hyperbolic plane in quadratic spaces
A nonsingular^{} (http://planetmath.org/NonDegenerateQuadraticForm) isotropic quadratic space $\mathscr{H}$ of dimension^{} 2 (over a field) is called a hyperbolic plane. In other words, $\mathscr{H}$ is a 2dimensional vector space^{} over a field equipped with a quadratic form^{} $Q$ such that there exists a nonzero vector $v$ with $Q(v)=0$.
Examples. Fix the ground field to be $\mathbb{R}$, and ${\mathbb{R}}^{2}$ be the twodimensional vector space over $\mathbb{R}$ with the standard basis $(0,1)$ and $(1,0)$.

1.
Let ${Q}_{1}(x,y)=xy$. Then ${Q}_{1}(a,0)={Q}_{1}(0,b)=0$ for all $a,b\in \mathbb{R}$. $({\mathbb{R}}^{2},{Q}_{1})$ is a hyperbolic plane. When ${Q}_{1}$ is written in matrix form, we have
${Q}_{1}(x,y)=\left(\begin{array}{cc}\hfill x\hfill & \hfill y\hfill \end{array}\right)\left(\begin{array}{cc}\hfill 0\hfill & \hfill \frac{1}{2}\hfill \\ \hfill \frac{1}{2}\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill x\hfill & \hfill y\hfill \end{array}\right)M({Q}_{1})\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right).$

2.
Let ${Q}_{2}(r,s)={r}^{2}{s}^{2}$. Then ${Q}_{2}(a,a)=0$ for all $a\in \mathbb{R}$. $({\mathbb{R}}^{2},{Q}_{2})$ is a hyperbolic plane. As above, ${Q}_{2}$ can be written in matrix form:
${Q}_{1}(x,y)=\left(\begin{array}{cc}\hfill x\hfill & \hfill y\hfill \end{array}\right)\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill x\hfill & \hfill y\hfill \end{array}\right)M({Q}_{2})\left(\begin{array}{c}\hfill x\hfill \\ \hfill y\hfill \end{array}\right).$
From the above examples, we see that the name “hyperbolic plane” comes from the fact that the associated quadratic form resembles the equation of a hyperbola^{} in a twodimensional Euclidean plane^{}.
It’s not hard to see that the two examples above are equivalent quadratic forms. To transform from the first form to the second, for instance, follow the linear substitutions $x=rs$ and $y=r+s$, or in matrix form:
$\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right)M({Q}_{1})\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right)\left(\begin{array}{cc}\hfill 0\hfill & \hfill \frac{1}{2}\hfill \\ \hfill \frac{1}{2}\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{cc}\hfill 1\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 1\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)=M({Q}_{2}).$
In fact, we have the following
Proposition^{}. Any two hyperbolic planes over a field $k$ of characteristic not 2 are isometric quadratic spaces.
Proof.
From the first example above, we see that the quadratic space with the quadratic form $xy$ is a hyperbolic plane. Conversely, if we can show that any hyperbolic plane $\mathscr{H}$ is isometric the example (with the ground field switched from $\mathbb{R}$ to $k$), we are done.
Pick a nonzero vector $u\in \mathscr{H}$ and suppose it is isotropic: $Q(u)=0$. Pick another vector $v\in \mathscr{H}$ so $\{u,v\}$ forms a basis for $\mathscr{H}$. Let $B$ be the symmetric bilinear form^{} associated with $Q$. If $B(u,v)=0$, then for any $w\in \mathscr{H}$ with $w=\alpha u+\beta v$, $B(u,w)=\alpha B(u,u)+\beta B(u,v)=0$, contradicting the fact that $\mathscr{H}$ is nonsingular. So $B(u,v)\ne 0$. By dividing $v$ by $B(u,v)$, we may assume that $B(u,v)=1$.
Suppose $\alpha =B(v,v)$. Then the matrix associated with the quadratic form $Q$ corresponding to the basis $\U0001d51f=\{u,v\}$ is
${M}_{\U0001d51f}(Q)=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill \alpha \hfill \end{array}\right).$
If $\alpha =0$ then we are done, since ${M}_{\U0001d51f}(Q)$ is equivalent^{} to ${M}_{\U0001d51f}({Q}_{1})$ via the isometry^{} $T:\mathscr{H}\to \mathscr{H}$ given by
$T=\left(\begin{array}{cc}\hfill \frac{1}{2}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\text{, so that}{T}^{t}\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right)T=\left(\begin{array}{cc}\hfill 0\hfill & \hfill \frac{1}{2}\hfill \\ \hfill \frac{1}{2}\hfill & \hfill 0\hfill \end{array}\right).$
If $\alpha \ne 0$, then the trick is to replace $v$ with an isotropic vector $w$ so that the bottom right cell is also 0. Let $w=\frac{\alpha}{2}u+v$. It’s easy to verify that $Q(w)=0$. As a result, the isometry $S$ required has the matrix form
$S=\left(\begin{array}{cc}\hfill 1\hfill & \hfill \frac{\alpha}{2}\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\text{, so that}{S}^{t}\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill \alpha \hfill \end{array}\right)S=\left(\begin{array}{cc}\hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill 0\hfill \end{array}\right).$
∎
Thus we may speak of the hyperbolic plane over a field without any ambiguity, and we may identify the hyperbolic plane with either of the two quadratic forms $xy$ or ${x}^{2}{y}^{2}$. Its notation, corresponding to the second of the forms, is $\u27e81\u27e9\perp \u27e81\u27e9$, or simply $\u27e81,1\u27e9$.
A hyperbolic space is a finite dimensional orthogonal direct sum of hyperbolic planes. It is always even dimensional and has the notation $\u27e81,1,1,1,\mathrm{\dots},1,1\u27e9$ or simply $n\u27e81\u27e9\perp n\u27e81\u27e9$, where $2n$ is the dimensional of the hyperbolic space.
Remarks.

•
The notion of the hyperbolic plane encountered in the theory of quadratic forms is different from the “hyperbolic plane”, a 2dimensional space of constant negative curvature^{} (Euclidean^{} signature^{}) that is commonly used in differential geometry^{}, and in nonEuclidean geometry.

•
Instead of being associated with a quadratic form, a hyperbolic plane is sometimes defined in terms of an alternating form. In any case, the two definitions of a hyperbolic plane coincide if the ground field has characteristic 2.
Title  hyperbolic plane in quadratic spaces 

Canonical name  HyperbolicPlaneInQuadraticSpaces 
Date of creation  20130322 15:41:47 
Last modified on  20130322 15:41:47 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  9 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 11E88 
Classification  msc 15A63 
Defines  hyperbolic plane 
Defines  hyperbolic space 