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 Introduction 1 Finite subset spaces Cell structures Homology Cell structures for finite subset spaces 2 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 exp k X = { nonempty subsets of X of size at most k } . Topology given by the quotient map ( x 1 , x 2 , . . . , x k ) �→ { x 1 , x 2 , . . . , x k } . = ⇒ 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 exp k X = { nonempty subsets of X of size at most k } . Topology given by the quotient map ( x 1 , x 2 , . . . , x k ) �→ { x 1 , x 2 , . . . , x k } . = ⇒ 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: Start with S 1 × S 1 1 Identify ( x , y ) and ( y , x ) 2 Result is a Möbius strip 3 Boundary is exp 1 S 1 (a circle) Glued edge is exp 2 ( S 1 , ∗ ) = { α ∈ exp 2 S 1 : ∗ ∈ 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: ( y , x ) Start with S 1 × S 1 1 Identify ( x , y ) and ( y , x ) 2 Result is a Möbius strip 3 ( x , y ) Boundary is exp 1 S 1 (a circle) Glued edge is exp 2 ( S 1 , ∗ ) = { α ∈ exp 2 S 1 : ∗ ∈ 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: { y , x } Start with S 1 × S 1 1 Identify ( x , y ) and ( y , x ) 2 Result is a Möbius strip 3 Boundary is exp 1 S 1 (a circle) Glued edge is exp 2 ( S 1 , ∗ ) = { α ∈ exp 2 S 1 : ∗ ∈ 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: Start with S 1 × S 1 1 Identify ( x , y ) and ( y , x ) 2 Result is a Möbius strip 3 ( x , x ) ∼ { x } Boundary is exp 1 S 1 (a circle) Glued edge is exp 2 ( S 1 , ∗ ) = { α ∈ exp 2 S 1 : ∗ ∈ 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: {∗ , y } Start with S 1 × S 1 1 Identify ( x , y ) and ( y , x ) 2 Result is a Möbius strip 3 Boundary is exp 1 S 1 (a circle) Glued edge is exp 2 ( S 1 , ∗ ) = { α ∈ exp 2 S 1 : ∗ ∈ 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: Start with S 1 × S 1 1 Identify ( x , y ) and ( y , x ) 2 Result is a Möbius strip 3 Boundary is exp 1 S 1 (a circle) Glued edge is exp 2 ( S 1 , ∗ ) = { α ∈ exp 2 S 1 : ∗ ∈ 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 exp 2 S 1 : Materials One 0-cell v Two 1-cells e 1 , e 2 One 2-cell f 1 Method Glue ends of e 1 , e 2 to v . Glue on boundary of f along e 1 e − 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 exp 2 S 1 : e 2 Materials One 0-cell v Two 1-cells e 1 , e 2 One 2-cell f 1 Method Glue ends of e 1 , e 2 to v . v Glue on boundary of f along e 1 e − 2 2 . e 1 f Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 5 / 15
Introduction Cell structures Cell structures: lego for topologists Instructions for exp 2 S 1 : e 2 Materials One 0-cell v Two 1-cells e 1 , e 2 One 2-cell f 1 Method Glue ends of e 1 , e 2 to v . v Glue on boundary of f along e 1 e − 2 2 . e 1 f Christopher Tuffley (Massey University) Cell structures for finite subset spaces ANZMC08 5 / 15
Introduction Cell structures Cell structures: lego for topologists Instructions for exp 2 S 1 : e 2 Materials One 0-cell v Two 1-cells e 1 , e 2 One 2-cell f 1 Method Glue ends of e 1 , e 2 to v . v Glue on boundary of f along e 1 e − 2 2 . e 1 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 i th 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 n th-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 H i ( 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 1 1 ∂ : boundary map from i -chains to ( i − 1 ) -chains 3 i -cycles: i -chains with boundary 0 − 1 2 boundaries: images of ∂ ∂ 2 = 0 so every boundary is a cycle = ⇒ can define H i ( 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 : − 1 i -chains: linear combinations of i -cells 1 1 ∂ : boundary map from i -chains − 2 to ( i − 1 ) -chains 1 3 i -cycles: i -chains with boundary 0 − 1 2 boundaries: images of ∂ 2 ∂ 2 = 0 so every boundary is a cycle = ⇒ can define H i ( 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 y x ∂ : boundary map from i -chains to ( i − 1 ) -chains x y i -cycles: i -chains with boundary 0 x y boundaries: images of ∂ ∂ 2 = 0 so every boundary is a cycle = ⇒ can define H i ( 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 x �������� �������� ∂ : boundary map from i -chains �������� �������� �������� �������� �������� �������� to ( i − 1 ) -chains x x �������� �������� �������� �������� i -cycles: i -chains with boundary 0 �������� �������� �������� �������� x boundaries: images of ∂ ∂ 2 = 0 so every boundary is a cycle = ⇒ can define H i ( 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
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