PlanetMath: hyperbolic plane in quadratic spaces

Examples. Fix the ground field to be $\mathbb{R}$ , and $\mathbb{R}^2$ be the two-dimensional vector space over $\mathbb{R}$ with the standard basis

and

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 two-dimensional 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

and

, or in matrix form:

Proof. From the first example above, we see that the quadratic space with the quadratic form

is a hyperbolic plane. Conversely, if we can show that any hyperbolic plane $\mathcal{H}$ is isometric the example (with the ground field switched from $\mathbb{R}$ to

), we are done.

Pick a non-zero vector $u\in\mathcal{H}$ and suppose it is isotropic: . Pick another vector $v\in\mathcal{H}$ so $\lbrace u,v\rbrace$ forms a basis for $\mathcal{H}$ . Let be the symmetric bilinear form associated with . If , then for any $w\in\mathcal{H}$ with $w=\alpha u+\beta v$ , $B(u,w)= \alpha B(u,u)+\beta B(u,v)=0$ , contradicting the fact that $\mathcal{H}$ is non-singular. So $B(u,v)\neq 0$ . By dividing by , we may assume that .

Suppose $\alpha=B(v,v)$ . Then the matrix associated with the quadratic form corresponding to the basis $\mathfrak{b}=\lbrace u,v\rbrace$ is

$M_{\mathfrak{b}}(Q)= \begin{pmatrix} 0 & 1 \ 1 & \alpha \end{pmatrix}.$

If $\alpha=0$ then we are done, since $M_{\mathfrak{b}}(Q)$ is equivalent to $M_{\mathfrak{b}}(Q_1)$ via the isometry $T:\mathcal{H}\to\mathcal{H}$ given by

$T= \begin{pmatrix} \frac{1}{2} & 0 \ 0 & 1 \end{pmatrix}$ , so that $T^t \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}T= \begin{pmatrix} 0 & \frac{1}{2} \ \frac{1}{2} & 0 \end{pmatrix}.$

If $\alpha\neq 0$ , then the trick is to replace with an isotropic vector so that the bottom right cell is also 0. Let $w=-\frac{\alpha}{2}u+v$ . It's easy to verify that . As a result, the isometry required has the matrix form

$S= \begin{pmatrix} 1 & -\frac{\alpha}{2} \ 0 & 1 \end{pmatrix}$ , so that $S^t \begin{pmatrix} 0 & 1 \ 1 & \alpha \end{pmatrix}S= \begin{pmatrix} 0 & 1 \ 1 & 0 \end{pmatrix}.$

$\qedsymbol$

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

. Its notation, corresponding to the second of the forms, is $\langle 1\rangle \bot\langle-1\rangle$ , or simply $\langle 1,-1\rangle$ .

A hyperbolic space is a finite dimensional orthogonal direct sum of hyperbolic planes. It is always even dimensional and has the notation $\langle 1,-1,1,-1,\ldots,1,-1\rangle$ or simply $n\langle 1\rangle\bot n\langle -1\rangle$ , where

is the dimensional of the hyperbolic space.