Some problems of geometric combinatorics

Pavle Blagojević

(based on the joint work with Günter M. Ziegler)

Theorem (Radon’s theorem)Let X ⊆ Rd with (at least) d + 2 elements. Then there are disjointsubsets P and N of X with the property that:

conv(P) ∩ conv(N) 6= ∅.

Theorem (Radon’s theorem)Let X ⊆ Rd with (at least) d + 2 elements. Then there are disjointsubsets P and N of X with the property that:

conv(P) ∩ conv(N) 6= ∅.

Theorem (Radon’s theorem)Let X ⊆ Rd with (at least) d + 2 elements. Then there are disjointsubsets P and N of X with the property that:

conv(P) ∩ conv(N) 6= ∅.

Theorem (Radon’s theorem)Let X ⊆ Rd with (at least) d + 2 elements. Then there are disjointsubsets P and N of X with the property that:

conv(P) ∩ conv(N) 6= ∅.

Theorem (Radon’s theorem)Let X ⊆ Rd with (at least) d + 2 elements. Then there are disjointsubsets P and N of X with the property that:

conv(P) ∩ conv(N) 6= ∅.

Theorem (Radon’s theorem)Let X ⊆ Rd with (at least) d + 2 elements. Then there are disjointsubsets P and N of X with the property that:

conv(P) ∩ conv(N) 6= ∅.

Theorem (Radon’s theorem)Let f : ∆d+1 −→ Rd be an affine map. Then there are disjointfaces P and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.

Theorem (Topological Radon’s theorem)Let f : ∆d+1 −→ Rd be an continuous map. Then there aredisjoint faces P and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.

Bajmóczy, Bárány: On a common generalization of Borsuk’s and Radon’s theorem, Acta Math. Acad.

Sci. hungar., 1979.

Theorem (Radon’s theorem)Let X ⊆ Rd with (at least) d + 2 elements. Then there are disjointsubsets P and N of X with the property that:

conv(P) ∩ conv(N) 6= ∅.

Theorem (Radon’s theorem)Let f : ∆d+1 −→ Rd be an affine map. Then there are disjointfaces P and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.

Theorem (Topological Radon’s theorem)Let f : ∆d+1 −→ Rd be an continuous map. Then there aredisjoint faces P and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.

Bajmóczy, Bárány: On a common generalization of Borsuk’s and Radon’s theorem, Acta Math. Acad.

Sci. hungar., 1979.

Theorem (Radon’s theorem)Let X ⊆ Rd with (at least) d + 2 elements. Then there are disjointsubsets P and N of X with the property that:

conv(P) ∩ conv(N) 6= ∅.

Theorem (Radon’s theorem)Let f : ∆d+1 −→ Rd be an affine map. Then there are disjointfaces P and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.

Theorem (Topological Radon’s theorem)Let f : ∆d+1 −→ Rd be an continuous map. Then there aredisjoint faces P and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.

Bajmóczy, Bárány: On a common generalization of Borsuk’s and Radon’s theorem, Acta Math. Acad.

Sci. hungar., 1979.

Groups action and the Borsuk–Ulam theoremDefinitionLet G be a group and X 6= ∅. A (left) G-action on X is a functionG × X → X, (g , x) 7−→ g · x with the property that:

g · (h · x) = (gh) · x and 1 · x = x .

A set X with a G-action is called a G-set.Examplea. Z/2 = 〈ε〉, X = Sd , ε · x = −xb. Sn, X = Y n, π · (y1, . . . , yn) = (yπ−1(1), . . . , yπ−1(n))

c. D8 = (Z/2)2oZ/2 = 〈ε1, ε2, ω : ε21 = ε22 = ω2 = 1, ε1ω = ωε2〉,X = Sd × Sd ,ε1·(x1, x2) = (−x1, x2), ε2·(x1, x2) = (x1,−x2), ω·(x1, x2) = (x2, x1)

DefinitionLet X be a G-set.1. A G-action on X is free if g · x = x implies that g = 1.2. The fixed point set of the G-action on X is the set:

XG = x ∈ X : g · x = x for every g ∈ G.

Groups action and the Borsuk–Ulam theoremDefinitionLet G be a group and X 6= ∅. A (left) G-action on X is a functionG × X → X, (g , x) 7−→ g · x with the property that:

g · (h · x) = (gh) · x and 1 · x = x .A set X with a G-action is called a G-set.

Examplea. Z/2 = 〈ε〉, X = Sd , ε · x = −xb. Sn, X = Y n, π · (y1, . . . , yn) = (yπ−1(1), . . . , yπ−1(n))

c. D8 = (Z/2)2oZ/2 = 〈ε1, ε2, ω : ε21 = ε22 = ω2 = 1, ε1ω = ωε2〉,X = Sd × Sd ,ε1·(x1, x2) = (−x1, x2), ε2·(x1, x2) = (x1,−x2), ω·(x1, x2) = (x2, x1)

DefinitionLet X be a G-set.1. A G-action on X is free if g · x = x implies that g = 1.2. The fixed point set of the G-action on X is the set:

XG = x ∈ X : g · x = x for every g ∈ G.

Groups action and the Borsuk–Ulam theoremDefinitionLet G be a group and X 6= ∅. A (left) G-action on X is a functionG × X → X, (g , x) 7−→ g · x with the property that:

g · (h · x) = (gh) · x and 1 · x = x .A set X with a G-action is called a G-set.Examplea. Z/2 = 〈ε〉, X = Sd , ε · x = −x

b. Sn, X = Y n, π · (y1, . . . , yn) = (yπ−1(1), . . . , yπ−1(n))

c. D8 = (Z/2)2oZ/2 = 〈ε1, ε2, ω : ε21 = ε22 = ω2 = 1, ε1ω = ωε2〉,X = Sd × Sd ,ε1·(x1, x2) = (−x1, x2), ε2·(x1, x2) = (x1,−x2), ω·(x1, x2) = (x2, x1)

DefinitionLet X be a G-set.1. A G-action on X is free if g · x = x implies that g = 1.2. The fixed point set of the G-action on X is the set:

XG = x ∈ X : g · x = x for every g ∈ G.

Groups action and the Borsuk–Ulam theoremDefinitionLet G be a group and X 6= ∅. A (left) G-action on X is a functionG × X → X, (g , x) 7−→ g · x with the property that:

g · (h · x) = (gh) · x and 1 · x = x .A set X with a G-action is called a G-set.Examplea. Z/2 = 〈ε〉, X = Sd , ε · x = −xb. Sn, X = Y n, π · (y1, . . . , yn) = (yπ−1(1), . . . , yπ−1(n))

c. D8 = (Z/2)2oZ/2 = 〈ε1, ε2, ω : ε21 = ε22 = ω2 = 1, ε1ω = ωε2〉,X = Sd × Sd ,ε1·(x1, x2) = (−x1, x2), ε2·(x1, x2) = (x1,−x2), ω·(x1, x2) = (x2, x1)

DefinitionLet X be a G-set.1. A G-action on X is free if g · x = x implies that g = 1.2. The fixed point set of the G-action on X is the set:

XG = x ∈ X : g · x = x for every g ∈ G.

Groups action and the Borsuk–Ulam theoremDefinitionLet G be a group and X 6= ∅. A (left) G-action on X is a functionG × X → X, (g , x) 7−→ g · x with the property that:

g · (h · x) = (gh) · x and 1 · x = x .A set X with a G-action is called a G-set.Examplea. Z/2 = 〈ε〉, X = Sd , ε · x = −xb. Sn, X = Y n, π · (y1, . . . , yn) = (yπ−1(1), . . . , yπ−1(n))

c. D8 = (Z/2)2oZ/2 = 〈ε1, ε2, ω : ε21 = ε22 = ω2 = 1, ε1ω = ωε2〉,

X = Sd × Sd ,ε1·(x1, x2) = (−x1, x2), ε2·(x1, x2) = (x1,−x2), ω·(x1, x2) = (x2, x1)

DefinitionLet X be a G-set.1. A G-action on X is free if g · x = x implies that g = 1.2. The fixed point set of the G-action on X is the set:

XG = x ∈ X : g · x = x for every g ∈ G.

Groups action and the Borsuk–Ulam theoremDefinitionLet G be a group and X 6= ∅. A (left) G-action on X is a functionG × X → X, (g , x) 7−→ g · x with the property that:

g · (h · x) = (gh) · x and 1 · x = x .A set X with a G-action is called a G-set.Examplea. Z/2 = 〈ε〉, X = Sd , ε · x = −xb. Sn, X = Y n, π · (y1, . . . , yn) = (yπ−1(1), . . . , yπ−1(n))

c. D8 = (Z/2)2oZ/2 = 〈ε1, ε2, ω : ε21 = ε22 = ω2 = 1, ε1ω = ωε2〉,X = Sd × Sd ,ε1·(x1, x2) = (−x1, x2),

ε2·(x1, x2) = (x1,−x2), ω·(x1, x2) = (x2, x1)

DefinitionLet X be a G-set.1. A G-action on X is free if g · x = x implies that g = 1.2. The fixed point set of the G-action on X is the set:

XG = x ∈ X : g · x = x for every g ∈ G.

Groups action and the Borsuk–Ulam theoremDefinitionLet G be a group and X 6= ∅. A (left) G-action on X is a functionG × X → X, (g , x) 7−→ g · x with the property that:

g · (h · x) = (gh) · x and 1 · x = x .A set X with a G-action is called a G-set.Examplea. Z/2 = 〈ε〉, X = Sd , ε · x = −xb. Sn, X = Y n, π · (y1, . . . , yn) = (yπ−1(1), . . . , yπ−1(n))

c. D8 = (Z/2)2oZ/2 = 〈ε1, ε2, ω : ε21 = ε22 = ω2 = 1, ε1ω = ωε2〉,X = Sd × Sd ,ε1·(x1, x2) = (−x1, x2), ε2·(x1, x2) = (x1,−x2),

ω·(x1, x2) = (x2, x1)

DefinitionLet X be a G-set.1. A G-action on X is free if g · x = x implies that g = 1.2. The fixed point set of the G-action on X is the set:

XG = x ∈ X : g · x = x for every g ∈ G.

Groups action and the Borsuk–Ulam theoremDefinitionLet G be a group and X 6= ∅. A (left) G-action on X is a functionG × X → X, (g , x) 7−→ g · x with the property that:

g · (h · x) = (gh) · x and 1 · x = x .A set X with a G-action is called a G-set.Examplea. Z/2 = 〈ε〉, X = Sd , ε · x = −xb. Sn, X = Y n, π · (y1, . . . , yn) = (yπ−1(1), . . . , yπ−1(n))

c. D8 = (Z/2)2oZ/2 = 〈ε1, ε2, ω : ε21 = ε22 = ω2 = 1, ε1ω = ωε2〉,X = Sd × Sd ,ε1·(x1, x2) = (−x1, x2), ε2·(x1, x2) = (x1,−x2), ω·(x1, x2) = (x2, x1)

DefinitionLet X be a G-set.1. A G-action on X is free if g · x = x implies that g = 1.2. The fixed point set of the G-action on X is the set:

XG = x ∈ X : g · x = x for every g ∈ G.

Groups action and the Borsuk–Ulam theoremDefinitionLet G be a group and X 6= ∅. A (left) G-action on X is a functionG × X → X, (g , x) 7−→ g · x with the property that:

g · (h · x) = (gh) · x and 1 · x = x .A set X with a G-action is called a G-set.Examplea. Z/2 = 〈ε〉, X = Sd , ε · x = −xb. Sn, X = Y n, π · (y1, . . . , yn) = (yπ−1(1), . . . , yπ−1(n))

c. D8 = (Z/2)2oZ/2 = 〈ε1, ε2, ω : ε21 = ε22 = ω2 = 1, ε1ω = ωε2〉,X = Sd × Sd ,ε1·(x1, x2) = (−x1, x2), ε2·(x1, x2) = (x1,−x2), ω·(x1, x2) = (x2, x1)

DefinitionLet X be a G-set.

1. A G-action on X is free if g · x = x implies that g = 1.2. The fixed point set of the G-action on X is the set:

XG = x ∈ X : g · x = x for every g ∈ G.

Groups action and the Borsuk–Ulam theoremDefinitionLet G be a group and X 6= ∅. A (left) G-action on X is a functionG × X → X, (g , x) 7−→ g · x with the property that:

g · (h · x) = (gh) · x and 1 · x = x .A set X with a G-action is called a G-set.Examplea. Z/2 = 〈ε〉, X = Sd , ε · x = −xb. Sn, X = Y n, π · (y1, . . . , yn) = (yπ−1(1), . . . , yπ−1(n))

c. D8 = (Z/2)2oZ/2 = 〈ε1, ε2, ω : ε21 = ε22 = ω2 = 1, ε1ω = ωε2〉,X = Sd × Sd ,ε1·(x1, x2) = (−x1, x2), ε2·(x1, x2) = (x1,−x2), ω·(x1, x2) = (x2, x1)

DefinitionLet X be a G-set.1. A G-action on X is free if g · x = x implies that g = 1.

2. The fixed point set of the G-action on X is the set:XG = x ∈ X : g · x = x for every g ∈ G.

Groups action and the Borsuk–Ulam theoremDefinitionLet G be a group and X 6= ∅. A (left) G-action on X is a functionG × X → X, (g , x) 7−→ g · x with the property that:

g · (h · x) = (gh) · x and 1 · x = x .A set X with a G-action is called a G-set.Examplea. Z/2 = 〈ε〉, X = Sd , ε · x = −xb. Sn, X = Y n, π · (y1, . . . , yn) = (yπ−1(1), . . . , yπ−1(n))

c. D8 = (Z/2)2oZ/2 = 〈ε1, ε2, ω : ε21 = ε22 = ω2 = 1, ε1ω = ωε2〉,X = Sd × Sd ,ε1·(x1, x2) = (−x1, x2), ε2·(x1, x2) = (x1,−x2), ω·(x1, x2) = (x2, x1)

DefinitionLet X be a G-set.1. A G-action on X is free if g · x = x implies that g = 1.2. The fixed point set of the G-action on X is the set:

XG = x ∈ X : g · x = x for every g ∈ G.

Groups action and the Borsuk–Ulam theoremDefinitionLet X and Y be G-sets. A map f : X −→ Y is a G-map if for everyg ∈ G and every x ∈ X:

f (g · x) = g · f (x).

Examplea. Inclusion Sd −→ Sd+1 is a Z/2-map.b. Radial retraction Rd\0 −→ Sd−1, x 7−→ x

‖x‖ is a Z/2-map.

Theorem (Borsuk–Ulam theorem)Let the spheres Sn and Sm be free Z/2-spaces. Then a continuousZ/2-map Sn −→ Sm exists if and only if n ≤ m.

Groups action and the Borsuk–Ulam theoremDefinitionLet X and Y be G-sets. A map f : X −→ Y is a G-map if for everyg ∈ G and every x ∈ X:

f (g · x) = g · f (x).

Examplea. Inclusion Sd −→ Sd+1 is a Z/2-map.

b. Radial retraction Rd\0 −→ Sd−1, x 7−→ x‖x‖ is a Z/2-map.

Theorem (Borsuk–Ulam theorem)Let the spheres Sn and Sm be free Z/2-spaces. Then a continuousZ/2-map Sn −→ Sm exists if and only if n ≤ m.

Groups action and the Borsuk–Ulam theoremDefinitionLet X and Y be G-sets. A map f : X −→ Y is a G-map if for everyg ∈ G and every x ∈ X:

f (g · x) = g · f (x).

Examplea. Inclusion Sd −→ Sd+1 is a Z/2-map.b. Radial retraction Rd\0 −→ Sd−1, x 7−→ x

‖x‖ is a Z/2-map.

Theorem (Borsuk–Ulam theorem)Let the spheres Sn and Sm be free Z/2-spaces. Then a continuousZ/2-map Sn −→ Sm exists if and only if n ≤ m.

Groups action and the Borsuk–Ulam theoremDefinitionLet X and Y be G-sets. A map f : X −→ Y is a G-map if for everyg ∈ G and every x ∈ X:

f (g · x) = g · f (x).

Examplea. Inclusion Sd −→ Sd+1 is a Z/2-map.b. Radial retraction Rd\0 −→ Sd−1, x 7−→ x

‖x‖ is a Z/2-map.

Theorem (Borsuk–Ulam theorem)Let the spheres Sn and Sm be free Z/2-spaces. Then a continuousZ/2-map Sn −→ Sm exists if and only if n ≤ m.

Deleted join of a simplicial complexDefinitionLet K be a simplicial complex. The n-fold 2-wise deleted join of K:

K ∗n∆(2) = λ1x1+· · ·+λnxn ∈ F1∗· · ·∗Fn ⊂ K ∗n : Fi∩Fj = ∅ for i 6= j

The symmetric group Sn acts on K ∗n∆(2).ExampleK = [0, 1],K∗2∆(2) = S1 K = [3] = 1, 2, 3,K∗2∆(2) = S1

Lemma(K ∗ L)∗n∆(2)

∼= K ∗n∆(2) ∗ L∗n∆(2)

Example(∆N)∗r∆(2)

∼= ([1]∗N+1)∗r∆(2)∼= ([1]∗r∆(2))∗N+1 ∼= [r ]∗N+1

Deleted join of a simplicial complexDefinitionLet K be a simplicial complex. The n-fold 2-wise deleted join of K:

K ∗n∆(2) = λ1x1+· · ·+λnxn ∈ F1∗· · ·∗Fn ⊂ K ∗n : Fi∩Fj = ∅ for i 6= j

The symmetric group Sn acts on K ∗n∆(2).

ExampleK = [0, 1],K∗2∆(2) = S1 K = [3] = 1, 2, 3,K∗2∆(2) = S1

Lemma(K ∗ L)∗n∆(2)

∼= K ∗n∆(2) ∗ L∗n∆(2)

Example(∆N)∗r∆(2)

∼= ([1]∗N+1)∗r∆(2)∼= ([1]∗r∆(2))∗N+1 ∼= [r ]∗N+1

Deleted join of a simplicial complexDefinitionLet K be a simplicial complex. The n-fold 2-wise deleted join of K:

K ∗n∆(2) = λ1x1+· · ·+λnxn ∈ F1∗· · ·∗Fn ⊂ K ∗n : Fi∩Fj = ∅ for i 6= j

The symmetric group Sn acts on K ∗n∆(2).ExampleK = [0, 1],K∗2∆(2) = S1

K = [3] = 1, 2, 3,K∗2∆(2) = S1

Lemma(K ∗ L)∗n∆(2)

∼= K ∗n∆(2) ∗ L∗n∆(2)

Example(∆N)∗r∆(2)

∼= ([1]∗N+1)∗r∆(2)∼= ([1]∗r∆(2))∗N+1 ∼= [r ]∗N+1

Deleted join of a simplicial complexDefinitionLet K be a simplicial complex. The n-fold 2-wise deleted join of K:

K ∗n∆(2) = λ1x1+· · ·+λnxn ∈ F1∗· · ·∗Fn ⊂ K ∗n : Fi∩Fj = ∅ for i 6= j

The symmetric group Sn acts on K ∗n∆(2).ExampleK = [0, 1],K∗2∆(2) = S1 K = [3] = 1, 2, 3,K∗2∆(2) = S1

Lemma(K ∗ L)∗n∆(2)

∼= K ∗n∆(2) ∗ L∗n∆(2)

Example(∆N)∗r∆(2)

∼= ([1]∗N+1)∗r∆(2)∼= ([1]∗r∆(2))∗N+1 ∼= [r ]∗N+1

Deleted join of a simplicial complexDefinitionLet K be a simplicial complex. The n-fold 2-wise deleted join of K:

K ∗n∆(2) = λ1x1+· · ·+λnxn ∈ F1∗· · ·∗Fn ⊂ K ∗n : Fi∩Fj = ∅ for i 6= j

The symmetric group Sn acts on K ∗n∆(2).ExampleK = [0, 1],K∗2∆(2) = S1 K = [3] = 1, 2, 3,K∗2∆(2) = S1

Lemma(K ∗ L)∗n∆(2)

∼= K ∗n∆(2) ∗ L∗n∆(2)

Example(∆N)∗r∆(2)

∼= ([1]∗N+1)∗r∆(2)∼= ([1]∗r∆(2))∗N+1 ∼= [r ]∗N+1

Deleted join of a simplicial complexDefinitionLet K be a simplicial complex. The n-fold 2-wise deleted join of K:

K ∗n∆(2) = λ1x1+· · ·+λnxn ∈ F1∗· · ·∗Fn ⊂ K ∗n : Fi∩Fj = ∅ for i 6= j

The symmetric group Sn acts on K ∗n∆(2).ExampleK = [0, 1],K∗2∆(2) = S1 K = [3] = 1, 2, 3,K∗2∆(2) = S1

Lemma(K ∗ L)∗n∆(2)

∼= K ∗n∆(2) ∗ L∗n∆(2)

Example(∆N)∗r∆(2)

∼= ([1]∗N+1)∗r∆(2)∼= ([1]∗r∆(2))∗N+1 ∼= [r ]∗N+1

Deleted join of a simplicial complexDefinitionLet K be a simplicial complex. The n-fold 2-wise deleted join of K:

K ∗n∆(2) = λ1x1+· · ·+λnxn ∈ F1∗· · ·∗Fn ⊂ K ∗n : Fi∩Fj = ∅ for i 6= j

The symmetric group Sn acts on K ∗n∆(2).ExampleK = [0, 1],K∗2∆(2) = S1 K = [3] = 1, 2, 3,K∗2∆(2) = S1

Lemma(K ∗ L)∗n∆(2)

∼= K ∗n∆(2) ∗ L∗n∆(2)

Example(∆N)∗r∆(2)

∼= ([1]∗N+1)∗r∆(2)

∼= ([1]∗r∆(2))∗N+1 ∼= [r ]∗N+1

Deleted join of a simplicial complexDefinitionLet K be a simplicial complex. The n-fold 2-wise deleted join of K:

K ∗n∆(2) = λ1x1+· · ·+λnxn ∈ F1∗· · ·∗Fn ⊂ K ∗n : Fi∩Fj = ∅ for i 6= j

The symmetric group Sn acts on K ∗n∆(2).ExampleK = [0, 1],K∗2∆(2) = S1 K = [3] = 1, 2, 3,K∗2∆(2) = S1

Lemma(K ∗ L)∗n∆(2)

∼= K ∗n∆(2) ∗ L∗n∆(2)

Example(∆N)∗r∆(2)

∼= ([1]∗N+1)∗r∆(2)∼= ([1]∗r∆(2))∗N+1

∼= [r ]∗N+1

Deleted join of a simplicial complexDefinitionLet K be a simplicial complex. The n-fold 2-wise deleted join of K:

K ∗n∆(2) = λ1x1+· · ·+λnxn ∈ F1∗· · ·∗Fn ⊂ K ∗n : Fi∩Fj = ∅ for i 6= j

The symmetric group Sn acts on K ∗n∆(2).ExampleK = [0, 1],K∗2∆(2) = S1 K = [3] = 1, 2, 3,K∗2∆(2) = S1

Lemma(K ∗ L)∗n∆(2)

∼= K ∗n∆(2) ∗ L∗n∆(2)

Example(∆N)∗r∆(2)

∼= ([1]∗N+1)∗r∆(2)∼= ([1]∗r∆(2))∗N+1 ∼= [r ]∗N+1

Topological Radon’s theoremLet f : ∆d+1 −→ Rd be an continuous map. Then there are disjoint facesP and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.

Proof:

I f : ∆d+1 −→ Rd fails the topological Radon’s theorem

I Ω := (∆d+1)∗2∆(2)∼=Lemma [2]∗d+2 ∼= (S0)∗d+2 ∼= Sd+1

Ω is a free Z/2-space and Z/2 = 〈ε〉

I F : Ω −→W2 ⊕ (Rd ⊕ Rd), W2 := (w1,w2) ∈ R2 : w1 + w2 = 0

F (λ1x1 + λ2x2) := (λ1 − 12 , λ2 − 1

2 )⊕(λ1f (x1)⊕ λ2f (x2)

)F is a Z/2-map

λ1x1 + λ2x2

·ε

f// (λ1 − 1

2 , λ2 − 12 )⊕

(λ1f (x1)⊕ λ2f (x2)

)·ε

λ2x2 + λ1x1f// (λ2 − 1

2 , λ1 − 12 )⊕

(λ2f (x2)⊕ λ1f (x1)

)

Topological Radon’s theoremLet f : ∆d+1 −→ Rd be an continuous map. Then there are disjoint facesP and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.Proof:

I f : ∆d+1 −→ Rd fails the topological Radon’s theorem

I Ω := (∆d+1)∗2∆(2)∼=Lemma [2]∗d+2 ∼= (S0)∗d+2 ∼= Sd+1

Ω is a free Z/2-space and Z/2 = 〈ε〉

I F : Ω −→W2 ⊕ (Rd ⊕ Rd), W2 := (w1,w2) ∈ R2 : w1 + w2 = 0

F (λ1x1 + λ2x2) := (λ1 − 12 , λ2 − 1

2 )⊕(λ1f (x1)⊕ λ2f (x2)

)F is a Z/2-map

λ1x1 + λ2x2

·ε

f// (λ1 − 1

2 , λ2 − 12 )⊕

(λ1f (x1)⊕ λ2f (x2)

)·ε

λ2x2 + λ1x1f// (λ2 − 1

2 , λ1 − 12 )⊕

(λ2f (x2)⊕ λ1f (x1)

)

Topological Radon’s theoremLet f : ∆d+1 −→ Rd be an continuous map. Then there are disjoint facesP and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.Proof:

I f : ∆d+1 −→ Rd fails the topological Radon’s theorem

I Ω := (∆d+1)∗2∆(2)

∼=Lemma [2]∗d+2 ∼= (S0)∗d+2 ∼= Sd+1

Ω is a free Z/2-space and Z/2 = 〈ε〉

I F : Ω −→W2 ⊕ (Rd ⊕ Rd), W2 := (w1,w2) ∈ R2 : w1 + w2 = 0

F (λ1x1 + λ2x2) := (λ1 − 12 , λ2 − 1

2 )⊕(λ1f (x1)⊕ λ2f (x2)

)F is a Z/2-map

λ1x1 + λ2x2

·ε

f// (λ1 − 1

2 , λ2 − 12 )⊕

(λ1f (x1)⊕ λ2f (x2)

)·ε

λ2x2 + λ1x1f// (λ2 − 1

2 , λ1 − 12 )⊕

(λ2f (x2)⊕ λ1f (x1)

)

Topological Radon’s theoremLet f : ∆d+1 −→ Rd be an continuous map. Then there are disjoint facesP and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.Proof:

I f : ∆d+1 −→ Rd fails the topological Radon’s theorem

I Ω := (∆d+1)∗2∆(2)∼=Lemma [2]∗d+2

∼= (S0)∗d+2 ∼= Sd+1

Ω is a free Z/2-space and Z/2 = 〈ε〉

I F : Ω −→W2 ⊕ (Rd ⊕ Rd), W2 := (w1,w2) ∈ R2 : w1 + w2 = 0

F (λ1x1 + λ2x2) := (λ1 − 12 , λ2 − 1

2 )⊕(λ1f (x1)⊕ λ2f (x2)

)F is a Z/2-map

λ1x1 + λ2x2

·ε

f// (λ1 − 1

2 , λ2 − 12 )⊕

(λ1f (x1)⊕ λ2f (x2)

)·ε

λ2x2 + λ1x1f// (λ2 − 1

2 , λ1 − 12 )⊕

(λ2f (x2)⊕ λ1f (x1)

)

Topological Radon’s theoremLet f : ∆d+1 −→ Rd be an continuous map. Then there are disjoint facesP and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.Proof:

I f : ∆d+1 −→ Rd fails the topological Radon’s theorem

I Ω := (∆d+1)∗2∆(2)∼=Lemma [2]∗d+2 ∼= (S0)∗d+2 ∼= Sd+1

Ω is a free Z/2-space and Z/2 = 〈ε〉

I F : Ω −→W2 ⊕ (Rd ⊕ Rd), W2 := (w1,w2) ∈ R2 : w1 + w2 = 0

F (λ1x1 + λ2x2) := (λ1 − 12 , λ2 − 1

2 )⊕(λ1f (x1)⊕ λ2f (x2)

)F is a Z/2-map

λ1x1 + λ2x2

·ε

f// (λ1 − 1

2 , λ2 − 12 )⊕

(λ1f (x1)⊕ λ2f (x2)

)·ε

λ2x2 + λ1x1f// (λ2 − 1

2 , λ1 − 12 )⊕

(λ2f (x2)⊕ λ1f (x1)

)

Topological Radon’s theoremLet f : ∆d+1 −→ Rd be an continuous map. Then there are disjoint facesP and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.Proof:

I f : ∆d+1 −→ Rd fails the topological Radon’s theorem

I Ω := (∆d+1)∗2∆(2)∼=Lemma [2]∗d+2 ∼= (S0)∗d+2 ∼= Sd+1

Ω is a free Z/2-space and Z/2 = 〈ε〉

I F : Ω −→W2 ⊕ (Rd ⊕ Rd), W2 := (w1,w2) ∈ R2 : w1 + w2 = 0

F (λ1x1 + λ2x2) := (λ1 − 12 , λ2 − 1

2 )⊕(λ1f (x1)⊕ λ2f (x2)

)F is a Z/2-map

λ1x1 + λ2x2

·ε

f// (λ1 − 1

2 , λ2 − 12 )⊕

(λ1f (x1)⊕ λ2f (x2)

)·ε

λ2x2 + λ1x1f// (λ2 − 1

2 , λ1 − 12 )⊕

(λ2f (x2)⊕ λ1f (x1)

)

Topological Radon’s theoremLet f : ∆d+1 −→ Rd be an continuous map. Then there are disjoint facesP and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.Proof:

I f : ∆d+1 −→ Rd fails the topological Radon’s theorem

I Ω := (∆d+1)∗2∆(2)∼=Lemma [2]∗d+2 ∼= (S0)∗d+2 ∼= Sd+1

Ω is a free Z/2-space and Z/2 = 〈ε〉

I F : Ω −→W2 ⊕ (Rd ⊕ Rd),

W2 := (w1,w2) ∈ R2 : w1 + w2 = 0

F (λ1x1 + λ2x2) := (λ1 − 12 , λ2 − 1

2 )⊕(λ1f (x1)⊕ λ2f (x2)

)F is a Z/2-map

λ1x1 + λ2x2

·ε

f// (λ1 − 1

2 , λ2 − 12 )⊕

(λ1f (x1)⊕ λ2f (x2)

)·ε

λ2x2 + λ1x1f// (λ2 − 1

2 , λ1 − 12 )⊕

(λ2f (x2)⊕ λ1f (x1)

)

Topological Radon’s theoremLet f : ∆d+1 −→ Rd be an continuous map. Then there are disjoint facesP and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.Proof:

I f : ∆d+1 −→ Rd fails the topological Radon’s theorem

I Ω := (∆d+1)∗2∆(2)∼=Lemma [2]∗d+2 ∼= (S0)∗d+2 ∼= Sd+1

Ω is a free Z/2-space and Z/2 = 〈ε〉

I F : Ω −→W2 ⊕ (Rd ⊕ Rd), W2 := (w1,w2) ∈ R2 : w1 + w2 = 0

F (λ1x1 + λ2x2) := (λ1 − 12 , λ2 − 1

2 )⊕(λ1f (x1)⊕ λ2f (x2)

)F is a Z/2-map

λ1x1 + λ2x2

·ε

f// (λ1 − 1

2 , λ2 − 12 )⊕

(λ1f (x1)⊕ λ2f (x2)

)·ε

λ2x2 + λ1x1f// (λ2 − 1

2 , λ1 − 12 )⊕

(λ2f (x2)⊕ λ1f (x1)

)

Topological Radon’s theoremLet f : ∆d+1 −→ Rd be an continuous map. Then there are disjoint facesP and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.Proof:

I f : ∆d+1 −→ Rd fails the topological Radon’s theorem

I Ω := (∆d+1)∗2∆(2)∼=Lemma [2]∗d+2 ∼= (S0)∗d+2 ∼= Sd+1

Ω is a free Z/2-space and Z/2 = 〈ε〉

I F : Ω −→W2 ⊕ (Rd ⊕ Rd), W2 := (w1,w2) ∈ R2 : w1 + w2 = 0

F (λ1x1 + λ2x2) := (λ1 − 12 , λ2 − 1

2 )⊕(λ1f (x1)⊕ λ2f (x2)

)

F is a Z/2-map

λ1x1 + λ2x2

·ε

f// (λ1 − 1

2 , λ2 − 12 )⊕

(λ1f (x1)⊕ λ2f (x2)

)·ε

λ2x2 + λ1x1f// (λ2 − 1

2 , λ1 − 12 )⊕

(λ2f (x2)⊕ λ1f (x1)

)

Topological Radon’s theoremLet f : ∆d+1 −→ Rd be an continuous map. Then there are disjoint facesP and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.Proof:

I f : ∆d+1 −→ Rd fails the topological Radon’s theorem

I Ω := (∆d+1)∗2∆(2)∼=Lemma [2]∗d+2 ∼= (S0)∗d+2 ∼= Sd+1

Ω is a free Z/2-space and Z/2 = 〈ε〉

I F : Ω −→W2 ⊕ (Rd ⊕ Rd), W2 := (w1,w2) ∈ R2 : w1 + w2 = 0

F (λ1x1 + λ2x2) := (λ1 − 12 , λ2 − 1

2 )⊕(λ1f (x1)⊕ λ2f (x2)

)F is a Z/2-map

λ1x1 + λ2x2

·ε

f// (λ1 − 1

2 , λ2 − 12 )⊕

(λ1f (x1)⊕ λ2f (x2)

)·ε

λ2x2 + λ1x1f// (λ2 − 1

2 , λ1 − 12 )⊕

(λ2f (x2)⊕ λ1f (x1)

)

Topological Radon’s theoremLet f : ∆d+1 −→ Rd be an continuous map. Then there are disjoint facesP and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.Proof:

I f : ∆d+1 −→ Rd fails the topological Radon’s theorem

I Ω := (∆d+1)∗2∆(2)∼=Lemma [2]∗d+2 ∼= (S0)∗d+2 ∼= Sd+1

Ω is a free Z/2-space and Z/2 = 〈ε〉

I F : Ω −→W2 ⊕ (Rd ⊕ Rd), W2 := (w1,w2) ∈ R2 : w1 + w2 = 0

F (λ1x1 + λ2x2) := (λ1 − 12 , λ2 − 1

2 )⊕(λ1f (x1)⊕ λ2f (x2)

)F is a Z/2-map

λ1x1 + λ2x2

·ε

f// (λ1 − 1

2 , λ2 − 12 )⊕

(λ1f (x1)⊕ λ2f (x2)

)·ε

λ2x2 + λ1x1f// (λ2 − 1

2 , λ1 − 12 )⊕

(λ2f (x2)⊕ λ1f (x1)

)

Topological Radon’s theoremLet f : ∆d+1 −→ Rd be an continuous map. Then there are disjoint facesP and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.Proof:

•F : (Ω ∼= Sd+1) −→Z/2 W2 ⊕ (Rd ⊕ Rd)

•F (λ1x1 + λ2x2) := (λ1 − 12 , λ2 − 1

2 )⊕(λ1f (x1)⊕ λ2f (x2)

)

I T := (w1,w2)⊕ (y1, y2) ∈W2 ⊕ (Rd ⊕ Rd) : w1 = w2 = 0, y1 = y2

=(W2 ⊕ (Rd ⊕ Rd)

)Z/2I F (λ1x1+λ2x2) ∈ T =⇒

λ1 − 1

2 = λ2 − 12 = 0

λ1f (x1) = λ2f (x2)=⇒

λ1 = λ2 = 1

2f (x1) = f (x2)

I f : ∆d+1 −→ Rd fails the theorem =⇒ F (Ω) ∩ T = ∅

Ω ∼= Sd+1 //

((

W2 ⊕ (Rd ⊕ Rd)

W2 ⊕ (Rd ⊕ Rd)\T

44

// T⊥\0 ' S(T⊥) ∼= Sd

Topological Radon’s theoremLet f : ∆d+1 −→ Rd be an continuous map. Then there are disjoint facesP and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.Proof:

•F : (Ω ∼= Sd+1) −→Z/2 W2 ⊕ (Rd ⊕ Rd)

•F (λ1x1 + λ2x2) := (λ1 − 12 , λ2 − 1

2 )⊕(λ1f (x1)⊕ λ2f (x2)

)I T := (w1,w2)⊕ (y1, y2) ∈W2 ⊕ (Rd ⊕ Rd) : w1 = w2 = 0, y1 = y2

=(W2 ⊕ (Rd ⊕ Rd)

)Z/2I F (λ1x1+λ2x2) ∈ T =⇒

λ1 − 1

2 = λ2 − 12 = 0

λ1f (x1) = λ2f (x2)=⇒

λ1 = λ2 = 1

2f (x1) = f (x2)

I f : ∆d+1 −→ Rd fails the theorem =⇒ F (Ω) ∩ T = ∅

Ω ∼= Sd+1 //

((

W2 ⊕ (Rd ⊕ Rd)

W2 ⊕ (Rd ⊕ Rd)\T

44

// T⊥\0 ' S(T⊥) ∼= Sd

Topological Radon’s theoremLet f : ∆d+1 −→ Rd be an continuous map. Then there are disjoint facesP and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.Proof:

•F : (Ω ∼= Sd+1) −→Z/2 W2 ⊕ (Rd ⊕ Rd)

•F (λ1x1 + λ2x2) := (λ1 − 12 , λ2 − 1

2 )⊕(λ1f (x1)⊕ λ2f (x2)

)I T := (w1,w2)⊕ (y1, y2) ∈W2 ⊕ (Rd ⊕ Rd) : w1 = w2 = 0, y1 = y2

=(W2 ⊕ (Rd ⊕ Rd)

)Z/2

I F (λ1x1+λ2x2) ∈ T =⇒λ1 − 1

2 = λ2 − 12 = 0

λ1f (x1) = λ2f (x2)=⇒

λ1 = λ2 = 1

2f (x1) = f (x2)

I f : ∆d+1 −→ Rd fails the theorem =⇒ F (Ω) ∩ T = ∅

Ω ∼= Sd+1 //

((

W2 ⊕ (Rd ⊕ Rd)

W2 ⊕ (Rd ⊕ Rd)\T

44

// T⊥\0 ' S(T⊥) ∼= Sd

Topological Radon’s theoremLet f : ∆d+1 −→ Rd be an continuous map. Then there are disjoint facesP and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.Proof:

•F : (Ω ∼= Sd+1) −→Z/2 W2 ⊕ (Rd ⊕ Rd)

•F (λ1x1 + λ2x2) := (λ1 − 12 , λ2 − 1

2 )⊕(λ1f (x1)⊕ λ2f (x2)

)I T := (w1,w2)⊕ (y1, y2) ∈W2 ⊕ (Rd ⊕ Rd) : w1 = w2 = 0, y1 = y2

=(W2 ⊕ (Rd ⊕ Rd)

)Z/2I F (λ1x1+λ2x2) ∈ T =⇒

λ1 − 1

2 = λ2 − 12 = 0

λ1f (x1) = λ2f (x2)=⇒

λ1 = λ2 = 1

2f (x1) = f (x2)

I f : ∆d+1 −→ Rd fails the theorem =⇒ F (Ω) ∩ T = ∅

Ω ∼= Sd+1 //

((

W2 ⊕ (Rd ⊕ Rd)

W2 ⊕ (Rd ⊕ Rd)\T

44

// T⊥\0 ' S(T⊥) ∼= Sd

Topological Radon’s theoremLet f : ∆d+1 −→ Rd be an continuous map. Then there are disjoint facesP and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.Proof:

•F : (Ω ∼= Sd+1) −→Z/2 W2 ⊕ (Rd ⊕ Rd)

•F (λ1x1 + λ2x2) := (λ1 − 12 , λ2 − 1

2 )⊕(λ1f (x1)⊕ λ2f (x2)

)I T := (w1,w2)⊕ (y1, y2) ∈W2 ⊕ (Rd ⊕ Rd) : w1 = w2 = 0, y1 = y2

=(W2 ⊕ (Rd ⊕ Rd)

)Z/2I F (λ1x1+λ2x2) ∈ T =⇒

λ1 − 1

2 = λ2 − 12 = 0

λ1f (x1) = λ2f (x2)=⇒

λ1 = λ2 = 1

2f (x1) = f (x2)

I f : ∆d+1 −→ Rd fails the theorem =⇒ F (Ω) ∩ T = ∅

Ω ∼= Sd+1 //

((

W2 ⊕ (Rd ⊕ Rd)

W2 ⊕ (Rd ⊕ Rd)\T

44

// T⊥\0 ' S(T⊥) ∼= Sd

Topological Radon’s theoremLet f : ∆d+1 −→ Rd be an continuous map. Then there are disjoint facesP and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.Proof:

•F : (Ω ∼= Sd+1) −→Z/2 W2 ⊕ (Rd ⊕ Rd)

•F (λ1x1 + λ2x2) := (λ1 − 12 , λ2 − 1

2 )⊕(λ1f (x1)⊕ λ2f (x2)

)I T := (w1,w2)⊕ (y1, y2) ∈W2 ⊕ (Rd ⊕ Rd) : w1 = w2 = 0, y1 = y2

=(W2 ⊕ (Rd ⊕ Rd)

)Z/2I F (λ1x1+λ2x2) ∈ T =⇒

λ1 − 1

2 = λ2 − 12 = 0

λ1f (x1) = λ2f (x2)=⇒

λ1 = λ2 = 1

2f (x1) = f (x2)

I f : ∆d+1 −→ Rd fails the theorem =⇒ F (Ω) ∩ T = ∅

Ω ∼= Sd+1 //

((

W2 ⊕ (Rd ⊕ Rd)

W2 ⊕ (Rd ⊕ Rd)\T

44

// T⊥\0 ' S(T⊥) ∼= Sd

Topological Radon’s theoremLet f : ∆d+1 −→ Rd be an continuous map. Then there are disjoint facesP and N of the simplex ∆d+1 with the property that:

f (P) ∩ f (N) 6= ∅.Proof:

•F : (Ω ∼= Sd+1) −→Z/2 W2 ⊕ (Rd ⊕ Rd)

•F (λ1x1 + λ2x2) := (λ1 − 12 , λ2 − 1

2 )⊕(λ1f (x1)⊕ λ2f (x2)

)I T := (w1,w2)⊕ (y1, y2) ∈W2 ⊕ (Rd ⊕ Rd) : w1 = w2 = 0, y1 = y2

=(W2 ⊕ (Rd ⊕ Rd)

)Z/2I F (λ1x1+λ2x2) ∈ T =⇒

λ1 − 1

2 = λ2 − 12 = 0

λ1f (x1) = λ2f (x2)=⇒

λ1 = λ2 = 1

2f (x1) = f (x2)

I f : ∆d+1 −→ Rd fails the theorem =⇒ F (Ω) ∩ T = ∅

Ω ∼= Sd+1 //

((

W2 ⊕ (Rd ⊕ Rd)

W2 ⊕ (Rd ⊕ Rd)\T

44

// T⊥\0 ' S(T⊥) ∼= Sd

The Complete bipartite graph K3,3 is not planarTheorem

There is no continuous injective map [3]∗2 −→ R2.

Proof:I f : [3]∗2 → R2 a continuous injection

I Ω := ([3]∗2)∗2∆(2)∼=Lemma ([3]∗2∆(2))

∗2 =Example (S1)∗2 = S3

Ω is a free Z/2-space

I F : Ω→ R3 ⊕ R3, F (λ1x1 + λ2x2) := (λ1, λ1f (x1))⊕ (λ2, λ2f (x2))

F is a Z/2-mapI T := (y1, y2) ∈ R3 ⊕ R3 : y1 = y2 = (R3 ⊕ R3)Z/2

I F (λ1x1+λ2x2) ∈ T =⇒λ1 = λ2λ1f (x1) = λ2f (x2)

=⇒λ1 = λ2 = 1

2f (x1) = f (x2)

I f : [3]∗2 → R2 a continuous injection =⇒ F (Ω) ∩ T = ∅S3 ∼= Ω

F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2

The Complete bipartite graph K3,3 is not planarTheorem

There is no continuous injective map [3]∗2 −→ R2.Proof:

I f : [3]∗2 → R2 a continuous injection

I Ω := ([3]∗2)∗2∆(2)∼=Lemma ([3]∗2∆(2))

∗2 =Example (S1)∗2 = S3

Ω is a free Z/2-space

I F : Ω→ R3 ⊕ R3, F (λ1x1 + λ2x2) := (λ1, λ1f (x1))⊕ (λ2, λ2f (x2))

F is a Z/2-mapI T := (y1, y2) ∈ R3 ⊕ R3 : y1 = y2 = (R3 ⊕ R3)Z/2

I F (λ1x1+λ2x2) ∈ T =⇒λ1 = λ2λ1f (x1) = λ2f (x2)

=⇒λ1 = λ2 = 1

2f (x1) = f (x2)

I f : [3]∗2 → R2 a continuous injection =⇒ F (Ω) ∩ T = ∅S3 ∼= Ω

F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2

The Complete bipartite graph K3,3 is not planarTheorem

There is no continuous injective map [3]∗2 −→ R2.Proof:I f : [3]∗2 → R2 a continuous injection

I Ω := ([3]∗2)∗2∆(2)∼=Lemma ([3]∗2∆(2))

∗2 =Example (S1)∗2 = S3

Ω is a free Z/2-space

I F : Ω→ R3 ⊕ R3, F (λ1x1 + λ2x2) := (λ1, λ1f (x1))⊕ (λ2, λ2f (x2))

F is a Z/2-mapI T := (y1, y2) ∈ R3 ⊕ R3 : y1 = y2 = (R3 ⊕ R3)Z/2

I F (λ1x1+λ2x2) ∈ T =⇒λ1 = λ2λ1f (x1) = λ2f (x2)

=⇒λ1 = λ2 = 1

2f (x1) = f (x2)

I f : [3]∗2 → R2 a continuous injection =⇒ F (Ω) ∩ T = ∅S3 ∼= Ω

F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2

The Complete bipartite graph K3,3 is not planarTheorem

There is no continuous injective map [3]∗2 −→ R2.Proof:I f : [3]∗2 → R2 a continuous injection

I Ω := ([3]∗2)∗2∆(2)

∼=Lemma ([3]∗2∆(2))∗2 =Example (S1)∗2 = S3

Ω is a free Z/2-space

I F : Ω→ R3 ⊕ R3, F (λ1x1 + λ2x2) := (λ1, λ1f (x1))⊕ (λ2, λ2f (x2))

F is a Z/2-mapI T := (y1, y2) ∈ R3 ⊕ R3 : y1 = y2 = (R3 ⊕ R3)Z/2

I F (λ1x1+λ2x2) ∈ T =⇒λ1 = λ2λ1f (x1) = λ2f (x2)

=⇒λ1 = λ2 = 1

2f (x1) = f (x2)

I f : [3]∗2 → R2 a continuous injection =⇒ F (Ω) ∩ T = ∅S3 ∼= Ω

F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2

The Complete bipartite graph K3,3 is not planarTheorem

There is no continuous injective map [3]∗2 −→ R2.Proof:I f : [3]∗2 → R2 a continuous injection

I Ω := ([3]∗2)∗2∆(2)∼=Lemma ([3]∗2∆(2))

∗2

=Example (S1)∗2 = S3

Ω is a free Z/2-space

I F : Ω→ R3 ⊕ R3, F (λ1x1 + λ2x2) := (λ1, λ1f (x1))⊕ (λ2, λ2f (x2))

F is a Z/2-mapI T := (y1, y2) ∈ R3 ⊕ R3 : y1 = y2 = (R3 ⊕ R3)Z/2

I F (λ1x1+λ2x2) ∈ T =⇒λ1 = λ2λ1f (x1) = λ2f (x2)

=⇒λ1 = λ2 = 1

2f (x1) = f (x2)

I f : [3]∗2 → R2 a continuous injection =⇒ F (Ω) ∩ T = ∅S3 ∼= Ω

F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2

The Complete bipartite graph K3,3 is not planarTheorem

There is no continuous injective map [3]∗2 −→ R2.Proof:I f : [3]∗2 → R2 a continuous injection

I Ω := ([3]∗2)∗2∆(2)∼=Lemma ([3]∗2∆(2))

∗2 =Example (S1)∗2 = S3

Ω is a free Z/2-space

I F : Ω→ R3 ⊕ R3, F (λ1x1 + λ2x2) := (λ1, λ1f (x1))⊕ (λ2, λ2f (x2))

F is a Z/2-mapI T := (y1, y2) ∈ R3 ⊕ R3 : y1 = y2 = (R3 ⊕ R3)Z/2

I F (λ1x1+λ2x2) ∈ T =⇒λ1 = λ2λ1f (x1) = λ2f (x2)

=⇒λ1 = λ2 = 1

2f (x1) = f (x2)

I f : [3]∗2 → R2 a continuous injection =⇒ F (Ω) ∩ T = ∅S3 ∼= Ω

F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2

The Complete bipartite graph K3,3 is not planarTheorem

There is no continuous injective map [3]∗2 −→ R2.Proof:I f : [3]∗2 → R2 a continuous injection

I Ω := ([3]∗2)∗2∆(2)∼=Lemma ([3]∗2∆(2))

∗2 =Example (S1)∗2 = S3

Ω is a free Z/2-space

I F : Ω→ R3 ⊕ R3, F (λ1x1 + λ2x2) := (λ1, λ1f (x1))⊕ (λ2, λ2f (x2))

F is a Z/2-mapI T := (y1, y2) ∈ R3 ⊕ R3 : y1 = y2 = (R3 ⊕ R3)Z/2

I F (λ1x1+λ2x2) ∈ T =⇒λ1 = λ2λ1f (x1) = λ2f (x2)

=⇒λ1 = λ2 = 1

2f (x1) = f (x2)

I f : [3]∗2 → R2 a continuous injection =⇒ F (Ω) ∩ T = ∅S3 ∼= Ω

F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2

The Complete bipartite graph K3,3 is not planarTheorem

There is no continuous injective map [3]∗2 −→ R2.Proof:I f : [3]∗2 → R2 a continuous injection

I Ω := ([3]∗2)∗2∆(2)∼=Lemma ([3]∗2∆(2))

∗2 =Example (S1)∗2 = S3

Ω is a free Z/2-space

I F : Ω→ R3 ⊕ R3,

F (λ1x1 + λ2x2) := (λ1, λ1f (x1))⊕ (λ2, λ2f (x2))

F is a Z/2-mapI T := (y1, y2) ∈ R3 ⊕ R3 : y1 = y2 = (R3 ⊕ R3)Z/2

I F (λ1x1+λ2x2) ∈ T =⇒λ1 = λ2λ1f (x1) = λ2f (x2)

=⇒λ1 = λ2 = 1

2f (x1) = f (x2)

I f : [3]∗2 → R2 a continuous injection =⇒ F (Ω) ∩ T = ∅S3 ∼= Ω

F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2

The Complete bipartite graph K3,3 is not planarTheorem

There is no continuous injective map [3]∗2 −→ R2.Proof:I f : [3]∗2 → R2 a continuous injection

I Ω := ([3]∗2)∗2∆(2)∼=Lemma ([3]∗2∆(2))

∗2 =Example (S1)∗2 = S3

Ω is a free Z/2-space

I F : Ω→ R3 ⊕ R3, F (λ1x1 + λ2x2) := (λ1, λ1f (x1))⊕ (λ2, λ2f (x2))

F is a Z/2-mapI T := (y1, y2) ∈ R3 ⊕ R3 : y1 = y2 = (R3 ⊕ R3)Z/2

I F (λ1x1+λ2x2) ∈ T =⇒λ1 = λ2λ1f (x1) = λ2f (x2)

=⇒λ1 = λ2 = 1

2f (x1) = f (x2)

I f : [3]∗2 → R2 a continuous injection =⇒ F (Ω) ∩ T = ∅S3 ∼= Ω

F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2

The Complete bipartite graph K3,3 is not planarTheorem

There is no continuous injective map [3]∗2 −→ R2.Proof:I f : [3]∗2 → R2 a continuous injection

I Ω := ([3]∗2)∗2∆(2)∼=Lemma ([3]∗2∆(2))

∗2 =Example (S1)∗2 = S3

Ω is a free Z/2-space

I F : Ω→ R3 ⊕ R3, F (λ1x1 + λ2x2) := (λ1, λ1f (x1))⊕ (λ2, λ2f (x2))

F is a Z/2-map

I T := (y1, y2) ∈ R3 ⊕ R3 : y1 = y2 = (R3 ⊕ R3)Z/2

I F (λ1x1+λ2x2) ∈ T =⇒λ1 = λ2λ1f (x1) = λ2f (x2)

=⇒λ1 = λ2 = 1

2f (x1) = f (x2)

I f : [3]∗2 → R2 a continuous injection =⇒ F (Ω) ∩ T = ∅S3 ∼= Ω

F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2

The Complete bipartite graph K3,3 is not planarTheorem

There is no continuous injective map [3]∗2 −→ R2.Proof:I f : [3]∗2 → R2 a continuous injection

I Ω := ([3]∗2)∗2∆(2)∼=Lemma ([3]∗2∆(2))

∗2 =Example (S1)∗2 = S3

Ω is a free Z/2-space

I F : Ω→ R3 ⊕ R3, F (λ1x1 + λ2x2) := (λ1, λ1f (x1))⊕ (λ2, λ2f (x2))

F is a Z/2-mapI T := (y1, y2) ∈ R3 ⊕ R3 : y1 = y2

= (R3 ⊕ R3)Z/2

I F (λ1x1+λ2x2) ∈ T =⇒λ1 = λ2λ1f (x1) = λ2f (x2)

=⇒λ1 = λ2 = 1

2f (x1) = f (x2)

I f : [3]∗2 → R2 a continuous injection =⇒ F (Ω) ∩ T = ∅S3 ∼= Ω

F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2

The Complete bipartite graph K3,3 is not planarTheorem

There is no continuous injective map [3]∗2 −→ R2.Proof:I f : [3]∗2 → R2 a continuous injection

I Ω := ([3]∗2)∗2∆(2)∼=Lemma ([3]∗2∆(2))

∗2 =Example (S1)∗2 = S3

Ω is a free Z/2-space

I F : Ω→ R3 ⊕ R3, F (λ1x1 + λ2x2) := (λ1, λ1f (x1))⊕ (λ2, λ2f (x2))

F is a Z/2-mapI T := (y1, y2) ∈ R3 ⊕ R3 : y1 = y2 = (R3 ⊕ R3)Z/2

I F (λ1x1+λ2x2) ∈ T =⇒λ1 = λ2λ1f (x1) = λ2f (x2)

=⇒λ1 = λ2 = 1

2f (x1) = f (x2)

I f : [3]∗2 → R2 a continuous injection =⇒ F (Ω) ∩ T = ∅S3 ∼= Ω

F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2

The Complete bipartite graph K3,3 is not planarTheorem

There is no continuous injective map [3]∗2 −→ R2.Proof:I f : [3]∗2 → R2 a continuous injection

I Ω := ([3]∗2)∗2∆(2)∼=Lemma ([3]∗2∆(2))

∗2 =Example (S1)∗2 = S3

Ω is a free Z/2-space

I F : Ω→ R3 ⊕ R3, F (λ1x1 + λ2x2) := (λ1, λ1f (x1))⊕ (λ2, λ2f (x2))

F is a Z/2-mapI T := (y1, y2) ∈ R3 ⊕ R3 : y1 = y2 = (R3 ⊕ R3)Z/2

I F (λ1x1+λ2x2) ∈ T

=⇒λ1 = λ2λ1f (x1) = λ2f (x2)

=⇒λ1 = λ2 = 1

2f (x1) = f (x2)

I f : [3]∗2 → R2 a continuous injection =⇒ F (Ω) ∩ T = ∅S3 ∼= Ω

F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2

The Complete bipartite graph K3,3 is not planarTheorem

There is no continuous injective map [3]∗2 −→ R2.Proof:I f : [3]∗2 → R2 a continuous injection

I Ω := ([3]∗2)∗2∆(2)∼=Lemma ([3]∗2∆(2))

∗2 =Example (S1)∗2 = S3

Ω is a free Z/2-space

I F : Ω→ R3 ⊕ R3, F (λ1x1 + λ2x2) := (λ1, λ1f (x1))⊕ (λ2, λ2f (x2))

F is a Z/2-mapI T := (y1, y2) ∈ R3 ⊕ R3 : y1 = y2 = (R3 ⊕ R3)Z/2

I F (λ1x1+λ2x2) ∈ T =⇒λ1 = λ2λ1f (x1) = λ2f (x2)

=⇒λ1 = λ2 = 1

2f (x1) = f (x2)

I f : [3]∗2 → R2 a continuous injection =⇒ F (Ω) ∩ T = ∅S3 ∼= Ω

F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2

The Complete bipartite graph K3,3 is not planarTheorem

There is no continuous injective map [3]∗2 −→ R2.Proof:I f : [3]∗2 → R2 a continuous injection

I Ω := ([3]∗2)∗2∆(2)∼=Lemma ([3]∗2∆(2))

∗2 =Example (S1)∗2 = S3

Ω is a free Z/2-space

I F : Ω→ R3 ⊕ R3, F (λ1x1 + λ2x2) := (λ1, λ1f (x1))⊕ (λ2, λ2f (x2))

F is a Z/2-mapI T := (y1, y2) ∈ R3 ⊕ R3 : y1 = y2 = (R3 ⊕ R3)Z/2

I F (λ1x1+λ2x2) ∈ T =⇒λ1 = λ2λ1f (x1) = λ2f (x2)

=⇒λ1 = λ2 = 1

2f (x1) = f (x2)

I f : [3]∗2 → R2 a continuous injection =⇒ F (Ω) ∩ T = ∅S3 ∼= Ω

F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2

The Complete bipartite graph K3,3 is not planarTheorem

There is no continuous injective map [3]∗2 −→ R2.Proof:I f : [3]∗2 → R2 a continuous injection

I Ω := ([3]∗2)∗2∆(2)∼=Lemma ([3]∗2∆(2))

∗2 =Example (S1)∗2 = S3

Ω is a free Z/2-space

I F : Ω→ R3 ⊕ R3, F (λ1x1 + λ2x2) := (λ1, λ1f (x1))⊕ (λ2, λ2f (x2))

F is a Z/2-mapI T := (y1, y2) ∈ R3 ⊕ R3 : y1 = y2 = (R3 ⊕ R3)Z/2

I F (λ1x1+λ2x2) ∈ T =⇒λ1 = λ2λ1f (x1) = λ2f (x2)

=⇒λ1 = λ2 = 1

2f (x1) = f (x2)

I f : [3]∗2 → R2 a continuous injection =⇒ F (Ω) ∩ T = ∅

S3 ∼= ΩF

//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2

The Complete bipartite graph K3,3 is not planarTheorem

There is no continuous injective map [3]∗2 −→ R2.Proof:I f : [3]∗2 → R2 a continuous injection

I Ω := ([3]∗2)∗2∆(2)∼=Lemma ([3]∗2∆(2))

∗2 =Example (S1)∗2 = S3

Ω is a free Z/2-space

I F : Ω→ R3 ⊕ R3, F (λ1x1 + λ2x2) := (λ1, λ1f (x1))⊕ (λ2, λ2f (x2))

F is a Z/2-mapI T := (y1, y2) ∈ R3 ⊕ R3 : y1 = y2 = (R3 ⊕ R3)Z/2

I F (λ1x1+λ2x2) ∈ T =⇒λ1 = λ2λ1f (x1) = λ2f (x2)

=⇒λ1 = λ2 = 1

2f (x1) = f (x2)

I f : [3]∗2 → R2 a continuous injection =⇒ F (Ω) ∩ T = ∅S3 ∼= Ω

F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2

The Complete bipartite graph K3,3 is not planarTheorem

There is no continuous injective map [3]∗2 −→ R2.Proof:I f : [3]∗2 → R2 a continuous injection

I Ω := ([3]∗2)∗2∆(2)∼=Lemma ([3]∗2∆(2))

∗2 =Example (S1)∗2 = S3

Ω is a free Z/2-space

I F : Ω→ R3 ⊕ R3, F (λ1x1 + λ2x2) := (λ1, λ1f (x1))⊕ (λ2, λ2f (x2))

F is a Z/2-mapI T := (y1, y2) ∈ R3 ⊕ R3 : y1 = y2 = (R3 ⊕ R3)Z/2

I F (λ1x1+λ2x2) ∈ T =⇒λ1 = λ2λ1f (x1) = λ2f (x2)

=⇒λ1 = λ2 = 1

2f (x1) = f (x2)

I f : [3]∗2 → R2 a continuous injection =⇒ F (Ω) ∩ T = ∅S3 ∼= Ω

F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2

Ham Sandwich theorem

DefinitionAn affine hyperplane H in Rd determined by v ∈ S(Rd ) and a ∈ Ris the set:

H = x ∈ Rd : 〈x, v〉 = a.

If H is accompanied with the defining vector v ∈ S(Rd ), then thepair (H, v) is called an oriented affine hyperplane.

An oriented affine hyperplane (H, v) determines two halfspaces:

H− = x ∈ Rd : 〈x, v〉 < a and H+ = x ∈ Rd : 〈x, v〉 > a.

Ham Sandwich theorem

DefinitionAn affine hyperplane H in Rd determined by v ∈ S(Rd ) and a ∈ Ris the set:

H = x ∈ Rd : 〈x, v〉 = a.

If H is accompanied with the defining vector v ∈ S(Rd ), then thepair (H, v) is called an oriented affine hyperplane.

An oriented affine hyperplane (H, v) determines two halfspaces:

H− = x ∈ Rd : 〈x, v〉 < a and H+ = x ∈ Rd : 〈x, v〉 > a.

Ham Sandwich theorem

DefinitionAn affine hyperplane H in Rd determined by v ∈ S(Rd ) and a ∈ Ris the set:

H = x ∈ Rd : 〈x, v〉 = a.

If H is accompanied with the defining vector v ∈ S(Rd ), then thepair (H, v) is called an oriented affine hyperplane.

An oriented affine hyperplane (H, v) determines two halfspaces:

H− = x ∈ Rd : 〈x, v〉 < a and H+ = x ∈ Rd : 〈x, v〉 > a.

Ham Sandwich theorem

DefinitionAn affine hyperplane H in Rd determined by v ∈ S(Rd ) and a ∈ Ris the set:

H = x ∈ Rd : 〈x, v〉 = a.

If H is accompanied with the defining vector v ∈ S(Rd ), then thepair (H, v) is called an oriented affine hyperplane.

An oriented affine hyperplane (H, v) determines two halfspaces:

H− = x ∈ Rd : 〈x, v〉 < a and H+ = x ∈ Rd : 〈x, v〉 > a.

Ham Sandwich theorem

TheoremLet d ≥ 1, and µ1, . . . , µd be finite Borel measures on Rd with theproperty that measure of every hyperplane in Rd is zero.

Then there exists an affine hyperplane H in Rd such that

µ1(H+) = µ1(H−), . . . , µd (H+) = µd (H−).

Ham Sandwich theorem

TheoremLet d ≥ 1, and µ1, . . . , µd be finite Borel measures on Rd with theproperty that measure of every hyperplane in Rd is zero.Then there exists an affine hyperplane H in Rd such that

µ1(H+) = µ1(H−), . . . , µd (H+) = µd (H−).

Ham Sandwich theorem (proof)

Ω =space of affine oriented hyperplanes in Rd

Ham Sandwich theorem (proof)

Ω =space of affine oriented hyperplanes in Rd

= oriented hyperlanes in Rd+1 withoutthe ones perpendicular to ±e := (0, . . . , 0,±1)

Ham Sandwich theorem (proof)

Ω =space of affine oriented hyperplanes in Rd

= oriented hyperlanes in Rd+1 withoutthe ones perpendicular to ±e := (0, . . . , 0,±1)

= S(Rd+1)\e,−e

Ω is a Z/2-space with respect to orientation change

Ham Sandwich theorem (proof)

Ω =space of affine oriented hyperplanes in Rd

= oriented hyperlanes in Rd+1 withoutthe ones perpendicular to ±e := (0, . . . , 0,±1)

= S(Rd+1)\e,−eΩ is a Z/2-space with respect to orientation change

Ham Sandwich theorem (proof)I Ω =space of affine oriented hyperplanes in Rd = S(Rd+1)\e,−e

I µ1, . . . , µd fail Ham Sandwich theorem

I F : Ω −→ Rd , F (v) :=(µi (H+

v ∩ Rd)− µi (H−v ∩ Rd))i=1,...,d

I F is a Z/2-mapv F

//

·ε

(µi (H+

v ∩ Rd )− µi (H−v ∩ Rd )

)i=1,...,d

·ε

−v F//(µi (H+

−v ∩ Rd )− µi (H−−v ∩ Rd )

)i=1,...,d

I T := 0 ⊂ Rd

I F (v) ∈ T =⇒ µi (H+v ∩ Rd) = µi (H−v ∩ Rd) for i = 1, . . . , d

=⇒ F : S(Rd+1)\e,−e −→Z/2 Rd\0

I F : S(Rd+1)\e,−e −→Z/2 Rd\0, F : S(Rd+1) −→Z/2 Rd\0

S(Rd+1) −→Z/2 Rd\0 −→Z/2 S(Rd)

Ham Sandwich theorem (proof)I Ω =space of affine oriented hyperplanes in Rd = S(Rd+1)\e,−e

I µ1, . . . , µd fail Ham Sandwich theorem

I F : Ω −→ Rd , F (v) :=(µi (H+

v ∩ Rd)− µi (H−v ∩ Rd))i=1,...,d

I F is a Z/2-mapv F

//

·ε

(µi (H+

v ∩ Rd )− µi (H−v ∩ Rd )

)i=1,...,d

·ε

−v F//(µi (H+

−v ∩ Rd )− µi (H−−v ∩ Rd )

)i=1,...,d

I T := 0 ⊂ Rd

I F (v) ∈ T =⇒ µi (H+v ∩ Rd) = µi (H−v ∩ Rd) for i = 1, . . . , d

=⇒ F : S(Rd+1)\e,−e −→Z/2 Rd\0

I F : S(Rd+1)\e,−e −→Z/2 Rd\0, F : S(Rd+1) −→Z/2 Rd\0

S(Rd+1) −→Z/2 Rd\0 −→Z/2 S(Rd)

Ham Sandwich theorem (proof)I Ω =space of affine oriented hyperplanes in Rd = S(Rd+1)\e,−e

I µ1, . . . , µd fail Ham Sandwich theorem

I F : Ω −→ Rd , F (v) :=(µi (H+

v ∩ Rd)− µi (H−v ∩ Rd))i=1,...,d

I F is a Z/2-mapv F

//

·ε

(µi (H+

v ∩ Rd )− µi (H−v ∩ Rd )

)i=1,...,d

·ε

−v F//(µi (H+

−v ∩ Rd )− µi (H−−v ∩ Rd )

)i=1,...,d

I T := 0 ⊂ Rd

I F (v) ∈ T =⇒ µi (H+v ∩ Rd) = µi (H−v ∩ Rd) for i = 1, . . . , d

=⇒ F : S(Rd+1)\e,−e −→Z/2 Rd\0

I F : S(Rd+1)\e,−e −→Z/2 Rd\0, F : S(Rd+1) −→Z/2 Rd\0

S(Rd+1) −→Z/2 Rd\0 −→Z/2 S(Rd)

Ham Sandwich theorem (proof)I Ω =space of affine oriented hyperplanes in Rd = S(Rd+1)\e,−e

I µ1, . . . , µd fail Ham Sandwich theorem

I F : Ω −→ Rd , F (v) :=(µi (H+

v ∩ Rd)− µi (H−v ∩ Rd))i=1,...,d

I F is a Z/2-mapv F

//

·ε

(µi (H+

v ∩ Rd )− µi (H−v ∩ Rd )

)i=1,...,d

·ε

−v F//(µi (H+

−v ∩ Rd )− µi (H−−v ∩ Rd )

)i=1,...,d

I T := 0 ⊂ Rd

I F (v) ∈ T =⇒ µi (H+v ∩ Rd) = µi (H−v ∩ Rd) for i = 1, . . . , d

=⇒ F : S(Rd+1)\e,−e −→Z/2 Rd\0

I F : S(Rd+1)\e,−e −→Z/2 Rd\0, F : S(Rd+1) −→Z/2 Rd\0

S(Rd+1) −→Z/2 Rd\0 −→Z/2 S(Rd)

Ham Sandwich theorem (proof)I Ω =space of affine oriented hyperplanes in Rd = S(Rd+1)\e,−e

I µ1, . . . , µd fail Ham Sandwich theorem

I F : Ω −→ Rd , F (v) :=(µi (H+

v ∩ Rd)− µi (H−v ∩ Rd))i=1,...,d

I F is a Z/2-mapv F

//

·ε

(µi (H+

v ∩ Rd )− µi (H−v ∩ Rd )

)i=1,...,d

·ε

−v F//(µi (H+

−v ∩ Rd )− µi (H−−v ∩ Rd )

)i=1,...,d

I T := 0 ⊂ Rd

I F (v) ∈ T =⇒ µi (H+v ∩ Rd) = µi (H−v ∩ Rd) for i = 1, . . . , d

=⇒ F : S(Rd+1)\e,−e −→Z/2 Rd\0

I F : S(Rd+1)\e,−e −→Z/2 Rd\0, F : S(Rd+1) −→Z/2 Rd\0

S(Rd+1) −→Z/2 Rd\0 −→Z/2 S(Rd)

Ham Sandwich theorem (proof)I Ω =space of affine oriented hyperplanes in Rd = S(Rd+1)\e,−e

I µ1, . . . , µd fail Ham Sandwich theorem

I F : Ω −→ Rd , F (v) :=(µi (H+

v ∩ Rd)− µi (H−v ∩ Rd))i=1,...,d

I F is a Z/2-mapv F

//

·ε

(µi (H+

v ∩ Rd )− µi (H−v ∩ Rd )

)i=1,...,d

·ε

−v F//(µi (H+

−v ∩ Rd )− µi (H−−v ∩ Rd )

)i=1,...,d

I T := 0 ⊂ Rd

I F (v) ∈ T

=⇒ µi (H+v ∩ Rd) = µi (H−v ∩ Rd) for i = 1, . . . , d

=⇒ F : S(Rd+1)\e,−e −→Z/2 Rd\0

I F : S(Rd+1)\e,−e −→Z/2 Rd\0, F : S(Rd+1) −→Z/2 Rd\0

S(Rd+1) −→Z/2 Rd\0 −→Z/2 S(Rd)

Ham Sandwich theorem (proof)I Ω =space of affine oriented hyperplanes in Rd = S(Rd+1)\e,−e

I µ1, . . . , µd fail Ham Sandwich theorem

I F : Ω −→ Rd , F (v) :=(µi (H+

v ∩ Rd)− µi (H−v ∩ Rd))i=1,...,d

I F is a Z/2-mapv F

//

·ε

(µi (H+

v ∩ Rd )− µi (H−v ∩ Rd )

)i=1,...,d

·ε

−v F//(µi (H+

−v ∩ Rd )− µi (H−−v ∩ Rd )

)i=1,...,d

I T := 0 ⊂ Rd

I F (v) ∈ T =⇒ µi (H+v ∩ Rd) = µi (H−v ∩ Rd) for i = 1, . . . , d

=⇒ F : S(Rd+1)\e,−e −→Z/2 Rd\0

I F : S(Rd+1)\e,−e −→Z/2 Rd\0, F : S(Rd+1) −→Z/2 Rd\0

S(Rd+1) −→Z/2 Rd\0 −→Z/2 S(Rd)

Ham Sandwich theorem (proof)I Ω =space of affine oriented hyperplanes in Rd = S(Rd+1)\e,−e

I µ1, . . . , µd fail Ham Sandwich theorem

I F : Ω −→ Rd , F (v) :=(µi (H+

v ∩ Rd)− µi (H−v ∩ Rd))i=1,...,d

I F is a Z/2-mapv F

//

·ε

(µi (H+

v ∩ Rd )− µi (H−v ∩ Rd )

)i=1,...,d

·ε

−v F//(µi (H+

−v ∩ Rd )− µi (H−−v ∩ Rd )

)i=1,...,d

I T := 0 ⊂ Rd

I F (v) ∈ T =⇒ µi (H+v ∩ Rd) = µi (H−v ∩ Rd) for i = 1, . . . , d

=⇒ F : S(Rd+1)\e,−e −→Z/2 Rd\0

I F : S(Rd+1)\e,−e −→Z/2 Rd\0, F : S(Rd+1) −→Z/2 Rd\0

S(Rd+1) −→Z/2 Rd\0 −→Z/2 S(Rd)

Ham Sandwich theorem (proof)I Ω =space of affine oriented hyperplanes in Rd = S(Rd+1)\e,−e

I µ1, . . . , µd fail Ham Sandwich theorem

I F : Ω −→ Rd , F (v) :=(µi (H+

v ∩ Rd)− µi (H−v ∩ Rd))i=1,...,d

I F is a Z/2-mapv F

//

·ε

(µi (H+

v ∩ Rd )− µi (H−v ∩ Rd )

)i=1,...,d

·ε

−v F//(µi (H+

−v ∩ Rd )− µi (H−−v ∩ Rd )

)i=1,...,d

I T := 0 ⊂ Rd

I F (v) ∈ T =⇒ µi (H+v ∩ Rd) = µi (H−v ∩ Rd) for i = 1, . . . , d

=⇒ F : S(Rd+1)\e,−e −→Z/2 Rd\0

I F : S(Rd+1)\e,−e −→Z/2 Rd\0,

F : S(Rd+1) −→Z/2 Rd\0

S(Rd+1) −→Z/2 Rd\0 −→Z/2 S(Rd)

Ham Sandwich theorem (proof)I Ω =space of affine oriented hyperplanes in Rd = S(Rd+1)\e,−e

I µ1, . . . , µd fail Ham Sandwich theorem

I F : Ω −→ Rd , F (v) :=(µi (H+

v ∩ Rd)− µi (H−v ∩ Rd))i=1,...,d

I F is a Z/2-mapv F

//

·ε

(µi (H+

v ∩ Rd )− µi (H−v ∩ Rd )

)i=1,...,d

·ε

−v F//(µi (H+

−v ∩ Rd )− µi (H−−v ∩ Rd )

)i=1,...,d

I T := 0 ⊂ Rd

I F (v) ∈ T =⇒ µi (H+v ∩ Rd) = µi (H−v ∩ Rd) for i = 1, . . . , d

=⇒ F : S(Rd+1)\e,−e −→Z/2 Rd\0

I F : S(Rd+1)\e,−e −→Z/2 Rd\0, F : S(Rd+1) −→Z/2 Rd\0

S(Rd+1) −→Z/2 Rd\0 −→Z/2 S(Rd)

Ham Sandwich theorem (proof)I Ω =space of affine oriented hyperplanes in Rd = S(Rd+1)\e,−e

I µ1, . . . , µd fail Ham Sandwich theorem

I F : Ω −→ Rd , F (v) :=(µi (H+

v ∩ Rd)− µi (H−v ∩ Rd))i=1,...,d

I F is a Z/2-mapv F

//

·ε

(µi (H+

v ∩ Rd )− µi (H−v ∩ Rd )

)i=1,...,d

·ε

−v F//(µi (H+

−v ∩ Rd )− µi (H−−v ∩ Rd )

)i=1,...,d

I T := 0 ⊂ Rd

I F (v) ∈ T =⇒ µi (H+v ∩ Rd) = µi (H−v ∩ Rd) for i = 1, . . . , d

=⇒ F : S(Rd+1)\e,−e −→Z/2 Rd\0

I F : S(Rd+1)\e,−e −→Z/2 Rd\0, F : S(Rd+1) −→Z/2 Rd\0

S(Rd+1) −→Z/2 Rd\0

−→Z/2 S(Rd)

Ham Sandwich theorem (proof)I Ω =space of affine oriented hyperplanes in Rd = S(Rd+1)\e,−e

I µ1, . . . , µd fail Ham Sandwich theorem

I F : Ω −→ Rd , F (v) :=(µi (H+

v ∩ Rd)− µi (H−v ∩ Rd))i=1,...,d

I F is a Z/2-mapv F

//

·ε

(µi (H+

v ∩ Rd )− µi (H−v ∩ Rd )

)i=1,...,d

·ε

−v F//(µi (H+

−v ∩ Rd )− µi (H−−v ∩ Rd )

)i=1,...,d

I T := 0 ⊂ Rd

I F (v) ∈ T =⇒ µi (H+v ∩ Rd) = µi (H−v ∩ Rd) for i = 1, . . . , d

=⇒ F : S(Rd+1)\e,−e −→Z/2 Rd\0

I F : S(Rd+1)\e,−e −→Z/2 Rd\0, F : S(Rd+1) −→Z/2 Rd\0

S(Rd+1) −→Z/2 Rd\0 −→Z/2 S(Rd)

Ham Sandwich theorem (proof)I Ω =space of affine oriented hyperplanes in Rd = S(Rd+1)\e,−e

I µ1, . . . , µd fail Ham Sandwich theorem

I F : Ω −→ Rd , F (v) :=(µi (H+

v ∩ Rd)− µi (H−v ∩ Rd))i=1,...,d

I F is a Z/2-mapv F

//

·ε

(µi (H+

v ∩ Rd )− µi (H−v ∩ Rd )

)i=1,...,d

·ε

−v F//(µi (H+

−v ∩ Rd )− µi (H−−v ∩ Rd )

)i=1,...,d

I T := 0 ⊂ Rd

I F (v) ∈ T =⇒ µi (H+v ∩ Rd) = µi (H−v ∩ Rd) for i = 1, . . . , d

=⇒ F : S(Rd+1)\e,−e −→Z/2 Rd\0

I F : S(Rd+1)\e,−e −→Z/2 Rd\0, F : S(Rd+1) −→Z/2 Rd\0

S(Rd+1) −→Z/2 Rd\0 −→Z/2 S(Rd)

Tverberg’s theoremTheorem (Tverberg’s theorem)Let d ≥ 1, r ≥ 1, N = (d + 1)(r − 1), and f : ∆N −→ Rd be an affinemap. Then there exist r pairwise disjoint faces F1, . . . ,Fr of the simplex∆N such that:

f (F1) ∩ · · · ∩ f (Fr ) 6= ∅.

The collection F1, . . . ,Fr is called an Tverberg partition for the map f .

Conjecture (Topological Tverberg’s conjecture)Let d ≥ 1, r ≥ 1, N = (d + 1)(r − 1), and f : ∆N −→ Rd be acontinuous map. Then there exist r pairwise disjoint faces F1, . . . ,Fr ofthe simplex ∆N such that:

f (F1) ∩ · · · ∩ f (Fr ) 6= ∅.

• r is a prime:Bárány, Shlosman, Szűcs: On a topological generalization of a theorem of Tverberg, J. London

Math. Soc., 1981.

• r is a prime power:Özaydin, Equivariant maps for the symmetric group, Preprint, 1987.

Tverberg’s theoremTheorem (Tverberg’s theorem)Let d ≥ 1, r ≥ 1, N = (d + 1)(r − 1), and f : ∆N −→ Rd be an affinemap. Then there exist r pairwise disjoint faces F1, . . . ,Fr of the simplex∆N such that:

f (F1) ∩ · · · ∩ f (Fr ) 6= ∅.

The collection F1, . . . ,Fr is called an Tverberg partition for the map f .

Conjecture (Topological Tverberg’s conjecture)Let d ≥ 1, r ≥ 1, N = (d + 1)(r − 1), and f : ∆N −→ Rd be acontinuous map. Then there exist r pairwise disjoint faces F1, . . . ,Fr ofthe simplex ∆N such that:

f (F1) ∩ · · · ∩ f (Fr ) 6= ∅.

• r is a prime:Bárány, Shlosman, Szűcs: On a topological generalization of a theorem of Tverberg, J. London

Math. Soc., 1981.

• r is a prime power:Özaydin, Equivariant maps for the symmetric group, Preprint, 1987.

Tverberg’s theoremTheorem (Tverberg’s theorem)Let d ≥ 1, r ≥ 1, N = (d + 1)(r − 1), and f : ∆N −→ Rd be an affinemap. Then there exist r pairwise disjoint faces F1, . . . ,Fr of the simplex∆N such that:

f (F1) ∩ · · · ∩ f (Fr ) 6= ∅.

The collection F1, . . . ,Fr is called an Tverberg partition for the map f .

Conjecture (Topological Tverberg’s conjecture)Let d ≥ 1, r ≥ 1, N = (d + 1)(r − 1), and f : ∆N −→ Rd be acontinuous map. Then there exist r pairwise disjoint faces F1, . . . ,Fr ofthe simplex ∆N such that:

f (F1) ∩ · · · ∩ f (Fr ) 6= ∅.

• r is a prime:Bárány, Shlosman, Szűcs: On a topological generalization of a theorem of Tverberg, J. London

Math. Soc., 1981.

• r is a prime power:Özaydin, Equivariant maps for the symmetric group, Preprint, 1987.

Tverberg’s theoremTheorem (Tverberg’s theorem)Let d ≥ 1, r ≥ 1, N = (d + 1)(r − 1), and f : ∆N −→ Rd be an affinemap. Then there exist r pairwise disjoint faces F1, . . . ,Fr of the simplex∆N such that:

f (F1) ∩ · · · ∩ f (Fr ) 6= ∅.

The collection F1, . . . ,Fr is called an Tverberg partition for the map f .

Conjecture (Topological Tverberg’s conjecture)Let d ≥ 1, r ≥ 1, N = (d + 1)(r − 1), and f : ∆N −→ Rd be acontinuous map. Then there exist r pairwise disjoint faces F1, . . . ,Fr ofthe simplex ∆N such that:

f (F1) ∩ · · · ∩ f (Fr ) 6= ∅.

• r is a prime:Bárány, Shlosman, Szűcs: On a topological generalization of a theorem of Tverberg, J. London

Math. Soc., 1981.

• r is a prime power:Özaydin, Equivariant maps for the symmetric group, Preprint, 1987.

Tverberg’s theoremTheorem (Tverberg’s theorem)Let d ≥ 1, r ≥ 1, N = (d + 1)(r − 1), and f : ∆N −→ Rd be an affinemap. Then there exist r pairwise disjoint faces F1, . . . ,Fr of the simplex∆N such that:

f (F1) ∩ · · · ∩ f (Fr ) 6= ∅.

The collection F1, . . . ,Fr is called an Tverberg partition for the map f .

Conjecture (Topological Tverberg’s conjecture)Let d ≥ 1, r ≥ 1, N = (d + 1)(r − 1), and f : ∆N −→ Rd be acontinuous map. Then there exist r pairwise disjoint faces F1, . . . ,Fr ofthe simplex ∆N such that:

f (F1) ∩ · · · ∩ f (Fr ) 6= ∅.

• r is a prime:Bárány, Shlosman, Szűcs: On a topological generalization of a theorem of Tverberg, J. London

Math. Soc., 1981.

• r is a prime power:Özaydin, Equivariant maps for the symmetric group, Preprint, 1987.

Grünbaum mass partition problem

I µ1, . . . , µj are finite Borel measures on Rd with the property thatmeasure of every hyperplane in Rd is zero.

I H1, . . . ,Hk affine hyperplanes in Rd

Orthant determined by H1, . . . ,Hk is a connected component of the

complement Rd\(H1 ∪ · · · ∪ Hk).

Orthant determined by H1, . . . ,Hk is an intersection Hσ11 ∩ · · · ∩ Hσk

k

where σi ∈ +,−.

ProblemDetermine all triples (d , j , k) with the property that for every collectionof j (proper) probability measures in Rd there exists k affine hyperplanesH1, . . . ,Hk such that each orthant determined by them contains the sameamount of every measure. Triples with this property are called admissible.

Grünbaum mass partition problemI µ1, . . . , µj are finite Borel measures on Rd with the property thatmeasure of every hyperplane in Rd is zero.

I H1, . . . ,Hk affine hyperplanes in Rd

Orthant determined by H1, . . . ,Hk is a connected component of the

complement Rd\(H1 ∪ · · · ∪ Hk).

Orthant determined by H1, . . . ,Hk is an intersection Hσ11 ∩ · · · ∩ Hσk

k

where σi ∈ +,−.

ProblemDetermine all triples (d , j , k) with the property that for every collectionof j (proper) probability measures in Rd there exists k affine hyperplanesH1, . . . ,Hk such that each orthant determined by them contains the sameamount of every measure. Triples with this property are called admissible.

Grünbaum mass partition problemI µ1, . . . , µj are finite Borel measures on Rd with the property thatmeasure of every hyperplane in Rd is zero.

I H1, . . . ,Hk affine hyperplanes in Rd

Orthant determined by H1, . . . ,Hk is a connected component of the

complement Rd\(H1 ∪ · · · ∪ Hk).

Orthant determined by H1, . . . ,Hk is an intersection Hσ11 ∩ · · · ∩ Hσk

k

where σi ∈ +,−.

ProblemDetermine all triples (d , j , k) with the property that for every collectionof j (proper) probability measures in Rd there exists k affine hyperplanesH1, . . . ,Hk such that each orthant determined by them contains the sameamount of every measure. Triples with this property are called admissible.

Grünbaum mass partition problemI µ1, . . . , µj are finite Borel measures on Rd with the property thatmeasure of every hyperplane in Rd is zero.

I H1, . . . ,Hk affine hyperplanes in Rd

Orthant determined by H1, . . . ,Hk is a connected component of the

complement Rd\(H1 ∪ · · · ∪ Hk).

Orthant determined by H1, . . . ,Hk is an intersection Hσ11 ∩ · · · ∩ Hσk

k

where σi ∈ +,−.

ProblemDetermine all triples (d , j , k) with the property that for every collectionof j (proper) probability measures in Rd there exists k affine hyperplanesH1, . . . ,Hk such that each orthant determined by them contains the sameamount of every measure. Triples with this property are called admissible.

Grünbaum mass partition problemI µ1, . . . , µj are finite Borel measures on Rd with the property thatmeasure of every hyperplane in Rd is zero.

I H1, . . . ,Hk affine hyperplanes in Rd

Orthant determined by H1, . . . ,Hk is a connected component of the

complement Rd\(H1 ∪ · · · ∪ Hk).

Orthant determined by H1, . . . ,Hk is an intersection Hσ11 ∩ · · · ∩ Hσk

k

where σi ∈ +,−.

ProblemDetermine all triples (d , j , k) with the property that for every collectionof j (proper) probability measures in Rd there exists k affine hyperplanesH1, . . . ,Hk such that each orthant determined by them contains the sameamount of every measure. Triples with this property are called admissible.

Grünbaum mass partition problemI µ1, . . . , µj are finite Borel measures on Rd with the property thatmeasure of every hyperplane in Rd is zero.

I H1, . . . ,Hk affine hyperplanes in Rd

Orthant determined by H1, . . . ,Hk is a connected component of the

complement Rd\(H1 ∪ · · · ∪ Hk).

Orthant determined by H1, . . . ,Hk is an intersection Hσ11 ∩ · · · ∩ Hσk

k

where σi ∈ +,−.

ProblemDetermine all triples (d , j , k) with the property that for every collectionof j (proper) probability measures in Rd there exists k affine hyperplanesH1, . . . ,Hk such that each orthant determined by them contains the sameamount of every measure. Triples with this property are called admissible.

Nandakumar & Ramana Rao problemProblem (Nandakumar & Ramana Rao 2006)Every convex polygon P in the plane can be partitioned into anyprescribed number r of convex pieces that have equal areas as well asequal perimeters.

Nandakumar & Ramana Rao problemProblem (Nandakumar & Ramana Rao 2006)Every convex polygon P in the plane can be partitioned into anyprescribed number r of convex pieces that have equal areas as well asequal perimeters.

Nandakumar & Ramana Rao problemProblem (Nandakumar & Ramana Rao 2006)Every convex polygon P in the plane can be partitioned into anyprescribed number r of convex pieces that have equal areas as well asequal perimeters.

Nandakumar & Ramana Rao problemProblem (Nandakumar & Ramana Rao 2006)Every convex polygon P in the plane can be partitioned into anyprescribed number r of convex pieces that have equal areas as well asequal perimeters.

Nandakumar & Ramana Rao problemProblem (Nandakumar & Ramana Rao 2006)Every convex polygon P in the plane can be partitioned into anyprescribed number r of convex pieces that have equal areas as well asequal perimeters.

Nandakumar & Ramana Rao problemProblem (Nandakumar & Ramana Rao 2006)Every convex polygon P in the plane can be partitioned into anyprescribed number r of convex pieces that have equal areas as well asequal perimeters.

Nandakumar & Ramana Rao problemProblem (Nandakumar & Ramana Rao 2006)Every convex polygon P in the plane can be partitioned into anyprescribed number r of convex pieces that have equal areas as well asequal perimeters.

Nandakumar & Ramana Rao problemProblem (Nandakumar & Ramana Rao 2006)Every convex polygon P in the plane can be partitioned into anyprescribed number r of convex pieces that have equal areas as well asequal perimeters.