Can anyone explain the persistent absolute cohomology bar codes? how are the indices defined in absolute persistent cohomology?

For example, enter image description here

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?

  • $\begingroup$ 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

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 $H^0(X_2)=\langle 1,2\rangle \simeq \Bbb{Z}^2$, 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.

  • $\begingroup$ 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

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