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The Setting

I am currently working on a project in Topological Data Analysis (TDA), where I aim to construct a density-robust spherical coordinate associated with a dataset $X$, sampled from a Riemannian manifold $\mathcal{M}$ with some probability density function (not necessarily uniform), using persistent cohomology. Specifically, the goal is to define a map $X \to S^2$ that depends only on the shape of the underlying manifold, not on the probability density function.

The key idea is that Brown representability implies that a $\mathbb{Z}$-cocycle of a topological space (in this case, a Vietoris-Rips complex $X_\varepsilon$ associated with the data) is equivalent to a map from this space to a $K(\mathbb{Z}, 2)$. This Eilenberg-MacLane space can be modeled by $\mathbb{C}\mathbb{P}^\infty$, meaning we obtain a map $X_\varepsilon \to \mathbb{C}\mathbb{P}^\infty$. This map can then be lifted to $X_\varepsilon \to S^2$ along the inclusion $S^2 \to \mathbb{C}\mathbb{P}^\infty$.

In practice, one computes the second-degree persistent cohomology associated with the Vietoris-Rips complex $X_\varepsilon$ for some well-chosen $\varepsilon$. Then, a cocycle $\alpha$ corresponding to a high-persistence feature is selected, and a map is constructed from this cocycle. This is achieved by mapping $0$- and $1$-simplices to a base point and, for triangles $T$, collapsing their boundaries and composing the result with the degree $n = \alpha(T)$ map. This process is described in detail in this paper.

The Problem

For my specific approach, it would be helpful to find a method to construct a map $X_\varepsilon \to S^2$ from a real cocycle $\alpha \in H^2(X_\varepsilon; \mathbb{R})$. This idea is not entirely new: in the one-dimensional case of computing circular coordinates, it is possible to use a real cocycle to define a map. Specifically, for a $1$-cocycle $\alpha \in H^1(X_\varepsilon; \mathbb{R})$, a map $X_\varepsilon \to S^1$ can be defined by winding a $1$-simplex $e$ around $S^1$ $d$-times, where $d = \alpha(e)$. This approach works because $\mathbb{R}/\mathbb{Z} \cong S^1$.

However, the two-dimensional case is more challenging. For $\mathbb{R}^2/\mathbb{Z}^2 \cong T^2$ (the torus), maps defined analogously often factor through $T^2$, making them unsuitable as spherical coordinates since they fail to map directly to $S^2$. The ideas for the one-dimensional case are drawn from this paper and this paper.

The Ideas

If there were a way to wrap a triangle or disk around $S^2$ $d \in \mathbb{R}$ times, one could apply a similar trick to the one-dimensional case. However, I have not yet found a viable method. Alternatively, there might exist a completely different approach to obtaining such a map that I am unaware of. Another possibility might involve constructing a map to $\mathbb{C}\mathbb{P}^\infty$ from the real cocycle and then restricting this map to $S^2$.

The Questions

  1. Does anyone know of a method or have a good idea for constructing a map $\Delta_2 \to S^2$ or $S^2 \to S^2$ of a real degree?
  2. Does anyone know of a method or idea for obtaining such a map from a real cocycle associated with a manifold in general?
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  • $\begingroup$ Cool question. Are you committed to using $S^2$ to get your coordinates? You could choose an $\mathbb{R}$-linear basis for $H^2(X_{\epsilon}; \mathbb{R})$ consisting of elements in the image of the map $H^2(X_{\epsilon}; \mathbb{Q}) \rightarrow H^2(X_{\epsilon}; \mathbb{R})$. Since those elements have rational coefficients and since $X$ is a finite point cloud, you can clear denominators to get a class in $H^2(X_{\epsilon}; \mathbb{Z}) \cong [X_{\epsilon}, CP^{\infty}]$. Since $X_{\epsilon}$ is a finite simp. cplx, the map $X_{\epsilon}\rightarrow CP^{\infty}$ lands in $CP^n$ for some $n$. $\endgroup$
    – user509184
    Commented yesterday
  • $\begingroup$ My point is that you could use the projective coordinates you get from that map $X_{\epsilon} \rightarrow CP^n$. This avoids having to move everything into $S^2$, and (because of taking the rational basis + clearing denominators) it avoids the strangeness you asked about, of having maps on $S^2$ of real degree. But maybe you already have a good reason to not take the path I described, and to do what you're doing instead. $\endgroup$
    – user509184
    Commented yesterday

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