absolute persistent cohomology bar codes

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

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?

• 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. – user113102 Dec 17 '18 at 0:34
• 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. – 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.
• 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? – user352102 Dec 17 '18 at 22:39