absolute persistent cohomology bar codes
$begingroup$
Can anyone explain the persistent absolute cohomology bar codes? how are the indices defined in absolute persistent cohomology?
For example, 
corresponds to the filtration $X_1 subset ... subset X_6$
Recall we have the persistent module:
$H^*(X_1) leftarrow ... leftarrow H^*(X_{5}) leftarrow H^*(X_6)$
there should thus be absolute cohomology barcodes:
${[1,infty)_0, [2,3)_0, [4,5)_1, [6,infty)_2}$ where the subscript refers to dimension of the generating cocycle.
How do we explain the barcodes for this example?
algebraic-topology homology-cohomology topological-data-analysis
asked Dec 16 '18 at 22:20
user352102user352102
38529
$endgroup$
add a comment |
$begingroup$
Can anyone explain the persistent absolute cohomology bar codes? how are the indices defined in absolute persistent cohomology?
For example,
corresponds to the filtration $X_1 subset ... subset X_6$
Recall we have the persistent module:
$H^*(X_1) leftarrow ... leftarrow H^*(X_{5}) leftarrow H^*(X_6)$
there should thus be absolute cohomology barcodes:
${[1,infty)_0, [2,3)_0, [4,5)_1, [6,infty)_2}$ where the subscript refers to dimension of the generating cocycle.
How do we explain the barcodes for this example?
algebraic-topology homology-cohomology topological-data-analysis
asked Dec 16 '18 at 22:20
user352102user352102
38529
$endgroup$
$begingroup$
Do you understand why those are the barcodes for the absolute homology? It's then a theorem that these barcodes agree. I guess I don't totally understand what you're asking.
$endgroup$
– user113102
Dec 17 '18 at 0:34
$begingroup$
You can write out the betti numbers of the 0,1,2-th homology of each complex in the filtration to check the barcodes for absolute homology. The issue is that I don't understand the result of the theorem. It looks like the indices are backwards or something.
$endgroup$
– user352102
Dec 17 '18 at 1:17
add a comment |
$begingroup$
Can anyone explain the persistent absolute cohomology bar codes? how are the indices defined in absolute persistent cohomology?
For example,
corresponds to the filtration $X_1 subset ... subset X_6$
Recall we have the persistent module:
$H^*(X_1) leftarrow ... leftarrow H^*(X_{5}) leftarrow H^*(X_6)$
there should thus be absolute cohomology barcodes:
${[1,infty)_0, [2,3)_0, [4,5)_1, [6,infty)_2}$ where the subscript refers to dimension of the generating cocycle.
How do we explain the barcodes for this example?
algebraic-topology homology-cohomology topological-data-analysis
asked Dec 16 '18 at 22:20
user352102user352102
38529
$endgroup$
Can anyone explain the persistent absolute cohomology bar codes? how are the indices defined in absolute persistent cohomology?
For example,
corresponds to the filtration $X_1 subset ... subset X_6$
Recall we have the persistent module:
$H^*(X_1) leftarrow ... leftarrow H^*(X_{5}) leftarrow H^*(X_6)$
there should thus be absolute cohomology barcodes:
${[1,infty)_0, [2,3)_0, [4,5)_1, [6,infty)_2}$ where the subscript refers to dimension of the generating cocycle.
How do we explain the barcodes for this example?
algebraic-topology homology-cohomology topological-data-analysis
algebraic-topology homology-cohomology topological-data-analysis
asked Dec 16 '18 at 22:20
user352102user352102
38529
asked Dec 16 '18 at 22:20
user352102user352102
38529
asked Dec 16 '18 at 22:20
user352102user352102
38529
asked Dec 16 '18 at 22:20
user352102user352102
38529
asked Dec 16 '18 at 22:20
user352102user352102
38529
38529
$begingroup$
Do you understand why those are the barcodes for the absolute homology? It's then a theorem that these barcodes agree. I guess I don't totally understand what you're asking.
$endgroup$
– user113102
Dec 17 '18 at 0:34
$begingroup$
You can write out the betti numbers of the 0,1,2-th homology of each complex in the filtration to check the barcodes for absolute homology. The issue is that I don't understand the result of the theorem. It looks like the indices are backwards or something.
$endgroup$
– user352102
Dec 17 '18 at 1:17
add a comment |
$begingroup$
Do you understand why those are the barcodes for the absolute homology? It's then a theorem that these barcodes agree. I guess I don't totally understand what you're asking.
$endgroup$
– user113102
Dec 17 '18 at 0:34
$begingroup$
You can write out the betti numbers of the 0,1,2-th homology of each complex in the filtration to check the barcodes for absolute homology. The issue is that I don't understand the result of the theorem. It looks like the indices are backwards or something.
$endgroup$
– user352102
Dec 17 '18 at 1:17
$begingroup$
Do you understand why those are the barcodes for the absolute homology? It's then a theorem that these barcodes agree. I guess I don't totally understand what you're asking.
$endgroup$
– user113102
Dec 17 '18 at 0:34
$begingroup$
Do you understand why those are the barcodes for the absolute homology? It's then a theorem that these barcodes agree. I guess I don't totally understand what you're asking.
$endgroup$
– user113102
Dec 17 '18 at 0:34
$begingroup$
You can write out the betti numbers of the 0,1,2-th homology of each complex in the filtration to check the barcodes for absolute homology. The issue is that I don't understand the result of the theorem. It looks like the indices are backwards or something.
$endgroup$
– user352102
Dec 17 '18 at 1:17
$begingroup$
You can write out the betti numbers of the 0,1,2-th homology of each complex in the filtration to check the barcodes for absolute homology. The issue is that I don't understand the result of the theorem. It looks like the indices are backwards or something.
$endgroup$
– user352102
Dec 17 '18 at 1:17
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The existence of a barcode of the form $[a,b)_n$ establishes that there is an $n$-cocycle with non-trivial cohomology class arising at time $a$ that persists up to time $b$.
For instance, the barcode $[2,3)_0$ corresponds to the $0$-cocycle associated to the point $2$. Note that $H^0(X_2)=langle 1,2rangle simeq Bbb{Z}^2$, where $1$ and $2$ denote the corresponding $0$-cocycles. The cochain $2$ is no longer a cocyle in $H^2(X_3)$, since its differential is not zero.
answered Dec 17 '18 at 21:32
F MF M
3,08652341
$endgroup$
$begingroup$
I do not understand. (let $sigma_i^*$ be the dual cochain associated with the basis chain $sigma_i$). How is cochain $sigma_2^*$ a cocycle on $X_3$? isn't the coboundary of cochain $sigma_2^*$ the cochain $sigma_3^*$? $delta$($sigma_2^*$)= $sigma_3^*$ $neq$ 0?
$endgroup$
– user352102
Dec 17 '18 at 22:39
$begingroup$
You are right - editing accordingly.
$endgroup$
– F M
Dec 17 '18 at 22:48
$begingroup$
It appears that the diagram was drawn with persistent homology rather than cohomology in mind (I was thinking about homology in my original answer as well). I now see why you're confused about the intervals been backwards - they are not if you think about it homologically, since the homology class of the point 2 ceases to be a non-trivial class at step 3, being homologous to 1.
$endgroup$
– F M
Dec 17 '18 at 22:51
add a comment |
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$begingroup$
The existence of a barcode of the form $[a,b)_n$ establishes that there is an $n$-cocycle with non-trivial cohomology class arising at time $a$ that persists up to time $b$.
For instance, the barcode $[2,3)_0$ corresponds to the $0$-cocycle associated to the point $2$. Note that $H^0(X_2)=langle 1,2rangle simeq Bbb{Z}^2$, where $1$ and $2$ denote the corresponding $0$-cocycles. The cochain $2$ is no longer a cocyle in $H^2(X_3)$, since its differential is not zero.
answered Dec 17 '18 at 21:32
F MF M
3,08652341
$endgroup$
$begingroup$
I do not understand. (let $sigma_i^*$ be the dual cochain associated with the basis chain $sigma_i$). How is cochain $sigma_2^*$ a cocycle on $X_3$? isn't the coboundary of cochain $sigma_2^*$ the cochain $sigma_3^*$? $delta$($sigma_2^*$)= $sigma_3^*$ $neq$ 0?
$endgroup$
– user352102
Dec 17 '18 at 22:39
$begingroup$
You are right - editing accordingly.
$endgroup$
– F M
Dec 17 '18 at 22:48
$begingroup$
It appears that the diagram was drawn with persistent homology rather than cohomology in mind (I was thinking about homology in my original answer as well). I now see why you're confused about the intervals been backwards - they are not if you think about it homologically, since the homology class of the point 2 ceases to be a non-trivial class at step 3, being homologous to 1.
$endgroup$
– F M
Dec 17 '18 at 22:51
add a comment |
$begingroup$
The existence of a barcode of the form $[a,b)_n$ establishes that there is an $n$-cocycle with non-trivial cohomology class arising at time $a$ that persists up to time $b$.
For instance, the barcode $[2,3)_0$ corresponds to the $0$-cocycle associated to the point $2$. Note that $H^0(X_2)=langle 1,2rangle simeq Bbb{Z}^2$, where $1$ and $2$ denote the corresponding $0$-cocycles. The cochain $2$ is no longer a cocyle in $H^2(X_3)$, since its differential is not zero.
answered Dec 17 '18 at 21:32
F MF M
3,08652341
$endgroup$
$begingroup$
I do not understand. (let $sigma_i^*$ be the dual cochain associated with the basis chain $sigma_i$). How is cochain $sigma_2^*$ a cocycle on $X_3$? isn't the coboundary of cochain $sigma_2^*$ the cochain $sigma_3^*$? $delta$($sigma_2^*$)= $sigma_3^*$ $neq$ 0?
$endgroup$
– user352102
Dec 17 '18 at 22:39
$begingroup$
You are right - editing accordingly.
$endgroup$
– F M
Dec 17 '18 at 22:48
$begingroup$
It appears that the diagram was drawn with persistent homology rather than cohomology in mind (I was thinking about homology in my original answer as well). I now see why you're confused about the intervals been backwards - they are not if you think about it homologically, since the homology class of the point 2 ceases to be a non-trivial class at step 3, being homologous to 1.
$endgroup$
– F M
Dec 17 '18 at 22:51
add a comment |
$begingroup$
The existence of a barcode of the form $[a,b)_n$ establishes that there is an $n$-cocycle with non-trivial cohomology class arising at time $a$ that persists up to time $b$.
For instance, the barcode $[2,3)_0$ corresponds to the $0$-cocycle associated to the point $2$. Note that $H^0(X_2)=langle 1,2rangle simeq Bbb{Z}^2$, where $1$ and $2$ denote the corresponding $0$-cocycles. The cochain $2$ is no longer a cocyle in $H^2(X_3)$, since its differential is not zero.
answered Dec 17 '18 at 21:32
F MF M
3,08652341
$endgroup$
The existence of a barcode of the form $[a,b)_n$ establishes that there is an $n$-cocycle with non-trivial cohomology class arising at time $a$ that persists up to time $b$.
For instance, the barcode $[2,3)_0$ corresponds to the $0$-cocycle associated to the point $2$. Note that $H^0(X_2)=langle 1,2rangle simeq Bbb{Z}^2$, where $1$ and $2$ denote the corresponding $0$-cocycles. The cochain $2$ is no longer a cocyle in $H^2(X_3)$, since its differential is not zero.
answered Dec 17 '18 at 21:32
F MF M
3,08652341
edited Dec 17 '18 at 22:49
answered Dec 17 '18 at 21:32
F MF M
3,08652341
answered Dec 17 '18 at 21:32
F MF M
3,08652341
answered Dec 17 '18 at 21:32
F MF M
3,08652341
3,08652341
$begingroup$
I do not understand. (let $sigma_i^*$ be the dual cochain associated with the basis chain $sigma_i$). How is cochain $sigma_2^*$ a cocycle on $X_3$? isn't the coboundary of cochain $sigma_2^*$ the cochain $sigma_3^*$? $delta$($sigma_2^*$)= $sigma_3^*$ $neq$ 0?
$endgroup$
– user352102
Dec 17 '18 at 22:39
$begingroup$
You are right - editing accordingly.
$endgroup$
– F M
Dec 17 '18 at 22:48
$begingroup$
It appears that the diagram was drawn with persistent homology rather than cohomology in mind (I was thinking about homology in my original answer as well). I now see why you're confused about the intervals been backwards - they are not if you think about it homologically, since the homology class of the point 2 ceases to be a non-trivial class at step 3, being homologous to 1.
$endgroup$
– F M
Dec 17 '18 at 22:51
add a comment |
$begingroup$
I do not understand. (let $sigma_i^*$ be the dual cochain associated with the basis chain $sigma_i$). How is cochain $sigma_2^*$ a cocycle on $X_3$? isn't the coboundary of cochain $sigma_2^*$ the cochain $sigma_3^*$? $delta$($sigma_2^*$)= $sigma_3^*$ $neq$ 0?
$endgroup$
– user352102
Dec 17 '18 at 22:39
$begingroup$
You are right - editing accordingly.
$endgroup$
– F M
Dec 17 '18 at 22:48
$begingroup$
It appears that the diagram was drawn with persistent homology rather than cohomology in mind (I was thinking about homology in my original answer as well). I now see why you're confused about the intervals been backwards - they are not if you think about it homologically, since the homology class of the point 2 ceases to be a non-trivial class at step 3, being homologous to 1.
$endgroup$
– F M
Dec 17 '18 at 22:51
$begingroup$
I do not understand. (let $sigma_i^*$ be the dual cochain associated with the basis chain $sigma_i$). How is cochain $sigma_2^*$ a cocycle on $X_3$? isn't the coboundary of cochain $sigma_2^*$ the cochain $sigma_3^*$? $delta$($sigma_2^*$)= $sigma_3^*$ $neq$ 0?
$endgroup$
– user352102
Dec 17 '18 at 22:39
$begingroup$
I do not understand. (let $sigma_i^*$ be the dual cochain associated with the basis chain $sigma_i$). How is cochain $sigma_2^*$ a cocycle on $X_3$? isn't the coboundary of cochain $sigma_2^*$ the cochain $sigma_3^*$? $delta$($sigma_2^*$)= $sigma_3^*$ $neq$ 0?
$endgroup$
– user352102
Dec 17 '18 at 22:39
$begingroup$
You are right - editing accordingly.
$endgroup$
– F M
Dec 17 '18 at 22:48
$begingroup$
You are right - editing accordingly.
$endgroup$
– F M
Dec 17 '18 at 22:48
$begingroup$
It appears that the diagram was drawn with persistent homology rather than cohomology in mind (I was thinking about homology in my original answer as well). I now see why you're confused about the intervals been backwards - they are not if you think about it homologically, since the homology class of the point 2 ceases to be a non-trivial class at step 3, being homologous to 1.
$endgroup$
– F M
Dec 17 '18 at 22:51
$begingroup$
It appears that the diagram was drawn with persistent homology rather than cohomology in mind (I was thinking about homology in my original answer as well). I now see why you're confused about the intervals been backwards - they are not if you think about it homologically, since the homology class of the point 2 ceases to be a non-trivial class at step 3, being homologous to 1.
$endgroup$
– F M
Dec 17 '18 at 22:51
add a comment |
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$begingroup$
Do you understand why those are the barcodes for the absolute homology? It's then a theorem that these barcodes agree. I guess I don't totally understand what you're asking.
$endgroup$
– user113102
Dec 17 '18 at 0:34
$begingroup$
You can write out the betti numbers of the 0,1,2-th homology of each complex in the filtration to check the barcodes for absolute homology. The issue is that I don't understand the result of the theorem. It looks like the indices are backwards or something.
$endgroup$
– user352102
Dec 17 '18 at 1:17