some problems of geometric combinatoricshomepages.warwick.ac.uk/~maskas/courses/blagojevic1.pdf ·...
TRANSCRIPT
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.