Quadric

 ( Projective Geometry ) 

Quadric surfaces have the nice property that they also transform under affine coordinate transformations. If a quadric surface, $\{\backslash bold\; M\}$, is moved to a coordinate location and orientation $C\_1$, its description as seen from the original coordinate system $C\_1$ is$$\{\backslash bold\; M\}\_\{moved\; \backslash ;\; to\}\; =\; \{\backslash bold\; \{P\_\{0\; 1\}\}\}^T\; \{\backslash bold\; M\}\; \backslash ;\; \{\backslash bold\; \{P\_\{0\; 1\}\}\}$$However, this needs to be distinguished from a quadric surface defined in coordinate system $C\_0$ as viewed from a coordinate system $C\_1$, where the transformation is

$\{\backslash bold\; M\}\_\{viewed\; \backslash ;\; from\}\; =\; \{\backslash bold\; \{P\_\{1\; 0\}\}\}^T\; \{\backslash bold\; M\}\; \backslash ;\; \{\backslash bold\; \{P\_\{1\; 0\}\}\}$

Note that since $\{\backslash bold\; M\}^T\; =\; \{\backslash bold\; M\}$, it does not matter whether or not this matrix is transposed, and one may see it either way in the literature.

When the Earth is better approximated as an ellipsoid, the longitudes are still the same planes as for the spherical Earth. Latitudes are still cones, but the points of the cones are no longer at the center of the Earth, except at the equator; the cone axis is still along the Earth's axis, but north or south of the Earth center. Altitudes are ellipsoids. In the World Geodetic System 1984 (WGS84), the equatorial radius of the Earth is ($G\_a$) is given as 6,378,137.0 meters and the polar radius ($G\_b$) as 6,356,752.3142 meters. The squared first eccentricity is then$$\{G\_e\}^2\; =\; \backslash frac\; \{1+\; t\_1^2+\; \backslash cdots\; +t\_\{n-1\}^2\}\backslash \backslash \; x\_n\; =\backslash frac\{1-\; t\_1^2-\; \backslash cdots\; -t\_\{n-1\}^2\}\{1+\; t\_1^2+\; \backslash cdots\; +t\_\{n-1\}^2\}.\; \backslash end\{cases\}$$

By homogenizing, one obtains the projective parametrization

$\backslash begin\{cases\}$

A straightforward verification shows that this induces a bijection between the points of the quadric such that $X\_n\backslash neq\; -X\_0$ and the points such that $T\_n\backslash neq\; 0$ in the projective space of the parameters. On the other hand, all values of $(T\_1,\backslash ldots,\; T\_n)$ such that $T\_n=0$ and $T\_1^2+\; \backslash cdots\; +T\_\{n-1\}^2\backslash neq\; 0$ give the point $A.$

In the case of conic sections ($n=2$), there is exactly one point with $T\_n=0.$ and one has a bijection between the circle and the projective line.

For $n>2,$ there are many points with $T\_n=0,$ and thus many parameter values for the point $A.$ On the other hand, the other points of the quadric for which $X\_n=-X\_0$ (and thus $x\_n=-1$) cannot be obtained for any value of the parameters. These points are the points of the intersection of the quadric and its tangent plane at $A.$ In this specific case, these points have nonreal complex coordinates, but it suffices to change one sign in the equation of the quadric for producing real points that are not obtained with the resulting parametrization.

A point of a quadric defined over a field $F$ is said rational point over $F$ if its coordinates belong to $F$ A rational point over the field $\backslash R$ of the real numbers, is called a real point.

A rational point over $\backslash Q$ is called simply a rational point. By clearing denominators, one can suppose and one supposes generally that the projective coordinates of a rational point (in a quadric defined over $\backslash Q$) are integers. Also, by clearing denominators of the coefficients, one supposes generally that all the coefficients of the equation of the quadric and the polynomials occurring in the parametrization are integers.

Finding the rational points of a projective quadric amounts thus to solving a Diophantine equation.

Given a rational point over a quadric over a field , the parametrization described in the preceding section provides rational points when the parameters are in , and, conversely, every rational point of the quadric can be obtained from parameters in , if the point is not in the tangent hyperplane at .

It follows that, if a quadric has a rational point, it has many other rational points (infinitely many if is infinite), and these points can be algorithmically generated as soon one knows one of them.

As said above, in the case of projective quadrics defined over $\backslash Q,$ the parametrization takes the form

$X\_i=F\_i(T\_1,\; \backslash ldots,\; T\_n)\backslash quad\; \backslash text\{for\; \}\; i=0,\backslash ldots,n,$

By choosing $A=(-1,\; 0,\; 1),$ the above method provides the parametrization

$\backslash begin\{cases\}$

If and are coprime integers such that $m>n>0,$ the resulting triple is a Pythagorean triple. If one of and is even and the other is odd, this resulting triple is primitive; otherwise, and are both odd, and one obtains a primitive triple by dividing by 2.

In summary, the primitive Pythagorean triples with $b$ even are obtained as

$a=m^2-n^2,\backslash quad\; b=2mn,\backslash quad\; c=\; m^2+n^2,$

$a=\backslash frac\{m^2-n^2\}\{2\},\backslash quad\; b=mn,\; \backslash quad\; c=\; \backslash frac\{m^2+n^2\}2,$

As the exchange of and transforms a Pythagorean triple into another Pythagorean triple, only one of the two cases is sufficient for producing all primitive Pythagorean triples up to the order of and .

(Q1) $\backslash ;q(\backslash lambda\backslash vec\; x)=\backslash lambda^2q(\backslash vec\; x\; )\backslash ;$ for any $\backslash lambda\backslash in\; K$ and $\backslash vec\; x\; \backslash in\; V$.

(Q2) $\backslash ;f(\backslash vec\; x,\backslash vec\; y\; ):=q(\backslash vec\; x+\backslash vec\; y)-q(\backslash vec\; x)-q(\backslash vec\; y)\backslash ;$ is a bilinear form.

In case of $\backslash operatorname\{char\}K\backslash ne2$ the bilinear form is $f(\backslash vec\; x,\backslash vec\; x)=2q(\backslash vec\; x)$, i.e. $f$ and $q$ are mutually determined in a unique way.
In case of $\backslash operatorname\{char\}K=2$ (that means: $1+1=0$) the bilinear form has the property $f(\backslash vec\; x,\backslash vec\; x)=0$, i.e. $f$ is symplectic.

For $V=K^n\backslash$ and $\backslash \; \backslash vec\; x=\backslash sum\_\{i=1\}^\{n\}x\_i\backslash vec\; e\_i\backslash quad$ ($\backslash \{\backslash vec\; e\_1,\backslash ldots,\backslash vec\; e\_n\backslash \}$ is a base of $V$) $\backslash \; q$ has the familiar form

$f(\backslash vec\; x,\backslash vec\; y)=\backslash sum\_\{1=i\backslash le\; k\}^\{n\}\; a\_\{ik\}(x\_iy\_k+x\_ky\_i)$.

For example:

$n=3,\backslash quad\; q(\backslash vec\; x)=x\_1x\_2-x^2\_3,\; \backslash quad\; f(\backslash vec\; x,\backslash vec\; y)=x\_1y\_2+x\_2y\_1-2x\_3y\_3\backslash ;\; .$

$V\_\{n+1\}$ an -dimensional vector space over the field $K,$

$\backslash langle\backslash vec\; x\backslash rangle$ the 1-dimensional linear span,

$\{\backslash mathcal\; P\}=\backslash \{\backslash langle\; \backslash vec\; x\backslash rangle\; \backslash mid\; \backslash vec\; x\; \backslash in\; V\_\{n+1\}\backslash \},\backslash$ the set of points ,

$\{\backslash mathcal\; G\}=\backslash \{\; \backslash text\{2-dimensional\; subspaces\; of\; \}\; V\_\{n+1\}\backslash \},\backslash$ the set of lines.

$P\_n(K)=(\{\backslash mathcal\; P\},\{\backslash mathcal\; G\})\backslash$ is the -dimensional projective space over $K$.

The set of points contained in a $(k+1)$-dimensional subspace of $V\_\{n+1\}$ is a $k$-dimensional subspace of $P\_n(K)$. A 2-dimensional subspace is a plane.

In case of $\backslash ;n>3\backslash ;$ a $(n-1)$-dimensional subspace is called hyperplane.

$\backslash mathcal\; Q=\backslash \{\backslash langle\backslash vec\; x\backslash rangle\; \backslash in\; \{\backslash mathcal\; P\}\; \backslash mid\; q(\backslash vec\; x)=0\backslash \}.$

Examples in $P\_2(K)$.:
(E1): For $\backslash ;q(\backslash vec\; x)=x\_1x\_2-x^2\_3\backslash ;$ one obtains a Conic section.
(E2): For $\backslash ;q(\backslash vec\; x)=x\_1x\_2\backslash ;$ one obtains the pair of lines with the equations $x\_1=0$ and $x\_2=0$, respectively. They intersect at point $\backslash langle(0,0,1)^\backslash text\{T\}\backslash rangle$;

For the considerations below it is assumed that $\backslash mathcal\; Q\backslash ne\; \backslash emptyset$.

$P^\backslash perp:=\backslash \{\backslash langle\backslash vec\; x\backslash rangle\backslash in\; \{\backslash mathcal\; P\}\; \backslash mid\; f(\backslash vec\; p,\backslash vec\; x)=0\backslash \}$

If $\backslash ;f(\backslash vec\; p,\backslash vec\; x)=0\backslash ;$ for all $\backslash vec\; x$, one obtains $P^\backslash perp=\backslash mathcal\; P$.

If $\backslash ;f(\backslash vec\; p,\backslash vec\; x)\backslash ne\; 0\backslash ;$ for at least one $\backslash vec\; x$, the equation $\backslash ;f(\backslash vec\; p,\backslash vec\; x)=0\backslash ;$is a non trivial linear equation which defines a hyperplane. Hence

$P^\backslash perp$ is either a hyperplane or $\{\backslash mathcal\; P\}$.

a) $g\backslash cap\; \backslash mathcal\; Q=\backslash emptyset\backslash ;$ and $g$ is called exterior line

b) $g\; \backslash subset\; \backslash mathcal\; Q\backslash ;$ and $g$ is called a line in the quadric

c) $|g\backslash cap\; \backslash mathcal\; Q|=1\backslash ;$ and $g$ is called tangent line

d) $|g\backslash cap\; \backslash mathcal\; Q|=2\backslash ;$ and $g$ is called secant line.

Proof: Let $g$ be a line, which intersects $\backslash mathcal\; Q$ at point $\backslash ;U=\backslash langle\backslash vec\; u\backslash rangle\backslash ;$ and $\backslash ;V=\; \backslash langle\backslash vec\; v\backslash rangle\backslash ;$ is a second point on $g$. From $\backslash ;q(\backslash vec\; u)=0\backslash ;$ one obtains
$q(x\backslash vec\; u+\backslash vec\; v)=q(x\backslash vec\; u)+q(\backslash vec\; v)+f(x\backslash vec\; u,\backslash vec\; v)=q(\backslash vec\; v)+xf(\backslash vec\; u,\backslash vec\; v)\backslash ;\; .$
I) In case of $g\backslash subset\; U^\backslash perp$ the equation $f(\backslash vec\; u,\backslash vec\; v)=0$ holds and it is $\backslash ;\; q(x\backslash vec\; u+\backslash vec\; v)=q(\backslash vec\; v)\backslash ;$ for any $x\backslash in\; K$. Hence either $\backslash ;q(x\backslash vec\; u+\backslash vec\; v)=0\backslash ;$ for any $x\backslash in\; K$ or $\backslash ;q(x\backslash vec\; u+\backslash vec\; v)\backslash ne\; 0\backslash ;$ for any $x\backslash in\; K$, which proves b) and b').
II) In case of $g\backslash not\backslash subset\; U^\backslash perp$ one obtains $f(\backslash vec\; u,\backslash vec\; v)\backslash ne\; 0$ and the equation $\backslash ;q(x\backslash vec\; u+\backslash vec\; v)=q(\backslash vec\; v)+xf(\backslash vec\; u,\backslash vec\; v)=\; 0\backslash ;$ has exactly one solution $x$. Hence: $|g\backslash cap\; \backslash mathcal\; Q|=2$, which proves c).

Additionally the proof shows:

A line $g$ through a point $P\backslash in\; \backslash mathcal\; Q$ is a tangent line if and only if $g\backslash subset\; P^\backslash perp$.

a) $\backslash mathcal\; R:=\backslash \{P\backslash in\{\backslash mathcal\; P\}\; \backslash mid\; P^\backslash perp=\backslash mathcal\; P\backslash \}$ is a projective subspace. $\backslash mathcal\; R$ is called f-radical of quadric $\backslash mathcal\; Q$.

b) $\backslash mathcal\; S:=\backslash mathcal\; R\backslash cap\backslash mathcal\; Q$ is called singular radical or $q$-radical of $\backslash mathcal\; Q$.

c) In case of $\backslash operatorname\{char\}K\backslash ne2$ one has $\backslash mathcal\; R=\backslash mathcal\; S$.

A quadric is called non-degenerate if $\backslash mathcal\; S=\backslash emptyset$.

Examples in $P\_2(K)$ (see above):
(E1): For $\backslash ;q(\backslash vec\; x)=x\_1x\_2-x^2\_3\backslash ;$ (conic) the bilinear form is $f(\backslash vec\; x,\backslash vec\; y)=x\_1y\_2+x\_2y\_1-2x\_3y\_3\backslash ;\; .$
In case of $\backslash operatorname\{char\}K\backslash ne2$ the polar spaces are never $\backslash mathcal\; P$. Hence $\backslash mathcal\; R=\backslash mathcal\; S=\backslash empty$.
In case of $\backslash operatorname\{char\}K=2$ the bilinear form is reduced to $f(\backslash vec\; x,\backslash vec\; y)=x\_1y\_2+x\_2y\_1\backslash ;$ and $\backslash mathcal\; R=\backslash langle(0,0,1)^\backslash text\{T\}\backslash rangle\backslash notin\; \backslash mathcal\; Q$. Hence $\backslash mathcal\; R\backslash ne\; \backslash mathcal\; S=\backslash empty\; \backslash ;\; .$ In this case the f-radical is the common point of all tangents, the so called knot.
In both cases $S=\backslash empty$ and the quadric (conic) ist non-degenerate.
(E2): For $\backslash ;q(\backslash vec\; x)=x\_1x\_2\backslash ;$ (pair of lines) the bilinear form is $f(\backslash vec\; x,\backslash vec\; y)=x\_1y\_2+x\_2y\_1\backslash ;$ and $\backslash mathcal\; R=\backslash langle(0,0,1)^\backslash text\{T\}\backslash rangle=\backslash mathcal\; S\backslash ;\; ,$ the intersection point.
In this example the quadric is degenerate.

For any point $P\backslash notin\; \backslash mathcal\; Q\backslash cup\; \{\backslash mathcal\; R\}\backslash ;$ there exists an involutorial central collineation $\backslash sigma\_P$ with center $P$ and $\backslash sigma\_P(\backslash mathcal\; Q)=\backslash mathcal\; Q$.

Proof: Due to $P\backslash notin\; \backslash mathcal\; Q\backslash cup\; \{\backslash mathcal\; R\}$ the polar space $P^\backslash perp$ is a hyperplane.

The linear mapping

$\backslash varphi:\; \backslash vec\; x\; \backslash rightarrow\; \backslash vec\; x-\backslash frac\{f(\backslash vec\; p,\backslash vec\; x)\}\{q(\backslash vec\; p)\}\backslash vec\; p$

induces an involutorial central collineation $\backslash sigma\_P$ with axis $P^\backslash perp$ and centre $P$ which leaves $\backslash mathcal\; Q$ invariant.
In the case of $\backslash operatorname\{char\}K\backslash ne2$, the mapping $\backslash varphi$ produces the familiar shape $\backslash ;\; \backslash varphi:\; \backslash vec\; x\; \backslash rightarrow\; \backslash vec\; x-2\backslash frac\{f(\backslash vec\; p,\backslash vec\; x)\}\{f(\backslash vec\; p,\backslash vec\; p)\}\backslash vec\; p\backslash ;$ with $\backslash ;\; \backslash varphi(\backslash vec\; p)=-\backslash vec\; p$ and $\backslash ;\; \backslash varphi(\backslash vec\; x)=\backslash vec\; x\backslash ;$ for any $\backslash langle\backslash vec\; x\backslash rangle\; \backslash in\; P^\backslash perp$.

Remark:

a) An exterior line, a tangent line or a secant line is mapped by the involution $\backslash sigma\_P$ on an exterior, tangent and secant line, respectively.

b) $\{\backslash mathcal\; R\}$ is pointwise fixed by $\backslash sigma\_P$.

For example: points on a sphere or ruled surface (s. below).

Any two maximal $q$-subspaces have the same dimension $m$.Beutelpacher/Rosenbaum, p.139

Let be $m$ the dimension of the maximal $q$-subspaces of $\backslash mathcal\; Q$ then

The integer $\backslash ;i:=m+1\backslash ;$ is called index of $\backslash mathcal\; Q$.

Theorem:F. Buekenhout: Ensembles Quadratiques des Espace Projective, Math. Teitschr. 110 (1969), p. 306-318.

For the index $i$ of a non-degenerate quadric $\backslash mathcal\; Q$ in $P\_n(K)$ the following is true:

:$i\backslash le\; \backslash frac\{n+1\}\{2\}$.

Let be $\backslash mathcal\; Q$ a non-degenerate quadric in $P\_n(K),\; n\backslash ge\; 2$, and $i$ its index.

In case of $i=1$ quadric $\backslash mathcal\; Q$ is called sphere (or oval conic if $n=2$).

In case of $i=2$ quadric $\backslash mathcal\; Q$ is called hyperboloid (of one sheet).

Examples:

a) Quadric $\backslash mathcal\; Q$ in $P\_2(K)$ with form $\backslash ;q(\backslash vec\; x)=x\_1x\_2-x^2\_3\backslash ;$ is non-degenerate with index 1.

b) If polynomial $\backslash ;p(\backslash xi)=\backslash xi^2+a\_0\backslash xi+b\_0\backslash ;$ is irreducible over $K$ the quadratic form $\backslash ;q(\backslash vec\; x)=x^2\_1+a\_0x\_1x\_2+b\_0x^2\_2-x\_3x\_4\backslash ;$ gives rise to a non-degenerate quadric $\backslash mathcal\; Q$ in $P\_3(K)$ of index 1 (sphere). For example: $\backslash ;p(\backslash xi)=\backslash xi^2+1\backslash ;$ is irreducible over $\backslash R$ (but not over $\backslash C$ !).

c) In $P\_3(K)$ the quadratic form $\backslash ;q(\backslash vec\; x)=x\_1x\_2+x\_3x\_4\backslash ;$ generates a hyperboloid.

A division ring $K$ is commutative ring if and only if any equation $x^2+ax+b=0,\; \backslash \; a,b\; \backslash in\; K$, has at most two solutions.

There are generalizations of quadrics: .Beutelspacher/Rosenbaum: p. 135 A quadratic set is a set of points of a projective space with the same geometric properties as a quadric: every line intersects a quadratic set in at most two points or is contained in the set.

• Klein quadric
• Rotation of axes
• Superquadrics
• Translation of axes

• M. Audin: Geometry, Springer, Berlin, 2002, , p. 200.
• M. Berger: Problem Books in Mathematics, ISSN 0941-3502, Springer New York, pp 79–84.
• A. Beutelspacher, U. Rosenbaum: Projektive Geometrie, Vieweg + Teubner, Braunschweig u. a. 1992, , p. 159.
• P. Dembowski: Finite Geometries, Springer, 1968, , p. 43.

• Interactive Java 3D models of all quadric surfaces
• Lecture Note Planar Circle Geometries, an Introduction to Moebius, Laguerre and Minkowski Planes, p. 117