Technische Universität München Zentrum Mathematik

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1 Technische Universität München Zentrum Mathematik Prof. r. r. Jürgen Richter-ebert, ernhard Werner Projective eometry SS www-m.ma.tum.de/projektiveeometriess Solutions for Worksheet (--) Question. Moulton plane lasswork The so-called Moulton plane is an affine plane, in which esargues theorem does not hold everywhere. Its projective closure can be described as follows: P = P R L = L R x a y I b z c { ax + by + cz = if x z > and a b > ax + by + cz = otherwise a) Verify, that the special case in the incidence relation I only applies to points to the right of the y axis and lines with finite negative slope. b) Show that the incidence relation is well defined. c) reate a descriptive picture of this plane. d) escribe a process by which joining lines and points of intersection can be computed. e) Verify that this projective Moulton plane satisfies all the axioms of a projective plane. f) reate a construction which demonstrates that Pappos theorem is not generally true in this plane. g) reate a construction which demonstrates that esargues theorem is not generally true in this plane. Solution: a) or points one can argue like this: ür Punkte kann man so argumentieren: x z > x z z endlich x z > x z > x > with x being the x-coordinate of the dehomogenised point. or lines it is best to rewrite the its linear equation: = ax + by + c by = (ax + c) y = a b x c b a b > a b b a b > a b > a b <

2 b) Once again, well-defined means idenpendent from the choice of the representative. I.e.: pil (λp)i(µl) x z > λx λz = λ (x z) > a b > µa µb = µ (a b) > a x + b y + c z = µa λx + µb λy + µc λz = µλ(a x + b y + c z) = a x + b y + c z = µa λx + µb λy + µc λz = µλ( a x + b y + c z) = It is important, that the squares of real numbers are always positive. So the criterion for the different cases as well as the computations in each case are independent of the choice of the representatives. c) The best way to think of the points of the Moulton plane is to see them as the points of the ordinary projective plane. nd the lines reprsented by the set of points incident to them. Lines of non-negative slope behave like normal lines in the plane i.e. they have the same points as in RP. This also holds for vertical lines, i.e. lines with infinite slope. Lines with negative slope get bent when crossing the y-axis: To the right of the axis the slope increases by the factor. The following picture shoes points and all lines between them. x = d) irst, the algorithm to determine the Join of p = (x, y, z ) T and p = (x, y, z ) T. or that we build the subsidiary line (a, b, c ) T := p p. When it has (finite) negative slope, i.e. when a b >, one has to translate the points to the right of the y-axis to the place where they would lie on the subsidiary line. I.e. to te place they would have to be when then line has no bent. { x := x falls a b > und x z > x sonst { x := x falls a b > und x z > x sonst rom p := (x, y, z ) and p := (x, y, z ) one gets the Join. p p := p p. The same ansatz works for Meets: One starts with the intersection of the lines without bents. nd if it lies to the right of the y-axis, one has to rewrite the lines, such that the fit the doubled slope to the right of the y-axis. x a y := l l = b b z c c { a i := a i falls x z > und a i b i > a i sonst a a l l := b b c c a für i {, } The calculations above result in the zero vector if and only if the given points are linearly dependent. I.e. the result can always be interpreted as an geometric object for different input objects.

3 e) The methods above always lead to a unique result. onsidering this, Join and Meet are already uniquely defined. ut strictly speaking, we showed only their existence. In theory, there might be other objects, fulfiling the incidence relations, not found by the algorithm. oth cases, two distinct Joins of given points and two distince Meets of given lines, can be formulated and discussed in equivalent ways: ssume we have p, p P; l, l L so dass p Il, p Il, p Il, p Il To contradict this, a proof by cases comes in handy. W.l.o.g. we assume that x z x z and a b a b. all : x z x z. oth points on the left. No special cases, just ordinary linear equations. So the properties of RP hold and not all incidences can be fulfilled. eide Punkte links. eine Sonderbehandlung, sondern normale eradengelichungen. aher gelten igenschaften von RP, und es können nicht alle vier Inzidenzen gleichzeitig auftreten. all : a b a b. oth lines have non-begative slope. Nothing special, ordinary linear equations. all : < x z x z ; < a b a b. ll objects get a special treatment. One can multiply all x i by and gets the ordinary linear equation. all : x z < x z ; < a b a b. oth lines with negative slope, but points on different sides. Special treatment for p, but nor for p. Multiply x by to get the ordinary linear equation. all : < x z x z ; a b < a b. oth points on the right, but line slopes have different signs. Special treatment for l. Multiply a by to get the ordinary linear equation. all : x z < x z ; a b < a b ; z. Points on different sides, slopes of lines with different signs, points finite. Moulton line does not change the sign of its slope while crossing the y-axis. Whence, this signs can be computed from the difference of the dehomogenised y-coordinates: y z y z a b y z > y z a b > oth situations contradict the assumption of this case. all : x z = < x z ; a b < a b ; z =. One point at infinity and one on the right, lines with differently signed slopes. ere, the point at infinity determines the slope. x y a b x y < a b > So, we have shown, that no diefferent points can be incident to different lines. With the existence of Join and Meet, the first two axioms hold. s a projective base we can choose. The negative sign in the last point guarantees that the special case of the incidence relation need not be considered. To prove that these are in general position, we can thus deduce from the proof in for RP. f) It is remarkably hard to find a construct in which the theorem is cleary depicted as wrong. In particular, when starting with a given collinear points as a violated conclusion. It is easier to start with given concurrent lines. The easiest way is to use this concurrency in a corner of the theorem as the conclusion and using only one bent line with a very visible bent. The rest of the configuarion should be completely on one side of the y-axis, as otherwise further bent lines complicate things.

4 x = g) esargue s Theorem as well is best depicted with one set of non-concurrent lines one side as the violated conclusion and the rest of the theorem on the other. x = Question. Symmetries of the ano plane onsider the ano plane consisting of the lines (,, ), (,, ), (,, ), (,, ), (,, ), (,, ), (,, ) Imagine the depicted triangular grid to extend infinitely in all directions.

5 a) nter the numbers... into the circles of the triangular grid in such a way that every yellow triangle corresponds to one line of the ano plane. urthermore, all diamonds should contain the same pattern of numbers, so they are related by a translation. very number should appear in every diamond exactly once. b) escribe several automorphisms τ of the ano plane which satisfy the equation τ = id. lso describe several automorphisms satisfying τ = id. c) What is the cardinality of the group of automorphisms of the ano plane? an you explain this number using the structure of the triangular grid? In order to answer this question, recall the arbitrary choices you made while filling in the triangular grid. Solution: a) You can see an appropriate pattern on the right. The was placed arbitralrily in the center of the diamond and the in one of the neighbouring fields. This determines the place of the. or the there are free circles inside the diamond and it fixes the position of the. The rest is then determined as well. The above the follows from the symmetry requirement. nd together they fix the. b) long the lines building the edges of the yellow triangles, every number occurs with period : efore one number occurs again, every other was entered into one of the circles. translation along these lines represents thus an automorphism whose seventh power is the identity. xamples for such permutations (in cycle notation) are: ( ) ( ) ( ) symmetry of order is given by the rotation by. s a center one can use either a circle or the center of one of the (yellow or white) triangles. The following examples are constructed by choosing the first cycle as the center of rotation and mapping the other numbers accordingly: ()( )( ) ( )( )() ( )( )() c) The number of automorphisms correspond to the number of labelings of the triangle lattice compliant to the combinatorics of the ano plane: The yellow triangles are the structure which has to be kept intact. nd the periodicity assures that mapping the labeling within one diamond to another is well-defined. There are ways to place the within the diamond and neighbouring cirlces for the or neighbouring triangles for the whole line and ways to orientate it. This leaves places for the to be. The rest is then determined. The total number is thus = 8. This matches the possible ways to choose points in the ano plane in general position, to use them as a projective base.

6 omework Question. 9 configurations (n r, n r ) configuration is often simply called a n r configuration. ere you see several drawings of 9 configurations. () () () () () () () (8) (9) a) Identify which of these images represent isomorphic incidence structures. Label the points in such a way that for isomorphic configurations, the labels of all collinear triples of points agree, so that the isomorphism is reafily apparent. Most of the subsequent subtasks can be addressed for all isomorphic configurations at once. b) hoose one of the given configurations which is not isomorphic to Pappos theorem. Try to reproduce this drawing (without measuring the original). Which parameters can be chosen arbitrarily during that process? c) onsider how you could compute coordinates for the points in the drawing you just created (or at least tried to create). d) or every configuration, determine the dual incidence structure, and decide which of the depicted primal configurations are isomorphic to that. Support this claimed isomorphism by labeling the lines of the original in a suitable way.

7 e) onstruct additional 9 configurations and investigate, which of the depicted configurations are isomporphic to yours. Solution: valuable tool to analyse such 9 -configuration is the not-adjacent-relation. It indicates which points are not next to each other. There are exactly points not-adjacent to a given point, as through every point run exactly lines and on every line there are additional points. One can use the non-adjacency to run through the graphs. eginning at an arbitrary point there are exactly two neighbours to go to. fter deciding for one (and actually gong there) only point is left, which was not already visited. This way, one can go through the graph until the return to the starting point. (graph-theoretical, not geometrical) circle is formed, containing at least some points of the graph. If some points were omitted, they can serve as starting points for new circles. a) The number of circles gained from the above heuristic is an invariant of the incidence structure it is independent from the embedding into the plane. So, if two graphs have a different number of cirlces, they cannot be isomorphic. aution: The opposite is, in general, false. raphs with the same number of circles need not be isomorphic. Only with the additional knowledge gained e.g. from books that there are only combinatorically different 9 -configuration and that they have different circle number one can use this number as a feature for classification. One can use the non-adjacency also for finding equivalent labelings. Though it is not enough to look at the circles alone. ut this, at least, limits the number of labelings one has to check. In total, there are the following classes, distinct by their circle numbers. Within the classes one can find equivalent labelings, like the ones given below.

8 circles with points each. This is the configuration of Pappos Theorem. () () ircles: ollinear triples of points: circle with 9 points. () (8) (), ( ), (),,,,,,,, ircles: ollinear triples of points: () () ( ),,,,,,,, 8

9 circle with and points, respectively. (9) () () ircles: ollinear triples of points: ( ), (),,,,,,,, b) In general, such a drawing will not work. Other than Pappos Theorem, the remaining 9 -configurations are no incindence-geometrical theorems. That means, that the last incidence drawn will not automattically be fulfilled. It works only for special starting positions. good example is configuration (). Starting with an equilateral triangle on the outside an positioning the corners of the middle triangle at the centers of the outside edges there is still one corner of the innermost triangle which can be choosen freely. nd only afterwards it will become cleaer whether this choice lead to a correct incidence structure. c) or the case above in configuration () one work with barycentric coordinates (i.e. in the drawing plane embedded in x + y + z = ): = = = = = = 9

10 = x x x = ( ) ( ) = x = x x x x x = ( ) ( ) = x = x x x x = ( ) ( ) = = x x x + (x )! = x x = x x, = ± So there are possibilities for to make a working configuration. d) The structures are all dual to themselves. One way to see this, is to specify a symmetric incidence matrix as dualising just means transposing this matrix. circle: a b c d e f g h k circles: a b c d e f g h k circles: a b c d e f g h k

11 aution: The existence of a symmetric incidence matrix is only a sufficient condition for an incidence structure to be self-dual. ut it is not clear that any self-dual configuration can be written as a symmetric incidence matrix. e) ombinatorically, there are only three different 9 -configurations. very configuration found can be assigned to one of the classes above by counting its circles. Question. initely Pappos a) What is the smalles finite projective plane in which a non-degenerate instance of Pappos theorem can be realized, and in which this theorem is generally true? b) reate a sketch of the corresponding projective plane, and in that emphasize one instance of the theorem. c) lso create a sketch of this theorem in RP. Label the points in both sketches correspondingly, so that the common combinatorics become clear. Solution: a) s Pappos Theorem needs 9 points, the ano plane only allows degenerated version of the theorem. Since Pappos Theorem holds in every projective plane constructed from a field, the plane P is the smallest plane in which the theorem appears in a non-degenerated way and in which it holds universally. That it holds follows from the the fact, that the algebraic proof from the lecture works over every field. b) The eassiest way is to start with the (known) depiction of of the plane P and marking an instance of the theorem in it. c) When using the same labels for corresponding points, it becomes clear the points in the middle line of the finite depiction appear reordered. That is an artefact from the embedding the combinatorics, i.e. the incidences, are all that matter.

12 Question. ollineations iven the following points in P : = = = = = i a) ind a projective transformation, which fixes the points, and and maps onto. Write down a function τ a : P P which describes the effect of this transformation for the points of the plane. b) ind a collineation which is not a projective transformation but still maps the points as stated above. escribe this second operation using a function τ b : P P c) onsider both the collineations you obtained in the previous subtasks, and apply each of them to the point = (i,, ) T. ompare the results. d) emonstrate that the map τ b is well defined. Solution: a) We want a matrix M fulfilling the following equations: M = α M = β M = γ M = δ s a scalar in front of the matrix does not alter the projective geometric interpretation of the matrix as a map, on of the scalars in the equations can be choosen arbitrarily, e.g. δ =. s of the pre-images are already unit vectors, the respective columns of M have to be the images of these unit vectors. With the fourth point we get the following linear system: α β = i γ α β = i γ Mit diesen Vorfaktoren ist die bbildungsmatrix bestimmt als amit ergibt sich die geforderte unktion als M = i τ a : p M p b) Mapping a number to its complex conjugate is an automorphism of. nd conjugating all entries of a vector cannot be described by a projective map. So we want a map gained from a matrix proudct after conjugating and still fulfilling the following equations: M Ā = α M = β M = γ M = δ s the pre-images are all real and the images do not get conjugated, the matrix M is the same as above. This results in the function τ b : p M p

13 c) Inserting the point yields i i τ a () = M = M =, i i τ b () = M Ē = M =. oth images really lie in different equivalence classes. s the projective map τ a was fully characterised by the pairs of pre-image and image points, τ b indeed cannot be a projective transformation. d) To show that τ b is well-defined we have to establish that the equivalence class of τ b ([p]) is independent from the representative of [p]. So let λp be a generic representative of the equivalence class [p]. Then we have τ b (λp) = M (λp) = M ( λ p) = λ(m p) = λ τ b (p) [τ b (p)]