2
$\begingroup$

Do higher groups classify the homotopy types of topological spaces? We may assume $\pi_n$ of the topological spaces are all finite and $\pi_n =0$ for large enough $n$.

For example, if only $\pi_1 \neq 0$ and finite, the homotopy types of topological spaces are classified by finite groups.

Added: the term "higher group" refers to the higher group defined here https://en.wikipedia.org/wiki/N-group_(category_theory)

Improve this question
$\endgroup$
12
  • 6
    $\begingroup$ Look up "Postnikov tower" and "k-invariants" $\endgroup$
    – Achim Krause
    Commented Sep 16, 2023 at 15:10
  • 7
    $\begingroup$ I think there is a miscommunication here. I believe the OP means "higher groups" in the sense of 2-groups etc - i.e. group objects in a homotopical context, NOT just the graded group given by the collection of homotopy groups. In this case the answer is I believe tautologically yes for connected spaces -- there's an equivalence of (oo-)categories between n-truncated connected pointed spaces (pi-finite) and n-groups given by taking loops / taking classifying spaces. $\endgroup$
    – David Ben-Zvi
    Commented Sep 16, 2023 at 21:54
  • 6
    $\begingroup$ @Xiao-GangWen I think the terminology "higher groups" is unambiguous, but familiar only to a relatively small community. In this general context people jumped to the conclusion you didn't mean them in the precise sense but rather meant "higher homotopy groups". $\endgroup$
    – David Ben-Zvi
    Commented Sep 16, 2023 at 23:29
  • 2
    $\begingroup$ Good catch, @DavidBen-Zvi! $\endgroup$
    – Omar Antolín-Camarena
    Commented Sep 18, 2023 at 18:16
  • 6
    $\begingroup$ The answer is supposed to be yes, @Xiao-GangWen (with a small adjustment about $\pi_0$): if a space has $\pi_k(X, x_0)=0$ for all $k>n$ and all $x_0 \in X$, then the homotopy type should be determined by the fundamental $n$-groupoid of $X$. If $\pi_0(X)=pt$, then this $n$-groupoid is an $n$-group. The difficulty is proving these claims for a given definition of $n$-group. There are definitions for which it can be proved, but those don't make $n$-groups feel very algebraic, and there are some more algebraic definitions of $n$-group for which this result can only be proved for small $n$. $\endgroup$
    – Omar Antolín-Camarena
    Commented Sep 18, 2023 at 18:23

0

Reset to default

You must log in to answer this question.