hyperbolic pairs and basis

Definition 1.

Given a reflexive non-degenerate sesquilinear form $b\mathrm{:}V\mathrm{\times }V\mathrm{\to }k$,a hyperbolic pair is a pair $e\mathrm{,}f\mathrm{\in }V$ such that

$$b(e,e)=0=b(f,f)\text{ and }b(e,f)=1.$$

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 subspace but a 1-dimensional projective space).

Definition 2.

A hyperbolic basis for a vector space $V$ with respect to a reflexivenon-degenerate sesquilinear form $b$ is a basis$\mathrm{\{}{e}_{\mathrm{1}}\mathrm{,}{f}_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{e}_m\mathrm{,}{f}_m\mathrm{\}}$ where

$$b(e_i,e_j)=0=b(f_i,f_j)\text{ and }b(e_i,f_j)={\delta }_{ij}.$$

Thus a hyperbolic basis is a basis composed of hyperbolic pairs. Furthermore,if $V$ has a hyperbolic basis then setting $H_i=⟨e_i,f_i⟩$ shows

$$V=H_1⟂H_2⟂\mathrm{\cdots }⟂H_m$$

where $X⟂Y=X\oplus Y$ 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 symmetric hyperbolic pair is a hyperbolic pair $e,f$ for which $b$ restricted to $L=⟨e,f⟩$ is a symmetric 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

When $1/2\in k$ we can consider the associated quadratic form

$$q(v)=\frac{1}{2}b(v,v)$$

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

Suppose the field is $ℝ$. Then we can associate a graph to theequations $c=q(xe+yf)=xy$ for any fixed $c\in ℝ$. If $c=0$ then$x=0$ or $y=0$ so the graph is the $x$ and $y$-axis – also called thedegenerate hyperbola. If $c\ne 0$ then $x\ne 0$ and so $y=\frac{c}{x}$.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 diagonalmatrices so that they reflect the content of Sylvester’s Law of Inertia.When $1/2\in k$ (characteristic of $k$ is not 2) we candiagonalize any symmetric hyperbolic pair as follows:

That is, we can change the basis to

$$e\mapsto u:=\frac{1}{2}e+f,f\mapsto v:=-\frac{1}{2}e+f.$$

Then $b(u,u)=1$, $b(v,v)=-1$, and $b(u,v)=b(v,u)=0$. Alternatively we findunder this basis we have the quadratic form $q(xu+yv)=x^2-y^2$ which is alsoseen as the standard equation of a hyperbola.

If we think of a quadratic form as generalizing norms – that is length, thenwe are observing that on a hyperbolic line length is not Euclidean, in fact,as the usual Euclidean length of $(x,y)$, $x^2+y^2$, gets large, the associated hyperbolic length get small: $x^2-y^2$ may get small, even 0 or negative. Thus the curvature of this space is negative (consider the graph of $z=x^2-y^2$ which is a saddle.)

All symmetric hyperbolic pairs are isometric so decomposing a bilinear form intothe radical plus hyperbolic pairs plus any left over anisotropic complement producesa 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 $b$ restrictedto $L=⟨e,f⟩$ is an alternating bilinear form. This requires theadditional 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 $v\in V$) we can still derive the equations ofa hyperbola. Specifically

$$b(xe,yf)=xy.$$

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

Alternating hyperbolic pairs cannot be diagonalized as every element$v\in V$ has $b(v,v)=0$.

All alternating bilinear forms decompose into hyperbolic lines and the radical andany two alternating hyperbolic lines are isometric and thus simply indicating the numberof hyperbolic pairs in an alternating bilinear form specifies the form uniquely. If wefurther insist the form is non-degenerate then the dimension of the vector spacespecifies 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 theadditional condition that $b(e,f)=1=b(f,e)$. The associated matrix does not revealmuch 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 isometricunless we further consider the automorphism of the two forms.

4 Characteristic 2

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

Thus the form cannot be diagonalized as it is alternating. Here it is generallymore useful to use a quadratic form then the bilinear form. Unfortunatelybecause we cannot recover the quadratic form from the bilinear formon account that $b(v,v)=0$, such a quadratic form must be provided externallyfrom some other method. Thus it is not always feasible.