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?

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    $\begingroup$ 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. $\endgroup$ – KCd Dec 21 '16 at 12:42
  • $\begingroup$ 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? $\endgroup$ – p Groups Dec 21 '16 at 12:51
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    $\begingroup$ The corresponding "elliptic" question. $\endgroup$ – 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.

  • $\begingroup$ 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! $\endgroup$ – p Groups Dec 28 '16 at 5:46

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