Midterm Questions 1. Prove that a system of points and lines satisfying the following conditions is an projective plane of order n (with n > 1): a. There are at least n2 + n + 1 lines. b. There are at least n + 1 lines through every point. c. Each pair of distinct lines meet at a unique point. d. Each line contains n + 1 points. We must show that the axioms for a projective plane are satisfied. Axiom 2' is c. Axiom 3 (each line has at least 3 points) follows from d since n > 1. Axiom 4 (there are at least two lines) follows from a since n > 1. Thus, we only have to prove Axiom 1: Two points determine a unique line. Let P and Q be two distinct points. If there were more than one line joining P and Q we would contradict c. So, assume that there is no line joining them. Take a line g through Q (there are at least 3 by b), and Midterm Questions 1. Prove that a system of points and lines satisfying the following conditions is an projective plane of order n (with n > 1): a. There are at least n2 + n + 1 lines. b. There are at least n + 1 lines through every point. c. Each pair of distinct lines meet at a unique point. d. Each line contains n + 1 points. Consider the lines through P. Each must intersect g by c. This gives at least n+1 points on g, not including Q, so g has at least n+2 points, contradicting d. Thus, P and Q are joined by a unique line. Midterm Questions 2. Let P = <u>, Q = <v>, and R = <w> be three noncollinear points of a projective plane P(V). Let P' = <v + aw> be a point on QR, Q' = <w + bu> a point on RP, and R' = <u + cv> a point on PQ. Prove the theorem of Menelaus: the points P', Q' and R' are collinear iff abc = -1. As P, Q and R are independent points in a 2 dimensional projective space, they form a basis. With respect to this ordered basis, the points P', Q' and R' have homogeneous coordinates P' = (0:1:a), Q' = (b:0:1) and R' = (1:c:0). These three points are collinear iff they are dependent iff the determinant of the matrix formed by the coordinates is zero. 0 1 a ∣b 0 1∣ = 1abc = 0. 1 c 0 Midterm Questions 3. A property that a set of axioms may or may not have is independence, i.e., that no axiom in the set can be proved from the remaining axioms. Independence of a set of axioms is usually proved by providing examples in which all except one of the axioms hold. By constructing 4 examples, prove that the axioms for a projective space given in the text are independent. That is, for each axiom, construct an example in which the other three axioms hold, but this one fails. Axiom 1 fails, others hold. Axiom 2 fails, others hold Any Affine plane, in particular the Euclidean plane. Midterm Questions 3. A property that a set of axioms may or may not have is independence, i.e., that no axiom in the set can be proved from the remaining axioms. Independence of a set of axioms is usually proved by providing examples in which all except one of the axioms hold. By constructing 4 examples, prove that the axioms for a projective space given in the text are independent. That is, for each axiom, construct an example in which the other three axioms hold, but this one fails. Axiom 3 fails, others hold. Axiom 4 fails, others hold Midterm Questions 4 a. Let F and K be fields. Suppose that f: F\{0} → K\{0} is a multiplicative isomorphism. Extend f to a map from F to K by defining f(0) = 0. Prove that this extended f is a field isomorphism if and only if f(a + 1) = f(a) + 1 for all a ∈ F. b. Let F be GF(16) constructed from the primitive polynomial x4 + x + 1 and let K be GF(16) constructed from the primitive polynomial x4 + x3 + 1. Prove that F and K are isomorphic by producing a specific isomorphism. a.) Since f is a multiplicative isomorphism, f(1) = 1. If f is a field automorphism then f(a+1) = f(a) + f(1) = f(a) + 1. In the other direction, the only thing that needs to be checked is that the extended f is additive, f(a+b) = f(a) + f(b). Now, f(a+b) = f(b(a/b + 1)) = f(b)f(a/b + 1) = f(b)(f(a/b) + 1) = f(b)f(a/b) + f(b) = f(a) + f(b), if b ≠ 0 and the statement is clearly true if b = 0. Midterm Questions f(ai) = b-i mod 15 a + 1 = a4 → b11 = b14 + 1 a = a b = b a2 + 1 = a8 → b7 = b13 + 1 a2 = a2 b2 = b2 a3 + 1 = a14 → b = b12 + 1 a3 = a3 b3 = b3 4 → 14 11 a4 = a + 1 b4 = b3 + 1 a + 1 = a b = b + 1 5 10 → 5 10 a5 = a2 + a b5 = b3 + b + 1 a + 1 = a b = b + 1 6 13 → 2 9 a6 = a3 + a2 b6 = b3 + b2 + b + 1 a + 1 = a b = b + 1 a7 = a3 + a + 1 b7 = b2 + b + 1 a7 + 1 = a9 → b6 = b8 + 1 a8 = a2 + 1 b8 = b3 + b2 + b a8 + 1 = a2 → b13 = b7 + 1 a9 = a3 + a b9 = b2 + 1 a9 + 1 = a7 → b8 = b6 + 1 a10 = a2 + a + 1 b10 = b3 + b a10 + 1 = a5 → b10 = b5 + 1 a11 = a3 + a2 + a b11 = b3 + b2 + 1 a11 + 1 = a12 → b3 = b4 + 1 a12 = a3 + a2 + a + 1 b12 = b + 1 a12 + 1 = a 11→ b4 = b3 + 1 a13 = a3 + a2 + 1 b13 = b2 + b a13 + 1 = a6 → b9 = b2 + 1 14 3 14 3 2 a = a + 1 b = b + b a14 + 1 = a3 → b12 = b + 1 15 15 a = 1 b = 1 a15 + 1 = 0 → 0 = 1 + 1 F K Midterm Questions 5. Prove that if U is a subspace of a projective space and if g is a line that intersects U in just one point, then there is a hyperplane containing U that does not contain g. Let the projective space be P of dimension d. Clearly, U ≠ P. If U is a hyperplane then we are done. If U is a subspace of dimension t, then t < d-1. By Lemma 1.3.10, there are d – t > 1 hyperplanes of P which intersect at U. Each of these contains U, and if they all contained the line g, their common intersection would be larger than U, a contradiction. So, at least one of them does not contain g. The Representation Theorems The Fundamental Theorem of Projective Geometry The Fundamental Theorem Now that we have determined the structure of essentially all projective geometries (projective planes still remain wildly different in their structures) we are in a position to classify the collineations (the bijections which preserve this structure). This classification is known as the Fundamental Theorem of Projective Geometry. We will assume throughout this section that P is a Desarguesian projective space, so there exists a division ring F and vector space V over F so that P = P(V). Let H be a hyperplane of P and A = P\H be the affine space determined by H. Also, let O be a fixed point of A. It will be convenient to work in the affine space A. Recall that the points of A can be identified with the vectors in V and that the lines of A are the cosets of the 1-dimensional subspaces of V. Decomposition Let T = T(H) be the translation group of A, i.e., the group of all elations of P with axis H. Let G be the group of all collineations of A, and let G be the subgroup of G consisting of all O collineations of A which fix the point O. [The facts that G is a group and that G is a subgroup of it are easy exercises.] O Lemma 3.5.1: Each g in G can be uniquely written as g = tg O where t in T and g in G . O O This lemma says that every collineation of A can be described as a product of a translation and an element of G . Thus, we need only O describe all the elements of these two subgroups in order to describe all collineations. Decomposition Lemma 3.5.1: Each g in G can be uniquely written as g = tg where t in T and g in G . O O O Pf: Let t be the unique translation that maps O to g(O). Define g = t-1g. O Note that g (O) = t-1g(O) = t-1(g(O)) = O, so g in G . Furthermore, O O O -1 tg = t(t g) = g, O so we can write any g as g = tg where t ∈ T and g ∈ G . O O O We now need to show that this expression is unique. Suppose that g = tg and g = t'g '. We then have tt'-1 = g 'g -1. But the map on the left O O O O is in T and the map on the right is in G , so this map is in T ∩ G = {id}. O O Therefore t = t' and g = g ', so the expression is unique.