Realisation of maps between spheres by simplicial maps

Question

Let $K^n_0 := \partial \Delta_{n+1}$ the simplicial set obtained by removing the $(n+1)$-simplex from the standard simplex. This gives a simplicial decomposition of the sphere $S^n$. More generally, let $K^n_m$ given by $m$-fold barycentric subdivision of $K^n_0$.

Given a map of simplicial sets $f: K^{n+k}_m \rightarrow K^n_0$, geometric realization induces a continuous map $F: S^{n+k} \rightarrow S^n$, and it is well-known that every such continuous map is homotopic to a simplicial map $f: K^{n+k}_m \rightarrow K^n_0$ for $m$ large enough.

For $k\geq 1$, the homotopy groups of sphere are finite, hence there exists a smallest $m$ such that all homotopy classes of maps $F: S^{n+k} \rightarrow S^n$ are represented by a simplicial map $f: K^{n+k}_m \rightarrow K^n_0$.

Question: Are there explicit bounds on this smallest $m$ in terms of $n$ and $k$?

Answer 1

First, $\pi_{4k-1}(S^{2k})$ has an infinite cyclic direct summand for every $k \geq 1$. As a simple example, you can think of $\pi_3(S^2) \cong \mathbb{Z}$, coming from the Hopf fibration.

We recently wrote a paper where we solved a problem, which is closely related to your question. Namely, we give an algorithm which for a given finite simply connected simplicial complex $Y$, and a given $d \geq 2$, under some technical conditions on $Y$ the algorithm outputs for every generator $\alpha \in \pi_d(Y)$, a simplicial map $f_\alpha : \Sigma^d_\alpha \rightarrow Y $, where $\Sigma^d_\alpha$ is a simplicial complex whose geometric realisation is homeomorphic to $S^d$. In particular this works for spheres. You can find the paper on archive (https://arxiv.org/pdf/1706.00380.pdf), and it will soon appear in the SODA 2018 proceedings. The complexity of the algorithm is singly exponential on the number of simplices of $Y$, and we show that this is optimal. The way we prove it is by constructing an example $X$ where the necessary size of the spheres $\Sigma^n_\alpha$ is exponential on the size of $X$. That shows that any general algorithm must necessarily have at least singly exponential complexity.

However, we do not work with barycentric subdivisions, but with a rather specific and non-canonical (if I might put it so) triangulations of the $d$-sphere. The construction of those was invented by Clemens Berger in his PhD dissertation (https://tel.archives-ouvertes.fr/tel-00339314/document - in French only). We believe that one could be able to fit those "non-canonical" models into an iterated barycentric subdivision of the boundary of a standard simplex. If this is true, it would mean that you need to subdivide $\partial \Delta[m+1]$ enough times to fit the exponentially large spheres produced by our algorithm, and you would be able to represent any homotopy element you wish.

On the other hand, even though our algorithm is exponential in the general case, it might happen that it runs in polynomial time for some very specific spaces. For example. It is not impossible that this turn out to be true for spheres, but I cannot say much more about it.

There is also another point of view. Spheres are compact Riemannian manifolds and you can rephrase your question in the language of Lipschitz constants. In this language, your question roughly translates to "what Lipschitz constants can one expect from representatives of homotopy groups of spheres". There is a connection between the Lipschitz constant of a Lipschitz function and the fineness of a triangulation required to homotope it to a simplicial map. This is somewhat expected because both notions measure some sort of geometric complexity of the function in mind. In particular, there is the following result by Gromov (http://www.ihes.fr/~gromov/PDF/4[19].pdf):

Theorem (Gromov): Let $X$ and $Y$ be compact simply connected Riemannian manifolds. Then $$\# \{ [f] \in [X,Y] \, : \, Lip(f) \leq L \} = O(L^\alpha)$$, where $\alpha$ depends only on the rational homotopy type of $X$ and $Y$.

You can check also this (http://www.pnas.org/content/110/48/19246.full) paper from Weinberger and Ferry. There is a long study of similar questions from that point of view.

I hope my comments are useful. In conclusion I would like to say that you question is very deep and very hard.