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Analytical Geometry
2.5 Normal Rational Curves
2.6 The Moulton Plane
2.7 Spacial Geometries are
Desarguesian
Normal Rational Curves
Def: Let P be a projective space of dimension d. We say
that a set S of at least d+1 points of P is in general position
if any d+1 points of S form a basis for P.
Examples:
a. A set of at least 3 points of a projective plane are in
general position iff no three are on a common line.
b. A set of at least 4 points of a 3-dimensional projective
space are in general position iff no 4 are in a common
plane.
Normal Rational Curves
Def: Let P = P(V) be a d-dimensional projective space
coordinatized over a field F. A normal rational curve is any set of
points projectively equivalent to the set
C = {(1:t:t2:t3: ... :td) | t∈ F} ∪{(0:0: ... :0:1)}.
Example:
In PG(3,4) (see last section) the following set of points form a
normal rational curve: (0:0:0:1), (1:0:0:0), (1:1:1:1), (1:a:a2:1) and
(1:a2:a:1). Note that no 4 of these points lie in a common plane.
This implies that no 3 of them lie on a common line.
Normal Rational Curves
Theorem 2.5.1: The points of a normal rational curve are in
general position.
Pf: We choose d+1 points of the curve and consider the matrix
whose rows are the coordinates of these points. By Theorem 2.3.1
the points are independent iff the determinant of the matrix is
different from zero.
Case I: The point (0:0: ... :0:1) is not in the set of points.
All the points have the form (1:ti:ti2: ... : tid) for i = 1, ..., d+1.
The determinant is:
2
d
1 t
t
⋯ t
∣
1
⋮
1
1
t2
⋮
td
1
1
t 22
⋮
2
td
1 t d 1 t d 1
⋯
⋱
⋯
2
∣
t 2d
⋮ .
d
td
⋯ t d 1
d
Normal Rational Curves
Theorem 2.5.1: The points of a normal rational curve are in
general position.
Pf(cont.): This is a Vandermonde determinant. It is not zero iff all
the ti are different. Since the points are different, the determinant
is not zero; hence the points are independent.
Case II: The point (0:0: ... :0:1) is in the set of points.
The determinant is now:
2
d
∣
1 t1
1
⋮
1
0
⋯ t1
t1
2
t2 t2
⋮ ⋮
2
td td
0 0
d
∣
⋯ t2
⋱ ⋮ .
d
⋯ td
⋯ 1
Develop the determinant with respect to the last row and get as in
case I a Vandermonde determinant, which is not zero since the
points are distinct.
❑
Normal Rational Curves
Corollary 2.5.2: Each hyperplane intersects a normal
rational curve in at most d points.
Pf: Since any d+1 points of the curve span a d dimensional
subspace, the intersection of this subspace with a hyperplane has
dimension d-1 and so contains at most d points of the curve. ❑
The Moulton Plane
We now provide an example of a nondesarguesian plane by
constructing an affine plane over the reals in which Desargues
theorem does not hold universally. This example is known as the
Moulton plane.
Def: Define the geometry M as follows:
Points
The pairs (x,y) with x,y ∈ℝ.
Lines
Described by the equations of the form x = c and y = mx + b with
c,m,b∈ℝ.
Incidence
(u,v) I (x=c) iff u = c. (u,v) I (y = mx + b) iff v = mu + b unless
u < 0 and m < 0, in which case (u,v) I (y = mx + b) iff v = 2mu + b.
The Moulton Plane
The incidence may be interpreted as follows: Vertical lines and
lines with non-negative slopes are the same as they are in the usual
Euclidean plane. Lines with negative slope are “bent” by a factor of
2 as they cross the y-axis. Some representative lines in this plane
look like this:
The Moulton Plane
Theorem 2.6.1: The Moulton plane is an affine plane in which the
theorem of Desargues is not true.
Pf: We first show that the Moulton plane M is an affine plane.
Axiom 1: Two distinct points determine a unique line.
Let the two points be (x0,y0) and (x1,y1) with x0 ≤ x1. In the cases
that the two points lie on the same side of the y-axis (x0 and
x1 > 0, or x0 and x1 < 0) or lie on a Euclidean line of non-negative
slope (y0 ≤ y1), the statement is obvious. Thus, we need only
consider the case where x0 < 0, x1 > 0 and y0 > y1. To find a “bent”
line joining these points we need to find m and b so that
y0 = mx0 + b and y1 = 2mx1 + b.
The Moulton Plane
Theorem 2.6.1: The Moulton plane is an affine plane in which the
theorem of Desargues is not true.
Pf(cont.): Now,
y0 – y1 = mx0 – 2mx1 = m(x0 – 2x1),
so m = (y0 – y1)/(x0 – 2x1) and b = (y1x0 – 2y0x1)/(x0 – 2x1). Thus,
m and b are uniquely determined.
Axiom 2: Playfair's Axiom.
The parallel classes are given by the vertical lines (x = c), and
the lines with fixed slope m (positive or negative). It is easy to see
that the axiom is satisfied.
Axiom 3: There exists a triangle.
Obvious.
Thus M is an affine plane.
The Moulton Plane
Theorem 2.6.1: The Moulton plane is an affine plane in which the
theorem of Desargues is not true.
Pf(cont.): Finally, we show that Desargues theorem is not
universally true. This is done by setting up a pair of triangles in a
special way so that we can compare them in the Euclidean plane
and in M.
Spacial Geometries are Desarguesian
Theorem 2.7.1: Let P be a projective space of dimension d. If
d ≥ 3, then the theorem of Desargues holds in P.
Pf: Let triangles A1, A2, A3 and B1, B2, B3 be perspective from the
point C. Let P12 = A1A2 ∩ B1B2 , P13 = A1A3 ∩ B1B3 , and P23 =
A2A3 ∩ B2B3 .
Case I: The planes π = < A1, A2, A3> and ψ = <B1, B2, B3 > are
distinct.
Since Ai and Bi are collinear with C, we have Bi ∈ <C, A1, A2, A3>
for i = 1,2,3. Therefore all points and lines are contained in the 3dimensional subspace U = <C, A1, A2, A3>. The points Pij all lie in
π ∩ ψ, and since this intersection is a line, the points are collinear.
Spacial Geometries are Desarguesian
Theorem 2.7.1: Let P be a projective space of dimension d. If
d ≥ 3, then the theorem of Desargues holds in P.
Pf(cont.):
Case II: The points C, A1, A2, A3, B1, B2, B3 are in the same plane π.
The idea is to “lift” this configuration into 3 space and then apply
case I. Let C' and C'' be two points not in π such that the line C'C''
meets π at C. Since C'C'' and A1B1 meet at C, these lines generate a
plane. The lines C'A1 and C''B1 intersect at a point D1 in this plane
and this point is not in π. In a similar way we construct the points
D2 = C'A2 ∩ C''B2 and D3 = C'A3 ∩ C''B3.
No three of C', D1, D2 and D3 are collinear. Suppose not, then dim of
<C',D1,D2,D3> ≤ 2. Since Ai is on C'Di, i = 1,2,3, we would have
Spacial Geometries are Desarguesian
C'
C''
D1
D3
C
A1
A3
D2
B1
A2
B3
B2
Spacial Geometries are Desarguesian
Theorem 2.7.1: Let P be a projective space of dimension d. If
d ≥ 3, then the theorem of Desargues holds in P.
Pf(cont.):
the points Ai lie in π ∩ <D1,D2,D3,C'>. Since the dimension of this
intersection is at most 1, it is a point or a line; therefore the Ai are
collinear, a contradiction.
By construction of C' it is clear that no three of C', A1, A2, A3 are
collinear. Similarly it follows that C'', D1, D2, D3, B1,B2, B3 also
satisfy the conditions of the theorem of Desargues.
Define ψ = <D1,D2,D3>. By Case I, the lines
D1D2 and A1A2, D2D3 and A2A3, D3D1 and A3A1
Spacial Geometries are Desarguesian
Theorem 2.7.1: Let P be a projective space of dimension d. If
d ≥ 3, then the theorem of Desargues holds in P.
Pf(cont.):
and
D1D2 and B1B2, D2D3 and B2B3, D3D1 and B3B1
intersect in points of the line g = π ∩ ψ. In particular, it follows that
AiAj ∩ BiBj is also a point of g (the point where DiDj meets g).
Therefore, the points Pij are all incident with g.
❑
Spacial Geometries are Desarguesian
C'
C''
D1
D3
C
A1
A3
D2
B1
A2
B3
B2
Spacial Geometries are Desarguesian
Corollary 2.7.2: Let P be a projective plane. The theorem of
Desargues holds in P iff P can be embedded as a plane in a
projective space of dimension ≥ 3.
Note: The converse of this corollary is also true, but requires more
work to prove.