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Introduction The Supertail Main Theorem Future Work

The complete characterization of the minimum size supertail

Esmeralda L. N˘ astase

Xavier University

Joint work with P. Sissokho

Illinois State University

May 18, 2019

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Introduction The Supertail Main Theorem Future Work Definitions Applications Motivation The Minimum Size

◮ V = V (n, q) the vector space of dimension n over GF(q). ◮ A subspace partition or partition P of V , is a collection of

subspaces {W1, . . . , Wk} s.t.

◮ V = W1 ∪ · · · ∪ Wk ◮ Wi ∩ Wj = {0} for i = j.

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Introduction The Supertail Main Theorem Future Work Definitions Applications Motivation The Minimum Size

◮ V = V (n, q) the vector space of dimension n over GF(q). ◮ A subspace partition or partition P of V , is a collection of

subspaces {W1, . . . , Wk} s.t.

◮ V = W1 ∪ · · · ∪ Wk ◮ Wi ∩ Wj = {0} for i = j.

◮ size of a subspace partition P is the number of subspaces in

P.

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Applications

◮ translation planes ◮ error-correcting codes ◮ orthogonal arrays ◮ designs ◮ subspace codes

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Let P be any partition of V .

◮ P has type d md1 1

. . . d

mdk k

, if for each i, there are mdi > 0 subspaces of dim di in P, and d1 < d2 < · · · < dk.

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Introduction The Supertail Main Theorem Future Work Definitions Applications Motivation The Minimum Size

Let P be any partition of V .

◮ P has type d md1 1

. . . d

mdk k

, if for each i, there are mdi > 0 subspaces of dim di in P, and d1 < d2 < · · · < dk.

Problem

◮ What are the necessary and sufficient conditions for the

existence of a partition of V of a given type?

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Every partition P of V satisfies:

◮ packing condition k

• i=1

mdi(qdi − 1) = qn − 1

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Every partition P of V satisfies:

◮ packing condition k

• i=1

mdi(qdi − 1) = qn − 1

◮ dimension condition

U, W ∈ P, U = W = ⇒ dim(U) + dim(W ) ≤ n

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Let P be a partition of V .

◮ σq(n, s) = the min size of any partition of V in which the

largest subspace has dim s.

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Theorem (Heden, Lehmann, N., and Sissokho, 2011, 2012). Let n, m, s, and r be integers such that 1 ≤ r < s, m ≥ 1, and n = ms + r. Then σq(n, s) =          qs + 1 for 3 ≤ n < 2s, qs+r

m−2

• i=0

qis + q⌈ s+r

2 ⌉ + 1 for n ≥ 2s.

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Let P be a partition of V of type d

md1 1

. . . d

mdk k

.

◮ For any s such that d1 < s ≤ dm, the set S of subspaces in P

• f dim less than s and with greatest subspace dim t is called

the st-supertail of P.

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Theorem (Heden, Lehmann, N., and Sissokho, 2013). Let P be a partition of V . If S is an st-supertail of P, then |S| ≥ σq(s, t).

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Theorem (Heden, Lehmann, N., and Sissokho, 2013). Let P be a partition of V . If S is an st-supertail of P, then |S| ≥ σq(s, t).

• Corollary. If s ≥ 2t and |S| = σq(s, t), then the union of the

subspaces in S forms a subspace of dim s.

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Conjecture (2013). If t < s < 2t and |S| = σq(s, t) = qt + 1,

then the union of the subspaces in S forms a subspace.

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Conjecture (2013). If t < s < 2t and |S| = σq(s, t) = qt + 1,

then the union of the subspaces in S forms a subspace.

Theorem (Heden, 2009). Let P be a partition of V of type

d

md1 1

. . . d

mdk k

. If S is the tail of P, i.e., all subspaces in S have the same dim d1 = t, s.t. |S| = qt + 1 and d2 = s < 2t, then the subspaces of S form a subspace of dim 2t.

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Theorem (N. and Sissokho, 2017). Let P be a partition of V of

type d

md1 1

. . . d

mdk k

, and let S be an st-supertail of P s.t. |S| = σq(s, t) and t < s < 2t. If one of the following conditions holds (i) S contains subspaces of at most 2 different dimensions (ii) s = 2t − 1 (iii) All the subspaces in P \ S have the same dimension s, then the union of the subspaces in S forms a subspace W , and either (a) d1 = t, md1 = qt + 1, and dim W = 2t, or (b) d1 = a and d2 = t, with md1 = qt and md2 = 1, and dim W = a + t.

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Theorem (N. and Sissokho, 2018). Let P be a partition of V .

Suppose S is an st-supertail of P such that t < s < 2t and |S| = σq(s, t) = qt + 1. Then (i) the union of the subspaces in S forms a subspace W (ii) S is a subspace partition of W whose type is either

◮ tqt+1, or ◮ t1aqt, for some 1 ≤ a < t.

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Let P be a partition of V .

• 1. Let Θk = (qk − 1)/(q − 1)

⇒ Θk is the the number of points, i.e., 1-subspaces, in an k-subspace.

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Let P be a partition of V .

• 1. Let Θk = (qk − 1)/(q − 1)

⇒ Θk is the the number of points, i.e., 1-subspaces, in an k-subspace.

• 2. Let H denote the set of all hyperplanes of V . For H ∈ H, and

any integer k ≥ 1,

◮ bH,k =the number of k-subspaces X ∈ S such that X ⊆ H. ◮ βH = t

k=a bH,kqk, where a ≤ dim X ≤ t for any X ∈ S.

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Lemma 1. The number of hyperplanes H ∈ H that contain a

given k-subspace of V is Θn−k = qn−k − 1 q − 1 . In particular, H contains Θn = qn − 1 q − 1 hyperplanes.

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Lemma 2. Let P be a partition of V . Suppose S is an st-supertail

• f P such that

|S| = σq(s, t) = qt + 1, and t < s < 2t. If H ∈ H, then βH ≥ qt and

t

• i=a

miΘi = cqs − 1 q − 1 for some integer c ≥ 1.

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Lemma 3. Let P be a partition of V . Suppose S is an st-supertail

• f P such that |S| = σq(s, t) = qt + 1 and t < s < 2t.

Let W =

X∈S X and let δ = δ(S) denote the number of points,

i.e., 1-subspaces, of W . For H ∈ H, let δH = δH(S) be the number of points in W ∩ H. Then (i) |S| − 1 = qt ≤ βH ≤ cqd + qt = δ(q − 1) + |S| (ii)

H δH = δΘn−1

(iii)

H δH(δH − 1) = δ(δ − 1)Θn−2.

(iv) βH = qδH − δ + |S|. (v)

H βH = |S|Θn − δ.

(vi)

H β2 H = Θn

• |S|2 + δ(q − 1)
• − δ2(q − 1) − δ(2|S| − 1).

(vii)

H (βH − (|S| − 1)) (βH − (δ(q − 1) + |S|)) = 0.

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Theorem (N. and Sissokho, 2018). Let P be a partition of V .

Suppose S is an st-supertail of P such that t < s < 2t and |S| = σq(s, t) = qt + 1. Then (i) the union of the subspaces in S forms a subspace W (ii) S is a subspace partition of W whose type is either

◮ tqt+1, or ◮ t1aqt, for some 1 ≤ a < t.

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Proof (i). Let βH = t

i=a bH,iqi. Then by Lemma 3 (i),(vii), for

any H ∈ H, we have βH = |S| − 1 = qt, or βH = δ(q − 1) + |S| = cqs + qt. Let

◮ x = # of hyperplanes H so that βH = qt ◮ y = # hyperplanes H so that βH = cqs + qt.

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Then

◮ By Lemma 1

x + y = Θn

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Then

◮ By Lemma 1

x + y = Θn

◮ By Lemma 2 and Lemma 3 (v),

xqt+y(cqs +qt) =

• H

βH = |S|Θn−δ = (qt+1)Θn− cqs − 1 q − 1

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Then

◮ By Lemma 1

x + y = Θn

◮ By Lemma 2 and Lemma 3 (v),

xqt+y(cqs +qt) =

• H

βH = |S|Θn−δ = (qt+1)Θn− cqs − 1 q − 1

◮ Solving the above system, yields

x = qn−s(cqs − 1) c(q − 1) and y = qn−s − c c(q − 1) .

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◮ gcd(c, cqs − 1) = 1, gcd(q − 1, qn−s) = 1, and x an integer

⇒ c | qn−s.

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◮ gcd(c, cqs − 1) = 1, gcd(q − 1, qn−s) = 1, and x an integer

⇒ c | qn−s.

◮ Thus, c = qj, for some j > 0, which implies that

x = qn−(s+j)(qs+j − 1) q − 1 and y = qn−(s+j) − 1 q − 1 .

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◮ gcd(c, cqs − 1) = 1, gcd(q − 1, qn−s) = 1, and x an integer

⇒ c | qn−s.

◮ Thus, c = qj, for some j > 0, which implies that

x = qn−(s+j)(qs+j − 1) q − 1 and y = qn−(s+j) − 1 q − 1 .

◮ Let W = X∈S X. Then for H ∈ H with W ⊆ H, we have

βH =

t

• i=a

bH,iqi =

t

• i=a

miqi > qt,

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Thus, the # of hyperplanes containing W is y = qn−(s+j) − 1 q − 1 .

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Thus, the # of hyperplanes containing W is y = qn−(s+j) − 1 q − 1 . For any H ∈ H, W ⊆ H ⇐ ⇒ W ⊆ H.

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Thus, the # of hyperplanes containing W is y = qn−(s+j) − 1 q − 1 . For any H ∈ H, W ⊆ H ⇐ ⇒ W ⊆ H.

◮ y = the # of hyperplanes containing W .

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Thus, the # of hyperplanes containing W is y = qn−(s+j) − 1 q − 1 . For any H ∈ H, W ⊆ H ⇐ ⇒ W ⊆ H.

◮ y = the # of hyperplanes containing W . ◮ dimW = s + j, by Lemma 1.

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Thus, the # of hyperplanes containing W is y = qn−(s+j) − 1 q − 1 . For any H ∈ H, W ⊆ H ⇐ ⇒ W ⊆ H.

◮ y = the # of hyperplanes containing W . ◮ dimW = s + j, by Lemma 1.

Since # of points in W is δ = cqs − 1 q − 1 = qs+j − 1 q − 1

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Thus, the # of hyperplanes containing W is y = qn−(s+j) − 1 q − 1 . For any H ∈ H, W ⊆ H ⇐ ⇒ W ⊆ H.

◮ y = the # of hyperplanes containing W . ◮ dimW = s + j, by Lemma 1.

Since # of points in W is δ = cqs − 1 q − 1 = qs+j − 1 q − 1 = the # of points in W

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Thus, the # of hyperplanes containing W is y = qn−(s+j) − 1 q − 1 . For any H ∈ H, W ⊆ H ⇐ ⇒ W ⊆ H.

◮ y = the # of hyperplanes containing W . ◮ dimW = s + j, by Lemma 1.

Since # of points in W is δ = cqs − 1 q − 1 = qs+j − 1 q − 1 = the # of points in W = ⇒ W = W is a subspace of dim s + j.

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Proof (ii). We consider the following cases:

• 1. mt = |S| = qt + 1, i.e., all the subspaces in S are t-subspaces
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Proof (ii). We consider the following cases:

• 1. mt = |S| = qt + 1, i.e., all the subspaces in S are t-subspaces
• 2. mt ≤ qt

◮ If mt > 1, then dim W ≥ 2t, by dim condition.

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Proof (ii). We consider the following cases:

• 1. mt = |S| = qt + 1, i.e., all the subspaces in S are t-subspaces
• 2. mt ≤ qt

◮ If mt > 1, then dim W ≥ 2t, by dim condition. ◮ By the packing condition, W has less than q2t − 1 vectors

⇒ contradicts dim W ≥ 2t. Thus, mt = 1.

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◮ Suppose S has 1 subspace of dim t and subspaces of dim

a1, . . . , aℓ, s.t. a1 < · · · < aℓ < t, for some ℓ ≥ 2.

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◮ Suppose S has 1 subspace of dim t and subspaces of dim

a1, . . . , aℓ, s.t. a1 < · · · < aℓ < t, for some ℓ ≥ 2.

◮ Since mt = 1 and W = X∈S X has dim s + j,

qt(qs+j−t−1) = (qs+j−1)−(qt−1) =

aℓ

• i=a1

mi(qi−1) < qt(qaℓ−1),

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◮ Suppose S has 1 subspace of dim t and subspaces of dim

a1, . . . , aℓ, s.t. a1 < · · · < aℓ < t, for some ℓ ≥ 2.

◮ Since mt = 1 and W = X∈S X has dim s + j,

qt(qs+j−t−1) = (qs+j−1)−(qt−1) =

aℓ

• i=a1

mi(qi−1) < qt(qaℓ−1), Thus, qs+j−t < qaℓ ⇒ j < t + aℓ − s.

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◮ dim W = s + j and S contains subspaces of dim t and aℓ.

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◮ dim W = s + j and S contains subspaces of dim t and aℓ. By

the dim condition s + j ≥ t + aℓ = ⇒ j ≥ t + aℓ − s,

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◮ dim W = s + j and S contains subspaces of dim t and aℓ. By

the dim condition s + j ≥ t + aℓ = ⇒ j ≥ t + aℓ − s, = ⇒ contradiction, i.e ℓ = 1. Let a = aℓ.

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◮ dim W = s + j and S contains subspaces of dim t and aℓ. By

the dim condition s + j ≥ t + aℓ = ⇒ j ≥ t + aℓ − s, = ⇒ contradiction, i.e ℓ = 1. Let a = aℓ.

◮ S has type t1aqt.

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Theorem

Let P be a subspace partition of V . If S is an st-supertail of P of size |S| = σq(s, t), then the set of points covered by the subspaces in S forms a subspace.

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• Fact. If P is a partition of V of type d

md1 1

. . . d

mdk k

and S is an st-supertail of P, then |S| ≥ σq(s, t) .

• Question. If |S| > σq(s, t), then what is the next size of S?
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Thank you!

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