Quadric

A quadric in an affine space of dimension n is the set of zeros of a polynomial of degree 2. That is, it is the set of the points whose coordinates satisfy an equation

${\displaystyle p(x_{1},\ldots ,x_{n})=0,}$

where the polynomial p has the form

${\displaystyle p(x_{1},\ldots ,x_{n})=\sum _{i=1}^{n}\sum _{j=1}^{n}a_{i,j}x_{i}x_{j}+\sum _{i=1}^{n}(a_{i,0}+a_{0,i})x_{i}+a_{0,0}\,,}$

for a matrix ${\displaystyle A=(a_{i,j})}$ with ${\displaystyle i}$ and ${\displaystyle j}$ running from 0 to ${\displaystyle n}$. When the characteristic of the field of the coefficients is not two, generally ${\displaystyle a_{i,j}=a_{j,i}}$ is assumed; equivalently ${\displaystyle A=A^{\mathsf {T}}}$. When the characteristic of the field of the coefficients is two, generally ${\displaystyle a_{i,j}=0}$ is assumed when ${\displaystyle j<i}$; equivalently ${\displaystyle A}$ is upper triangular.

The equation may be shortened, as the matrix equation

${\displaystyle \mathbf {x} ^{\mathsf {T}}A\mathbf {x} =0\,,}$

with

${\displaystyle \mathbf {x} ={\begin{pmatrix}1&x_{1}&\cdots &x_{n}\end{pmatrix}}^{\mathsf {T}}\,.}$

The equation of the projective completion is almost identical:

${\displaystyle \mathbf {X} ^{\mathsf {T}}A\mathbf {X} =0,}$

with

${\displaystyle \mathbf {X} ={\begin{pmatrix}X_{0}&X_{1}&\cdots &X_{n}\end{pmatrix}}^{\mathsf {T}}.}$

These equations define a quadric as an algebraic hypersurface of dimension n – 1 and degree two in a space of dimension n.

A quadric is said to be non-degenerate if the matrix ${\displaystyle A}$ is invertible.

A non-degenerate quadric is non-singular in the sense that its projective completion has no singular point (a cylinder is non-singular in the affine space, but it is a degenerate quadric that has a singular point at infinity).

The singular points of a degenerate quadric are the points whose projective coordinates belong to the null space of the matrix A.

A quadric is reducible if and only if the rank of A is one (case of a double hyperplane) or two (case of two hyperplanes).

Let consider the quadric of equation

${\displaystyle x_{1}^{2}+x_{2}^{2}+\cdots x_{n}^{2}-1=0.}$

For ${\displaystyle n=2,}$ this is the unit circle; for ${\displaystyle n=3}$ this is the unit sphere; in higher dimension, this is the unit hypersphere.

The point ${\displaystyle \mathbf {a} =(0,\ldots ,0,-1)}$ belongs to the quadric (the choice of this point among other similar points is only a question of convenience). So, the equation ${\displaystyle q(\mathbf {a} +\lambda \mathbf {t} )=0}$ of the preceding section becomes

${\displaystyle (\lambda t_{1}^{2})+\cdots +(\lambda t_{n-1})^{2}+(1-\lambda )^{2}-1=0.}$

By expanding the squares, simplifying the constant terms, dividing by ${\displaystyle \lambda ,}$ and solving in ${\displaystyle \lambda ,}$ one obtains

${\displaystyle \lambda ={\frac {2}{1+t_{1}^{2}+\cdots +t_{n-1}^{2}}}.}$

Substituting this into ${\displaystyle \mathbf {x} =\mathbf {a} +\lambda \mathbf {t} }$ and simplifying the expression of the last coordinate, one obtains the parametric equation

${\displaystyle {\begin{cases}x_{1}={\frac {2t_{1}}{1+t_{1}^{2}+\cdots +t_{n-1}^{2}}}\\\vdots \\x_{n-1}={\frac {2t_{n-1}}{1+t_{1}^{2}+\cdots +t_{n-1}^{2}}}\\x_{n}={\frac {1-t_{1}^{2}-\cdots -t_{n-1}^{2}}{1+t_{1}^{2}+\cdots +t_{n-1}^{2}}}.\end{cases}}}$

By homogenizing, one obtains the projective parametrization

${\displaystyle {\begin{cases}X_{0}=T_{1}^{2}+\cdots +T_{n}^{2}\\X_{1}=2T_{1}T_{n}\\\vdots \\X_{n-1}=2T_{n-1}T_{n}\\X_{n}=T_{n}^{2}-T_{1}^{2}-\cdots -T_{n-1}^{2}.\end{cases}}}$

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

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

For ${\displaystyle n>2,}$ there are many points with ${\displaystyle T_{n}=0,}$ and thus many parameter values for the point ${\displaystyle A.}$ On the other hand, the other points of the quadric for which ${\displaystyle X_{n}=-X_{0}}$ (and thus ${\displaystyle x_{n}=-1}$) 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 ${\displaystyle A.}$ 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.

This is well illustrated by Pythagorean triples. A Pythagorean triple is a triple ${\displaystyle (a,b,c)}$ of positive integers such that ${\displaystyle a^{2}+b^{2}=c^{2}.}$ A Pythagorean triple is primitive if ${\displaystyle a,b,c}$ are setwise coprime, or, equivalently, if any of the three pairs ${\displaystyle (a,b),}$ ${\displaystyle (b,c)}$ and ${\displaystyle (a,c)}$ is coprime.

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

${\displaystyle {\begin{cases}a=m^{2}-n^{2}\\b=2mn\\c=m^{2}+m^{2}\end{cases}}}$

for the quadric of equation ${\displaystyle a^{2}+b^{2}-c^{2}=0.}$ (The names of variables and parameters are being changed from the above ones to those that are common when considering Pythagorean triples).

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

In summary, the primitive Pythagorean triples with ${\displaystyle b}$ even are obtained as

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

with m and n coprime integers such that one is even and ${\displaystyle m>n>0}$ (this is Euclid's formula). The primitive Pythagorean triples with ${\displaystyle b}$ odd are obtained as

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

with m and n coprime odd integers such that ${\displaystyle m>n>0.}$

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

Let ${\displaystyle K}$ be a field and ${\displaystyle V}$ a vector space over ${\displaystyle K}$. A mapping ${\displaystyle q}$ from ${\displaystyle V}$ to ${\displaystyle K}$ such that

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

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

is called quadratic form. The bilinear form ${\displaystyle f}$ is symmetric.

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

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

${\displaystyle q({\vec {x}})=\sum _{1=i\leq k}^{n}a_{ik}x_{i}x_{k}\ {\text{ with }}\ a_{ik}:=f({\vec {e}}_{i},{\vec {e}}_{k})\ {\text{ for }}\ i\neq k\ {\text{ and }}\ a_{ii}:=q({\vec {e}}_{i})\ }$ and

${\displaystyle f({\vec {x}},{\vec {y}})=\sum _{1=i\leq k}^{n}a_{ik}(x_{i}y_{k}+x_{k}y_{i})}$.

For example:

${\displaystyle n=3,\quad q({\vec {x}})=x_{1}x_{2}-x_{3}^{2},\quad f({\vec {x}},{\vec {y}})=x_{1}y_{2}+x_{2}y_{1}-2x_{3}y_{3}\;.}$

Let ${\displaystyle K}$ be a field, ${\displaystyle 2\leq n\in \mathbb {N} }$,

${\displaystyle V_{n+1}}$ an (n + 1)-dimensional vector space over the field ${\displaystyle K,}$

${\displaystyle \langle {\vec {x}}\rangle }$ the 1-dimensional subspace generated by ${\displaystyle {\vec {0}}\neq {\vec {x}}\in V_{n+1}}$,

${\displaystyle {\mathcal {P}}=\{\langle {\vec {x}}\rangle \mid {\vec {x}}\in V_{n+1}\},\ }$ the set of points ,

${\displaystyle {\mathcal {G}}=\{{\text{2-dimensional subspaces of }}V_{n+1}\},\ }$ the set of lines.

${\displaystyle P_{n}(K)=({\mathcal {P}},{\mathcal {G}})\ }$ is the n-dimensional projective space over ${\displaystyle K}$.

The set of points contained in a ${\displaystyle (k+1)}$-dimensional subspace of ${\displaystyle V_{n+1}}$ is a ${\displaystyle k}$-dimensional subspace of ${\displaystyle P_{n}(K)}$. A 2-dimensional subspace is a plane.

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

A quadratic form ${\displaystyle q}$ on a vector space ${\displaystyle V_{n+1}}$ defines a quadric ${\displaystyle {\mathcal {Q}}}$ in the associated projective space ${\displaystyle {\mathcal {P}},}$ as the set of the points ${\displaystyle \langle {\vec {x}}\rangle \in {\mathcal {P}}}$ such that ${\displaystyle q({\vec {x}})=0}$. That is,

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

Examples in ${\displaystyle P_{2}(K)}$.:
(E1): For ${\displaystyle \;q({\vec {x}})=x_{1}x_{2}-x_{3}^{2}\;}$ one obtains a conic.
(E2): For ${\displaystyle \;q({\vec {x}})=x_{1}x_{2}\;}$ one obtains the pair of lines with the equations ${\displaystyle x_{1}=0}$ and ${\displaystyle x_{2}=0}$, respectively. They intersect at point ${\displaystyle \langle (0,0,1)^{\text{T}}\rangle }$;

For the considerations below it is assumed that ${\displaystyle {\mathcal {Q}}\neq \emptyset }$.

For point ${\displaystyle P=\langle {\vec {p}}\rangle \in {\mathcal {P}}}$ the set

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

is called polar space of ${\displaystyle P}$ (with respect to ${\displaystyle q}$).

If ${\displaystyle \;f({\vec {p}},{\vec {x}})=0\;}$ for any ${\displaystyle {\vec {x}}}$, one obtains ${\displaystyle P^{\perp }={\mathcal {P}}}$.

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

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

For the intersection of an arbitrary line ${\displaystyle g}$ with a quadric ${\displaystyle {\mathcal {Q}}}$, the following cases may occur:

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

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

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

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

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

Additionally the proof shows:

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

In the classical cases ${\displaystyle K=\mathbb {R} }$ or ${\displaystyle \mathbb {C} }$ there exists only one radical, because of ${\displaystyle \operatorname {char} K\neq 2}$ and ${\displaystyle f}$ and ${\displaystyle q}$ are closely connected. In case of ${\displaystyle \operatorname {char} K=2}$ the quadric ${\displaystyle {\mathcal {Q}}}$ is not determined by ${\displaystyle f}$ (see above) and so one has to deal with two radicals:

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

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

c) In case of ${\displaystyle \operatorname {char} K\neq 2}$ one has ${\displaystyle {\mathcal {R}}={\mathcal {S}}}$.

A quadric is called non-degenerate if ${\displaystyle {\mathcal {S}}=\emptyset }$.

Examples in ${\displaystyle P_{2}(K)}$ (see above):
(E1): For ${\displaystyle \;q({\vec {x}})=x_{1}x_{2}-x_{3}^{2}\;}$ (conic) the bilinear form is ${\displaystyle f({\vec {x}},{\vec {y}})=x_{1}y_{2}+x_{2}y_{1}-2x_{3}y_{3}\;.}$
In case of ${\displaystyle \operatorname {char} K\neq 2}$ the polar spaces are never ${\displaystyle {\mathcal {P}}}$. Hence ${\displaystyle {\mathcal {R}}={\mathcal {S}}=\emptyset }$.
In case of ${\displaystyle \operatorname {char} K=2}$ the bilinear form is reduced to ${\displaystyle f({\vec {x}},{\vec {y}})=x_{1}y_{2}+x_{2}y_{1}\;}$ and ${\displaystyle {\mathcal {R}}=\langle (0,0,1)^{\text{T}}\rangle \notin {\mathcal {Q}}}$. Hence ${\displaystyle {\mathcal {R}}\neq {\mathcal {S}}=\emptyset \;.}$ In this case the f-radical is the common point of all tangents, the so called knot.
In both cases ${\displaystyle S=\emptyset }$ and the quadric (conic) ist non-degenerate.
(E2): For ${\displaystyle \;q({\vec {x}})=x_{1}x_{2}\;}$ (pair of lines) the bilinear form is ${\displaystyle f({\vec {x}},{\vec {y}})=x_{1}y_{2}+x_{2}y_{1}\;}$ and ${\displaystyle {\mathcal {R}}=\langle (0,0,1)^{\text{T}}\rangle ={\mathcal {S}}\;,}$ the intersection point.
In this example the quadric is degenerate.

A quadric is a rather homogeneous object:

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

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

The linear mapping

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

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

Remark:

a) An exterior line, a tangent line or a secant line is mapped by the involution ${\displaystyle \sigma _{P}}$ on an exterior, tangent and secant line, respectively.

b) ${\displaystyle {\mathcal {R}}}$ is pointwise fixed by ${\displaystyle \sigma _{P}}$.

A subspace ${\displaystyle \;{\mathcal {U}}\;}$ of ${\displaystyle P_{n}(K)}$ is called ${\displaystyle q}$-subspace if ${\displaystyle \;{\mathcal {U}}\subset {\mathcal {Q}}\;}$

For example: points on a sphere or lines on a hyperboloid (s. below).

Any two maximal ${\displaystyle q}$-subspaces have the same dimension ${\displaystyle m}$.[5]

Let be ${\displaystyle m}$ the dimension of the maximal ${\displaystyle q}$-subspaces of ${\displaystyle {\mathcal {Q}}}$ then

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

Theorem: (BUEKENHOUT)[6]

For the index ${\displaystyle i}$ of a non-degenerate quadric ${\displaystyle {\mathcal {Q}}}$ in ${\displaystyle P_{n}(K)}$ the following is true:

${\displaystyle i\leq {\frac {n+1}{2}}}$.

Let be ${\displaystyle {\mathcal {Q}}}$ a non-degenerate quadric in ${\displaystyle P_{n}(K),n\geq 2}$, and ${\displaystyle i}$ its index.

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

In case of ${\displaystyle i=2}$ quadric ${\displaystyle {\mathcal {Q}}}$ is called hyperboloid (of one sheet).

Examples:

a) Quadric ${\displaystyle {\mathcal {Q}}}$ in ${\displaystyle P_{2}(K)}$ with form ${\displaystyle \;q({\vec {x}})=x_{1}x_{2}-x_{3}^{2}\;}$ is non-degenerate with index 1.

b) If polynomial ${\displaystyle \;p(\xi )=\xi ^{2}+a_{0}\xi +b_{0}\;}$ is irreducible over ${\displaystyle K}$ the quadratic form ${\displaystyle \;q({\vec {x}})=x_{1}^{2}+a_{0}x_{1}x_{2}+b_{0}x_{2}^{2}-x_{3}x_{4}\;}$ gives rise to a non-degenerate quadric ${\displaystyle {\mathcal {Q}}}$ in ${\displaystyle P_{3}(K)}$ of index 1 (sphere). For example: ${\displaystyle \;p(\xi )=\xi ^{2}+1\;}$ is irreducible over ${\displaystyle \mathbb {R} }$ (but not over ${\displaystyle \mathbb {C} }$ !).

c) In ${\displaystyle P_{3}(K)}$ the quadratic form ${\displaystyle \;q({\vec {x}})=x_{1}x_{2}+x_{3}x_{4}\;}$ generates a hyperboloid.

It is not reasonable to formally extend the definition of quadrics to spaces over genuine skew fields (division rings). Because one would obtain secants bearing more than 2 points of the quadric which is totally different from usual quadrics.[7][8][9] The reason is the following statement.

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

There are generalizations of quadrics: quadratic sets.[10] 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.