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Both cohomology and homotopy groups capture global topological information of a manifold $X$. It is interesting to ask if they can be computed from local data. A triangulation $T$ is a natural presentation of a manifold.

The cohomology by definition can be computed from $T$. However, the problem seems much harder for homotopy groups. In fact, humans struggle even for the simplest case $X=S^n$, at least for $\pi_{k>n}(X)$.

Hope is not lost, as we know the answer to $k \leq n$ for spheres. Therefore my question:

Question: Given a triangulated (compact) manifold $X$ of dimension $n$, is there an algorithm that computes $\pi_{k \leq n}(X)$ (up to isomorphism, or in the form of generators and relations)?

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  • $\begingroup$ Your question is a little imprecise: how do you want the output (e.g., generators and relations)? Certainly somethings around this problem are undecidable so I suspect the answer to your question (once it is made more precise) will be no $\endgroup$
    – Sam Hopkins
    Commented Apr 18, 2022 at 1:06
  • $\begingroup$ See e.g. mathoverflow.net/questions/304481/… $\endgroup$
    – Sam Hopkins
    Commented Apr 18, 2022 at 1:20
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    $\begingroup$ Another previous MO question you may be interested in: mathoverflow.net/questions/31004/… $\endgroup$
    – Sam Hopkins
    Commented Apr 18, 2022 at 1:21
  • $\begingroup$ Thanks for pointing out the nuance. I have edited accordingly. I was excited about that MO question too, but it seems that the answer there wasn't so clear.. $\endgroup$
    – Student
    Commented Apr 18, 2022 at 1:26
  • $\begingroup$ Note that for $k\ge 2$ for a closed manifold the $\pi_k$ is abelian, but however can fail to be finitely generated. So, it is unclear what would be meant by "outputting a presentation in terms of generators and relations". $\endgroup$
    – YCor
    Commented Apr 18, 2022 at 8:45

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Being a manifold or the dimension restriction $k\leq n$ doesn't matter, the following applies to finite simplicial complexes in general:

As others have explained, if the fundamental group is not finite, the higher homotopy groups might be infinitely generated, so in that case even the format of the answer is unclear. Also, it is known that deciding whether a group given by a finite presentation (the simplicial complex gives you such a presentation for $\pi_1$) is actually finite is not possible algorithmically (the halting problem can be reduced to this problem, so if this is problem would admit an algorithmic solution, the halting problem would, too). This problem arises even if you restrict attention to manifolds: Any finite presentation of a group can be turned into an explicit manifold of dimension $4$ or greater via handlebodies.

However, there are algorithms that will produce the list of elements in finite time if $\pi_1$ happens to be finite (and simply never terminate otherwise). From this, it is possible to produce a finite simplicial complex describing the universal cover. For simply-connected finite simplicial complexes, the problem of computing any given homotopy group is known to be be algorithmic, in fact, there exist actual implementations. (I do think they scale pretty badly with degree though).

EDIT: For the $\pi_1$ statement, let me give you an easy but horribly inefficient algorithm. For a finitely presented group $G$, and a finite group $H$ given by a multiplication table, a map $G\to H$ can be specified by finite data: Pick an image for each generator, such that the relations are satisfied. Conversely, a map $H\to G$ can be described by choosing for each element of $H$ a word in the generators of $G$, and for each pair of elements $h_1,h_2\in H$, a finite sequence of rewritings using the relations of $G$, identifying the word associated to $h_1h_2$ with the product of words associated to $h_1,h_2$. For two such maps, a proof witnessing that the composite $G\to H\to G$ is the identity can similarly be specified in a finite amount of data: For each generator, a finite sequence of rewrites identifying it with its image. Similarly for the other composite. So a proof identifying a finitely presented group with a concrete finite group $H$ can be given as a finite bundle of data, now just enumerate all such bundles by size until you found one.

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    $\begingroup$ For algorithms, note that these seem to available now also in Sage (via the linked Kenzo program): mdpi.com/2227-7390/9/7/722 --- while they scale badly, it seems that we are now at a point, where they can provide non-trivial (and sometimes unknown) computations in reasonable time. $\endgroup$
    – Lennart Meier
    Commented Nov 28, 2022 at 6:50
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    $\begingroup$ Actually, French doesn't seem too different from English, which allowed me to locate the magic. The key to the implementation is given in section 6.3: Assume $X$ is $(n-1)$-connected (and "effective" and "reduced", for computation reasons), where $n \geq 2$. It creates other "effective" and "reduced" spaces $E, E', E'', \ldots, E^{(n)}, \ldots$ from the information of $X$. Finally, it claims that $\pi_{n+k}(X) \simeq H_{n+k}(E^{(k-1)})$! And since homology is easier it's done. Could someone fill in the details of why the homologies of such $E$'s are equivalent of higher homotopy groups of $X$? $\endgroup$
    – Student
    Commented Dec 3, 2022 at 14:23
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    $\begingroup$ It's just Hurewicz. More precisely, the $E^{(n)} $ are constructed by killing lower homotopy groups of $X$ without changing the higher ones, and once you're $n-1$-connected $\pi_n$ and $H_n$ agree by Hurewicz. Minus the constructive/effective description of the spaces, this is precisely Serre's method. $\endgroup$
    – Achim Krause
    Commented Dec 3, 2022 at 17:27
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    $\begingroup$ @AchimKrause Note that the undecidability of the finiteness problem is distinct from the word problem, the former being a consequence of the Adian-Rabin theorem (it is also a consequence of the weaker form of the Adian-Rabin theorem proved by Adian some years before he proved the full one). $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Dec 4, 2022 at 8:49
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    $\begingroup$ @Carl-FredrikNybergBrodda: We're on the same page! But since these issues are subtle, I wanted to point out that "RH is true" is not really a great example of the phenomenon under discussion. Distinguishing "unknown", "undecidable" and "computationally intractable" is exactly the kind of difficulty that people new to these considerations have difficulty with. $\endgroup$
    – HJRW
    Commented Dec 5, 2022 at 10:14
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Collins and Miller proved that it is algorithmically undecidable whether or not $\pi_2(X)$ is trivial, for $X$ a finite 2-complex. As Achim Krause points out, in general $\pi_2(X)$ is only a module over $\pi_1(X)$, and so it's not clear what kind of a description of it you might want. But the Collins--Miller result shows that any such description that works in general won't be good enough to determine whether or not $\pi_2(X)$ is trivial (much as knowing the generators and relations for $\pi_1(X)$ doesn't enable you to decide whether or not $\pi_1(X)$ is trivial).

You might also like to bear the Whitehead asphericity conjecture in mind.

Whitehead's conjecture: If a 2-complex $X$ is aspherical and $Y$ is a subcomplex of $X$, then $Y$ is also aspherical.

This has been open for 81 years, so is often reckoned to be one of the hardest questions in topology.

For finite $X$, the question can be rephrased as:

If $X$ is obtained by attaching a 2-cell to $Y$ and $\pi_2(Y)$ is non-trivial, must $\pi_2(X)$ be non-trivial?

As this makes clear, we don't really understand the effect that attaching a 2-cell can have on $\pi_2$.

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  • $\begingroup$ Wow, good to know. Just to double-check, this issue goes away completely when $\pi_1(X)$ is finite, right (as shown in Achim Krause's answer and the comments below it)? $\endgroup$
    – Student
    Commented Dec 3, 2022 at 23:42
  • $\begingroup$ That’s correct. There are some other conditions on $\pi_1$ that can also be useful, but I’ll leave them be unless anyone asks. $\endgroup$
    – HJRW
    Commented Dec 4, 2022 at 6:46
  • $\begingroup$ If you feel that this thread is a good place to expand. Feel free. Personally I am interested in knowing more about it. Besides finiteness, what sort of conditions would allow higher homotopy groups get computed? References to related papers are also appreciated. $\endgroup$
    – Student
    Commented Dec 4, 2022 at 14:10
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    $\begingroup$ Ok! There are various results about 2-complexes saying that if $\pi_1(X)$ satisfies various properties and $H_2(X)=0$ then $\pi_2(X)=0$. Of course, Hurewicz gives this when $\pi_1(X)$ is trivial. Cockcroft proved this when $\pi_1(X)$ is free, and Howie generalised it further, to when $\pi_1(X)$ is locally indicable, meaning that every non-trivial finitely generated subgroup has infinite abelianisation. I don’t know any such results when $H_2(X)$ is non-trivial. $\endgroup$
    – HJRW
    Commented Dec 4, 2022 at 17:07

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