Old and new results on the MDS-conjecture J. De Beule ( joint work - - PowerPoint PPT Presentation

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Old and new results on the MDS-conjecture J. De Beule ( joint work - - PowerPoint PPT Presentation

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Introduction old(er) results Lemma of tangents Beyond k p ? Old and new results on the MDS-conjecture J. De Beule ( joint work with Simeon Ball) Department of Mathematics Ghent University February 9, 2012 Incidence Geometry and Buildings

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university-logo Introduction

  • ld(er) results

Lemma of tangents Beyond k ≤ p?

Old and new results on the MDS-conjecture

  • J. De Beule

(joint work with Simeon Ball)

Department of Mathematics Ghent University

February 9, 2012 Incidence Geometry and Buildings 2012

Jan De Beule MDS-conjecture

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university-logo Introduction

  • ld(er) results

Lemma of tangents Beyond k ≤ p?

Definitions

Definition An arc of a projective space PG(k − 1, q) is a set K of points such that no k points of K are incident with a common

  • hyperplane. An arc K is also called a n-arc if |K| = n.

Definition A linear [n, k, d] code C over Fq is an MDS code if it satisfies k = n − d + 1.

Jan De Beule MDS-conjecture

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Lemma of tangents Beyond k ≤ p?

Lemma Suppose that C is a linear [n, k, d] over Fq with parity check matrix H. Then C is an MDS-code if and only if every collection

  • f n − k columns of H is linearly indepent.

Corollary Linear MDS codes are equivalent with arcs in projective spaces.

Jan De Beule MDS-conjecture

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Lemma of tangents Beyond k ≤ p?

Lemma Suppose that C is a linear [n, k, d] over Fq with parity check matrix H. Then C is an MDS-code if and only if every collection

  • f n − k columns of H is linearly indepent.

Corollary Linear MDS codes are equivalent with arcs in projective spaces.

Jan De Beule MDS-conjecture

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Lemma of tangents Beyond k ≤ p?

fundamental questions

What is the largest size of an arc in PG(k − 1, q)? For which values of k − 1, q, q > k, is each (q + 1)-arc in PG(k − 1, q) a normal rational curve? {(1, t, . . . , tk−1) | t ∈ Fq} ∪ {(0, . . . , 0, 1)} For a given k − 1, q, q > k, which arcs of PG(k − 1, q) are extendable to a (q + 1)-arc?

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Lemma of tangents Beyond k ≤ p?

Early results

In the following list, q = ph, and we consider an l-arc in PG(k − 1, q). Bose (1947): l ≤ q + 1 if p ≥ k = 3. Segre (1955): a (q + 1)-arc in PG(2, q), q odd, is a conic. Lemma (Bush, 1952) An arc in PG(k − 1, q), k ≥ q, has size at most k + 1. An arc attaining this bound is equivalent to a frame of PG(k − 1, q). q = 2, k = 3: hyperovals are (q + 2)-arcs.

Jan De Beule MDS-conjecture

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Lemma of tangents Beyond k ≤ p?

MDS-conjecture

Conjecture An arc of PG(k − 1, q), k ≤ q, has size at most q + 1, unless q is even and k = 3 or k = q − 1, in which case it has size at most q + 2.

Jan De Beule MDS-conjecture

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Lemma of tangents Beyond k ≤ p?

more (recent) results

Conjecture is known to be true for all q ≤ 27, for all k ≤ 5 and k ≥ q − 3 and for k = 6, 7, q − 4, q − 5, see overview paper of J. Hirschfeld and L. Storme, pointing to results of Segre, J.A. Thas, Casse, Glynn, Bruen, Blokhuis, Voloch, Storme, Hirschfeld and Korchmáros. many examples of hyperovals, see e.g. Cherowitzo’s hyperoval page, pointing to examples of Segre, Glynn, Payne, Cherowitzo, Penttila, Pinneri, Royle and O’Keefe.

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Lemma of tangents Beyond k ≤ p?

more (recent) results

An example of a (q + 1)-arc in PG(4, 9), different from a normal rational curve, (Glynn): K = {(1, t, t2+ηt6, t3, t4) | t ∈ F9, η4 = −1}∪{(0, 0, 0, 0, 1)} An example of a (q + 1)-arc in PG(3, q), q = 2h, gcd(r, h) = 1, different from a normal rational curve, (Hirschfeld): K = {(1, t, t2r , t2r +1) | t ∈ Fq} ∪ {(0, 0, 0, 1)}

Jan De Beule MDS-conjecture

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Lemma of tangents Beyond k ≤ p?

arcs in PG(2, q)

tangent lines through p1 = (1, 0, 0): X1 = aiX2 p2 = (0, 1, 0): X2 = biX0 p3 = (0, 0, 1): X0 = ciX1 Lemma (B. Segre)

t

  • i=1

aibici = −1

Jan De Beule MDS-conjecture

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Lemma of tangents Beyond k ≤ p?

arcs in PG(2, q)

tangent lines through p1 = (1, 0, 0): X1 = aiX2 p2 = (0, 1, 0): X2 = biX0 p3 = (0, 0, 1): X0 = ciX1 Lemma (B. Segre)

t

  • i=1

aibici = −1

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Lemma of tangents Beyond k ≤ p?

coordinate free version

T{p1} := (X1 − aiX2) T{p2} := (X2 − biX0) T{p3} := (X0 − ciX1) Lemma T{p1}(p2)T{p2}(p3)T{p3}(p1) = (−1)t+1T{p1}(p3)T{p2}(p1)T{p3}(p2)

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Lemma of tangents Beyond k ≤ p?

coordinate free version in PG(k − 1, q)

Lemma (S. Ball) Choose S ⊂ K, |S| = k − 3, choose p1, p2, p3 ∈ K \ S. TS∪{p1}(p2)TS∪{p2}(p3)TS∪{p3}(p1) = (−1)t+1TS∪{p1}(p3)TS∪{p2}(p1)TS∪{p3}(p2)

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Lemma of tangents Beyond k ≤ p?

Interpolation

Lemma (S. Ball) Let |K| ≥ k + t > k. Choose Y = {y1, . . . , yk−2} ⊂ K and E ⊂ K \ Y, |E| = t + 2. Then 0 =

  • a∈E

TY(a)

  • z∈E\{a}

det(a, z, y1, . . . , yk−2)−1

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Lemma of tangents Beyond k ≤ p?

Exploiting interpolation and Segre’s lemma

Let |K| ≥ k + t > k. Choose Y = {y1, . . . , yk−2} ⊂ K and E ⊂ K \ Y, |E| = t + 2, r ≤ min(k − 1, t + 2). Let θi = (a1, . . . , ai−1, yi, . . . , yk−2) denote an ordered sequence, for the elements a1, . . . , ai−1 ∈ E Lemma (S. Ball) 0 =

  • a1,...,ar∈E

r−1

  • i=1

Tθi(ai) Tθi+1(yi)

  • Tθr(ar)
  • z∈(E∪Y)\(θr ∪{ar})

det(ar, z, θr)−1 , The r! terms in the sum for which {a1, . . . , ar} = A, A ⊂ E, |A| = r, are the same.

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university-logo Introduction

  • ld(er) results

Lemma of tangents Beyond k ≤ p?

Exploiting interpolation and Segre’s lemma

Let |K| ≥ k + t > k. Choose Y = {y1, . . . , yk−2} ⊂ K and E ⊂ K \ Y, |E| = t + 2, r ≤ min(k − 1, t + 2). Let θi = (a1, . . . , ai−1, yi, . . . , yk−2) denote an ordered sequence, for the elements a1, . . . , ai−1 ∈ E Lemma (S. Ball) 0 = r!

  • a1<...<ar∈E

r−1

  • i=1

Tθi(ai) Tθi+1(yi)

  • Tθr(ar)
  • z∈(E∪Y)\(θr ∪{ar })

det(ar, z, θr)−1 .

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Lemma of tangents Beyond k ≤ p?

avoiding some restriction

Lemma Suppose that K is an arc in PG(k − 1, q), then one can construct an arc K′ in PG(|K| − k − 1, q), with |K| = |K′|.

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Lemma of tangents Beyond k ≤ p?

Segre product

Let A = (a1, . . . , an) and B = (b0, . . . , bn−1) be two subsequences of K of the same length n and let D be a subset

  • f K \ (A ∪ B) of size k − n − 1.

Definition PD(A, B) =

n

  • i=1

TD∪{a1,...,ai−1,bi,...,bn−1}(ai) TD∪{a1,...,ai−1,bi,...,bn−1}(bi−1) and PD(∅, ∅) = 1.

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Using Segre’s lemma again

Lemma PD(A∗, B) = (−1)t+1PD(A, B), PD(A, B∗) = (−1)t+1PD(A, B), where the sequence X ∗ is obtained from X by interchanging two elements.

Jan De Beule MDS-conjecture

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Lemma of tangents Beyond k ≤ p?

Interpolation again

Suppose that |K| = q + 2. Let L of size p − 1, Ω of size p − 2, X and Y both of size k − p be disjoint ordered sequences of K. Let Sτ denote the sequence (sτ(i) | i ∈ τ), τ ⊆ {1, 2, . . . , |S|} for any sequence S. Let σ(Xτ, X) denote the number of transpositions needed to map X onto Xτ. M = {1, . . . , k − p} Lemma 0 =

  • τ⊆M

(−1)|τ|+σ(Xτ ,X)PL∪XM\τ (Yτ, Xτ)

  • z∈Ω∪Xτ ∪YM\τ

det(z, XM\τ , Yτ, L)−1

Jan De Beule MDS-conjecture

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Lemma of tangents Beyond k ≤ p?

Interpolation again

Let E ⊂ Ω, |E| = 2p − k − 2. Let W = (w1, . . . , w2n) be an ordered subssequence of K disjoint from L ∪ X ∪ Y ∪ E. Corollary 0 =

n

  • i=1

det(yn+1−i, X, L)

  • z∈E∪Y∪W2n

det(z, X, L)−1 . . . which is a contradiction

Jan De Beule MDS-conjecture

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Lemma of tangents Beyond k ≤ p?

Corollary (Ball and DB) An arc in PG(k − 1, q), q = ph, p prime, h > 1, k ≤ 2p − 2 has size at most q + 1.

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Lemma of tangents Beyond k ≤ p?

  • B. Cherowitzo.

Bill Cherowtizo’s Hyperoval Page. http://www-math.cudenver.edu/~wcherowi/research/h 1999.

  • J. W. P

. Hirschfeld and L. Storme, The packing problem in statistics, coding theory and finite projective spaces: update 2001, in Developments in Mathematics, 3, Kluwer Academic Publishers. Finite Geometries, Proceedings of the Fourth Isle of Thorns Conference, pp. 201–246.

Jan De Beule MDS-conjecture