Cell structures for finite subset spaces Christopher Tuffley - - PowerPoint PPT Presentation

Cell structures for finite subset spaces

Christopher Tuffley

Institute of Fundamental Sciences Massey University, Palmerston North

7th Australia — New Zealand Mathematics Convention December 2008

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 1 / 15

Outline

1

Introduction Finite subset spaces Cell structures Homology

2

Cell structures for finite subset spaces Goals The construction Results

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 2 / 15

Introduction Finite subset spaces

Finite subset spaces

—spaces whose points are finite subsets of a fixed space X. The kth finite subset space of X is expkX = {nonempty subsets of X of size at most k}. Topology given by the quotient map (x1, x2, . . . , xk) → {x1, x2, . . . , xk}. = ⇒ α, β close if each point close to a member of the other subset

✈ ✈ ✈ ✈ ✈

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 3 / 15

Introduction Finite subset spaces

Finite subset spaces

—spaces whose points are finite subsets of a fixed space X. The kth finite subset space of X is expkX = {nonempty subsets of X of size at most k}. Topology given by the quotient map (x1, x2, . . . , xk) → {x1, x2, . . . , xk}. = ⇒ α, β close if each point close to a member of the other subset

✈ ✈ ✈ ✈ ✈

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 3 / 15

Introduction Finite subset spaces

Example: the circle

The second finite subset space:

1

Start with S1 × S1

2

Identify (x, y) and (y, x)

3

Result is a Möbius strip Boundary is exp1S1 (a circle) Glued edge is exp2(S1, ∗) = {α ∈ exp2S1 : ∗ ∈ α} (another circle)

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 4 / 15

Introduction Finite subset spaces

Example: the circle

The second finite subset space:

1

Start with S1 × S1

2

Identify (x, y) and (y, x)

3

Result is a Möbius strip (x, y) (y, x) Boundary is exp1S1 (a circle) Glued edge is exp2(S1, ∗) = {α ∈ exp2S1 : ∗ ∈ α} (another circle)

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 4 / 15

Introduction Finite subset spaces

Example: the circle

The second finite subset space:

1

Start with S1 × S1

2

Identify (x, y) and (y, x)

3

Result is a Möbius strip {y, x} Boundary is exp1S1 (a circle) Glued edge is exp2(S1, ∗) = {α ∈ exp2S1 : ∗ ∈ α} (another circle)

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 4 / 15

Introduction Finite subset spaces

Example: the circle

The second finite subset space:

1

Start with S1 × S1

2

Identify (x, y) and (y, x)

3

Result is a Möbius strip (x, x) ∼ {x} Boundary is exp1S1 (a circle) Glued edge is exp2(S1, ∗) = {α ∈ exp2S1 : ∗ ∈ α} (another circle)

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 4 / 15

Introduction Finite subset spaces

Example: the circle

The second finite subset space:

1

Start with S1 × S1

2

Identify (x, y) and (y, x)

3

Result is a Möbius strip {∗, y} Boundary is exp1S1 (a circle) Glued edge is exp2(S1, ∗) = {α ∈ exp2S1 : ∗ ∈ α} (another circle)

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 4 / 15

Introduction Finite subset spaces

Example: the circle

The second finite subset space:

1

Start with S1 × S1

2

Identify (x, y) and (y, x)

3

Result is a Möbius strip Boundary is exp1S1 (a circle) Glued edge is exp2(S1, ∗) = {α ∈ exp2S1 : ∗ ∈ α} (another circle)

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 4 / 15

Introduction Cell structures

Cell structures: lego for topologists

Instructions for exp2S1: Materials One 0-cell v Two 1-cells e1, e2 One 2-cell f1 Method Glue ends of e1, e2 to v. Glue on boundary of f along e1e−2

2 .

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 5 / 15

Introduction Cell structures

Cell structures: lego for topologists

Instructions for exp2S1: Materials One 0-cell v Two 1-cells e1, e2 One 2-cell f1 Method Glue ends of e1, e2 to v. Glue on boundary of f along e1e−2

2 .

v e1 e2 f

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 5 / 15

Introduction Cell structures

Cell structures: lego for topologists

Instructions for exp2S1: Materials One 0-cell v Two 1-cells e1, e2 One 2-cell f1 Method Glue ends of e1, e2 to v. Glue on boundary of f along e1e−2

2 .

v e1 e2 f

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 5 / 15

Introduction Cell structures

Cell structures: lego for topologists

Instructions for exp2S1: Materials One 0-cell v Two 1-cells e1, e2 One 2-cell f1 Method Glue ends of e1, e2 to v. Glue on boundary of f along e1e−2

2 .

v e1 e2 f

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 5 / 15

Introduction Cell structures

Cell structures: more formally speaking

A cell structure builds X inductively from simple pieces: Start with some vertices (the 0-skeleton) At ith step, glue on i-dimensional balls (i-cells) using attaching maps defined on their boundaries. Result is the i-skeleton. X is an n-complex if process stops at nth-step.

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 6 / 15

Introduction Homology

Cellular homology: counting the holes

Given a cell structure for a space X: i-chains: linear combinations of i-cells ∂: boundary map from i-chains to (i − 1)-chains i-cycles: i-chains with boundary 0 boundaries: images of ∂ ∂2 = 0 so every boundary is a cycle = ⇒ can define Hi(X) = i-chains mod i-boundaries = ker ∂i/ image ∂i+1. (Depends on choice of co-efficient group, but not cell structure)

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 7 / 15

Introduction Homology

Cellular homology: counting the holes

Given a cell structure for a space X: i-chains: linear combinations of i-cells ∂: boundary map from i-chains to (i − 1)-chains i-cycles: i-chains with boundary 0 boundaries: images of ∂ − 1 2 1 1 3 ∂2 = 0 so every boundary is a cycle = ⇒ can define Hi(X) = i-chains mod i-boundaries = ker ∂i/ image ∂i+1. (Depends on choice of co-efficient group, but not cell structure)

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 7 / 15

Introduction Homology

Cellular homology: counting the holes

Given a cell structure for a space X: i-chains: linear combinations of i-cells ∂: boundary map from i-chains to (i − 1)-chains i-cycles: i-chains with boundary 0 boundaries: images of ∂ − 1 2 − 2 1 − 1 2 1 1 3 ∂2 = 0 so every boundary is a cycle = ⇒ can define Hi(X) = i-chains mod i-boundaries = ker ∂i/ image ∂i+1. (Depends on choice of co-efficient group, but not cell structure)

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 7 / 15

Introduction Homology

Cellular homology: counting the holes

Given a cell structure for a space X: i-chains: linear combinations of i-cells ∂: boundary map from i-chains to (i − 1)-chains i-cycles: i-chains with boundary 0 boundaries: images of ∂ x y x y y x ∂2 = 0 so every boundary is a cycle = ⇒ can define Hi(X) = i-chains mod i-boundaries = ker ∂i/ image ∂i+1. (Depends on choice of co-efficient group, but not cell structure)

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 7 / 15

Introduction Homology

Cellular homology: counting the holes

Given a cell structure for a space X: i-chains: linear combinations of i-cells ∂: boundary map from i-chains to (i − 1)-chains i-cycles: i-chains with boundary 0 boundaries: images of ∂

• x

x x x ∂2 = 0 so every boundary is a cycle = ⇒ can define Hi(X) = i-chains mod i-boundaries = ker ∂i/ image ∂i+1. (Depends on choice of co-efficient group, but not cell structure)

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 7 / 15

Introduction Homology

Cellular homology: counting the holes

Given a cell structure for a space X: i-chains: linear combinations of i-cells ∂: boundary map from i-chains to (i − 1)-chains i-cycles: i-chains with boundary 0 boundaries: images of ∂

• x

x x x ∂2 = 0 so every boundary is a cycle = ⇒ can define Hi(X) = i-chains mod i-boundaries = ker ∂i/ image ∂i+1. (Depends on choice of co-efficient group, but not cell structure)

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 7 / 15

Introduction Homology

Example: H1(exp2S1)

cycles: generated by e1, e2 boundaries: generated by ∂f = e1 − 2e2 = ⇒ e1 = 2e2 in homology = ⇒ H1(exp2S1) ∼ = Z, generated by e2

• e1

e2 f exp2(S1, ∗) = e2 represents 1 in H1; exp1S1 = e1 represents 2

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 8 / 15

Introduction Homology

Example: H1(exp2S1)

cycles: generated by e1, e2 boundaries: generated by ∂f = e1 − 2e2 = ⇒ e1 = 2e2 in homology = ⇒ H1(exp2S1) ∼ = Z, generated by e2

• e1

e2 f exp2(S1, ∗) = e2 represents 1 in H1; exp1S1 = e1 represents 2

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 8 / 15

Introduction Homology

Example: H1(exp2S1)

cycles: generated by e1, e2 boundaries: generated by ∂f = e1 − 2e2 = ⇒ e1 = 2e2 in homology = ⇒ H1(exp2S1) ∼ = Z, generated by e2

• e1

e2 f exp2(S1, ∗) = e2 represents 1 in H1; exp1S1 = e1 represents 2

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 8 / 15

Cell structures for finite subset spaces Goals

Goals

Find recipe for cell structure for expkX from cell structure for X Use cell structure to study H∗(expkX) Relate homology to that of other spaces related to X

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 9 / 15

Cell structures for finite subset spaces Goals

Re-inventing the wheel, and making it square?

Theorem (CT, 2003) For X a 2-complex, expkX has a “lexicographic” cell structure.

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 10 / 15

Cell structures for finite subset spaces Goals

Re-inventing the wheel, and making it square?

Theorem (CT, 2003) For X a 2-complex, expkX has a “lexicographic” cell structure. Recipe also provided by the machinery of simplicial sets (Jacob Mostovoy, private communication) Applicable to complexes of arbitrary dimension Appear difficult to work with:

S2 requires only two cells, but resulting structure for exp3S2 has 77 cells exp4S2 1039

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 10 / 15

Cell structures for finite subset spaces Goals

Re-inventing the wheel, and making it square?

Theorem (CT, 2003) For X a 2-complex, expkX has a “lexicographic” cell structure. Recipe also provided by the machinery of simplicial sets (Jacob Mostovoy, private communication) Applicable to complexes of arbitrary dimension Appear difficult to work with:

S2 requires only two cells, but resulting structure for exp3S2 has 77 cells exp4S2 1039

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 10 / 15

Cell structures for finite subset spaces Goals

Re-inventing the wheel, and making it square?

Theorem (CT, 2003) For X a 2-complex, expkX has a “lexicographic” cell structure. Recipe also provided by the machinery of simplicial sets (Jacob Mostovoy, private communication) Applicable to complexes of arbitrary dimension Appear difficult to work with:

S2 requires only two cells, but resulting structure for exp3S2 has 77 cells exp4S2 1039

Lexicographic approach: exp3S2 has 11 cells exp4S2 23

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 10 / 15

Cell structures for finite subset spaces The construction

Putting things in order

Idea Systematically choose a preferred order for the points in each subset of X Order all the cells Within each edge, order points linearly Within each 2-cell, order the points lexicographically

→ get cells labelled by ordered partitions

1 1 1 1 2 2 2 2 3 3 3 3 (1, 1, 1) (2, 1) (1, 2) (3) Take products of such cells to get a cell structure for expkX.

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 11 / 15

Cell structures for finite subset spaces The construction

Hitting those boundaries

Four kinds of contributions to the boundary of a basic cell:

• Christopher Tuffley (Massey University)

Cell structures for finite subset spaces ANZMC08 12 / 15

Cell structures for finite subset spaces The construction

Hitting those boundaries

Four kinds of contributions to the boundary of a basic cell:

• Christopher Tuffley (Massey University)

Cell structures for finite subset spaces ANZMC08 12 / 15

Cell structures for finite subset spaces The construction

Hitting those boundaries

Four kinds of contributions to the boundary of a basic cell:

• Christopher Tuffley (Massey University)

Cell structures for finite subset spaces ANZMC08 12 / 15

Cell structures for finite subset spaces The construction

Hitting those boundaries

Four kinds of contributions to the boundary of a basic cell:

• Christopher Tuffley (Massey University)

Cell structures for finite subset spaces ANZMC08 12 / 15

Cell structures for finite subset spaces The construction

Hitting those boundaries

Four kinds of contributions to the boundary of a basic cell:

• Technical issues arise in this case in higher dimensions

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 12 / 15

Cell structures for finite subset spaces Results

A direct sum decomposition theorem

Theorem (CT, 2008) If X is a connected 2-complex, then Hi(expkX) ∼ = Hi(expk+1(X, ∗)) ⊕ Hi(expk(X, ∗)) for i ≥ 1. Algebraically, expkX looks like

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 13 / 15

Cell structures for finite subset spaces Results

A direct sum decomposition theorem

Theorem (CT, 2008) If X is a connected 2-complex, then Hi(expkX) ∼ = Hi(expk+1(X, ∗)) ⊕ Hi(expk(X, ∗)) for i ≥ 1. Algebraically, expkX looks like expk+1(X, ∗) expk(X, ∗)

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 13 / 15

Cell structures for finite subset spaces Results

The two-sphere

Theorem (CT, 2003) The rational homology of expkS2 is Hi(expkS2) ∼ =

• Q

i = 0, 2k − 2, 2k;

• therwise.

Integer homology appears to be very complicated Mod-2 homology appears to be tractable, using results on the braid groups due to Fuchs (work in progress) In principle, the mod-p homology could be found using a similar method, and results of Vainshtein

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 14 / 15

Cell structures for finite subset spaces Results

The two-sphere

Theorem (CT, 2003) The rational homology of expkS2 is Hi(expkS2) ∼ =

• Q

i = 0, 2k − 2, 2k;

• therwise.

Integer homology appears to be very complicated Mod-2 homology appears to be tractable, using results on the braid groups due to Fuchs (work in progress) In principle, the mod-p homology could be found using a similar method, and results of Vainshtein

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 14 / 15

Future directions

Future directions

Extend construction to cell complexes of arbitrary dimension Extend direct-sum decomposition theorem to arbitrary cell complexes (perhaps via simplicial set construction?)

Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 15 / 15