Are the results from persistent homology complete?

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Do the homology groups for each Betti number as produced by persistent homology completely capture all the topological characteristics of the space? If not, what topological features are not captured?

I.e. will two topologically distinct spaces always have distinct persistent homology (e.g. barcodes or persistence diagrams) and will two topologically indistinguishable spaces always have identical results from persistent homology?

algebraic-topology homology-cohomology data-analysis topological-data-analysis

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edited Aug 3 at 23:17

Alexander Gruber♦

20.1k24101169

asked Jul 22 at 20:56

Brandon Brown

1546

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  First of all, persistent homology is defined for metric spaces. It is not an invariant of topological spaces. Second, it is definitely very far from being a complete invariant. For example all the persistent homology for $mathbb R^n$ (with, say, the Euclidean metric) is essentially trivial, but these spaces are not homeomorphic to each other.
  – Cheerful Parsnip
  Jul 22 at 21:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @CheerfulParsnip My feeling is that persistent homology is designed for finite metric spaces (at least, that is were people want to apply it). Do you know of any examples of two finite metric spaces that have the same barcodes, but are not isometric?
  – Lorenzo
  Jul 22 at 21:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Lorenzo: you can definitely cook up finite examples too. I think the following works. (ctd)
  – Cheerful Parsnip
  Jul 22 at 22:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Let the first space have points $a,b,c,d$ where the distances between a,b and a,c are $1$. The distance between $a,c$ is $2$. Let the distance from $d$ to the three other points also be $2$. For the second space, let the distance between a,b and c,d be 1, and let the four distances a,c; a,d; b,c; b,d be $2$. Now the persistent homology in degree $0$ starts at rank 4, then jumps to rank $2$ at distance $1$, then jumps to rank $1$ at distance $2$. The components in each case are contractible, so the higher Betti numbers are all $0$.
  – Cheerful Parsnip
  Jul 22 at 22:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BrandonBrown: PH is a useful and powerful invariant, but there's no reason to expect it to classify metric spaces.
  – Cheerful Parsnip
  Jul 24 at 2:34
  

  
  
  
  

 | 
show 2 more comments

up vote
4
down vote

favorite

1

Do the homology groups for each Betti number as produced by persistent homology completely capture all the topological characteristics of the space? If not, what topological features are not captured?

I.e. will two topologically distinct spaces always have distinct persistent homology (e.g. barcodes or persistence diagrams) and will two topologically indistinguishable spaces always have identical results from persistent homology?

algebraic-topology homology-cohomology data-analysis topological-data-analysis

share|cite|improve this question

edited Aug 3 at 23:17

Alexander Gruber♦

20.1k24101169

asked Jul 22 at 20:56

Brandon Brown

1546

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  First of all, persistent homology is defined for metric spaces. It is not an invariant of topological spaces. Second, it is definitely very far from being a complete invariant. For example all the persistent homology for $mathbb R^n$ (with, say, the Euclidean metric) is essentially trivial, but these spaces are not homeomorphic to each other.
  – Cheerful Parsnip
  Jul 22 at 21:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @CheerfulParsnip My feeling is that persistent homology is designed for finite metric spaces (at least, that is were people want to apply it). Do you know of any examples of two finite metric spaces that have the same barcodes, but are not isometric?
  – Lorenzo
  Jul 22 at 21:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Lorenzo: you can definitely cook up finite examples too. I think the following works. (ctd)
  – Cheerful Parsnip
  Jul 22 at 22:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Let the first space have points $a,b,c,d$ where the distances between a,b and a,c are $1$. The distance between $a,c$ is $2$. Let the distance from $d$ to the three other points also be $2$. For the second space, let the distance between a,b and c,d be 1, and let the four distances a,c; a,d; b,c; b,d be $2$. Now the persistent homology in degree $0$ starts at rank 4, then jumps to rank $2$ at distance $1$, then jumps to rank $1$ at distance $2$. The components in each case are contractible, so the higher Betti numbers are all $0$.
  – Cheerful Parsnip
  Jul 22 at 22:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BrandonBrown: PH is a useful and powerful invariant, but there's no reason to expect it to classify metric spaces.
  – Cheerful Parsnip
  Jul 24 at 2:34
  

  
  
  
  

 | 
show 2 more comments

up vote
4
down vote

favorite

1

up vote
4
down vote

favorite

1

1

Do the homology groups for each Betti number as produced by persistent homology completely capture all the topological characteristics of the space? If not, what topological features are not captured?

I.e. will two topologically distinct spaces always have distinct persistent homology (e.g. barcodes or persistence diagrams) and will two topologically indistinguishable spaces always have identical results from persistent homology?

algebraic-topology homology-cohomology data-analysis topological-data-analysis

share|cite|improve this question

edited Aug 3 at 23:17

Alexander Gruber♦

20.1k24101169

asked Jul 22 at 20:56

Brandon Brown

1546

Do the homology groups for each Betti number as produced by persistent homology completely capture all the topological characteristics of the space? If not, what topological features are not captured?

I.e. will two topologically distinct spaces always have distinct persistent homology (e.g. barcodes or persistence diagrams) and will two topologically indistinguishable spaces always have identical results from persistent homology?

algebraic-topology homology-cohomology data-analysis topological-data-analysis

share|cite|improve this question

edited Aug 3 at 23:17

Alexander Gruber♦

20.1k24101169

asked Jul 22 at 20:56

Brandon Brown

1546

share|cite|improve this question

share|cite|improve this question

edited Aug 3 at 23:17

Alexander Gruber♦

20.1k24101169

edited Aug 3 at 23:17

Alexander Gruber♦

20.1k24101169

edited Aug 3 at 23:17

Alexander Gruber♦

20.1k24101169

20.1k24101169

asked Jul 22 at 20:56

Brandon Brown

1546

asked Jul 22 at 20:56

Brandon Brown

1546

asked Jul 22 at 20:56

Brandon Brown

1546

1546

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  First of all, persistent homology is defined for metric spaces. It is not an invariant of topological spaces. Second, it is definitely very far from being a complete invariant. For example all the persistent homology for $mathbb R^n$ (with, say, the Euclidean metric) is essentially trivial, but these spaces are not homeomorphic to each other.
  – Cheerful Parsnip
  Jul 22 at 21:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @CheerfulParsnip My feeling is that persistent homology is designed for finite metric spaces (at least, that is were people want to apply it). Do you know of any examples of two finite metric spaces that have the same barcodes, but are not isometric?
  – Lorenzo
  Jul 22 at 21:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Lorenzo: you can definitely cook up finite examples too. I think the following works. (ctd)
  – Cheerful Parsnip
  Jul 22 at 22:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Let the first space have points $a,b,c,d$ where the distances between a,b and a,c are $1$. The distance between $a,c$ is $2$. Let the distance from $d$ to the three other points also be $2$. For the second space, let the distance between a,b and c,d be 1, and let the four distances a,c; a,d; b,c; b,d be $2$. Now the persistent homology in degree $0$ starts at rank 4, then jumps to rank $2$ at distance $1$, then jumps to rank $1$ at distance $2$. The components in each case are contractible, so the higher Betti numbers are all $0$.
  – Cheerful Parsnip
  Jul 22 at 22:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BrandonBrown: PH is a useful and powerful invariant, but there's no reason to expect it to classify metric spaces.
  – Cheerful Parsnip
  Jul 24 at 2:34
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  First of all, persistent homology is defined for metric spaces. It is not an invariant of topological spaces. Second, it is definitely very far from being a complete invariant. For example all the persistent homology for $mathbb R^n$ (with, say, the Euclidean metric) is essentially trivial, but these spaces are not homeomorphic to each other.
  – Cheerful Parsnip
  Jul 22 at 21:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @CheerfulParsnip My feeling is that persistent homology is designed for finite metric spaces (at least, that is were people want to apply it). Do you know of any examples of two finite metric spaces that have the same barcodes, but are not isometric?
  – Lorenzo
  Jul 22 at 21:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Lorenzo: you can definitely cook up finite examples too. I think the following works. (ctd)
  – Cheerful Parsnip
  Jul 22 at 22:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Let the first space have points $a,b,c,d$ where the distances between a,b and a,c are $1$. The distance between $a,c$ is $2$. Let the distance from $d$ to the three other points also be $2$. For the second space, let the distance between a,b and c,d be 1, and let the four distances a,c; a,d; b,c; b,d be $2$. Now the persistent homology in degree $0$ starts at rank 4, then jumps to rank $2$ at distance $1$, then jumps to rank $1$ at distance $2$. The components in each case are contractible, so the higher Betti numbers are all $0$.
  – Cheerful Parsnip
  Jul 22 at 22:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @BrandonBrown: PH is a useful and powerful invariant, but there's no reason to expect it to classify metric spaces.
  – Cheerful Parsnip
  Jul 24 at 2:34
  

  
  
  
  

1

1

First of all, persistent homology is defined for metric spaces. It is not an invariant of topological spaces. Second, it is definitely very far from being a complete invariant. For example all the persistent homology for $mathbb R^n$ (with, say, the Euclidean metric) is essentially trivial, but these spaces are not homeomorphic to each other.
– Cheerful Parsnip
Jul 22 at 21:01

First of all, persistent homology is defined for metric spaces. It is not an invariant of topological spaces. Second, it is definitely very far from being a complete invariant. For example all the persistent homology for $mathbb R^n$ (with, say, the Euclidean metric) is essentially trivial, but these spaces are not homeomorphic to each other.
– Cheerful Parsnip
Jul 22 at 21:01

@CheerfulParsnip My feeling is that persistent homology is designed for finite metric spaces (at least, that is were people want to apply it). Do you know of any examples of two finite metric spaces that have the same barcodes, but are not isometric?
– Lorenzo
Jul 22 at 21:51

@CheerfulParsnip My feeling is that persistent homology is designed for finite metric spaces (at least, that is were people want to apply it). Do you know of any examples of two finite metric spaces that have the same barcodes, but are not isometric?
– Lorenzo
Jul 22 at 21:51

@Lorenzo: you can definitely cook up finite examples too. I think the following works. (ctd)
– Cheerful Parsnip
Jul 22 at 22:19

@Lorenzo: you can definitely cook up finite examples too. I think the following works. (ctd)
– Cheerful Parsnip
Jul 22 at 22:19

1

1

Let the first space have points $a,b,c,d$ where the distances between a,b and a,c are $1$. The distance between $a,c$ is $2$. Let the distance from $d$ to the three other points also be $2$. For the second space, let the distance between a,b and c,d be 1, and let the four distances a,c; a,d; b,c; b,d be $2$. Now the persistent homology in degree $0$ starts at rank 4, then jumps to rank $2$ at distance $1$, then jumps to rank $1$ at distance $2$. The components in each case are contractible, so the higher Betti numbers are all $0$.
– Cheerful Parsnip
Jul 22 at 22:19

Let the first space have points $a,b,c,d$ where the distances between a,b and a,c are $1$. The distance between $a,c$ is $2$. Let the distance from $d$ to the three other points also be $2$. For the second space, let the distance between a,b and c,d be 1, and let the four distances a,c; a,d; b,c; b,d be $2$. Now the persistent homology in degree $0$ starts at rank 4, then jumps to rank $2$ at distance $1$, then jumps to rank $1$ at distance $2$. The components in each case are contractible, so the higher Betti numbers are all $0$.
– Cheerful Parsnip
Jul 22 at 22:19

1

1

@BrandonBrown: PH is a useful and powerful invariant, but there's no reason to expect it to classify metric spaces.
– Cheerful Parsnip
Jul 24 at 2:34

@BrandonBrown: PH is a useful and powerful invariant, but there's no reason to expect it to classify metric spaces.
– Cheerful Parsnip
Jul 24 at 2:34

 | 
show 2 more comments

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