2024-10-21

I may have found a non $q$-tame example in a bounded space.

Let our data set be $X=[0,2]$. Construct the cover $\mathcal{U}=\{U_n{\}}_{n=0}^{\infty }$ by $U_n=[a_n,b_n]$ where

$$a_n=\{\begin{array}{ll}0& \text{if }n=0\\ a_{n-1}+2^{-(n-1)}-2^{-(n+1)}& \text{if }n>0\end{array}$$

and $b_n=a_n+2^{-n}$. Notice that $a_n\to 2$ and $b_n\to 2$. Moreover, under the Lebesgue measure $\ell$,

$$\begin{array}{rl}\ell (U_n)& =b_n-a_n=\frac{1}{2^n}\\ \ell (U_n\cap U_{n+1})& =a_n-b_{n-1}=\frac{1}{2^{n+1}}\\ \ell (U_n\cup U_{n+1})& =\ell (U_n)+\ell (U_{n+1})-\ell (U_n\cap U_{n+1})=\frac{1}{2^n}.\end{array}$$

Thus, in the Steinhaus filtration the edge $\{n,n+1\}$ is born at time

$$d_{St}(U_n,U_{n+1})=1-\frac{\ell (U_n\cap U_{n+1})}{\ell (U_n\cup U_{n+1})}=1-\frac{2^n}{2^{n+1}}=\frac{1}{2}.$$

So all the edges are born at the same time, which is not what we want.

Either way, the first-homology group of this complex is the trivial group as there are no 1-cycles.

However, we might be able to take this idea to the non $q$-tame example from 2024-10-10. The idea is to make each diamond have a height of $2^{-n}$.

We remove the bottom half of the diamonds and compress the remaining top half into a $24\times 2$ box. The boxes should have the following sizes:

$$\begin{array}{rl}{R}_n& =11\times \frac{1}{2^n}\\ G_n& =6\times \frac{11}{14\cdot 2^n}\\ B_n& =11\times \frac{12}{14\cdot 2^n}\end{array}$$

The overlap $R_n\cap G_n$, $R_n\cap G_{n+1}$, $B_n\cap G_n$ and $B_n\cap G_{n+1}$ should be as follows:

$$\begin{array}{rl}{R}_n\cap G_n& =2.5\times \frac{4}{14\cdot 2^n}\\ R_n\cap G_{n+1}& =2.5\times \frac{4}{14\cdot 2^{n+1}}\\ B_n\cap G_n& =2.5\times \frac{3}{14\cdot 2^n}\\ B_n\cap G_{n+1}& =2.5\times \frac{3}{14\cdot 2^{n+1}}.\end{array}$$

Assuming this is even possible, we get the Steinhaus distances of

$$1-\frac{\ell (B_n\cap G_n)}{\ell (B_n\cup G_n)}=1-\frac{\frac{7.5}{14\cdot 2^n}}{\frac{132}{14\cdot 2^n}+\frac{66}{14\cdot 2^n}-\frac{7.5}{14\cdot 2^n}}=\frac{361}{381}\approx 0.948$$ $$1-\frac{\ell (B_n\cap G_{n+1})}{\ell (B_n\cup G_{n+1})}=1-\frac{\frac{7.5}{14\cdot 2^{n+1}}}{\frac{132}{14\cdot 2^n}+\frac{66}{14\cdot 2^{n+1}}-\frac{7.5}{14\cdot 2^{n+1}}}=1-\frac{7.5}{264+66-7.5}=\frac{630}{645}\approx 0.977$$ $$1-\frac{\ell (R_n\cap G_n)}{\ell (R_n\cup G_n)}=1-\frac{\frac{10}{14\cdot 2^n}}{\frac{154}{14\cdot 2^n}+\frac{66}{14\cdot 2^n}-\frac{10}{14\cdot 2^n}}=\frac{200}{210}\approx 0.952$$

and

$$1-\frac{\ell (R_n\cap G_{n+1})}{\ell (R_n\cup G_n)}=1-\frac{\frac{10}{14\cdot 2^{n+1}}}{\frac{154}{14\cdot 2^n}+\frac{66}{14\cdot 2^{n+1}}-\frac{10}{14\cdot 2^n}}=\frac{354}{364}\approx \mathrm{0.973.}$$

Therefore, in the first-homology persistence module the map from $354/364$ to $630/645$ will have infinite rank. Unsure if this is true.