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To compute the simplicial homotopy group of a space $X$, we find a Kan fibrant replacement $X \to Y$ and calculate for that for $Y$, which can be implemented in a computer program.

Computing homotopy groups for spheres are fundamentally hard, and I believe the problem lies in the difficulty of finding their Kan fibrant replacement.

Could you please show why it's hard, for the easiest possible nontrivial case?

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Computing homotopy groups for spheres are fundamentally hard, and I believe the problem lies in the difficulty of finding their Kan fibrant replacement.

Computing the fibrant replacement for simplicial sets is quite easy: it is given by the Kan fibrant replacement functor Ex^∞. Explicitly, n-simplices in the fibrant replacement of a simplicial set X are maps Sd^k Δ^n → X, for some k≥0. Here Sd^k denotes the k-fold barycentric subdivision of a simplicial set. We allow to increase k by further subdividing, this does not change the simplex.

This description is very simple and can be easily programmed into a computer.

The problem is, however, is that the number of simplices grows exponentially with k, and we also do not have an efficient way to get an a priori upper bound for k. So some problems are bound to be computationally undecidable, such as the problem of computing whether π_1 of a simplicial set is trivial or not.

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    $\begingroup$ Unless I'm confused, there is no finite $k$ that works for all maps into the space. $\endgroup$
    – Phil Tosteson
    Commented May 2, 2020 at 17:34
  • $\begingroup$ @PhilTosteson: No part of the answer claims that k is fixed at any point. As I write in my answer, "We allow to increase k by further subdividing, this does not change the simplex." $\endgroup$
    – Dmitri Pavlov
    Commented May 2, 2020 at 17:37
  • $\begingroup$ An n-simplex of Ex^∞ X is a simplicial map Sd^k Δ^n → X for some k≥0. Composing this map with the last vertex map Sd^l Δ^n → Sd^k Δ^n for some l≥k produces a map Sd^l Δ^n → X, which defines exactly the same n-simplex of Ex^∞ X. $\endgroup$
    – Dmitri Pavlov
    Commented May 2, 2020 at 17:42
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    $\begingroup$ @Student: Concerning 1): computers can perform computations with integers, yet the set of integers is infinite. The story for simplicial sets is analogous. Concerning 2): yes, just postulate that the boundary of Sd^k Δ^n maps to the given simplicial subset instead of the basepoint. 3) Yes. Presentations are easy to produce, but extracting invariants from these presentations (e.g., the cardinality of a group) is hard or even undecidable. $\endgroup$
    – Dmitri Pavlov
    Commented May 2, 2020 at 21:20
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    $\begingroup$ Perhaps I am missing something, but I don't think it is straightforward use $Ex^{\infty} X$ to obtain a finite presentation of $\pi_n X$ for $n > 1$. You need an effective bound on how far you need to subdivide to get generators and relations. Just because you can make computations locally, it does not mean that you can compute globally-- and this is what is required to actually compute $\pi_n X$. Note, if you had a presentation, it would be trivial to determine the isomorphism class of $\pi_n X$, because $\pi_n X$ is abelian. $\endgroup$
    – Phil Tosteson
    Commented May 2, 2020 at 21:59
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Any fibrant replacement for $S^n$, $n \geq 1$ is going to have infinitely many non-degenerate simplices. This is simply because there are infinitely many elements of $\pi_nS^n$. So, even though to a mathematician it seems that we can "compute" a fibrant replacement, it is not actually easy to program it in such a way that we can determine the homotopy groups.

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  • $\begingroup$ Can the fibrant replacement be finite at each level? $\endgroup$
    – Student
    Commented May 2, 2020 at 15:05
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    $\begingroup$ No, that's what the argument in the answer shows. It has to have infinitely many non degenerate $n$-simplices, at the very least $\endgroup$
    – Maxime Ramzi
    Commented May 2, 2020 at 15:41
  • $\begingroup$ Maxime is right. However, there must be workarounds-- to represent any given map, one only has to subdivide $Y$ finitely many times. I don't know the literature, but I found this on the arxiv arxiv.org/abs/1706.00380. $\endgroup$
    – Phil Tosteson
    Commented May 2, 2020 at 17:31
  • $\begingroup$ But note that if $X$ is a simply connected simplicial set each of whose homotopy groups is finite (e.g., a Moore space), then it does have a fibrant replacement with finitely many simplices in each dimension. Unfortunately, the number of simplices still grows exponentially by dimension, so this is not very helpful. $\endgroup$
    – Charles Rezk
    Commented May 5, 2020 at 23:12
  • $\begingroup$ @Charles That's a good point. Also you can construct this replacement (up to some level) by constructing a Postnikov tower for $X$-- which is how Brown originally showed that homotopy groups are computable. $\endgroup$
    – Phil Tosteson
    Commented May 8, 2020 at 13:43

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