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 = , Q =
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 tg = t(t-1g) = 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. ❑ O O Translations Lemma 3.5.2: Let t ∈ T be an arbitrary translation. Choose an arbitrary point P of A and define P' = t(P). Viewing P and P' as vectors of V, we can describe t by: t(X) = X + P' - P for all points X ∈ A.
Pf: The map X → X + P' – P is a translation of A. This translation maps P to P', but t maps P to P' and since there is a unique translation mapping P to P', t must be this map. ❑ Semilinear Automorphisms
We now turn our attention to G . O
Def: Let V be a vector space over the division ring F, and let λ be an automorphism of F. A map g of V into itself is called a semilinear map with companion automorphism λ if for all v, w in V and all a in F we have: g(v+w) = g(v) + g(w), and g(av) = λ(a)g(v).
If λ is the identity automorphism, then g is just a linear map. Example Recall that in a finite field GF(q), with q = pe, the map x → xp is an automorphism of the field. Let V be a 3-dimensional vector space over GF(2e). The map g: V → V given by g (x,y,z) = (y2, z2, x2) is a semilinear map with companion automorphism being the squaring automorphism.
To see this, let v = (x,y,z) and w = (x',y',z'), then g(v) =(y2,z2,x2) and g(w) = (y'2,z'2,x'2). Now, v + w = (x+x', y+y', z+z') and we have, g(v+w) = ((y+y')2,(z+z')2,(x+x')2) = (y2 + y'2, z2 + z'2, x2+x'2) = g(v) + g(w), and g(av) = ((ay)2, (az)2, (ax)2) = (a2y2, a2z2, a2x2) = a2(y2,z2,x2) = a2 g(v). Additive Maps Lemma 3.5.3: Every element of G is an additive map, that is, O s ∈ G implies that s(P+Q) = s(P) + s(Q), for all P,Q ∈ A. O Pf: Recall the definition of addition of points. P + Q = t (Q), where P t is the unique translation mapping O to P. Geometrically, we can P see this in the projective space P. If P, Q and O are not collinear: O P* is the center of the translation Q P P+Q t (as an elation of P). So, t (Q) P P
P* Q* lies on QP*. OQ* is mapped to PQ* since Q* is a fixed point. Thus, t (Q) lies on both PQ* and P QP*. Additive Maps Lemma 3.5.3: Every element of G is an additive map, that is, O s ∈ G implies that s(P+Q) = s(P) + s(Q), for all P,Q ∈ A. O Pf(cont.): Since s is a collineation, we have P+Q = P*Q ∩ Q*P ⇒ s(P+Q) = s(P*)s(Q) ∩ s(Q*)s(P). By the geometric definition of addition we have s(P) + s(Q) = s(P)*s(Q) ∩ s(Q)*s(P).
O Since s has center O and stabilizes H, we see that s(P) P s(P)* = s(P*) (and the same for Q), so we have P* s(P)* s(P*) s(P+Q) = s(P) + s(Q). Additive Maps Lemma 3.5.3: Every element of G is an additive map, that is, O s ∈ G implies that s(P+Q) = s(P) + s(Q), for all P,Q ∈ A. O Pf(cont.): Furthermore, since P – Q , Q and O are not collinear, we have s(P) = s(P – Q + Q) = s(P-Q) + s(Q), i.e., s(P-Q) = s(P) – s(Q). But we also have s(P-Q) = s(P + (-Q)) = s(P) + s(-Q). Comparing these equations gives us s(-Q) = -s(Q) for all points Q.
Now suppose that P, Q and O are collinear, so Q = aP. Let R be a point not on this line. Then s(P+Q) = s(P+Q + R – R) = s(P + Q + R) – s(R) = s(P) + s(Q + R) – s(R) = s(P) + s(Q) + s(R) – s(R) = s(P) + s(Q). ❑ Semilinear Maps Theorem 3.5.4 : Every element of G is a semilinear map of the O vector space V. Pf: Lemma 3.5.3 shows that each map of G is additive, so we need to O show the existence of a companion automorphism. For any non-zero element a ∈ F, and any point X, not O, of A, the points O, X and aX are collinear. Since s ∈ G is a collineation which O fixes O, the points O, s(X) and s(aX) are also collinear, thus s(aX) is a scalar multiple of s(X), i.e., s(aX) = λ (a)s(X), where the notation X indicates that the scalar may depend on X. However, this is not the case, as we now show. Suppose that O, X and Y are distinct and noncollinear points. Now, s(a(X+Y)) = λ (a) s(X+Y) = λ (a) [s(X) + s(Y)] X+Y X+Y = λ (a) s(X) + λ (a) s(Y). X+Y X+Y On the other hand, Semilinear Maps
Theorem 3.5.4 : Every element of G is a semilinear map of the vector space V. O Pf(cont.): s(a(X+Y)) = s(aX + aY) = s(aX) + s(aY) = λ (a)s(X) + λ (a)s(Y). X Y Therefore, since s(X) and s(Y) are linearly independent vectors, λ (a) = λ (a) = λ (a). X X+Y Y Now suppose that O, X and Y are distinct and collinear points. Consider a point Z of A not on OX. By the above we obtain: λ (a) = λ (a) = λ (a). X Ζ Y Now define λ (0) = 0, for all X in A and λ (a) = 0 for all a ∈ F, and we X O have a map from F to F which satisfies, s(aX) = λ(a)s(X) for all a ∈ F and all X ∈ A. We now need to show that this λ is an automorphism of F. We first show that it is a homomorphism. Semilinear Maps Theorem 3.5.4 : Every element of G is a semilinear map of the vector space V. O Pf(cont.): λ(a+b) s(X) = s((a+b)X) = s(aX + bX) = s(aX) + s(bX) = λ(a)s(X) + λ(b)s(X) = (λ(a) + λ(b)) s(X) so λ(a+b) = λ(a) + λ(b). Also, we have λ(ab) s(X) = s((ab)X) = s(a(bX)) = λ(a)s(bX) = λ(a)λ(b)s(X), and so, λ(ab) = λ(a)λ(b). We now need only show that λ is a bijection to finish the proof. If λ(a) = λ(b), then s(aX) = λ(a)s(X) = λ(b)s(X) = s(bX), but s is a bijection so this implies aX = bX, and so, a = b. Now consider any a ∈ F, and let X be any point other than O of A. The point as(X) is a point on the line through O and s(X). Thus, the pre-image Semilinear Maps
Theorem 3.5.4 : Every element of G is a semilinear map of the vector space V. O Pf(cont.): of as(X) is a point on the line through O and X, namely it is a point Y with Y = bX. Therefore, s(Y) = s(bX) = λ(b)s(X) = as(X), so λ(b) = a and λ is surjective. We have now shown that λ is an automorphism of F, completing the proof. ❑ The Fundamental Theorem
We now turn to the projective space. It is easy to see that each collineation of A induces a collineation of P, but it is remarkable that the converse statement holds and this is the hard part of the following:
Theorem 3.5.8 : [The Fundamental Theorem of Projective Geometry] If P is a Desarguesian projective space and V a vector space such that P = P(V) then α is a collineation of P if and only if there is a bijective semilinear map of V which induces α.
We will not prove this result. Homographies
We can now refine our view of the group of collineations of the projective space P, and see what role the central collineations play in this group.
Def: A collineation of the projective space P = P(V) is called a projective collineation (also known as a homography) if it is induced by a linear collineation of V (and so can be represented by a matrix with respect to some basis of V). Central Collineations
Theorem 3.6.1 : Every central collineation is a projective collineation.
Theorem 3.6.7 : Every projective collineation of P is a product of central collineations.
We will not prove these theorems, but shall give a few examples. Homographies We will examine some collineations in PG(2,4). GF(4) = {0,1,a,a2}. We will work with homogeneous coordinates (x:y:z). Consider the map given by: 1 0 0 x : y : z x : y : z0 0 1= x : z : y= x ' : y ' : z ' . 0 1 0 The fixed points of this map are those with y = z. The line at infinity (x = 0) is stabilized with (0:1:1) being a fixed point. The points of the line kx + y + z = 0 have coordinates (1:y:y+k) or (0:1:1). Under the map, these points become (1:y+k:y) and they satisfy the same equation, so these lines through (0:1:1) are fixed lines. This map is the elation with axis y = z and center (0:1:1), a central collineation. Homographies
Now consider the homology with axis x = 0, center (1:0:0) which maps the point P = (1:1:a) to P' = (1:a:a2) [note that P,P' and the center are collinear on the line z = ay.] Observe that a homography having the form: c 0 0 x : y : z x : y : z :0 b 0 = cx : by : bz 0 0 b
Will fix (1:0:0) and any point on x = 0 (which have coordinates (0:y:z) ). The point (1:1:a) is mapped to (c:b:ba). We can get the map we want by setting c = 1 and b = a. Homographies Recall that a rotation in the Euclidean plane with center at the origin can be given by a matrix. For instance, a rotation by 90o is given by (in affine coordinates): 0 1 X ,Y X ,Y = Y ,− X . −1 0 If we write this in homogenous coordinates, the matrix becomes: 1 0 0 0 0 1. 0 −1 0 The corresponding homography has only one fixed point, so it can not be a central collineation. However, we know that a rotation can be written as a product of two reflections – which are central collineations. Group Names
The group of collineations of P is denoted by PΓ L(n,F) or PΓ L(n,q) if F is a finite field of order q. The subgroup of homographies (projective collineations) is denoted by PGL(n,F) or PGL(n,q).