Feb. 12 -- intro to arcs and ovals; intersection triangles You're already had to look up "oval" on page 17 to construct a (3,6,22) Steiner system from the square Paley 2-(11,5,2) design. An arc in a 2-design is a set S of points no three of which are contained in a block. (The book uses "n-arc" for an arc with n points, and limits the definition to square designs.) Equivalently, the intersection of S with any block B has size 0, 1, or 2. For geometrically obvious reasons (at least in the context of projective planes where the blocks are called "lines"), B is said to be "tangent" to S if the intersection has size 1, and "secant" if it has size 2; a block disjoint from S is said to be "passant" to S. Why do we care? It turns out to be a useful notion in several contexts when \D is a square design; in particular it will let us classify and analyze projective planes of small order. For starters (see p.18), suppose we ask whether a projective plane \D = (X,\B) is extendable: is there a 3-design \E such that \D is \E_p? If so, we already know the points of \E -- namely X u {p} -- and some of the blocks, namely B u {p} for any B in \B. Since B is square, any two blocks meet in \lam=1 point. For each x in X, \E_x has the same parameters as \E_p, so is also square with the same value of \lam; hence any two of its blocks meet in a unique point too. Thus any two blocks of \E are either disjoint or meet in 2 points. (That's a special case of the argument at the beginning of the proof of Cameron's Theorem 1.35 .) So, any block of \E that does not contain p is an arc. Indeed we'll see it's an arc of maximal size, which we'll call a "hyperoval". How big can an arc be? Prop. 1.46: A point of an n-arc in a square 2-(v,k,\lam) design lies on (n-1)\lam secants and k-(n-1)\lam tangents. In particular n <= 1 + (k/\lam) [because the count of tangents is nonnegative]. Proof: let S be an arc and p be one of its points. There are n-1 other points of S, hence n-1 pairs {p,q} in S, each lying in \lam blocks B; each such B is a tangent, which determines q uniquely, so there are indeed (n-1)\lam such B's. In a square design each point is in a total of k blocks, and if such a block is not one of the (n-1)\lam secants then it's a tangent. QED An _oval_ is an arc for which the tangent count is either 1 or 0; that is, for which n is either 1 + (k-1)/\lam (an oval of _Type I_) or 1 + k\lam (_Type II_) respectively. You must know the distinction but need not memorize which is which. You should, however, remember that ovals that have no tangents (and thus intersect each block in either 0 or 2 points) are called _hyperovals_. Example: a line complement in P^2 is a hyperoval; removing any of its 4 points yields a Type I oval. In the Paley 2-(11,5,2) design an oval has Type I and 1 + (5-1)/2 = 3 points; there are 11*10*3/3! = 55 ovals. Type I is geometrically familiar -- and we'll see (though not fully prove in class) that this familiarity extends to algebraic projective planes P^2(k). Observe that Type I exists ==> k is 1 mod \lam Type II exists ==> k is 0 mod \lam Both exist ==> \lam = 1 (so projective plane). As usual, none of these is <==> ! For example: Prop. 1.47 (p.18) If a square 2-(v,k,\lam) design has a hyperoval then k == \lam mod 2. Pf.: Fix a point p outside S -- such must exist, else S=X so k=2 and we have the trivial 2-(3,2,1) design (which indeed is an exception). The number of secants containing p is n \lam / 2 (why?), which is (\lam + k) / 2, so this must be an integer, QED. Example: if a projective plane of order q has a hyperoval then 2|q. In fact we can say more. If there's a hyperoval in a projective plane then removing any point yields an oval. Conversely, an oval yields a hyperoval iff all its tangents meet in a point. Prop. 1.48 (p.18): If k == \lam mod 2 and there's a Type I oval S then any point on the design lies on either one or all tangents. Cor.: in a projective plane with q even, all the tangents meet in a unique point p (proof: intersect any two of them), called its "center" (the book says "nucleus", which feels rarer to me, or "knot", which I don't think I've seen elsewhere). So we recover a hyperoval S u {p}. Proof: We have k(k-1) = (v-1) \lam (because it's a square 2-design) and n = 1 + (k-1)/\lam. Claim: k, \lam, n are odd. Proof: if \lam were even then k-1 would be odd, contradicting \lam | k - 1. Since \lam is odd, k-1 is even, so k is odd and so is n = 1 + (even)/(odd). It follows that the number of tangents through each point is odd (so in particular not zero). Apply "variance trick" again, computing the first and second moments of the numbers of tangents going through the points of X; this time it will turn out that the variance must be as large as possible. Let N_i be the number of points that lie on exactly i tangents. [NB the book calls it n_i, which would lead to the monstrosity n_n.] Then the N_i sum to v = #X; i N_i sums to (# tangents) * (# points per tangent) = n k; and i (i-1) N_i sums to (# pairs of tangents) * (# points per pair) = (n^2-n) \lam. Rearranging yields sum_i (i-1) (i-n) N_i = 0. Since N_i = 0 for i=0 and i>n it follows that the only nonzero N_i are N_1 and N_n, QED. Exercise: use these counts to verify that N_n = \lam, so one might say that in general there are \lam "nuclei". Examples: in P^2, "conics" = zeros of "nonsingular quadratic forms" = zeros of homogeneous quadratic polynomials P(x,y,z) over k that are irreducible even over the algebraic closure. [That's a necessary proviso: consider x^2 + xy + y^2 mod 2, or x^2 + y^2 mod 3.] Note/recall that the set of zeros of a homogeneous P is well-defined even though its value at a point of P^2 in general is not. A conic is clearly an arc because restricting to a line yields a homogeneous quadratic polynomial in two variables that is not identically zero (else P had a linear factor). In fact it is known -- though we will not prove this (yet another possible paper topic) -- that: (i) all conics are equivalent under PGL_3(k); (ii) all conics are Type I ovals (that is, they have q+1 points); and remarkably, one can prove in general that (iii) if q is odd then every oval is a conic (Segre [1954]). Given (i), we can prove (ii) by checking it for one conic, and it is convenient to use the conic xz-y^2; indeed the general point is (r^2:rs:s^2) for some (r:s) in P^1(k) (check that this makes sense; algebraic geometers recognize this as the first nontrivial example of a "rational normal curve", or the image of the "degree-2 Veronese map" P^1 --> P^2). So a conic is identified with of the q+1 points of P^1. When q is even: (iii) is true for q=2 and q=4 (we'll say more about the case q=4 soon); but for q>4 we can start with a conic, add the center, and remove some other point to get a non-conic oval; and for q>8 (except q=32) there are even hyperovals not of the form "conic + center". Finally I owe you some words about "intersection triangles" (p.21). That's a convenient way to do a bunch of inclusion-exclusion calculations at once without an explicit sum of binomial coefficients etc. The general setup is: we have a subset Y of X such that, for each s up to l=#Y, the number of blocks containing any s-element subset S of Y depends only on s; and we want to count the blocks whose intersection with Y is exactly S. The hypotheses always hold if \D is a t-design with l <= t, and in some other cases, e.g. if Y is an arc, or notably a block of a Steiner system, plus one other case explored in problems 6 and 7. The method is as follows, and illustrates the paradox that sometimes a problem is made easier by generalizing it (even though one might expect that proving a more general result should be at least as hard). Here we ask: given nonnegative i and j with j <= i <= l, an i-point subset I of Y, and a j-point subset J of I, how many blocks intersect I in J? (The problem we started with is the special case I=Y.) This depends only on i and j; the count \nu_{i,j} can be computed recursively by starting from the known values \nu_{i,i} (e.g. \nu_{i,i} = \lam_i for i <= s) and then using \nu_{i,j} = \nu_{i-1,j} - \nu_{i,j+1} to fill each row of a triangle from right to left. See the example on page 21 (Table 1.1); yes, the Gross responsible for Theorem 1.55 (1974) on page 22 (which gives necessary conditions for the bottom row of such a triangle in the Steiner case to contain a zero) is our Benedict Gross.