Persistent homology has to be free, right? I've been convinced that the homology groups you get when computing the persistent homology of a data cloud have to be free. But now I'm second guessing myself. Can we quickly say why this has to be? It's not just because it's the homology of a simplicial complex, because those can have torsion.
 A: Here are a couple of different possible takes on the question.
Persistent homology is essentially always computed with coefficients in a field. And homology with coefficients in a field is always a vector space. But this is an incomplete answer for several reasons.
Equation (5) of Computing Persistent Homology explains how when persistent homology is computed over a field $F$, then the persistence module decomposes as a sum
$$ \left(\bigoplus_{i=1}^n \Sigma^{\alpha_i} F[t]\right) \oplus\left( \bigoplus_{j=1}^m \Sigma^{\gamma_j} F[t]/(t^{n_j})\right). $$
From this you can read off the persistent homology barcode: you have $n$ semi-infinite intervals of the form $[\alpha_i,\infty)$, and $m$ finite intervals of the form $[\gamma_j,\gamma_j+n_j)$. Note that the semi-infinite intervals correspond to free summands $F[t]$ (shifted forwards in time by $\alpha_i$), whereas the finite intervals correspond to torsion summands $F[t]/(t^{n_j})$ (shifted forwards in time by $\gamma_j$). So in this sense, every finite-length bar in persistent homology exhibits torsion "in time."
But probably you are looking for torsion in space more than torsion in time. Vietoris-Rips and Čech simplicial complexes can certainly have torsion in them. As @MaximeRamzi and @JasonDeVito explain, take any non-orientable manifold you want, embed it in some high-dimensional Euclidean space, sample data points finely on this manifold, and then the Vietoris-Rips or Čech complex on this data (at appropriate scale parameters) will exhibit the torsion of this manifold.
Another way to see this is as follows. Take any simplicial complex $K$, say one with lots of torsion. The barycentric subdivision $sd(K)$ is a clique (or flag) simplicial complex. The one-skeleton of this barycentric subdivision $sd(K)$ is a graph, $sd(K)^{(1)}$. Let $X=V(sd(K)^{(1)})$ be the metric space whose set of points is the vertex set of this graph $sd(K)^{(1)}$, equipped with the shortest path metric in the graph. If you build the Vietoris-Rips complex on $X$ at scale parameter 1, then you see that you obtain $VR(X;1)=sd(K)$ as a Vietoris-Rips complex of a metric space. Since $sd(K)$ is homeomorphic to our original simplicial complex $K$, we see that we can realize the torsion of any simplicial complex as the torsion of some Vietoris-Rips complex.
There is active research on trying to do persistent homology with coefficients instead in the integers, which would allow this "torsion in space" to be measured even from the lens of persistent homology. See for example Generalized Persistence Diagrams.
