Prismatic Maps for the Topological Tverberg Conjecture Isaac - PowerPoint PPT Presentation

Prismatic Maps for the Topological Tverberg Conjecture Isaac Mabillard Joint work with Uli Wagner

Geometry → Algebra

Hope ∼ Geometry → Algebra =

General Problem: Let K be a simplicial complex and r ≥ 2 . Does there exist a continuous map f : K → R d without r -fold intersections?

General Problem: Let K be a simplicial complex and r ≥ 2 . Does there exist a continuous map f : K → R d without r -fold intersections? A point p ∈ R d is an r -fold intersection if there exit x 1 , ..., x r ∈ | K | distinct such that p = fx 1 = · · · = fx r K x 2 f f ( K ) − → p x 1

General Problem: Let K be a simplicial complex and r ≥ 2 . Does there exist a continuous map f : K → R d without r -fold intersections? A point p ∈ R d is an r -fold intersection if there exit x 1 , ..., x r ∈ | K | distinct such that p = fx 1 = · · · = fx r K x 2 f f ( K ) − → p x 1 A map f : K → R d without r -fold intersection is called r -embedding

Example: f : K 2 → R 3

Example: f : K 2 → R 3 K = real projective plane R P 2

Example: f : K 2 → R 3 K = real projective plane R P 2 f R P 2 − → Boy’s Surface

Example: f : K 2 → R 3 K = real projective plane R P 2 f R P 2 − → 2 -fold intersection Boy’s Surface

Example: f : K 2 → R 3 K = real projective plane R P 2 (unique) 3 -fold intersection f R P 2 − → 2 -fold intersection Boy’s Surface

Example: f : K 2 → R 3 K = real projective plane R P 2 (unique) 3 -fold intersection f R P 2 − → 2 -fold intersection Boy’s Surface f : R P 2 → R 3 is a 4 -embedding (no 4 -fold intersections)

Classical Case: Maps without 2 -fold intersections

Classical Case: Maps without 2 -fold intersections Goal: Find f : K → R d continuous & injective (i.e., f is an embedding )

Classical Case: Maps without 2 -fold intersections Goal: Find f : K → R d continuous & injective (i.e., f is an embedding ) Theorem (van Kampen–Shapiro–Wu): ∃ f : K m ֒ ∃ � → R 2 m f : K × 2 → S 2 S 2 m − 1 ⇔ δ provided m � = 2 .

Classical Case: Maps without 2 -fold intersections Goal: Find f : K → R d continuous & injective (i.e., f is an embedding ) Theorem (van Kampen–Shapiro–Wu): ∃ f : K m ֒ ∃ � → R 2 m f : K × 2 → S 2 S 2 m − 1 ⇔ δ provided m � = 2 . → S 2 S 2 m − 1 is ‘easy’ Proposition The existence of K × 2 δ algorithmically solvable .

Classical Case: Maps without 2 -fold intersections Goal: Find f : K → R d continuous & injective (i.e., f is an embedding ) Theorem (van Kampen–Shapiro–Wu): ∃ f : K m ֒ ∃ � → R 2 m f : K × 2 → S 2 S 2 m − 1 ⇔ δ provided m � = 2 . → S 2 S 2 m − 1 is ‘easy’ Proposition The existence of K × 2 δ algorithmically solvable . → R 2 m is Corollary. The existence of an embedding K m ֒ algorithmically solvable , provided m � = 2 .

What about maps without r -fold intersections?

What about maps without r -fold intersections? Goal: Find f : K → R d continuous & r = 3 without r -fold intersection (i.e., f is an r -embedding ) f ( K ) ⊂ R 3

What about maps without r -fold intersections? Goal: Find f : K → R d continuous & r = 3 without r -fold intersection (i.e., f is an r -embedding ) An necessary condition for the existence of f : f ( K ) ⊂ R 3 1) Define the r -fold deleted product of K : K × r := { σ 1 × · · · × σ r | σ i ∈ K and σ i ∩ σ j = ∅} ⊂ K × r δ

What about maps without r -fold intersections? Goal: Find f : K → R d continuous & r = 3 without r -fold intersection (i.e., f is an r -embedding ) An necessary condition for the existence of f : f ( K ) ⊂ R 3 1) Define the r -fold deleted product of K : K × r := { σ 1 × · · · × σ r | σ i ∈ K and σ i ∩ σ j = ∅} ⊂ K × r δ 2) Given an r -embedding f : K → R d , define � K × r R d × r → f : δ ( x 1 , . . . , x r ) �→ ( fx 1 , . . . , fx r )

Two properties of � f � K × r R d × r → f : δ �→ ( x 1 , . . . , x r ) ( fx 1 , . . . , fx r )

Two properties of � f � K × r R d × r → f : δ �→ ( x 1 , . . . , x r ) ( fx 1 , . . . , fx r ) and R d × r by A) The symmetric group S r acts on both K × r δ permutation of the coordinates f is compatible with both actions (i.e., � � f is S r -equivariant): For all ρ ∈ S r f ◦ ρ = ρ ◦ � � f

Two properties of � f � K × r R d × r → f : δ �→ ( x 1 , . . . , x r ) ( fx 1 , . . . , fx r ) and R d × r by A) The symmetric group S r acts on both K × r δ permutation of the coordinates f is compatible with both actions (i.e., � � f is S r -equivariant): For all ρ ∈ S r f ◦ ρ = ρ ◦ � � f B) ( x i ∈ σ i ∈ K and σ i ∩ σ j = ∅ ) ⇒ all the x i are distinct f is an r -embedding ⇒ ¬ ( fx 1 = · · · = fx r )

Two properties of � f � K × r R d × r → f : δ �→ ( x 1 , . . . , x r ) ( fx 1 , . . . , fx r ) and R d × r by A) The symmetric group S r acts on both K × r δ permutation of the coordinates f is compatible with both actions (i.e., � � f is S r -equivariant): For all ρ ∈ S r f ◦ ρ = ρ ◦ � � f B) ( x i ∈ σ i ∈ K and σ i ∩ σ j = ∅ ) ⇒ all the x i are distinct f is an r -embedding ⇒ ¬ ( fx 1 = · · · = fx r ) Hence: � f : K × r → S r R d × r \{ ( x, . . . , x ) | x ∈ R d } δ

Two properties of � f � K × r R d × r → f : δ �→ ( x 1 , . . . , x r ) ( fx 1 , . . . , fx r ) and R d × r by A) The symmetric group S r acts on both K × r δ permutation of the coordinates f is compatible with both actions (i.e., � � f is S r -equivariant): For all ρ ∈ S r f ◦ ρ = ρ ◦ � � f B) ( x i ∈ σ i ∈ K and σ i ∩ σ j = ∅ ) ⇒ all the x i are distinct f is an r -embedding ⇒ ¬ ( fx 1 = · · · = fx r ) Hence: � f : K × r → S r R d × r \{ ( x, . . . , x ) | x ∈ R d } ≃ S ( r − 1) d − 1 δ

f : K m → R d such that for all σ 1 , . . . , σ r ∈ K with σ i ∩ σ j = ∅ fσ 1 ∩ · · · ∩ fσ r = ∅

f : K m → R d such that for all σ 1 , . . . , σ r ∈ K with σ i ∩ σ j = ∅ fσ 1 ∩ · · · ∩ fσ r = ∅ ⇓ ∃ � → S r S ( r − 1) d − 1 f : K × r δ

f : K m → R d such that for all σ 1 , . . . , σ r ∈ K with σ i ∩ σ j = ∅ fσ 1 ∩ · · · ∩ fσ r = ∅ ? ⇑ ⇓ ∃ � → S r S ( r − 1) d − 1 f : K × r δ

f : K m → R d such that for all σ 1 , . . . , σ r ∈ K with σ i ∩ σ j = ∅ fσ 1 ∩ · · · ∩ fσ r = ∅ ⇑ ⇓ ∃ � → S r S ( r − 1) d − 1 f : K × r δ yes! provided m = ( r − 1) k, d = rk and k ≥ 3

f is an almost r -embedding f : K m → R d such that for all σ 1 , . . . , σ r ∈ K with σ i ∩ σ j = ∅ fσ 1 ∩ · · · ∩ fσ r = ∅ ⇑ ⇓ ∃ � → S r S ( r − 1) d − 1 f : K × r δ yes! provided m = ( r − 1) k, d = rk and k ≥ 3

Theorem: ∃ f : K ( r − 1) k → R rk almost r -embedding ⇔ ∃ � f : K × r → S r S ( r − 1) rk − 1 δ provided k ≥ 3 .

Theorem: ∃ f : K ( r − 1) k → R rk almost r -embedding ⇔ ∃ � f : K × r → S r S ( r − 1) rk − 1 δ provided k ≥ 3 . algebraic problem geometric problem ⇔ (equivariant map) (map without intersection)

Theorem: ∃ f : K ( r − 1) k → R rk almost r -embedding ⇔ ∃ � f : K × r → S r S ( r − 1) rk − 1 δ provided k ≥ 3 . algebraic problem geometric problem ⇔ (equivariant map) (map without intersection) → S r S ( r − 1) rk − 1 is easy Proposition The existence of K × r δ algorithmically solvable .

Theorem: ∃ f : K ( r − 1) k → R rk almost r -embedding ⇔ ∃ � f : K × r → S r S ( r − 1) rk − 1 δ provided k ≥ 3 . algebraic problem geometric problem ⇔ (equivariant map) (map without intersection) → S r S ( r − 1) rk − 1 is easy Proposition The existence of K × r δ algorithmically solvable . Corollary. The existence of f : K ( r − 1) k → R rk almost r -embedding is algorithmically solvable , provided k ≥ 3 .

Our Main Tool: an r -fold analogue of the Whitney Trick

Our Main Tool: an r -fold analogue of the Whitney Trick Classical Whitney Trick with two balls σ p and τ q in R p + q : p, q ≥ 3 σ p τ q x y R p + q

Our Main Tool: an r -fold analogue of the Whitney Trick Classical Whitney Trick with two balls σ p and τ q in R p + q : p, q ≥ 3 σ p τ q x y − 1 +1 R p + q

Our Main Tool: an r -fold analogue of the Whitney Trick Classical Whitney Trick with two balls σ p and τ q in R p + q : p, q ≥ 3 σ p τ q x y − 1 +1 R p + q Whitney Disk D 2

Our Main Tool: an r -fold analogue of the Whitney Trick Classical Whitney Trick with two balls σ p and τ q in R p + q : p, q ≥ 3 � σ p σ p τ q x y − 1 +1 R p + q Whitney Disk D 2 push σ p along the Whitney Disk

What happens with more than two balls? τ 6 µ 6 σ 6 − 1 y x +1 R 9

What happens with more than two balls? σ ∩ τ σ 6 − 1 y σ ∩ µ x +1 R 9

What happens with more than two balls? σ ∩ τ σ 6 − 1 y � σ ∩ µ σ ∩ µ x +1 Whitney trick for two balls R 9