# 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 $(0,1)$ and $(1,0)$.

1. 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. 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:

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

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

. 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 $\mathcal{H}$ 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 $xy$ 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.

• 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 HyperbolicPlaneInQuadraticSpaces 2013-03-22 15:41:47 2013-03-22 15:41:47 CWoo (3771) CWoo (3771) 9 CWoo (3771) Definition msc 11E88 msc 15A63 hyperbolic plane hyperbolic space