ is symmetric with respect to  $n$.    $\blacksquare$

6.2. Lemma.   *Let  $n\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{N}\hskip 0.50003pt(V)$,  $n\hskip 1.99997pt\neq\hskip 1.99997pt0$.    If  $f\hskip 1.00006pt\colon\hskip 1.00006ptV\longrightarrow V$  is anti-linear ,   then  $f\cdot n$  is defined and  $f\cdot n\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{A}\hskip 0.50003pt(V)$.    Conversely ,   if  $q\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{A}\hskip 0.50003pt(V)$,    then  $f\hskip 1.99997pt=\hskip 1.99997ptq/n$  is anti-linear .   *

Proof**.**     If  $f\hskip 1.00006pt\colon\hskip 1.00006ptV\longrightarrow V$  is anti-linear ,   then by Lemma  6. Anti-linear  maps  $f$  is symmetric with respect to  $n$  and hence  $f\cdot n$  is defined.   Let  $q\hskip 1.99997pt=\hskip 1.99997ptf\cdot n$.   If  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$  and  $v\hskip 0.50003pt,\hskip 3.00003ptw\hskip 1.00006pt\in\hskip 1.00006ptV$,   then

[TABLE]

by anti-linearity of  $f$  and the definition of  $q$,   and

[TABLE]

by Lemma  3. Quadratic  forms  in  dimension  $1$  over  $\mathbold{K}$  and the definition of  $q$.   It follows that   $\displaystyle(\hskip 0.50003ptz\hskip 1.00006ptv\hskip 0.50003pt,\hskip 3.00003ptw\hskip 0.50003pt)_{q}\hskip 3.99994pt=\hskip 3.99994pt(\hskip 0.50003ptv\hskip 0.50003pt,\hskip 3.00003ptz\hskip 1.00006ptw\hskip 0.50003pt)_{q}$   for all  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$  and all  $v\hskip 0.50003pt,\hskip 3.00003ptw\hskip 1.00006pt\in\hskip 1.00006ptV$.   By Lemma  3. Quadratic  forms  in  dimension  $1$  over  $\mathbold{K}$  this implies that  $f\cdot n\hskip 1.99997pt=\hskip 1.99997ptq\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{A}\hskip 0.50003pt(V)$.

Suppose now that  $q\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{A}\hskip 0.50003pt(V)$.   Since $n$ is non-degenerate,   the map   $\displaystyle f\hskip 1.99997pt=\hskip 1.99997ptq/n\hskip 1.00006pt\colon\hskip 1.00006ptV\longrightarrow V$   is defined and is symmetric with respect to  $n$.    If  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$  and  $v\hskip 0.50003pt,\hskip 3.00003ptw\hskip 1.00006pt\in\hskip 1.00006ptV$,    then

[TABLE]

by Lemma  3. Quadratic  forms  in  dimension  $1$  over  $\mathbold{K}$  and hence   $\displaystyle(\hskip 1.00006ptf\hskip 0.50003pt(z\hskip 1.00006ptv)\hskip 0.50003pt,\hskip 3.00003ptw\hskip 1.00006pt)_{n}\hskip 3.99994pt=\hskip 3.99994pt(\hskip 1.00006ptf\hskip 0.50003pt(v)\hskip 0.50003pt,\hskip 3.00003ptz\hskip 1.00006ptw\hskip 1.00006pt)_{n}$   by the definition of  $f$.   Since

[TABLE]

by Lemma  3. Quadratic  forms  in  dimension  $1$  over  $\mathbold{K}$,   it follows that

[TABLE]

for all  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$  and all  $v\hskip 0.50003pt,\hskip 3.00003ptw\hskip 1.00006pt\in\hskip 1.00006ptV$.   Since  $n$  is non-degenerate and  $w$  is arbitrary ,   it follows that  $f\hskip 0.50003pt(z\hskip 1.00006ptv)\hskip 1.99997pt=\hskip 1.99997pt\overline{z}\hskip 1.99997ptf\hskip 0.50003pt(v)$  for all  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$,  $v\hskip 1.00006pt\in\hskip 1.00006ptV$,    i.e.  that  $f$  is anti-linear .    $\blacksquare$

6.3. Lemma.   Let  $n\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{N}\hskip 0.50003pt(V)$,  $n\hskip 1.99997pt\neq\hskip 1.99997pt0$.    If   $f\hskip 1.00006pt\in\hskip 1.00006pt\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$   and   $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$,    then

[TABLE]

Proof**.**     Let  $q\hskip 1.99997pt=\hskip 1.99997ptf\cdot n$  and  $p\hskip 1.99997pt=\hskip 1.99997pt(z\hskip 1.00006ptf)\cdot n$.   Let  $v\hskip 0.50003pt,\hskip 3.00003ptw\hskip 1.00006pt\in\hskip 1.00006ptV$.    Then

[TABLE]

by the definition of the product  $z\hskip 0.50003ptq$.   On the other hand,  $(z\hskip 1.00006ptf)\hskip 0.50003pt(v)\hskip 1.99997pt=\hskip 1.99997ptz\hskip 1.00006ptf\hskip 0.50003pt(v)\hskip 1.99997pt=\hskip 1.99997ptf\hskip 0.50003pt(\hskip 1.00006pt\overline{z}\hskip 1.99997ptv)$  by the definition of the product  $z\hskip 0.50003ptf$  and the anti-linearity of  $f$  and hence

[TABLE]

by the definition of  $p$  and  $q$.    It follows that  $p\hskip 1.99997pt=\hskip 1.99997ptz\hskip 0.50003ptq$,   i.e.  $(z\hskip 1.00006ptf)\cdot n\hskip 3.99994pt=\hskip 3.99994ptz\hskip 1.00006pt(f\cdot n)$.    $\blacksquare$

7. The  structure  of  $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$

The structure of a vector space over  $\mathbold{K}$  on  $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$.   The product  $z\hskip 0.50003ptf$  of an element  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$  with  $f\hskip 1.00006pt\in\hskip 1.00006pt\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$  is defined by the formula

[TABLE]

Obviously ,   this operation turns $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$ into a vector space over  $\mathbold{K}$.   Since the dimension of  $V$  over  $\mathbold{K}$  is  $1$,   an anti-linear map  $V\longrightarrow V$  is determined by its value on any given non-zero  $v\hskip 1.00006pt\in\hskip 1.00006ptV$.   It follows that the dimension of  $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$ over  $\mathbold{K}$  is also  $1$,   and hence any non-zero  $g\hskip 1.00006pt\in\hskip 1.00006pt\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$  is a basis of  $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$.

The composition  $(f\hskip 0.50003pt,\hskip 3.00003ptg)\hskip 1.00006pt\longmapsto\hskip 1.00006ptf\circ g$  is a  hermitian map

[TABLE]

in the sense that it is bilinear and   $\displaystyle(z\hskip 0.50003ptf)\circ(u\hskip 0.50003ptg)\hskip 3.99994pt=\hskip 3.99994ptz\hskip 1.00006pt\overline{u}\hskip 3.00003pt(f\circ g)\hskip 1.00006pt$   for all  $z\hskip 0.50003pt,\hskip 3.00003ptu\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$.

A canonical bilinear form on  $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$.   Given  $f\hskip 0.50003pt,\hskip 3.00003ptg\hskip 1.00006pt\in\hskip 1.00006pt\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$,    let

[TABLE]

where the trace is taken over $\mathbold{k}$.   Clearly ,  $\langle\hskip 0.50003ptf\hskip 0.50003pt,\hskip 3.00003ptg\hskip 0.50003pt\rangle$  is a bilinear form and since

[TABLE]

for all maps  $f\hskip 0.50003pt,\hskip 3.00003ptg\hskip 1.00006pt\colon\hskip 1.00006ptV\longrightarrow V$  linear over $\mathbold{k}$,   this bilinear form is symmetric.   Therefore

[TABLE]

is a quadratic form on  $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$  having the form  $\langle\hskip 0.50003ptf\hskip 0.50003pt,\hskip 3.00003ptg\hskip 0.50003pt\rangle$  as its polarization.

Reflections.   Let  $\sigma$  be the map $z\longrightarrow\overline{z}$.   Then  $\sigma$  is an anti-linear map  $\mathbold{K}\longrightarrow\mathbold{K}$  and  $\sigma\circ\sigma\hskip 1.99997pt=\hskip 1.99997pt\mathop{\mbox{id}}\nolimits_{\hskip 0.70004pt\mathbold{K}\hskip 0.35002pt}$.   Since  $\sigma\hskip 0.50003pt(a)\hskip 1.99997pt=\hskip 1.99997pta$  for  $a\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{k}$  and  $\sigma\hskip 0.50003pt(\rho)\hskip 1.99997pt=\hskip 1.99997pt-\hskip 1.99997pt\rho$  for  $\rho\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{k}^{\hskip 0.70004pt\perp}$,   the determinant  $\det\sigma$  over  $\mathbold{k}$  is equal to  $-\hskip 1.99997pt1$.

Since  $V$  is isomorphic to  $\mathbold{K}$  as a vector space over  $\mathbold{K}$,   it follows that there exist anti-linear maps  $g\hskip 1.00006pt\colon\hskip 1.00006ptV\longrightarrow V$  such that   $\displaystyle g\circ g\hskip 3.99994pt=\hskip 3.99994pt\mathop{\mbox{id}}\nolimits_{\hskip 0.70004ptV\hskip 0.35002pt}$   and   $\det\hskip 1.00006ptg\hskip 3.99994pt=\hskip 3.99994pt-\hskip 1.99997pt1$.   We will call any such anti-linear map  $V\longrightarrow V$  a  reflection.

7.1. Lemma.   *If   $f\hskip 1.00006pt\in\hskip 1.00006pt\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$,    then   $\displaystyle f\circ f\hskip 1.99997pt\hskip 3.99994pt=\hskip 3.99994pt\mathcal{D}\hskip 1.00006pt(\hskip 1.00006ptf\hskip 1.00006pt)\hskip 1.00006pt\mathop{\mbox{id}}\nolimits_{\hskip 0.70004ptV\hskip 0.35002pt}$   and   $\displaystyle\det\hskip 1.00006ptf\hskip 3.99994pt=\hskip 3.99994pt-\hskip 1.99997pt\mathcal{D}\hskip 1.00006pt(\hskip 1.00006ptf\hskip 1.00006pt)$,   where the determinant is taken over  $\mathbold{k}$.   *

Proof**.**     Let  $g\hskip 1.00006pt\in\hskip 1.00006pt\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$  be a reflection.   Then  $g\hskip 1.99997pt\neq\hskip 1.99997pt0$  and hence  $g$  is a basis of  $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$.    It follows that   $f\hskip 3.99994pt=\hskip 3.99994ptz\hskip 1.00006ptg$   for some  $z\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{K}$  and  hence

[TABLE]

Therefore   $\mathcal{D}\hskip 1.00006pt(\hskip 1.00006ptf\hskip 1.00006pt)\hskip 3.99994pt=\hskip 3.99994pt\operatorname{T{\hskip 0.24994pt}r}\hskip 0.50003pt\hskip 1.00006pt(\hskip 1.00006ptf\circ f\hskip 1.00006pt)/2\hskip 3.99994pt=\hskip 3.99994ptN\hskip 0.50003pt(z)$.    It follows that   $\displaystyle f\circ f\hskip 1.99997pt\hskip 3.99994pt=\hskip 3.99994pt\mathcal{D}\hskip 1.00006pt(\hskip 1.00006ptf\hskip 1.00006pt)\hskip 1.00006pt\mathop{\mbox{id}}\nolimits_{\hskip 0.70004ptV\hskip 0.35002pt}$.

On the other hand,   $f\hskip 3.99994pt=\hskip 3.99994ptm_{\hskip 0.70004ptz}\circ g$   and  hence Lemma  4. The  structure  of  a  $\mathbold{K}$-vector  space  on  $\mathbold{A}\hskip 0.50003pt(V)$  implies that

[TABLE]

Since we already proved that   $\mathcal{D}\hskip 1.00006pt(\hskip 1.00006ptf\hskip 1.00006pt)\hskip 3.99994pt=\hskip 3.99994ptN\hskip 0.50003pt(z)$,   it follows that   $\displaystyle\det\hskip 1.00006ptf\hskip 3.99994pt=\hskip 3.99994pt-\hskip 1.99997pt\mathcal{D}\hskip 1.00006pt(\hskip 1.00006ptf\hskip 1.00006pt)$.    $\blacksquare$

8. The  conformal  dilatation  and  Beltrami  forms

Pull-backs of quadratic forms.   Let $U\hskip 0.50003pt,\hskip 3.00003ptU^{\prime}$ be vector spaces over $\mathbold{k}$ and $f\hskip 1.00006pt\colon\hskip 1.00006ptU\longrightarrow U^{\prime}$ be a linear map.    Let  $q\hskip 1.00006pt\in\hskip 1.00006ptQ\hskip 0.50003pt(U^{\prime})$.    The  pull-back  $f^{\hskip 0.70004pt*}\hskip 0.50003ptq\hskip 1.00006pt\in\hskip 1.00006ptQ\hskip 0.50003pt(U)$  is the quadratic form

[TABLE]

The pull-backs  $f^{\hskip 0.70004pt*}\hskip 0.50003ptq$  lead to a linear map   $\displaystyle f^{\hskip 0.70004pt*}\hskip 1.99997pt\colon\hskip 1.99997ptQ\hskip 0.50003pt(U^{\prime})\hskip 1.99997pt\longrightarrow\hskip 1.99997ptQ\hskip 0.50003pt(U)$.   If $U^{\prime\prime}$ is one more vector space over $\mathbold{k}$ and  $g\hskip 1.00006pt\colon\hskip 1.00006ptU^{\prime}\longrightarrow U^{\prime\prime}$  is a linear map,   then  $\displaystyle(g\circ\hskip 0.50003ptf)^{*}\hskip 1.99997pt=\hskip 1.99997ptf^{\hskip 0.70004pt*}\circ\hskip 1.99997ptg^{\hskip 0.35002pt*}$.   If $f$ is an isomorphism,   then  $f^{\hskip 0.70004pt*}$  is an isomorphism also.

If  $p\hskip 1.99997pt\in\hskip 1.00006ptQ\hskip 0.50003pt(U^{\prime})$  and  $f^{\hskip 0.70004pt*}\hskip 1.00006ptp\hskip 1.99997pt\neq\hskip 1.99997pt0$,   then the conformal structure $[\hskip 1.00006ptf^{\hskip 0.70004pt*}\hskip 1.00006ptp\hskip 1.00006pt]$ depends only on $[\hskip 1.00006ptp\hskip 1.00006pt]$.   It is called the  pull-back  of  $p$  by  $f$  and is denoted by  $f^{\hskip 0.70004pt*}\hskip 1.00006pt[\hskip 1.00006ptp\hskip 1.00006pt]$.

The conformal dilatation.   Let $W$ be another vector space of dimension $1$ over $\mathbold{K}$ and let  $f\hskip 1.00006pt\colon\hskip 1.00006ptV\longrightarrow W$  be a non-zero map linear over $\mathbold{k}$.   The pull-back

[TABLE]

of the canonical conformal structure  $c_{\hskip 0.35002ptW}$  on  $W$  is called the  conformal dilatation  of  $f$.

The map  $f$  is called  exceptional  if the conformal structure  $\mathbold{c}\hskip 0.50003pt(f)$  is exceptional ,   and  regular  otherwise.   If  $f$  is regular ,   then we consider  $\mathbold{c}\hskip 0.50003pt(f)$  as an element of  $\mathbb{M}\hskip 0.50003pt(V)$.

If the field extension  $\mathbold{K}/\hskip 0.50003pt\hskip 0.24994pt\mathbold{k}$  resembles enough the classical case  $\mathbold{C}/\hskip 0.50003pt\mathbold{R}$,   then all non-zero $f$ are regular .   For example,   by using Theorem  10. Comparing   $\mathbold{c}\hskip 0.50003pt(f)$   and   $\mu_{\hskip 0.70004ptf}$  it is not hard to prove that this is the case  if  $\mathbold{k}$  is an ordered field and  $\mathbold{K}\hskip 3.00003pt=\hskip 3.00003pt\mathbold{k}\hskip 1.00006pt(\rho)$  with  $\rho^{\hskip 0.35002pt2}\hskip 1.99997pt<\hskip 1.99997pt0$.

The map  $f$  is called  conformal  if  the pull-back by  $f$  of the canonical conformal structure on  $W$  is equal to the canonical conformal structure on  $V$,   i.e.  if

[TABLE]

Since  $c_{\hskip 0.35002ptV}$  serves as the zero of  $\mathbb{M}\hskip 0.50003pt(V)$,   the map  $f$  is conformal if and only if  $f$  is regular and  $\mathbold{c}\hskip 0.50003pt(f)\hskip 1.99997pt=\hskip 1.99997pt0\hskip 1.00006pt\in\hskip 1.00006pt\mathbb{M}\hskip 0.50003pt(V)$.   In general,   the conformal dilatation of  $f$  is a natural measure of the distortion of the canonical conformal structure by $f$.

Linear maps over  $\mathbold{k}$  as sums of linear and anti-linear over  $\mathbold{K}$  maps.   Every map  $f\hskip 1.00006pt\colon\hskip 1.00006ptV\longrightarrow W$  linear over  $\mathbold{k}$  admits a unique presentation as a sum

[TABLE]

of a map  $Lf$  linear over  $\mathbold{K}$  and a map  $Af$  anti-linear over  $\mathbold{K}$.   Indeed,   let  $\rho\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{k}^{\hskip 0.70004pt\perp}$,  $\rho\hskip 1.99997pt\neq\hskip 1.99997pt0$.    If  $f\hskip 3.99994pt=\hskip 3.99994ptLf\hskip 1.99997pt+\hskip 1.99997ptAf$   is  such a presentation of  $f$,   then

[TABLE]

and hence

[TABLE]

for all  $v\hskip 1.00006pt\in\hskip 1.00006ptV$.   This proves uniqueness.   Conversely ,   one can define  $Lf$  and  $Af$  by the last displayed formulas.   Since  $\mathbold{k}^{\hskip 0.70004pt\perp}$  has dimension $1$ over $\mathbold{k}$,   this definition does not depend on the choice of $\rho$.   We leave to the reader the verification that so defined maps  $Lf$  and  $Af$  are linear and anti-linear over $\mathbold{K}$ respectively .

Beltrami forms.   Let  $f\hskip 1.00006pt\colon\hskip 1.00006ptV\longrightarrow W$  is a map linear over $\mathbold{k}$ such that  $Lf\hskip 1.99997pt\neq\hskip 1.99997pt0$.   Then

[TABLE]

is called the  Beltrami form  of  $f$.   The map  $f$  is linear over  $\mathbold{K}$  if and only if  $\mu_{\hskip 0.70004ptf}\hskip 1.99997pt=\hskip 1.99997pt0$.   In general,   the Beltrami form of  $f$  is a natural measure of deviation of  $f$  from being linear over  $\mathbold{K}$.   It is a direct generalization of the classical Beltrami forms from the theory of quasi-conformal mappings.   See,   for example,   [H],   Section  4.8.

9. An  identification  of   $\mathbb{M}\hskip 0.50003pt(V)$  with  $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$

An identification of  $\mathbb{M}\hskip 0.50003pt(V)$  with  $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$.   Every conformal structure in  $\mathbb{M}\hskip 0.50003pt(V)$  has the form  $[\hskip 0.50003ptn\hskip 1.99997pt+\hskip 1.99997ptq\hskip 0.50003pt]$  for some non-zero  $n\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{N}(V)$  and some  $q\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{A}\hskip 0.50003pt(V)$.    Let

[TABLE]

be the map defined by the rule   $\displaystyle[\hskip 0.50003ptn\hskip 1.99997pt+\hskip 1.99997ptq\hskip 0.50003pt]\hskip 1.99997pt\longmapsto\hskip 1.99997ptq/n$.   Since,   obviously ,   $\displaystyle a\hskip 0.50003ptq/a\hskip 0.50003ptn\hskip 3.99994pt=\hskip 3.99994ptq/n$   for all non-zero  $a\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{k}$,    this map is well defined.

Let us choose some  $n\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{N}(V)$,  $n\hskip 1.99997pt\neq\hskip 1.99997pt0$.    Let

[TABLE]

be the map defined by the rule   $\displaystyle f\hskip 1.99997pt\longmapsto\hskip 1.99997pt[\hskip 0.50003ptn\hskip 1.99997pt+\hskip 1.99997ptf\cdot n\hskip 0.50003pt]$.    Since,   obviously ,   $\displaystyle f\cdot(a\hskip 0.50003ptn)\hskip 3.99994pt=\hskip 3.99994pta\hskip 0.50003pt(f\cdot n)$   and hence   $\displaystyle[\hskip 0.50003pta\hskip 0.50003ptn\hskip 1.99997pt+\hskip 1.99997ptf\cdot(a\hskip 0.50003ptn)\hskip 0.50003pt]\hskip 3.99994pt=\hskip 3.99994pt[\hskip 0.50003pta\hskip 1.00006pt(n\hskip 1.99997pt+\hskip 1.99997ptf\cdot n)\hskip 0.50003pt]\hskip 3.99994pt=\hskip 3.99994pt[\hskip 0.50003ptn\hskip 1.99997pt+\hskip 1.99997ptf\cdot n\hskip 0.50003pt]$   for all non-zero  $a\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{k}$,   this map in fact does not depend on the choice of $n$.

Since the operations  $q\hskip 1.00006pt\longmapsto q/n$  and $f\hskip 1.00006pt\longmapsto f\cdot n$ are mutually inverse,   these two maps between  $\mathbb{M}\hskip 0.50003pt(V)$  and  $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$  are mutually inverse bijections.   Lemma  6. Anti-linear  maps  implies that the second map is linear over $\mathbold{K}$.   It follows that the first map is linear over $\mathbold{K}$ also,   and hence both of these maps are isomorphisms of vector spaces over $\mathbold{K}$.   We will identify  $\mathbb{M}\hskip 0.50003pt(V)$  and  $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$  by these isomorphisms.

The identification of the quadratic forms  $\mathbb{D}$  and  $\mathcal{D}$.   Let  $\det\hskip 1.00006pt\colon\hskip 1.00006ptQ\hskip 0.50003pt(V)\longrightarrow\mathbold{k}$  be a determinant map.   Let  $n\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{N}(V)$,  $n\hskip 1.99997pt\neq\hskip 1.99997pt0$.   Suppose that  $f\hskip 1.00006pt\in\hskip 1.00006pt\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$  and let  $q\hskip 1.99997pt=\hskip 1.99997ptf\cdot n$.   Then  $f$  is identified with the conformal class   $[\hskip 0.50003ptn\hskip 1.99997pt+\hskip 1.99997ptf\cdot n\hskip 0.50003pt]\hskip 3.99994pt=\hskip 3.99994pt[\hskip 0.50003ptn\hskip 1.99997pt+\hskip 1.99997ptq\hskip 0.50003pt]$.

By applying  (1.2)  to  $f$  and to  $n$  in the role of  $q$,   we see that   $\det\hskip 1.00006ptq\hskip 3.99994pt=\hskip 3.99994pt\det\hskip 1.00006ptf\cdot\det\hskip 1.00006ptn$.   Since   $\displaystyle\det\hskip 1.00006ptf\hskip 3.99994pt=\hskip 3.99994pt-\hskip 1.99997pt\mathcal{D}\hskip 1.00006pt(\hskip 1.00006ptf\hskip 1.00006pt)$   by Lemma  7. The  structure  of  $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$,    it follows that

[TABLE]

On the other hand,   by the definition  (5.1)  the quotient  $-\hskip 1.99997pt\det\hskip 1.00006ptq\hskip 1.00006pt/\det\hskip 1.00006ptn$  is nothing else but the value of  $\mathbb{D}$  on the conformal class  $[\hskip 0.50003ptn\hskip 1.99997pt+\hskip 1.99997ptq\hskip 0.50003pt]$  identified with  $f$.

It follows that the pull-back of the quadratic form  $\mathcal{D}$  by the map  (9.2)  is equal to  $\mathbb{D}$.   Since the map  (9.1)  is equal to the inverse of  (9.2),   this implies that also the pull-back of the quadratic form  $\mathbb{D}$  by the map  (9.1)  is equal to  $\mathcal{D}$.

In other words,   the identification of  $\mathbb{M}\hskip 0.50003pt(V)$  and  $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$  identifies the canonical quadratic forms  $\mathbb{D}$  and  $\mathcal{D}$  on these spaces.

10. Comparing   $\mathbold{c}\hskip 0.50003pt(f)$   and   $\mu_{\hskip 0.70004ptf}$

The framework.   Let  $V\hskip 0.50003pt,\hskip 3.00003ptW$  be two vector spaces over  $\mathbold{K}$,   and  let  $f\hskip 1.00006pt\colon\hskip 1.00006ptV\longrightarrow W$  be a map linear over  $\mathbold{k}$.   Suppose that  $Lf\hskip 1.99997pt\neq\hskip 1.99997pt0$  and let  $\mu\hskip 1.99997pt=\hskip 1.99997pt\mu_{\hskip 0.70004ptf}$  be the Beltrami form of $f$.

10.1. Lemma.   The conformal dilatation  $\mathbold{c}\hskip 0.50003pt(f)$  of  $f$  is equal to the conformal class of

[TABLE]

*for any non-zero  $n\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{N}(V)$.   *

Proof**.**     Since the conformal class of  $(\hskip 0.50003pt1\hskip 1.99997pt+\hskip 1.99997pt\mathcal{D}\hskip 0.50003pt(\mu)\hskip 0.50003pt)\hskip 1.00006ptn\hskip 3.00003pt+\hskip 3.00003pt2\hskip 1.00006pt\mu\cdot n$  does not depend on the choice of  $n\hskip 1.99997pt\neq\hskip 1.99997pt0$,   it is sufficient to prove the claim only for one particular  $n\hskip 1.99997pt\neq\hskip 1.99997pt0$.

Let  $m\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{N}(W)$,  $m\hskip 1.99997pt\neq\hskip 1.99997pt0$.   By the definition,  $\mathbold{c}\hskip 0.50003pt(f)$  is equal to the conformal class of  $f^{\hskip 0.70004pt*}\hskip 1.00006ptm$.   Let  $n\hskip 1.99997pt=\hskip 1.99997pt(\hskip 0.50003ptLf\hskip 0.50003pt)^{\hskip 0.35002pt*}\hskip 0.50003pt(m)$.   Then  $n\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{N}(V)$  because  $Lf$  is linear over  $\mathbold{K}$  and  $n\hskip 1.99997pt\neq\hskip 1.99997pt0$  because  $Lf\hskip 1.99997pt\neq\hskip 1.99997pt0$  and  $m$  is anisotropic.    It follows that   $\displaystyle(\hskip 0.50003ptLf^{\hskip 1.39998pt-\hskip 0.70004pt1}\hskip 0.50003pt)^{\hskip 0.35002pt*}\hskip 1.00006ptn\hskip 3.99994pt=\hskip 3.99994ptm$   and  hence

[TABLE]

[TABLE]

[TABLE]

Let  $p\hskip 1.99997pt=\hskip 1.99997ptf^{\hskip 0.70004pt*}\hskip 1.00006ptm$.    By the above calculation  $p\hskip 1.99997pt=\hskip 1.99997pt(\hskip 0.50003pt\mathop{\mbox{id}}\nolimits\hskip 1.99997pt+\hskip 1.99997pt\mu\hskip 0.50003pt)^{*}\hskip 1.00006ptn$  and  hence

[TABLE]

[TABLE]

[TABLE]

By Lemma  6. Anti-linear  maps  the map  $\mu$  is symmetric with respect to  $n$.   It follows that

[TABLE]

Lemma  7. The  structure  of  $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$  implies that   $\displaystyle\mu\circ\mu\hskip 1.99997pt\hskip 3.99994pt=\hskip 3.99994pt\mathcal{D}\hskip 1.00006pt(\hskip 1.00006pt\mu\hskip 1.00006pt)\hskip 1.00006pt\mathop{\mbox{id}}\nolimits_{\hskip 0.70004ptV\hskip 0.35002pt}$   and  hence

[TABLE]

It follows that   $\displaystyle f^{\hskip 0.70004pt*}\hskip 1.00006ptm\hskip 3.99994pt=\hskip 3.99994ptp\hskip 3.99994pt=\hskip 3.99994pt(\hskip 0.50003pt1\hskip 1.99997pt+\hskip 1.99997pt\mathcal{D}\hskip 0.50003pt(\mu)\hskip 0.50003pt)\hskip 1.00006ptn\hskip 3.00003pt+\hskip 3.00003pt2\hskip 1.00006pt\mu\cdot n$.   Since  $\mathbold{c}\hskip 0.50003pt(f)$  is equal to the conformal class of  $f^{\hskip 0.70004pt*}\hskip 1.00006ptm$,   this proves the lemma.    $\blacksquare$

10.2. Theorem.   If   $\displaystyle 1\hskip 1.99997pt+\hskip 1.99997pt\mathcal{D}\hskip 0.50003pt(\mu)\hskip 3.99994pt\neq\hskip 3.99994pt0$,    then  $f$  is regular ,   $\mathbold{c}\hskip 0.50003pt(f)\hskip 1.00006pt\in\hskip 1.99997pt\mathbb{M}\hskip 0.50003pt(V)$,    and

[TABLE]

*after the identification of  $\mathbb{M}\hskip 0.50003pt(V)$  with  $\operatorname{Hom}_{\hskip 0.35002pta}\hskip 0.50003pt(V,\hskip 1.99997ptV)$.   *

Proof**.**     If   $\displaystyle 1\hskip 1.99997pt+\hskip 1.99997pt\mathcal{D}\hskip 0.50003pt(\mu)\hskip 3.99994pt\neq\hskip 3.99994pt0$,   then the quadratic form  (10.1)  does not belong to  $\mathbold{A}\hskip 0.50003pt(V)$  and the conformal class of the form  (10.1)  is equal to the conformal class of

[TABLE]

In view of Lemma  10. Comparing   $\mathbold{c}\hskip 0.50003pt(f)$   and   $\mu_{\hskip 0.70004ptf}$,   this implies that   $\mathbold{c}\hskip 0.50003pt(f)\hskip 1.00006pt\in\hskip 1.99997pt\mathbb{M}\hskip 0.50003pt(V)$   and  $\mathbold{c}\hskip 0.50003pt(f)$ is equal to the conformal class of  (10.2).   It remains to notice that conformal class is identified with

[TABLE]

10.3. Theorem.   If   $\displaystyle 1\hskip 1.99997pt+\hskip 1.99997pt\mathcal{D}\hskip 0.50003pt(\mu)\hskip 3.99994pt=\hskip 3.99994pt0$,    then  $f$  is exceptional ,   $\mathbold{c}\hskip 0.50003pt(f)\hskip 1.00006pt\in\hskip 1.99997pt\mathbb{P}\mathbold{A}\hskip 0.50003pt(V)$,    and

[TABLE]

*for any non-zero  $n\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{N}(V)$.   *

Proof**.**     Since  $2\hskip 1.00006pt\mu\cdot n\hskip 1.00006pt\in\hskip 1.00006pt\mathbold{A}\hskip 0.50003pt(V)$  by Lemma  6. Anti-linear  maps,   this follows from Lemma  10. Comparing   $\mathbold{c}\hskip 0.50003pt(f)$   and   $\mu_{\hskip 0.70004ptf}$.    $\blacksquare$