Some problems of geometric combinatoricshomepages.warwick.ac.uk/~maskas/courses/blagojevic1.pdf · Some problems of geometric combinatorics Pavle Blagojević (based on the joint work

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 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.




f (P) ∩ f (N) 6= ∅.


f (P) ∩ f (N) 6= ∅.


Sci. hungar., 1979.




f (P) ∩ f (N) 6= ∅.


f (P) ∩ f (N) 6= ∅.


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.


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))





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))





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)





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)





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:





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.


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.



f (g · x) = g · f (x).





f (g · x) = g · f (x).




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



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


∼= K ∗n∆(2) ∗ L∗n∆(2)


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



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


∼= K ∗n∆(2) ∗ L∗n∆(2)


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





∼= K ∗n∆(2) ∗ L∗n∆(2)


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





∼= K ∗n∆(2) ∗ L∗n∆(2)


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





∼= K ∗n∆(2) ∗ L∗n∆(2)


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





∼= K ∗n∆(2) ∗ L∗n∆(2)


∼= ([1]∗N+1)∗r∆(2)

∼= ([1]∗r∆(2))∗N+1 ∼= [r ]∗N+1





∼= K ∗n∆(2) ∗ L∗n∆(2)


∼= ([1]∗N+1)∗r∆(2)∼= ([1]∗r∆(2))∗N+1

∼= [r ]∗N+1





∼= K ∗n∆(2) ∗ L∗n∆(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)

)


f (P) ∩ f (N) 6= ∅.Proof:


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


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)

)


f (P) ∩ f (N) 6= ∅.Proof:


I Ω := (∆d+1)∗2∆(2)

∼=Lemma [2]∗d+2 ∼= (S0)∗d+2 ∼= Sd+1


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)

)


f (P) ∩ f (N) 6= ∅.Proof:


I Ω := (∆d+1)∗2∆(2)∼=Lemma [2]∗d+2

∼= (S0)∗d+2 ∼= Sd+1


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)

)


f (P) ∩ f (N) 6= ∅.Proof:


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


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)

)


f (P) ∩ f (N) 6= ∅.Proof:


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


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)

)


f (P) ∩ f (N) 6= ∅.Proof:


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


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)

)


f (P) ∩ f (N) 6= ∅.Proof:


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


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)

)


f (P) ∩ f (N) 6= ∅.Proof:


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


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)

)


f (P) ∩ f (N) 6= ∅.Proof:


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


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)

)


f (P) ∩ f (N) 6= ∅.Proof:


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


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)

)


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


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)


Ω ∼= Sd+1 //

((

W2 ⊕ (Rd ⊕ Rd)

W2 ⊕ (Rd ⊕ Rd)\T

44

// T⊥\0 ' S(T⊥) ∼= Sd


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)


=(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)


Ω ∼= Sd+1 //

((

W2 ⊕ (Rd ⊕ Rd)

W2 ⊕ (Rd ⊕ Rd)\T

44

// T⊥\0 ' S(T⊥) ∼= Sd


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)


=(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)


Ω ∼= Sd+1 //

((

W2 ⊕ (Rd ⊕ Rd)

W2 ⊕ (Rd ⊕ Rd)\T

44

// T⊥\0 ' S(T⊥) ∼= Sd


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)


=(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)


Ω ∼= Sd+1 //

((

W2 ⊕ (Rd ⊕ Rd)

W2 ⊕ (Rd ⊕ Rd)\T

44

// T⊥\0 ' S(T⊥) ∼= Sd


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)


=(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)


Ω ∼= Sd+1 //

((

W2 ⊕ (Rd ⊕ Rd)

W2 ⊕ (Rd ⊕ Rd)\T

44

// T⊥\0 ' S(T⊥) ∼= Sd


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)


=(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)


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


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





=⇒λ1 = λ2 = 1

2f (x1) = f (x2)


F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2


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





=⇒λ1 = λ2 = 1

2f (x1) = f (x2)


F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2



I Ω := ([3]∗2)∗2∆(2)

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





=⇒λ1 = λ2 = 1

2f (x1) = f (x2)


F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2



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

∗2

=Example (S1)∗2 = S3





=⇒λ1 = λ2 = 1

2f (x1) = f (x2)


F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2



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

∗2 =Example (S1)∗2 = S3





=⇒λ1 = λ2 = 1

2f (x1) = f (x2)


F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2



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

∗2 =Example (S1)∗2 = S3





=⇒λ1 = λ2 = 1

2f (x1) = f (x2)


F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2



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

∗2 =Example (S1)∗2 = S3


I F : Ω→ R3 ⊕ R3,

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



=⇒λ1 = λ2 = 1

2f (x1) = f (x2)


F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2



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

∗2 =Example (S1)∗2 = S3





=⇒λ1 = λ2 = 1

2f (x1) = f (x2)


F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2



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

∗2 =Example (S1)∗2 = S3



F is a Z/2-map

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


=⇒λ1 = λ2 = 1

2f (x1) = f (x2)


F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2



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

∗2 =Example (S1)∗2 = S3



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

= (R3 ⊕ R3)Z/2


=⇒λ1 = λ2 = 1

2f (x1) = f (x2)


F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2



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

∗2 =Example (S1)∗2 = S3





=⇒λ1 = λ2 = 1

2f (x1) = f (x2)


F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2



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

∗2 =Example (S1)∗2 = S3




I F (λ1x1+λ2x2) ∈ T

=⇒λ1 = λ2λ1f (x1) = λ2f (x2)

=⇒λ1 = λ2 = 1

2f (x1) = f (x2)


F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2



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

∗2 =Example (S1)∗2 = S3





=⇒λ1 = λ2 = 1

2f (x1) = f (x2)


F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2



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

∗2 =Example (S1)∗2 = S3





=⇒λ1 = λ2 = 1

2f (x1) = f (x2)


F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2



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

∗2 =Example (S1)∗2 = S3





=⇒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



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

∗2 =Example (S1)∗2 = S3





=⇒λ1 = λ2 = 1

2f (x1) = f (x2)


F//

&&

R3 ⊕ R3

(R3 ⊕ R3)\T

55

// T⊥\0 ' S(T⊥) ∼= S2



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

∗2 =Example (S1)∗2 = S3





=⇒λ1 = λ2 = 1

2f (x1) = f (x2)


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.



H = x ∈ Rd : 〈x, v〉 = a.






H = x ∈ Rd : 〈x, v〉 = a.






H = x ∈ Rd : 〈x, v〉 = a.





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−).


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



= 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




= 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)



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


=⇒ F : S(Rd+1)\e,−e −→Z/2 Rd\0


S(Rd+1) −→Z/2 Rd\0 −→Z/2 S(Rd)



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


=⇒ F : S(Rd+1)\e,−e −→Z/2 Rd\0


S(Rd+1) −→Z/2 Rd\0 −→Z/2 S(Rd)



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


=⇒ F : S(Rd+1)\e,−e −→Z/2 Rd\0


S(Rd+1) −→Z/2 Rd\0 −→Z/2 S(Rd)



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


=⇒ F : S(Rd+1)\e,−e −→Z/2 Rd\0


S(Rd+1) −→Z/2 Rd\0 −→Z/2 S(Rd)



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


S(Rd+1) −→Z/2 Rd\0 −→Z/2 S(Rd)



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


=⇒ F : S(Rd+1)\e,−e −→Z/2 Rd\0


S(Rd+1) −→Z/2 Rd\0 −→Z/2 S(Rd)



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


=⇒ F : S(Rd+1)\e,−e −→Z/2 Rd\0


S(Rd+1) −→Z/2 Rd\0 −→Z/2 S(Rd)



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


=⇒ 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)



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


=⇒ F : S(Rd+1)\e,−e −→Z/2 Rd\0


S(Rd+1) −→Z/2 Rd\0 −→Z/2 S(Rd)



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


=⇒ F : S(Rd+1)\e,−e −→Z/2 Rd\0


S(Rd+1) −→Z/2 Rd\0

−→Z/2 S(Rd)



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


=⇒ F : S(Rd+1)\e,−e −→Z/2 Rd\0


S(Rd+1) −→Z/2 Rd\0 −→Z/2 S(Rd)



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


=⇒ F : S(Rd+1)\e,−e −→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.


f (F1) ∩ · · · ∩ f (Fr ) 6= ∅.



f (F1) ∩ · · · ∩ f (Fr ) 6= ∅.


Math. Soc., 1981.



f (F1) ∩ · · · ∩ f (Fr ) 6= ∅.



f (F1) ∩ · · · ∩ f (Fr ) 6= ∅.


Math. Soc., 1981.



f (F1) ∩ · · · ∩ f (Fr ) 6= ∅.



f (F1) ∩ · · · ∩ f (Fr ) 6= ∅.


Math. Soc., 1981.



f (F1) ∩ · · · ∩ f (Fr ) 6= ∅.



f (F1) ∩ · · · ∩ f (Fr ) 6= ∅.


Math. Soc., 1981.


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.





k








k








k








k








k



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.








Some problems of geometric combinatoricshomepages.warwick.ac.uk/~maskas/courses/blagojevic1.pdf · Some problems of geometric combinatorics Pavle Blagojević (based on the joint work

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