Technische Universität München Zentrum Mathematik

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1 Technische Universität München Zentrum Mathematik Prof. Dr. Dr. Jürgen Richter-Gebert, Bernhard Werner Projective Geometry SS 8 Solutions for Worksheet 9 (8-6-8) Question. Intersecting conics Classwork Let the following two matrices describe two conics: 3 A = 3 B = 8 4 The goal of this task is to determine all points of intersection between these two conics. a) Determine a linear combination C = A + λb which represents a degenerate conic, and compute its matrix. b) Transform this matrix C into an equivalent matrix C of rank. c) Read the constituent lines g and h from this matrix. d) Determine a matrix M g in such a way that for every vector v R 3 the equation M g v = g v holds. Also compute M h in the same way. If you don t remember the structure of such a matrix by heart, try to derive it instead of looking it up in your lecture notes. e) Find the matrix D of a degenerate dual conic that has all those lines as tangents which pass through either of the points of intersection between g and A. f) Find a matrix D equivalent to D but of rank. Make use of the fact that you know the homogeneous coordinates of g. Read the coordinates of two points of intersection from this matrix. g) In the same fashion, compute the points of intersection between line h and conic B. h) Write down a list of all the points during this computation where you either made an arbitrary choice, or where the problem statement made an arbitrary choice, even though a computation based on a different choice would have lead to the same final results. i) Identify the other degenerate conics which could result from subtask a), and split at least one of them into lines k and l. j) Compute all those points of intersection between the lines g, h, k, l which are also points of intersection between the conics. Use this to verify your previous results.

2 Solution: a) A conic is degenerated if the determinant of its symmetric matrix is zero. This leads to: 3 = det(a + λb) = det 3 + λ λ λ = 3 λ λ λ λ 4(λ ) 3 λ λ = (λ ) 3 λ λ 4 At this point we already see that λ = has to be a root of the determinant, which is enough for this part. C = A + B = b) Because of the very easy structure of C one can spot a suitable rank--matrix basically immediately: C = C + = c) A non-zero row or column of the matrix is a representative of one of the factor lines. C = = (,, ) g = h = d) If you do not know the structure of such a matrix M g by heart, reconstruct it from an example: v v g v = v = v 3 = v = M g v M g = v 3 v v 3 v v 3 v h v = v = = v = M h v M h = v 3 v v 3 e) A line m is tangential to D iff the intersection of m and g lies on A. (g m) T A (g m) = m T M T g A M g m We can deduce the matrix of the conic from this: D = M T g A M g = = 8 = 8 3

3 f) The matrix D must result as a linear combination of D and M g, as g is the connecting line of the two points described by the dual conic to D. D = D + λm g = 8 + λ In this easy case, one can see without computing the determinant that λ = ± leads to a rank--matrix. D = D + M g = 8 + = 8 4 D = 4 (,, ) This leads to the two intersections of g and A: p = p = g) Analogously we get ther intersections of h and B: E = M T h B M h = = = 4 E = E + M h = = (,, ) p 3 = h) At the following points we made an arbitrary choice: p 4 = In the linear combination in part a): Which conic has the coefficient. Another possibility would have been C = B + λa. In part a): Which of three roots of the determinant to use. These roots correspond to the three degenerated conics through the four points. In part b) and f), we could have used the transposed matrix or could have done without the scaling. The assignment of the letters g an h to rows and columns is arbitrary. The other assignment leads to swapping the lines just as transposing the matrix does. As we explicitly use g in the succeeding computations, this choice influences the next steps, but not the entirety of the intersection points. It does not matter which line and which conic to intersect. One can intersect both with A, both with B or g with B ans h with A. 3

4 i) To determine the other degenerated conics, we have to expand the determinant from part a). 3 λ λ det(a + λb) = (λ ) 3 λ λ 4 ( = (λ ) (λ 3) 3 λ λ ) 4 3 λ λ λ = (λ ) (λ 3) ((3 λ) 4 ( λ) ( ) (λ ) ) = (λ ) (λ 3) ( 4λ + λ + λ ) = (λ ) (λ 3) (8 λ) = 4 (λ ) (λ 3) (λ 4) With this, we know the other roots. And the respective conics unfold as linear combinations: A + 3B = = (,, ) 4 4 A + 4B = 4 = (,, ) k = l = j) We can the four intersection points also by the following computation: g = h = k = l = g k = = p g l = = p h k = = p 4 h l = = p 3 This ansatz yields the same four intersection points (up to scalars) as above. To close this exercise, a picture to visualise the conics and their finite and infinite intersections: A B 4

5 Question. Decomposing a conic Decompose the following conic into its constituent lines: 4 A = Solution: First, we determine the dual conic B = A = = From this matrix we can infer (by considering any non-zero row or column) that the intersection of the two lines is p = (3,, ) T. So it would have been enough to compute a single row or column. When looking a the structure of the adjugate matrix, we see that a single column is computed via -determinants with alternating sign. We recall this structure from the cross product. So, computing directly from A: = 8 = p 7 9 The anti-symmetric matrix we have to add to A to get rank is therefore a multiple of C = M p = 3 3 Now we have to find a suitable multiple. When A + αc has rank, every -determinant will be zero. Using the bottom right sub-matrix, we get α 7 3α 8 = (3α) = α = ± = ± = ±3 3 3 Of course, we could have used any of the other two -determinants with the top left one we would even have avoided division. What becomes apparent with the above solution is, however, that the respective coordinate of p appears in the denominator of the solution for α. The corresponding determinant appears in the numerator. This determinant we could also read from the adjugate. With the proportion factor α = 3 we found, we can compute the rank--matrix: D = A + 3C = = 4 = (,, ) So, the two lines have the homogeneous coordinates. g = h = 4

6 Question 3. From where was a picture taken Given the following points in RP Homework A = B = 3 C = D = E = you are asked to compute the point F RP (distinct from A through E) which satisfies (A, B; C, D) F = (B, C; D, E) F = 3 a) Each of these cross ratios defines a conic. The value has special significance for cross ratios. What is special about the corresponding conic, and how can you exploit this fact in subsequent tasks? b) Find matrices which represent the two conics for the given cross ratios. c) Find the point F which satisfies the above condition. Solution: a) A cross-ratio of, or in RP means that at least two of the points are the same. Concretely, for (A, B; C, D) = we have A B or C D. This is obvious from the definition of the cross-ratio. Transferring this to the plane viewed from a fifth point F, this means that the respective connecting lines collapse. I.e. either ABF are collinear or CDF. So the conic degenerates into two lines. b) The conic corresponding to the first equation can be directly determined from the two lines. (A B) (C D) T = 3 A suitable symmetric representation would be 6 6 Q + Q T = 6 T 3 3 = (,, ) = = Q 6 6 For the conic with cross-ratio 3 it is best to orientate oneself towards exercise on sheet 8. First, we have to determine the conics with cross-ratio, and. (B D) (C E) T = (,, ) = R 4 (B C) (D E) T = (,, ) = = R 4 4 (B E) (C D) T = (,, ) = = R 7 7 6

7 These conics should be symmetrised to be able to handle the next step. S = R + R T = 8 S = R + R T = 8 7 S = R + R T = 7 The next step is to adjust the scale defined by these three conics we need representatives for S and S such that their sum is (a multiple of) S. Because of the zeros we can read the respective factors directly from the top left and bottom right corner: S + S = S ( So we found the basis matrices: S takes on the role of and S ) the role of ( 3 this scale we get by a point. ) ( ). The cross-ratio of 3 for S + S = 3 + = 6 = K 7 6 c) In theory, we would now have to intersect the conics Q and K. We know, however, that Q is degenerated and we even know its components. Furthermore, we know that Q and K have the points B, C and D in common. It thus suffices to intersect the line AB with the conic K. 3 A B = 6 6 M A B = L = M T A B K M A B = = = This matrix describes the two intersection points of the line AB with the conic K. To decompose it into these points we have to add an anti-symmetric matrix to get a rank--matric. To be precise, we have to add a suitable multiple of M A B. We can get the proportion factor for example from the minor obtained by deleting the middle row and middle column. This has the advantage that the middle coordinate of A B is and we therefore do not have to divide by it. α = = = 4 + = 3 L + α M A B = 3 8 = 3 (, 3, ) So, the conic K intersects the line AB in the known point B and the desired point F. It is therefore determined as F = 3 3 7

8 The computation to get this final result was quite long and some intermediate result are still uncomfortably large. Moreover, we had to take care of many signs. This makes computational errors very likely. So it makes sense to verify the result at this point. Computing the cross-ratios given in the instructions with the computed F does result in the given values. The result is correct. Question 4. Degrees of freedom of conics a) How many real degrees of freedom does one have in choosing an arbitrary conic in RP? b) Suppose you have chosen such a conic. Now we are looking for projective transformations which map the unit circle to this conic. How many real degrees of freedom are there to choose such a transformation? In subsequent subtasks this figure shall be denoted as k. c) Suppose you are given k pairwise distinct points P i on the unit circle, and also k pairwise distinct points P i on a conic represented by some matrix B. Describe how one could obtain the matrix of the projective transformation which maps the unit circle onto B. d) Execute this computation using specific numbers. You are free to choose coordinates in such a way that the computations become as simple as possible. The conic B shall not be degenerate, though, and not a circle either. Furthermore, you should perform all steps as you described them in c), even though your specific choice of coordinates could make some steps more simple. You may of course use these easier solutions to verify your results. Solution: a) A conic is described by an equation with 6 coefficients. This corresponds to the 6 entries of a triangular matrix, from which the other other entries of a symmetric matrix yield. A multiple of such an equation or matrix gives, however, the same conic. This leaves degrees of freedom. An alternative approach is the following: A conic can be described by points in the plane. Each point gives real degrees of freedom. But each of these points can be shifted along the conic without changing the conic. This eliminates degree of freedom. This leaves degrees, too. b) A conic is parametrised by RP. Hence, three points on the conic fix a projective scale. All other points are uniquely determined by their cross-ratio w.r.t. these three points. Each of these three base points can be chosen on the conic, giving degree of freedom each. I.e. 3 in total. c) One possibility is to construct the point P 4 which is in harmonic position w.r.t. P through P 3 i.e. satisfying (P, P ; P 3, P 4 ) =. It can be determined, for example, by Hesse s transfer principle. Analogously, we can determine a point P 4 as its image. Then we have four points and their images which determine a unique projective transformation, as we know. d) There are many reasonable choices for suitable numbers. Furthermore, there is a certain trade-of between easy computations and a high learning effect. The more complicated the coordinates are the better one has to understand the general approach. This solution treats a rather easy case: The transformation which is the final result will be given in advance. For example, one could choose the transformation that scales every x-coordinate by. The image of the unit circle would then be an ellipse with its center of symmetry at the origin. As defining points one could use points on the coordinate axes. The corresponding details would like this: A = B = P = P = P = P = P 3 = P 3 = 8

9 With this we can compute the points P 4 and P 4. To do so, we take the tangents at the conic in P and P, connect their intersection with P 3 and intersect this line with the conic. Starting with the intersection of the tangents: Q (A P ) (A P ) = = = Q 8 Q (B P ) (B P ) = = = Q And then... P 4 = Q T AQ P 3 P3 T AQ Q = P 3 Q = P 4 = Q T BQ T P 3 P 3 T BQ Q = P 3 Q = Now we need transformations mapping the standard basis of RP onto the corresponding point quadruple. λ λ µ = µ = τ τ λ λ µ = µ = τ τ M = M = Of these maps, M will gets executed in the opposite direction and afterwards M in forward direction. The compound map is therefore 4 M = M M = = That is exactly the transformation we chose in the beginning. This means the computation is correct. It can be applied to other points and conics by using other numbers. The method is sufficiently general. 9