hyperbolic plane in quadratic spaces

A non-singular (http://planetmath.org/NonDegenerateQuadraticForm) isotropic quadratic space $\mathscr{H}$ of dimension2 (over a field) is called a hyperbolic plane. In otherwords, $\mathscr{H}$ is a 2-dimensional vector space over a fieldequipped with a quadratic form $Q$ such that there exists a non-zerovector $v$ with $Q(v)=0$.

Examples. Fix the ground field to be $ℝ$, and${ℝ}^2$ be the two-dimensional vector space over $ℝ$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 $a,b\in ℝ$. $({ℝ}^2,Q_1)$ is a hyperbolicplane. When $Q_1$ is written in matrix form, we have

2. Let $Q_2(r,s)=r^2-s^2$. Then $Q_2(a,a)=0$ for all $a\in ℝ$. $({ℝ}^2,Q_2)$ is a hyperbolicplane. As above, $Q_2$ can be written in matrix form:

From the above examples, we see that the name “hyperbolic plane”comes from the fact that the associated quadratic form resembles theequation of a hyperbola in a two-dimensional Euclidean plane.

It’s not hard to see that the two examples above are equivalentquadratic 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:

In fact, we have the following

Proposition. Any two hyperbolic planes over a field $k$ ofcharacteristic 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 $ℝ$ to $k$), we are done.

Pick a non-zero vector $u\in \mathscr{H}$ and suppose it isisotropic: $Q(u)=0$. Pick another vector $v\in \mathscr{H}$ so$\{u,v\}$ forms a basis for $\mathscr{H}$. Let $B$ bethe 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 non-singular. 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 $𝔟=\{u,v\}$ is

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

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

∎

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 $⟨1⟩\perp ⟨-1⟩$, or simply $⟨1,-1⟩$.

A hyperbolic space is a finite dimensional orthogonal direct sum of hyperbolic planes. It is always even dimensional and has the notation$⟨1,-1,1,-1,\mathrm{\dots },1,-1⟩$ or simply $n⟨1⟩\perp n⟨-1⟩$, 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.