Notes from February 1

Transcription

1 Notes from February 1 Math 130 Course web site: 1. The order of an affine plane A theorem Throughout, we will suppose that (, ) is a finite affine plane: in particular, it should satisfy the axioms (A1), (A2) and (A3). Lemma. Let l 1 and m be two lines that are not parallel. Then the number of points on m is equal to the number of lines parallel to l 1. Proof. The lines l 1 and m are not parallel, so no line can be parralel to both of them, by Theorem 2 of the previous handout. Thus, a line l that is parallel to l 1 is not parallel to m, and will therefore meet m in exactly one point, (by Theorem 1 from last time). Consider now the set of all lines parallel to l 1. Each of these lines meets m in one point; and furthermore, by axiom (A2) (the parralel postulate ), each point of m lies on exactly one of these lines. This establishes a one-to-one correspondence between the points of m and the lines parallel to l 1 ; and there are therefore the same number of each. The diagram illustrates the proof of the lemma.

2 Using the lemma, we can prove: Theorem 3. In a finite affine plane, any two lines have the same number of points. Proof. Let l and m be any two lines. We will show #l = #m. According to axiom (A3), our affine plane has three non-collinear points. Joining these in pairs, we obtain three lines l 1, l 2 and l 3 no two of which are parallel. Each of l and m can be parallel to at most one of these three lines. Therefore, there is one the these three (we may as well call it l 1 ) which is parallel to neither l nor m. If we apply the lemma to the lines l 1 and m, we learn that #m is equal to the number of lines parallel to l 1. If we apply the lemma to the lines l 1 and l, we learn that #l is also equal to the number of lines parallel to l 1. It follows that #l = #m. Order The standard terminology is to say that an affine plane as order N if every line in the affine plane has exactly N points. The theorem just proved tells us that this is a sensible notion. The previous handout gave diagrams representing affine planes of order 2 and 3 (these are the 4-point plane and the 9-point plane respectively. It is not hard to prove that the number of points in an affine plane of order N must be N 2.

3 The question for which N does there exist an affine plane of order N is wide open. It is known that there is an affine plane of order N whenever N is a power of a prime. There is no known example of a finite affine plane whose order is not a prime power. On the other hand, they may exist. The only other two known facts about this question are: (1) If there exists an affine plane whose order N is 1 or 2 mod 4, then N must be a sum of two squares. 1 (2) There does not exist an affine plane of order 10. The first of these statements rules out, for example, the orders 6 and 14. The second statement was proved using a lot of computer time in the late 1980 s. Does there exist a plane of order 12? No one knows. Does there exist a plane of order 34? Unless there is a new breakthrough, we will never know. 2. Addition in affine planes Given a pair of vectors v and w in 2, we can form their sum v + w using vector addition. 1 A natural number N is the sum of two squares provided only that every prime p which is equal to 3 mod 4 divides N an even number of times. This is the sort of thing that is proved in Math 124, for example.

4 Although we most often think of vector addition in terms of coordinates, the point v + w above can be obtained from v and w geometrically, by constructing a parallelogram with vertices at the origin, v and w. Thought of this way, the construction can be carried out in any affine plane (, ). To carry out the construction in this way, we need two points P and Q in an affine plane, and a third point O to play the role of the origin. We will require that O, P and Q are not collinear. (In particular, this means that O, P and Q are three distinct points, because any two points lie on a line.) Under these conditions, we can construct a point P + Q as follows: (1) take l to be the line OP (which exists and is unique by axiom (A1)); (2) take m to be the line OQ, similarly; (3) take l to be the unique line through Q parallel to l, using axiom (A2); (4) take m to be the unique line through P parallel to m, again using axiom (A2); (5) define P + Q to the unique point of intersection of l and m. At the last step, we need to be assured that l and m cannot be parallel, to ensure that they do meet in a single point. To see that they are not parallel, we can argue as follows. If they were parallel, then l and m would be parallel (by repeated application of Theorem 2 from last time); and then we would deduce that l = m, because l and m are parallel and meet at O; and from this we would conclude that O, P and Q all lie on l and are therefore collinear, which is contrary to our assumption.

5 In class, we did a worked example, using the affine plane of order 3. In the first homework, there is another example to do. There are two points to remember here. First, the point we end up with as P +Q here does depend on our choice of origin O. Second, the construction breaks down if O, P and Q are collinear. Of course, in the coordinate plane 2, we are quite used to being able to add vectors v and w whether or not they are collinear with the origin; but in a general affine plane we cannot do this. Later, we will see what is needed to make addition really work in an affine plane when O, P and Q are collinear. 3. A larger affine plane For later reference, we record here an affine plane of order 4. There are 16 points in this plane, which we can label A, B,..., P. To indicate which sets of four points constitute the lines, we arrange the 16 points in a 4-by-4 array, There are 20 lines, and they are indicated by 4 red dots each in the following picture:

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