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.
See main article: Oval (projective plane).
ako
(1) Any line
g
ako
|g\capako|=0
g
|g\capako|=1
|g\capako|=2
(2) For any point
P\inako
t
t\capako=\{P\}
For finite planes (i.e. the set of points is finite) we have a more convenient characterization:
ako
|ako|=n+1
ako
\ne2
ako
(P3): Let be
P1,P2,P3
ako
\overline{PiPi}
Pi
ako
P4:=\overline{P1P1}\cap\overline{P2P3}, P5:=\overline{P2P2}\cap\overline{P1P3}, P6:=\overline{P3P3}\cap\overline{P1P2}
are collinear.[1]
K
P3=(0), ginfty
P3, (0,0)\inako
(0,0)
ako
(1,1)
P1=(x1,y1), P2=(x2,y2) .
ako
f:K\mapstoK
ako=\{(x,y)\inK2 | y=f(x)\} \cup\{(infty)\} .
(x0,f(x0))
f'
y=f'(x0)(x-x0)+f(x0)
P5=(x1,f'(x2)(x1-x2)+f(x2))
P4=(x2,f'(x1)(x2-x1)+f(x1)) .
ako
f(x)=x2
f'(x)=2x
P4,P5,P6
II: If
ako
\overline{P4P5}
\overline{P1P2}
f'(x2)+f'(x1)-
f(x2)-f(x1) | = | |
x2-x1 |
f(x2)-f(x1) | |
x2-x1 |
(i):
(f'(x2)+f'(x1))(x2-x1)=2(f(x2)-f(x1))
x1,x2\inK
f(0)=f'(0)=0
(ii):
f'(x2)x2=2f(x2)
f(1)=1
(iii):
f'(1)=2 .
(iv):
f'(x2)x1=f'(x1)x2
(v):
f'(x2)=2x2
x2\inK
f(x2)=x
2, | |
2 |
x2\inK
ako
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]
ako
Let be
P1,P2,P3
ako
P4,P5,P6
K
P3=(infty), P2=(0), P1=(1,1)
(0,0)
P2
P3
ako
f:K*:=K\cup\setminus\{0\}\mapstoK*
ako=\{(x,y)\inK2 | y=f(x), x\ne0\} \cup \{(0),(infty)\} .
P=(x,y), x\inK\setminus\{0,1\}
m(x)=\tfrac{f(x)-1}{x-1}
\overline{PP1} .
x\mapstof(x)-1
x\mapstox-1
K\setminus\{0,1\}
K\setminus\{0,-1\}
x\mapstom(x)
K\setminus\{0,1\}
K\setminus\{0,m1\}
m1
P1
K**:=K\setminus\{0,1\} :
\prod | |
x\inK** |
(f(x)-1)=\prod | |
x\inK** |
(x-1)=1 und m1 ⋅ \prod
x\inK** |
f(x)-1 | |
x-1 |
=-1 .
K*:=K\setminus\{0\}
\displaystyle
\prod | |
k\inK* |
k=-1 .
-1=m1 ⋅ \prod
x\inK** |
f(x)-1 | |
x-1 |
= m | ||||||||||||||||
|
=m1 .
\overline{P5P6}
\overline{P1P1}
-1
\overline{P1P1}\cap\overline{P2P3}=P4\in\overline{P5P6}
P1,P2,P3\inako
So: (P3) of the 3-point Pascal theorem holds and the oval is a non degenerate conic.