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hyperbolic plane in quadratic spaces
A nonsingular isotropic quadratic space $\mathcal{H}$ of dimension 2 (over a field) is called a hyperbolic plane. In other words, $\mathcal{H}$ 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
$Q_{1}(x,y)=\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}0&\frac{1}{2}\\ \frac{1}{2}&0\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}x&y\end{pmatrix}M(Q_{1})\begin{pmatrix}x\\ y\end{pmatrix}.$
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 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:
$\begin{pmatrix}1&1\\ 1&1\end{pmatrix}M(Q_{1})\begin{pmatrix}1&1\\ 1&1\end{pmatrix}=\begin{pmatrix}1&1\\ 1&1\end{pmatrix}\begin{pmatrix}0&\frac{1}{2}\\ \frac{1}{2}&0\end{pmatrix}\begin{pmatrix}1&1\\ 1&1\end{pmatrix}=\begin{pmatrix}1&0\\ 0&1\end{pmatrix}=M(Q_{2}).$
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 $xy$ 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 nonzero 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 nonsingular. 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 $\langle 1\rangle\bot\langle1\rangle$, 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 $n\langle 1\rangle\bot n\langle1\rangle$, 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 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.
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