Technische Universität München
Zentrum Mathematik
Prof. Dr. Dr. Jürgen Richter-Gebert, Bernhard Werner
Projective Geometry SS 2016
www-m10.ma.tum.de/ProjektiveGeometrieSS16
Solutions for Worksheet 11 (2016-06-30)
— Classwork—
Question 1. Projective reflection
This task will generalize euclidean reflections to other geometries.
a) On a projective line RP1 there are three given points ∞, a and (a + b). How can you construct the point (a − b),
which is in a certain sense the mirror image of (a + b) in a?
b) The mirror image P 0 of a point P after reflection in a line g can in the Euclidean plane be constructed using the
following steps:
1. Determine a line h which is perpendicular to g and passes through P .
2. Fins the point P 0 on h which lies at the same distance from the intersection of these two lines as P , but on
the opposite side.
Expand these construction steps so it will become clear exactly how each step can be achieved. Restrict your
description to tools which can be transferred to other geometries: meet, join, harmonic set as well as tangents,
polars and poles with respect to a fundamental conic. Don’t refer to the fundamental conic in any way except
for the last three operations just mentioned.
c) Formulate a corresponding construction for point reflections. Again restrict your set of tools like in the previous
subtask.
d) Describe a general set of transformations (which will subsequently be called “projective reflections”) which contains Euclidean reflections both in a line and in a point as special cases. Every element of this class of transformations shall be described using one point and one line. The transformations from this class should not refer to
any fundamental conic.
e) Provide a short argument as to why a projective reflection is an involution, i.e. why executing the transformation
twice results in the identity transformation.
f) How many pairs of preimage and image points are required to define such a projective reflection in general?
g) Show that every projective reflection is a projective transformation.
h) Which relation between the defining line and point must hold with respect to some fundamental conic K if the
transformation should describe a reflection in the stricter sense of a given geometry? Choose your condition in
such a way that for the Euclidean plane, reflections in a point or in a line are the only two permissible situations.
To formalize the required relation, you may want to adapt ideas from the definition of primal-dual pairs. A
projective reflection which satisfies this extra condition shall henceforth be called a “K-reflection”.
i) Describe the structure of a K-reflection in hyperbolic geometry. Can you again distinguish between reflections
in a line nd reflections in a point?
j) Also investigate the situation in elliptic geometry. Imagine that geometry on the surface of a sphere. How does
this affect the distinction between reflections in a line and in a point?
k) Show at least for hyperbolic geometry that every K-reflection preserves distances and angles. For which other
geometries can your proof be generalized?
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Solution:
a) We have (a, ∞; a + b, a − b) = −1. So we can use the standard construction of harmonic points.
b) A line reflection can be described as follows:
(a) Q = Bg, the pole of the reflection line
(b) h = Q ∨ P
(c) M = g ∧ h
(d) (P, P 0 ; M, Q) = −1 on h
The pole of a line is the point through which all perpendicular lines have to go. In Euclidean geometry, this can
easily be checked considering the matrix multiplication.
c) Here, we let Q be the point at which we reflect—in order to guarantee consistency with the subsequent parts of
the exercise. There are several ways to formulate reflections at a point, e.g. translating it into line reflections.
For the following parts, this formulations is the most handy one
(a) g = AQ, die Polare zu Q
(b) h = Q ∨ P
(c) M = g ∧ h
(d) (P, P 0 ; M, Q) = −1 auf h
In Euclidean geometry, the polar of any finite point is the line at infinity. So, g is just the line at infinity and
M is the point at infinity of the line h. Thus, we could have computed AP instead of AQ. But that would have
been harder to generalise.
d) Starting with a point Q and a line g, we do the following:
(a) h = Q ∨ P
(b) M = g ∨ h
(c) (P, P 0 ; M, Q) = −1
When g is a line and Q its orthogonal point at infinity, we get a Euclidean line reflection. When Q is a finite
point and g the line at infinity, we get a point reflection. Due to the uniform (but, maybe, unintuitive) choice of
labelling above, this construction correspond exactly to the steps 2 to 4 of the preceding parts.
e) As Q, P and P 0 all lie on h, we get this line both by Q ∨ P and Q ∨ P 0 . Hence, M does not change, when
we interchange P and P 0 . And as the harmonic position is also invariant under swapping one pair, the whole
construction is symmetric w.r.t. the interchange of P and P 0 . So, the image of the image is the start point again.
f) In general, two pairs are enough—e.g. P1 7→ P10 and P2 7→ P20 . We get Q = (P1 ∨ P10 ) ∧ (P2 ∨ P20 ). With this, we
can construct Mi on both lines by (Pi , Pi0 ; Mi , Q). From this we get g = M1 ∨ M2 .
g) It is well-known that the Euclidean line reflection, e.g. at the x-axis, can be written as a matrix multiplication
(say with a matrix S). Thus, it is a projective transformation. Ever instance of a projective reflection can be
reduced to this case via a projective transformation T : we can always map g to the x-axis and Q to the point
(0, 1, 0)T . As the projective reflection only uses incidence geometry in its construction, it is projectively invariant.
So we can write every projective reflection as the multiplication with a matrix T −1 · S · T . I.e. it is a projective
transformation.
h) Q and g have to form a pair of pole and polar. In the case of a degenerated fundamental object, this connection
is not always unique. So we define (Q, g) to be a pole/polar pair iff
AQ = λg
Bg = µQ
When we face a degenerated fundamental object, λ or µ is zero.
i) In hyperbolic geometry, line and point reflections are quite similar, when we consider all of RP 2 . When the line
intersects the conic, its pole is outside and for the inside we get a line reflection. When the point is in the inside
of the conic, the polar lies fully on the outside and in the inside we have a point reflection.
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j) In elliptic geometry, both concept collapse. The elliptic pole of the equator of a sphere is represented by the
north and south pole, which are identified with each other as antipodal points. The analogue holds for other
great circles. A rotation around such a pole by 180° corresponds to the reflection at the great circle. The reason
is that P and P 0 are also represented by antipodal points on the sphere.
k) To show that lengths and angles are preserved, we have to show that the conic is invariant under projective
reflections. To do so, it is handy to work with the hyperbolic geometry of the unit circle. The following considerations should be visualised over the real numbers. But they work perfectly fine, when intermediate results are
complex.
The line g intersects K in X and Y . The tangents x and y in these points intersect in M . Another line h intersects
g in Q and K in the two points P and P 0 . What remains is to show that (P, P 0 ; M, Q) indeed is −1.
The construction so far is known from the lecture—from the the context of Hesse’s transfer principle. It verifies
that (P, P 0 ; X, Y )K = −1. With this we have
−1 = lim (P, P 0 ; X, Y )Z
Z→Y
= (Y ∨ P, Y ∨ P 0 ; Y ∨ X, AY ) = (Y ∨ P, Y ∨ P 0 ; Y ∨ M, Y ∨ Q)
= (P, P 0 ; M, Q)Y = (P, P 0 ; M, Q)
Consequently, P 0 is indeed the image of P under the K-reflection. So, points on the fundamental object get
mapped to points of the fundamental object. I.e. it is a proper K-transformation.
— Homework—
Question 2. Angle bisectors
In this task you will show that the angle bisectors in a triangle will meet in a single point in any Cayley-Klein geometry.
You may assume that the triangle is non-degenerate, i.e. that the three corners are not collinear.
Note: This question has also been discussed in the lecture. Try to follow the arguments suggested by the individual
subtasks, supplemented by what you remember from the lecture. Simply reading and reproducing the proof from the
lecture is not the intended way.
a) How many ways are there to construct an angle bisector for a given pair of lines in Euclidean geometry?
b) Draw a sufficiently generic triangle, and draw in all the angle bisectors between its edges. How many points are
there – apart from the triangle corners – where two or more of these bisectors meet? And how many bisectors
meet in these points?
c) Consider two lines which intersect in the center of a circle. Describe how you can construct a (Euclidean) bisector
using only the operations tangent, meet and join. Utilize the symmetry of the situation.
d) Given two lines which intersect in any point inside the fundamental conic. Describe a construction which can be
used to construct a hyerbolic angle bisector for the given pair of lines.
e) Reformulate your construction in such a way that the point of intersection between the two original lines is not
used at all.
f) The circle depicted below, with center M , is to be used as the fundamental conic of hyperbolic geometry. Using a
ruler and possibly a compass, construct the interior angle bisectors of the triangle ABC, i.e. those angle bisectors
which pass through the interior of the triangle.
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B
M
C
A
g) Verify that these three angle bisectors indeed meet in a single point.
h) Prove that these bisectors must meet in a single point. Recall a theorem which has six tangents to a conic as
part of its premise. If you fail to recall such a theorem, dualizing the situation might help.
i) How far can this construction of angle bisectors be generalized to other geometries? Does the proof still hold?
Solution:
a) There are two possibilities.
b) There are 4 intersection point of three angle bisectors each. In one, the three inner angle bisectors intersect. In
the others, two outer and one inner bisector are involved.
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c) For every line take one intersection point with the circle. Then, construct the tangents in this points at the circle.
Finally, connect their intersection point with the intersection point of the original lines.
d) In the center of the circle, Euclidean and hyperbolic angle coincide. Hence, there the construction is the same,
too. The construction is projectively invariant and ever point can be translated to the center by an hyperbolic
transformation. Therefore, the construction can be used for any point to get an hyperbolic angle bisector.
e) Starting with two lines, we can consider the other pair of intersection points with the fundamental object.
Intersecting the corresponding tangents results in a point which also lies on the same angle bisector. So, the
angle bisector can alternatively be described as the connecting line of two tangent intersection.
f) The complete construction looks like this. The center of the circle can be used to draw the tangents, as they are
perpendicular to the radii.
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B
M
C
A
g) True, as can be seen in the drawing.
h) The construction corresponds to the theorem of Brianchon—i.e. the dualised theorem of Pascal.
i) In elliptic geometry, intermediate results would be complex, so the construction could only be done with complex
tools. The proof holds, however, as the algebraic relations are the same. For other geometries, the construction
fails due to the fact that some degenerated objects collapse and that the resulting object are not uniquely
determined anymore.
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