Segre's theorem explained

In projective geometry, Segre's theorem, named after the Italian mathematician Beniamino Segre, is the statement:

This statement was assumed 1949 by the two Finnish mathematicians G. Järnefelt and P. Kustaanheimo and its proof was published in 1955 by B. Segre.

A finite pappian projective plane can be imagined as the projective closure of the real plane (by a line at infinity), where the real numbers are replaced by a finite field . Odd order means that is odd. An oval is a curve similar to a circle (see definition below): any line meets it in at most 2 points and through any point of it there is exactly one tangent. The standard examples are the nondegenerate projective conic sections.

In pappian projective planes of even order greater than four there are ovals which are not conics. In an infinite plane there exist ovals, which are not conics. In the real plane one just glues a half of a circle and a suitable ellipse smoothly.

The proof of Segre's theorem, shown below, uses the 3-point version of Pascal's theorem and a property of a finite field of odd order, namely, that the product of all the nonzero elements equals -1.

Definition of an oval

See main article: Oval (projective plane).

ako

of points is called oval, if:

(1) Any line

g

meets

ako

in at most two points.If

|g\capako|=0

the line

g

is an exterior (or passing) line; in case

|g\capako|=1

a tangent line and if

|g\capako|=2

the line is a secant line.

(2) For any point

P\inako

there exists exactly one tangent

t

at, i.e.,

t\capako=\{P\}

.

For finite planes (i.e. the set of points is finite) we have a more convenient characterization:

ako

of points is an oval if and only if

|ako|=n+1

and no three points are collinear (on a common line).

Pascal's 3-point version

Theorem:Let be

ako

an oval in a pappian projective plane of characteristic

\ne2

.

ako

is a nondegenerate conic if and only if statement (P3)holds:

(P3): Let be

P1,P2,P3

any triangle on

ako

and

\overline{PiPi}

the tangent at point

Pi

to

ako

, then the points

P4:=\overline{P1P1}\cap\overline{P2P3},P5:=\overline{P2P2}\cap\overline{P1P3},P6:=\overline{P3P3}\cap\overline{P1P2}

are collinear.[1]

Proof:Let the projective plane be coordinatized inhomogeneously over a field

K

such that

P3=(0),ginfty

is the tangent at

P3,(0,0)\inako

, the x-axis is the tangent at the point

(0,0)

and

ako

contains the point

(1,1)

. Furthermore, we set

P1=(x1,y1),P2=(x2,y2).

(s. image)
The oval

ako

can be described by a function

f:K\mapstoK

such that:

ako=\{(x,y)\inK2|y=f(x)\}\cup\{(infty)\}.

The tangent at point

(x0,f(x0))

will be described using a function

f'

such that its equation is

y=f'(x0)(x-x0)+f(x0)

Hence (s. image)

P5=(x1,f'(x2)(x1-x2)+f(x2))

and

P4=(x2,f'(x1)(x2-x1)+f(x1)).

I: if

ako

is a non degenerate conic we have

f(x)=x2

and

f'(x)=2x

and one calculates easily that

P4,P5,P6

are collinear.

II: If

ako

is an oval with property (P3), the slope of the line

\overline{P4P5}

is equal to the slope of the line

\overline{P1P2}

, that means:

f'(x2)+f'(x1)-

f(x2)-f(x1)=
x2-x1
f(x2)-f(x1)
x2-x1
and hence

(i):

(f'(x2)+f'(x1))(x2-x1)=2(f(x2)-f(x1))

for all

x1,x2\inK

. With

f(0)=f'(0)=0

one gets

(ii):

f'(x2)x2=2f(x2)

and from

f(1)=1

we get

(iii):

f'(1)=2.

(i) and (ii) yield

(iv):

f'(x2)x1=f'(x1)x2

and with (iii) at least we get

(v):

f'(x2)=2x2

for all

x2\inK

. A consequence of (ii) and (v) is

f(x2)=x

2,
2

x2\inK

. Hence

ako

is a nondegenerate conic.

Remark:Property (P3) is fulfilled for any oval in a pappian projective plane of characteristic 2 with a nucleus (all tangents meet at the nucleus). Hence in this case (P3) is also true for non-conic ovals.[2]

Segre's theorem and its proof

Theorem:Any oval

ako

in a finite pappian projective plane of odd order is a nondegenerate conic section.
Proof:[3] For the proof we show that the oval has property (P3) of the 3-point version of Pascal's theorem.

Let be

P1,P2,P3

any triangle on

ako

and

P4,P5,P6

defined as described in (P3). The pappian plane will be coordinatized inhomogeneously over a finite field

K

, such that

P3=(infty),P2=(0),P1=(1,1)

and

(0,0)

is the common point of the tangents at

P2

and

P3

. The oval

ako

can be described using a bijective function

f:K*:=K\cup\setminus\{0\}\mapstoK*

:

ako=\{(x,y)\inK2 |y=f(x),x\ne0\}\cup\{(0),(infty)\}.

For a point

P=(x,y),x\inK\setminus\{0,1\}

, the expression

m(x)=\tfrac{f(x)-1}{x-1}

is the slope of the secant

\overline{PP1}.

Because both the functions

x\mapstof(x)-1

and

x\mapstox-1

are bijections from

K\setminus\{0,1\}

to

K\setminus\{0,-1\}

, and

x\mapstom(x)

a bijection from

K\setminus\{0,1\}

onto

K\setminus\{0,m1\}

, where

m1

is the slope of the tangent at

P1

, for

K**:=K\setminus\{0,1\}:

we get
\prod
x\inK**
(f(x)-1)=\prod
x\inK**

(x-1)=1und m1 ⋅ \prod

x\inK**
f(x)-1
x-1

=-1 .

(Remark: For

K*:=K\setminus\{0\}

we have:

\displaystyle

\prod
k\inK*

k=-1 .

)
Hence

-1=m1 ⋅ \prod

x\inK**
f(x)-1
x-1
= m
1 ⋅
\displaystyle
\prod
x\inK**
(f(x)-1)
\displaystyle
\prod
x\inK**
(x-1)

=m1 .

Because the slopes of line

\overline{P5P6}

and tangent

\overline{P1P1}

both are

-1

, it follows that

\overline{P1P1}\cap\overline{P2P3}=P4\in\overline{P5P6}

.This is true for any triangle

P1,P2,P3\inako

.

So: (P3) of the 3-point Pascal theorem holds and the oval is a non degenerate conic.

Sources

External links

Notes and References

  1. E. Hartmann: Planar Circle Geometries, an Introduction to Moebius-, Laguerre- and Minkowski Planes. Skript, TH Darmstadt (PDF; 891 kB), p. 34.
  2. E. Hartmann: Planar Circle Geometries, an Introduction to Moebius-, Laguerre- and Minkowski Planes. Skript, TH Darmstadt (PDF; 891 kB), p. 35.
  3. E. Hartmann: Planar Circle Geometries, an Introduction to Moebius-, Laguerre- and Minkowski Planes. Skript, TH Darmstadt (PDF; 891 kB), p. 41.