THE BORSUK CONJECTURE Saleem Watson California State University, Long Beach
What is the Borsuk Conjecture? The Borsuk Conjecture is about cutting objects into smaller pieces. Smaller means smaller diameter (not smaller area or volume). The Borsuk Cojecture is still an open problem, in many of its cases.
What is the Borsuk Conjecture? Karol Borsuk (1905 1982) Geometric Topology
The Diameter of a Set The diameter of a set is the largest distance between any two points of the set.
The Diameter of a Set The diameter of a set is the largest distance between any two points of the set. Diameter
The Diameter of a Set Quiz: Find the diameter of the set.
The Diameter of a Set Quiz: Find the diameter of the set. 1
The Diameter of a Set Quiz: Find the diameter of the set. 1
The Diameter of a Set Quiz: Find the diameter of each set.
The Diameter of a Set Quiz: Find the diameter of each set. 3
The Diameter of a Set Quiz: Find the diameter of each set. 1
The Borsuk Conjecture Every bounded set in n-dimensional space is the union of n+1 sets of smaller diameter. For example, every set in the plane (dimension 2) can be expressed as a union of three subsets of smaller diameter.
The Borsuk Conjecture Cutting up a square into pieces of smaller diameter. 1
The Borsuk Conjecture Cutting up a square into pieces of smaller diameter. 1
The Borsuk Conjecture Cutting up a circle into pieces of smaller diameter.
The Borsuk Conjecture Cutting up a circle into pieces of smaller diameter.
The Borsuk Conjecture Cutting up a triangle into pieces of smaller diameter.
The Borsuk Conjecture Cutting up a triangle into pieces of smaller diameter. (Note: no piece can contain two vertices.)
The Borsuk Conjecture Question 1: Can we get by with fewer than n+1 pieces? Question 2: Do we maybe, perhaps, sometimes need more than n+1 pieces?
The Borsuk Conjecture Question 1: Can we get by with fewer than n+1 pieces? NO 1 1-simplex 2-simplex 3-simplex n-simplex
The Borsuk Conjecture Question 2: Do we maybe, perhaps, sometimes need more than n+1 pieces? We ll see later.
Proof for dimension 1 The conjecture states: Every bounded set in a line can be expressed as the union of two sets of smaller diameter. Proof: Here is a typical set.
Proof for dimension 1 Step1: Inscribe the set in an interval. Theorem: Every set in the line of diameter 1 can be inscribed in an interval of diameter D, where D = 1
Proof for dimension 1 Step 2: Divide the interval in two pieces (each of smaller diameter).
Proof for dimension 2 The conjecture states: Every bounded set in the plane can be expressed as the union of 3 sets of smaller diameter.
Proof for dimension 2 Proof: Start with a bounded set in the plane.
Proof for dimension 2 Step 1 Inscribe the set in an equilateral triangle. Theorem: Any set of diameter 1 in the plane can be inscribed in an equilateral triangle of diameter D where D 3
Proof for dimension 2 D 3
Proof for dimension 2 Step 2 Divide the triangle into three pieces of smaller diameter as shown.
Proof for dimension 2 Step 2 Each piece has diameter D where 3 D < 1 2
Proof for dimension 3 The conjecture states: Every bounded set in the space can be expressed as the union of 4 sets of smaller diameter.
Proof for dimension 3 Proof: Start with a bounded set in space.
Proof for dimension 3 Step 1: Inscribe the set in a regular tetrahedron. Theorem: Any set of diameter 1 can be inscribed in a regular tetrahedron of diameter D where D 6
Proof for dimension 3
Proof for dimension 3 Step 2: Truncate tetrahedron to form an octahedron in which the set is inscribed.
Proof for dimension 3 The set is now inscribed in an octahedron. A 1 d 2 x r A' 3 u A 2 v 1 2 e 2 w A 3 A' 2 A 1 '
Proof for dimension 3 Step 3: Truncate the tetrahedron as shown. A 1 C 3 C 2 ' A 2 C 3 ' B 3 B 2 ' C 2 A 3 B 3 ' C 1 B 2 C 1 ' B 1 B 1 '
Proof for dimension 3 The 4 regions.
Proof for dimension 3 Step 4: This shape can be divided into four regions each of diameter D where D 2 2 2 1231 3 409 118 3 409 3 1 0.9887 1 3036 1012 + + < 759 1012 4 2
What are higher dimensions? The story begins with the ancient Greeks. They studied geometric shapes using ruler and compass constructions. The Greeks found formulas for the area of any shape bounded by straight line segments.
What are higher dimensions? The curves that the Greeks knew about were ellipses, parabolas, and hyperbolas (the conic sections). They could not find areas of regions bounded by such curves.
What are higher dimensions? Rene Descartes 1596-1650
What are higher dimensions? The invention of the coordinate plane is.. the greatest single step ever made in the progress of the exact sciences. -- John Stewart Mill
What are higher dimensions? Why? Many more curves (every equation in two variables corresponds to a curve). Analytic description of curves. Makes it possible to express the area bounded by a curve analytically (and this lead to the invention of calculus). Allows us to peer into higher dimensions (using analysis, not geometry).
What are higher dimensions? Points Dimension 1: Dimension 2: Dimension 3: Dimension 4: Dimension n: x 1 ( x1, x2) ( x, x, x ) 1 2 3 ( x1, x2, x3, x4) (,,, ) x1 x2 x n
What are higher dimensions? Distance between points: Dimension 2: ( x1, x2) ( y1, y2) Dimension 3: Dimension n: d = ( x y ) + ( x y ) ( x, x, x ) 1 2 3 (,,, ) x1 x2 x n 2 2 1 1 2 2 ( y1, y2, y3) d = ( x y ) + ( x y ) + ( x y ) 2 2 2 1 1 2 2 3 3 (,,, ) y1 y2 y n d = ( x y ) + ( x y ) + + ( x y ) 2 2 2 1 1 2 2 n n
What are higher dimensions? Example 1: The diagonal of a unit cube in dimension 100.
What are higher dimensions? Example 1: The diagonal of a unit cube in dimension 100. Length of diagonal is distance between (0,0,,0) and (1,1,,1) d = + + + = = 2 2 2 (1 0) (1 0) (1 0) 100 10
What are higher dimensions? Example 2: Unit balls tangent to the coordinate planes. Dimension: 2 Number of balls: 4 Distance to origin: 2 1.41 2 2 d = (1 0) + (1 0) 1 = 2 1
What are higher dimensions? Example 3: Unit balls tangent to the coordinate planes. Dimension: 3 Number of balls: Distance to origin: 2 2 2 d = (1 0) + (1 0) + (1 0) 1 = 3 1
What are higher dimensions? Example 3: Unit balls tangent to the coordinate planes. Dimension: 3 Number of balls: 8 Distance to origin: 3 1.73 2 2 2 d = (1 0) + (1 0) + (1 0) 1 = 3 1
What are higher dimensions? Example 4: Unit balls tangent to the coordinate planes. Dimension: 100 Number of balls: Distance to origin: d = + + + = = 2 2 2 (1 0) (1 0) (1 0) 1 100 1 9 d
What are higher dimensions? Example 4: Unit balls tangent to the coordinate planes. Dimension: 100 100 Number of balls: 2 Distance to origin: 100 1 = 9 d = + + + = = 2 2 2 (1 0) (1 0) (1 0) 1 100 1 9 d
What are higher dimensions? Example 5: In n dimensions The volume of the ball. The volume of the superscribed cube.
What are higher dimensions? n Cube ball 1 2 2 0 [ ] 1 2 4 3.14 3 8 4.19 4 16 4.93 5 32 5.26 6 64 5.16 7 128 4.72 8 256 4.06 9 512 3.30 10 11 12 1024 2048 4096 2.55 1.88 1.33 V2 k k π = k! V2 k 1 k+ 11 k = 2 π + 13 (2k + 1)
What are higher dimensions? n Cube ball 13 8,192.91 0 [ ] 1 14 16,384.60 15 32,768.38 16 65,536.23 17 131,072.14 18 262,144.08 19 524,288.05 20 1,048,576.03 21 2,097,152.01 22 23 24 4,194,304 8,388,608 16,777,216.007.004.002 V2 k k π = k! V2 k 1 k+ 11 k = 2 π + 13 (2k + 1)
What are higher dimensions? Example 6: Any set of diameter 1 in n- dimensional space can be inscribed in an n-simplex of diameter D where D nn ( + 1) 2 1 1-simplex 2-simplex 3-simplex n-simplex
Counterexample to the Borsuk conjecture The Borsuk conjecture is FALSE in dimension 1326. (Kalai 1983).
Counterexample to the Borsuk conjecture The counterexample is a set X with the property that any representation of the set as a union of sets of smaller diameter contains more than 1326+1=1327 elements.
Counterexample to the Borsuk conjecture The number 1326 is chosen as follows: 52 = 1326 2
Counterexample to the Borsuk conjecture The set X consists of some of the corners of the unit cube in R 1326. (there are 2 1326 corners)
Counterexample to the Borsuk conjecture The set X contains the following number of these corners. 51 = 25 247,959,266,474,052
Counterexample to the Borsuk conjecture Any diameter-reducing subset M of X contains at most 51 = 158,753,389,900 12
Counterexample to the Borsuk conjecture So how many diameter-reducing subset are needed to cover all of X?
Counterexample to the Borsuk conjecture So how many diameter-reducing subset are needed to cover all of X? The answer is k, where k 51 Number of points in X 25 = = 1561 Number of points in each M 51 12
Counterexample to the Borsuk conjecture But 1561 is greater than 1326+1, so the set X is a counterexample to the conjecture. k 51 25 = 1561 51 12
Counterexample to the Borsuk conjecture Questions: 1. Does the Borsuk Conjecture fail in dimension 4? 2. What is the dimension for which the Borsuk conjecture first fails?
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