hyperbolic pairs and basis

Definition 1.

Given a reflexive non-degenerate sesquilinear form b:V×Vk, a hyperbolic pair is a pair e,fV such that

b(e,e)=0=b(f,f) and b(e,f)=1.

The span of a hyperbolic pair is a hyperbolic line (recall that a line refers to the projective dimension thus we have a 2-dimensional subspacePlanetmathPlanetmathPlanetmath but a 1-dimensional projective space).

Definition 2.

A hyperbolic basis for a vector spaceMathworldPlanetmath V with respect to a reflexive non-degenerate sesquilinear form b is a basis {e1,f1,,em,fm} where

b(ei,ej)=0=b(fi,fj) and b(ei,fj)=δij.

Thus a hyperbolic basis is a basis composed of hyperbolic pairs. Furthermore, if V has a hyperbolic basis then setting Hi=ei,fi shows


where XY=XY with the added condition b(X,Y)=0.

Hyperbolic bases are the foundation of a “standard basis” for a vector spaces V equipped with a reflexive non-degenerate sesquilinear form.

1 Symmetric pairs

A symmetricPlanetmathPlanetmath hyperbolic pair is a hyperbolic pair e,f for which b restricted to L=e,f is a symmetric bilinear formMathworldPlanetmath. This requires the additional condition that b(e,f)=1=b(f,e).

This means that the form restricted the hyperbolic line L=e,f can be represented by the matrix


When 1/2k we can consider the associated quadratic formMathworldPlanetmath


so if v=xe+yf we arrive at the polynomial


Suppose the field is . Then we can associateMathworldPlanetmath a graph to the equations c=q(xe+yf)=xy for any fixed c. If c=0 then x=0 or y=0 so the graph is the x and y-axis – also called the degenerate hyperbola. If c0 then x0 and so y=cx. This is the graph of a a hyperbola, hence the title of a hyperbolic pair.

Symmetric bilinear maps are often preferred to be presented as diagonal matricesMathworldPlanetmath so that they reflect the content of Sylvester’s Law of Inertia. When 1/2k (characteristicPlanetmathPlanetmath of k is not 2) we can diagonalize any symmetric hyperbolic pair as follows:


That is, we can change the basis to


Then b(u,u)=1, b(v,v)=-1, and b(u,v)=b(v,u)=0. Alternatively we find under this basis we have the quadratic form q(xu+yv)=x2-y2 which is also seen as the standard equation of a hyperbola.

If we think of a quadratic form as generalizing norms – that is length, then we are observing that on a hyperbolic line length is not Euclidean, in fact, as the usual Euclidean length of (x,y), x2+y2, gets large, the associated hyperbolic length get small: x2-y2 may get small, even 0 or negative. Thus the curvature of this space is negative (consider the graph of z=x2-y2 which is a saddle.)

All symmetric hyperbolic pairs are isometric so decomposing a bilinear formPlanetmathPlanetmath into the radicalPlanetmathPlanetmathPlanetmath plus hyperbolic pairs plus any left over anisotropic complement produces a standard basis which allows for easy comparison of one symmetric bilinear form to another.

2 Alternating pairs

An alternating hyperbolic pair is a hyperbolic pair e,f for which b restricted to L=e,f is an alternating bilinear form. This requires the additional condition that b(e,f)=1=-b(f,e).

This means that the form restricted the hyperbolic line L=e,f can be represented by the matrix


Although we do not associate a quadratic form with an alternating bilinear (since b(v,v)=0 for all vV) we can still derive the equations of a hyperbola. Specifically


So again setting c=b(xe,yf)=xy we observe the various hyperbola graphs.

Alternating hyperbolic pairs cannot be diagonalized as every element vV has b(v,v)=0.

All alternating bilinear forms decompose into hyperbolic lines and the radical and any two alternating hyperbolic lines are isometric and thus simply indicating the number of hyperbolic pairs in an alternating bilinear form specifies the form uniquely. If we further insist the form is non-degenerate then the dimensionPlanetmathPlanetmath of the vector space specifies the form completely.

3 Hermitian pairs

A is a hyperbolic pair e,f for which b restricted to L=e,f is an Hermitian bilinear form. This requires the additional condition that b(e,f)=1=b(f,e). The associated matrix does not reveal much difference from the symmetric as we still obtain


What is different is how the matrix is used to compute the bilinear products:


So if we compute b(v,v) we find:


We see from this that two hyperbolic pairs of Hermitian type need not be isometric unless we further consider the automorphismPlanetmathPlanetmathPlanetmath of the two forms.

4 Characteristic 2

Hyperbolic pairs over fields of characteristic 2 are a special breed because they are at the same time symmetric and alternating. That is, the form is the matrix:


Thus the form cannot be diagonalized as it is alternating. Here it is generally more useful to use a quadratic form then the bilinear form. Unfortunately because we cannot recover the quadratic form from the bilinear form on account that b(v,v)=0, such a quadratic form must be provided externally from some other method. Thus it is not always feasible.

Title hyperbolic pairs and basis
Canonical name HyperbolicPairsAndBasis
Date of creation 2013-03-22 15:50:59
Last modified on 2013-03-22 15:50:59
Owner Algeboy (12884)
Last modified by Algeboy (12884)
Numerical id 13
Author Algeboy (12884)
Entry type Definition
Classification msc 15A99
Classification msc 15A63
Synonym symplectic pair
Defines hyperbolic pair
Defines hyperbolic basis