# Geometry behind calling the pair hyperbolic in bilinear space

The name hyperbola is originally comes from an object in geometry, but several other objects in mathematics wear partly the name hyperbola or hyperbolic; for example, hyperbolic Mobius transformation. There are some geometric reasons for calling them hyperbolic.

In the study of symplectic spaces (i.e. a vector space $V$ with a biliear form $(\cdot,\cdot)$ with $(u,u)=0$), a pair of vectors $\{u,v\}$ is called hyperbolic if $(u,v)=1$.

However, the books (or notes or sites) which give this definition of hyperbolic pair do not mention the geometric reason behind hyperbolic.

I tried to search in the books with title Linear algebra and geometry (lot of books are there with this title) the reason for it, but I didn't succeed! [I didn't even find it in some books on Linear algebra written by famous geometers - Sahafarevich, Dieudonne,...]

Can one explain a little the reason behind calling the pair $\{u,v\}$ with $(u,v)=1$ hyperbolic?

• The term is motivated by the case of symmetric bilinear forms. If $B$ is a symmetric bilinear form and there are vectors $u$ and $v$ such that $B(u,u) = 0$, $B(v,v) = 0$, and $B(u,v) = 1$ then $u$ and $v$ are linearly independent, so they span a 2-dimensional subspace. For scalars $x$ and $y$, $B(xu+yv,xu+yv) = x^2B(u,u) + xy(B(u,v) + B(v,u)) + y^2B(v,v) = 2xy$, and that formula looks like the expression for a hyperbola. (Changing $v$ to $v/2$ makes the formula $xy$, and by another linear change of variables the formula becomes $x^2-y^2$.) Such a 2-dim. subspace is called a hyperbolic plane. – KCd Dec 21 '16 at 12:42
• Ok; so a little modification: Consider the set $H=\{(v_1,v_2)\in V\times V: B(v_1,v_2)=1\}$; this is non-empty, since it contains $u,v$ and the possible linear combination of $u,v$ which are in $H$ give locus of hyperbola, so the pair $\{u,v\}$ is hyperbolic pair; am I right? – p Groups Dec 21 '16 at 12:51
• The corresponding "elliptic" question. – J. M. is a poor mathematician Dec 27 '16 at 18:14

There are three types of second order curves in $\Bbb R^2$:

1. ellips (in particular, circle), e.g. $x^2+y^2=1$,
2. hyperbola, e.g. $x^2-y^2=1$,
3. parabola, e.g. $y=x^2$.

In general, all three types can be represented as level curves $f(x,y)=\text{const}$ of a quadratic function $$f(x,y)=\begin{bmatrix}x\\y\end{bmatrix}^TH\begin{bmatrix}x\\y\end{bmatrix}+c^T\begin{bmatrix}x\\y\end{bmatrix}.$$ One gets (see also discriminant of a conic section)

1. ellips iff $H$ has two eigenvalues of the same sign,
2. hyperbola iff $H$ has two eigenvalues of opposite signs,
3. parabola iff $H$ has one zero eigenvalue and $\operatorname{rank}([H\ c])>\operatorname{rank}(H)$.

Traditionally, mathematical objects that are described by a quadratic form inherit the same names depending on the signature of the corresponding matrix $H$ (elliptic/hyperbolic/parabolic PDE, elliptic/hyperbolic geometry etc)

In your case, a hyperbolic pair are two linear independent vectors $u$, $v$ with those particular properties. If the "distance" is defined by the associated quadratic form $q(w)=B(w,w)=w^THw$ then the unit "circle" $q(xu+yv)=1$ becomes $2xy=1$, which is a hyperbola. Note that the matrix $H$ must be indefinite, in particular, in the plane $uv$ the quadratic form with the matrix $\begin{bmatrix}0 &1\\1 & 0\end{bmatrix}$ having one positive and one negative eigenvalues, i.e. the subspace $uv$ with the metric $q$ is hyperbolic.

• Very good, and thanks for explanation; I still didn't get a single explanation statement in any book whether it is on linear algebra; or linear algebra and geometry, or quadratic forms or bilinear forms; now many things are clear! – p Groups Dec 28 '16 at 5:46