I was recently reading with interest the following paper:https://arxiv.org/pdf/2102.07835.pdf and, going to appendix to retrieve some general notions of TDA, I've been stuck for a while trying to understand how the persistence diagram $\mathcal{D}$ is retrieved for the specific example presented. I'm posting here the picture with filtrations of graph and the relative persistence diagram. If anyone could help me understanding this, it would be much appreciated.
"since all edges are inserted at $a^{(3)} = 3$ we obtain the following $0$-th dimensional persistence diagram":
$\mathcal{D}_0 = \{(1, \infty), (1,3), (2,3), (3,3), (3,3)\}$
To answer the question it is not needed to read the paper, but knowing TDA i.e. topics presented in appendix.
Many thanks,
James
The idea behind a persistence diagram is you look at the 'birth' and the 'death' of generators of the $n$th homology. There are $5$ vertices in the leftmost diagram, corresponding to the $5$ elements of $\mathcal{D}_0$. Two of them appear at $G^{(1)}$, so you have $(1, something), (1, something)$ to be in your final diagram. One more vertex is added at $G^{(2)}$, giving you $(2, something)$, and so on. The 'something' will be when those components go away. However, in the rightmost diagram, there is only one component. So, everything 'dies' at $G^{(3)}$, giving you the '3' for the 'death' values.