hyperbolic plane in quadratic spaces

A non-singular (http://planetmath.org/NonDegenerateQuadraticForm) isotropic quadratic space $\mathcal{H}$ of dimension 2 (over a field) is called a hyperbolic plane. In other words, $\mathcal{H}$ is a 2-dimensional vector space over a field equipped with a quadratic form $Q$ such that there exists a non-zero vector $v$ with $Q(v)=0$ .

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 $(0,1)$ 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

$Q_{1}(x,y)=\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}0&\frac{1}{2}\\ \frac{1}{2}&0\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}x&y\end{pmatrix}M(Q_{1})\begin{pmatrix}x\\ y\end{pmatrix}.$
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:

$Q_{1}(x,y)=\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}x&y\end{pmatrix}M(Q_{2})\begin{pmatrix}x\\ y\end{pmatrix}.$

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 $x=r-s$ and $y=r+s$ , or in matrix form:

$\begin{pmatrix}1&1\\ -1&1\end{pmatrix}M(Q_{1})\begin{pmatrix}1&-1\\ 1&1\end{pmatrix}=\begin{pmatrix}1&1\\ -1&1\end{pmatrix}\begin{pmatrix}0&\frac{1}{2}\\ \frac{1}{2}&0\end{pmatrix}\begin{pmatrix}1&-1\\ 1&1\end{pmatrix}=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}=M(Q_{2}).$

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 $x y$ 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 $k$ ), we are done.

Pick a non-zero vector $u\in\mathcal{H}$ and suppose it is isotropic: $Q(u)=0$ . Pick another vector $v\in\mathcal{H}$ so $\{u,v\}$ forms a basis for $\mathcal{H}$ . Let $B$ be the symmetric bilinear form associated with $Q$ . If $B(u,v)=0$ , 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 $v$ 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 $\mathfrak{b}=\{u,v\}$ 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

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

If $\alpha\neq 0$ , then the trick is to replace $v$ 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

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

∎

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 $x y$ or $x^{2}-y^{2}$ . 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 $2n$ 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 2-dimensional space of constant negative curvature (Euclidean signature) that is commonly used in differential geometry, and in non-Euclidean 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	2013-03-22 15:41:47
Last modified on	2013-03-22 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