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SETr and AG(4, 3)Combinatorics Seminar
Robert Won
(Joint with M. Follett, K. Kalail, E. McMahon, C. Pelland)
“Partitions of AG(4, 3) into maximal caps”, Discrete Mathematics (2014)
October 19, 2016 1 / 41
The card game SETr
• SETr is played with 81 cards. Each card is characterized by 4attributes:
• Number: 1, 2 or 3 symbols.• Color: Red, purple or green.• Shading: Empty, striped or solid.• Shape: Ovals, diamonds or squiggles.
• A set is three cards where each attribute is independently eitherall the same or all different
October 19, 2016 An introduction to SETr 2 / 41
The card game SETr
The number of attributes that are the same can vary.
Shape and shading are the same,color and number are different.
All attributes are different.
October 19, 2016 An introduction to SETr 3 / 41
The card game SETr
• To start the game, twelve SETr cards are dealt face up.• If a player finds a set, he takes it and three new cards are dealt.• If there are no sets, three more cards are dealt. The three cards are
not replaced on the next set, reducing the number back to twelve.• The player who finds the most sets is the winner.
Image adapted from Davis and Maclagan“The Card Game SET”
October 19, 2016 An introduction to SETr 4 / 41
The card game SETr
Can you find a set?
October 19, 2016 An introduction to SETr 5 / 41
Finite affine geometry
• For affine geometry on a plane, there are three axioms:
1 There exist (at least) 3 non-collinear points.2 Any two points determine a unique line.3 Given a line ` and a point P not on `, there is a unique line through
P parallel to `.
• The order of a finite geometry is the number of points on each line.• Using these axioms, we can draw AG(2, 3), the affine plane of
order 3.
October 19, 2016 Finite affine geometry 6 / 41
Finite affine geometry
October 19, 2016 Finite affine geometry 7 / 41
Seeing geometry in SETr
• A deck of SETr cards is a finite affine geometry. The cards are thepoints; three points are on a line if those three cards form a set.
• This works because any two cards uniquely determine a thirdcard that completes the set.
October 19, 2016 Connecting SETr to geometry 8 / 41
Seeing geometry in SETr
• A deck of SETr cards is a finite affine geometry. The cards are thepoints; three points are on a line if those three cards form a set.
• This works because any two cards uniquely determine a thirdcard that completes the set.
October 19, 2016 Connecting SETr to geometry 8 / 41
Parallel sets• We can also see parallel lines as parallel sets.• If the original set has any attribute all the same, the parallel set
will also have the same attribute all the same.• If any attributes are different in the set, you can lay the cards of
the parallel set down so that each of those attributes cycle in thesame way as in the original.
_______________________________________________________
October 19, 2016 Connecting SETr to geometry 9 / 41
Finite affine planes of SETr Cards
October 19, 2016 Connecting SETr to geometry 10 / 41
Finite affine planes of SETr Cards
October 19, 2016 Connecting SETr to geometry 10 / 41
Finite affine planes of SETr Cards
October 19, 2016 Connecting SETr to geometry 10 / 41
Finite affine planes of SETr Cards
October 19, 2016 Connecting SETr to geometry 10 / 41
Finite affine planes of SETr Cards
This picture is sometimes called a magic square. Find all the sets in it!
There are 12 sets: one for each line in an affine plane of order 3.
October 19, 2016 Connecting SETr to geometry 11 / 41
Finite affine planes of SETr Cards
This picture is sometimes called a magic square. Find all the sets in it!
There are 12 sets: one for each line in an affine plane of order 3.October 19, 2016 Connecting SETr to geometry 11 / 41
A finite affine hyperplane
Select any remaining card and construct two more magic squares. Thiscreates a hyperplane.
October 19, 2016 Connecting SETr to geometry 12 / 41
A finite affine hyperplane
A line in AG(3, 3):AG(3, 3) is represented by three side-by-side 3 ! 3 grids. Again, three points are collinear if their sum
mod 3 is the point (0, 0, 0). In Figure 2, three collinear points are shown.
! !!
Figure 2. AG(3, 3) with one set of collinear points shown.
AG(4, 3) is represented by a 9 ! 9 grid, consisting of nine 3 ! 3 grids. A line will be three points thatappear either in the same subgrid, or in three subgrids that correspond to a line in AG(2, 3). Figure 3 showsa complete cap in AG(4, 3) whose anchor point is in the upper left. You can verify that the cap does consistof 10 pairs of points, where the third point completing the line for each pair is the point in the upper left.
Figure 3. A complete cap in AG(4, 3) whose anchor point is in the upper left.
This grid can also be realized as a way of viewing points with 4 coordinates, where the first two coordinatesgive the subgrid, and the second two give the point within that subgrid. In this case, the cap S picturedabove is the first complete cap lexicographically.
Let S1 be an arbitrary cap with anchor point a. Since the coordinates of a line sum to (0, 0, 0, 0), if thecoordinates for the points in S1 are summed with 10a, the result must be (0, 0, 0, 0), since this is sum of 10lines. Thus, the sum of the points in S1 is "a mod 3.
A computer search shows that any two caps with di!erent anchor points necessarily intersect. This factis interesting; it would be useful to have a geometric proof.
Another computer search verifies that there are 8424 complete caps with anchor a. These are all a"nelyequivalent, and the a"ne transformation that take one to another is actually a linear transformation, sincethe anchor point (0, 0, 0, 0) must go to itself. This reduces the transformations we must consider to GL(4, 3).
It is possible to decompose all of AG(4, 3) into 4 mutually disjoint complete caps with anchor a. Onesuch decomposition, where one of the complete caps is S, pictured above, is shown in Figure 4
Figure 4. A complete cap in AG(4, 3) whose anchor point is in the upper left.
2
October 19, 2016 Connecting SETr to geometry 13 / 41
The entire deck (AG(4, 3))
October 19, 2016 Connecting SETr to geometry 14 / 41
5-Attribute SETr
October 19, 2016 Connecting SETr to geometry 15 / 41
Some easy counting
• How many cards (points) are there?
81 = 34
• How many sets (lines) are there?1080 = (81× 80)/3! =
(812
)/3
• How many sets through a given card are there?40 = 80/2
October 19, 2016 Connecting SETr to geometry 16 / 41
Some easy counting
• How many cards (points) are there?81 = 34
• How many sets (lines) are there?1080 = (81× 80)/3! =
(812
)/3
• How many sets through a given card are there?40 = 80/2
October 19, 2016 Connecting SETr to geometry 16 / 41
Some easy counting
• How many cards (points) are there?81 = 34
• How many sets (lines) are there?
1080 = (81× 80)/3! =(81
2
)/3
• How many sets through a given card are there?40 = 80/2
October 19, 2016 Connecting SETr to geometry 16 / 41
Some easy counting
• How many cards (points) are there?81 = 34
• How many sets (lines) are there?1080 = (81× 80)/3! =
(812
)/3
• How many sets through a given card are there?40 = 80/2
October 19, 2016 Connecting SETr to geometry 16 / 41
Some easy counting
• How many cards (points) are there?81 = 34
• How many sets (lines) are there?1080 = (81× 80)/3! =
(812
)/3
• How many sets through a given card are there?
40 = 80/2
October 19, 2016 Connecting SETr to geometry 16 / 41
Some easy counting
• How many cards (points) are there?81 = 34
• How many sets (lines) are there?1080 = (81× 80)/3! =
(812
)/3
• How many sets through a given card are there?40 = 80/2
October 19, 2016 Connecting SETr to geometry 16 / 41
SETr and Fn3
• We can also think of a deck of SETr cards as the vector space F43.
• Each attribute corresponds to a coordinate, which can take on oneof three possible values
Number Color Shading Shape1→1 red→1 empty→1 oval→12→2 green→2 striped→2 diamond→23→0 purple→0 solid→0 squiggle→0
October 19, 2016 Caps and partitions in Fn3 17 / 41
SETr and Fn3
• With this choice of coordinates:
• The first set has coordinates (1, 0, 2, 1), (2, 1, 2, 1) and (0, 2, 2, 1).• The second set has coordinates (2, 1, 0, 0), (1, 2, 1, 1), (0, 0, 2, 2).• Three cards form a set if and only if
• the vector sum is ~0 mod 3• they are of the form ~x, ~x +~a, ~x + 2~a for some~a 6= ~0
October 19, 2016 Caps and partitions in Fn3 18 / 41
SETr and Fn3
• In Fn3 , we define a line (a.k.a. an algebraic line) to be
• three points that sum to ~0 mod 3• three points of the form ~x, ~x +~a, ~x + 2~a for some~a 6= ~0
• Now we have linear (well, affine) algebra! The maps Fn3 → Fn
3taking lines to lines are precisely the affine transformations
~x 7→ A~x +~b
for A ∈ GL(n, 3)
October 19, 2016 Caps and partitions in Fn3 19 / 41
Caps
• A k-cap is a collection of k points with no three collinear.• A complete cap is a cap for which any other point in the space
makes a line with a subset of points from the complete cap.• A maximal cap is a cap of maximum size.• In F2
3, maximal caps contain four points.
w w ww w ww w w
y yy y
October 19, 2016 Caps and partitions in Fn3 20 / 41
Caps
• A k-cap is a collection of k points with no three collinear.• A complete cap is a cap for which any other point in the space
makes a line with a subset of points from the complete cap.• A maximal cap is a cap of maximum size.• In F2
3, maximal caps contain four points.
w w ww w ww w w
y y
y yOctober 19, 2016 Caps and partitions in Fn
3 20 / 41
Caps
A complete 8-cap:t t tt t ttt tt
t t tt t ttt tt
t t tt t tt t tA complete (and maximal) 9-cap:t ttt t tt tt
t t tt ttt t tt ttt ttt tt
October 19, 2016 Caps and partitions in Fn3 21 / 41
Caps
• Two caps c1, c2 are called equivalent if there exists an affinetransformation mapping c1 to c2.
• Fact: All maximal caps in Fn3 are equivalent for n ≤ 6.
n = 4, the Pellegrino cap (Hill – 1983)n = 5, the Hill cap (Edel, Ferret, Landjev, Storme – 2002)n = 6, (Potechin – 2008)
• Open question: n > 6?
October 19, 2016 Caps and partitions in Fn3 22 / 41
An integer sequence
Denote by r3(Fn3) the size of a maximal cap in Fn
3
October 19, 2016 Caps and partitions in Fn3 23 / 41
An integer sequence
• Terry Tao’s blog: “Open question: best bounds for cap sets”( http://terrytao.wordpress.com/2007/02/23/open-question-best-bounds-for-cap-sets/)
• “Perhaps my favourite open question is the problem on themaximal size of a cap set — a subset of Fn
3 which contains nolines...”
n = 1 2 3 4 5 6 7 8 9r3(Fn
3) ≥ 2 4 9 20 45 112 236 472 1008r3(Fn
3) ≤ 2 4 9 20 45 112 292 773 2075
October 19, 2016 Caps and partitions in Fn3 24 / 41
An integer sequence(Almost) trivially, 2n ≤ r3(Fn
3) < 3n.
Asymptotically...• Edel (2004): r3(Fn
3) = Ω(2.2174n).• Brown-Buhler (1982, JCTA): r3(Fn
3) = o(3n).• Meshulam (1995, JCTA): r3(Fn
3) = O(3n/n).• Bateman-Katz (2012, J AMS): r3(Fn
3) = O(3n/n1+ε)
• Ellenberg-Gijswijt (to appear in Annals) “On large subsets of Fnq
with no three-term arithmetic progression”: r3(Fn3) = O(2.756n)
(!!!)• Ellenberg-Gijswijt adapt a method of Croot-Lev-Pach (to appear
in Annals).The punchline: asymptotically,
2.2174n < r3(Fn3) < 2.756n
October 19, 2016 Caps and partitions in Fn3 25 / 41
Maximal caps in AG(4, 3)
Remember this cap? Consider this card.
October 19, 2016 Caps and partitions in Fn3 26 / 41
Maximal caps in AG(4, 3)Theorem (F, K, M, P, -, 2014 )(First observed by Gary Gordon) Every 20-cap in AG(4, 3) consists of tenlines intersecting at one point with the point of intersection removed. We callthis point the anchor point.
October 19, 2016 Caps and partitions in Fn3 27 / 41
Maximal caps in AG(4, 3)Theorem (F, K, M, P, -, 2014 )Any two maximal caps in AG(4, 3) with different anchor points intersect
0
37
October 19, 2016 Caps and partitions in Fn3 28 / 41
Partitioning AG(4, 3)
(Tony Forbes) AG(4, 3) can be partitioned into 4 disjoint 20-caps andtheir anchor point.
October 19, 2016 Caps and partitions in Fn3 29 / 41
Partitioning AG(4, 3)
October 19, 2016 Caps and partitions in Fn3 30 / 41
Partitioning AG(4, 3)
Are all partitions of AG(4, 3) equivalent?
October 19, 2016 Caps and partitions in Fn3 31 / 41
Linear transformations
When the anchor point is fixed at ~0, affine transformations are lineartransformations. Here’s one:
October 19, 2016 Equivalence classes of partitions 32 / 41
Spot the difference
Consider these two caps with respect to our favorite cap, S.
October 19, 2016 Equivalence classes of partitions 33 / 41
Spot the difference
• The first is “1-completable”.• The second is “2-completable”.
October 19, 2016 Equivalence classes of partitions 34 / 41
Spot the difference
6-completables, too!
October 19, 2016 Equivalence classes of partitions 35 / 41
Completability
• 198 caps disjoint from S.
• With respect to our favorite cap, S:• 36 1-completable caps• 90 2-completable caps• 72 6-completable caps
• Every partition consists of:• S, 1-completable, 6-completable, 6-completable• S, 2-completable, 6-completable, 6-completable
October 19, 2016 Equivalence classes of partitions 36 / 41
Linear transformations
Theorem (F, M, K, P, -, 2014)Let T be an affine transformation fixing S:
T(n-comp) is an n-comp, n ∈ 1, 2, 6
No affine transformations exist between 1-completables and2-completables.
October 19, 2016 Equivalence classes of partitions 37 / 41
Partition classes
• 216 different partitions of AG(4, 3) with S.
36× 1 + 90× 2 = 216
• Two equivalence classes (no affine transformations):
• E1: 36 partitions S, 1-comp, 6-comp, 6-comp• E2: 180 partitions S, 2-comp, 6-comp, 6-comp
• Each 6-completable once in E1 and five times in E2.
October 19, 2016 Equivalence classes of partitions 38 / 41
Linear transformations of E2
Suppose D2 ∈ E2, and let S2 be the 2-completable of D2.
• 8 linear transformations fix D2 cap-wise (∼= Z4 × Z2).
• 8 linear transformations fix S and S2 and switch 6-completables.• Thus, a group of order 16 fixes D2 set-wise (∼= Z4 o Z4).
• Another set of 16 linear transformations fix S and S2 but send6-completables to two new 6-completables.
• Thus, group of order 32 fixing S and S2 (∼= (Z8 × Z2) o Z2).
October 19, 2016 Equivalence classes of partitions 39 / 41
Linear transformations of E1
Suppose D1 ∈ E1, and let S1 be the 1-completable of D1.
• 40 transformations fix D1 cap-wise (∼= Z4 ×D5).
• Also, 40 transformations fix S and S1 and switch 6-completables.
• Thus, group of order 80 fixing S and S1.• Isomorphic to Z20 o Z4.
October 19, 2016 Equivalence classes of partitions 40 / 41
Summary
• Every maximal cap in AG(4, 3) consists of ten lines intersecting atan anchor point.
• AG(4, 3) can be partitioned into four disjoint maximal caps andtheir anchor point.
• There are two equivalence classes of partitions.
Also interesting• Building complete caps (Jordan Awan):
http://webbox.lafayette.edu/~mcmahone/capbuilder.html
October 19, 2016 Equivalence classes of partitions 41 / 41