# On minimal Kan simplicial sets having finite number of simplexes in each dimension

What are the examples of “tame” minimal Kan simplicial sets having finite number of simplexes in each dimension besides simplicial point and $$B(G)\approx K(G,1)$$ for a finite group $$G$$? I believe that Alain Connes’ simplicial circle is also minimal Kan.

A tame minimal Kan complex has finite homotopy groups. The converse is also true: if homotopy groups of a minimal Kan complex are finite, then there are finitely many $$n$$-simplices. The proof is by induction on $$n$$. First, $$X_0 = \pi_0(X)$$ is finite. If $$n > 0$$, there are finitely many possible choices of boundaries for $$n$$-simplices by induction hypothesis. For every such boundary, the set of its fillers is either empty or isomorphic to $$\pi_n(X,x_0)$$ for some $$x_0$$. This completes the proof. This fact implies that, for every Kan complex with finite homotopy group, there is an equivalent tame minimal Kan complex.
• Tank you Valery. I should understand better why the tame complex has finite homotopy groups. It should be some very basic knowledge which i am missing for the moment. Therefore Connes’ circle (which is $K(Z,1)$ can’t be minimal Kan? Good to know. – Nikolai Mnev Jul 7 '19 at 7:33
• @NikolaiMnev This is true for any Kan complex $X$. The $n$-th homotopy group at $x_0$ could be defined as the set of homotopy classes of $n$-cells of $X$ with the degenerate at $x_0$ boundary. So, there is a surjection from a subset of $X_n$ onto $\pi_n(X,x_0)$. As a side note, if $X$ is minimal, then the surjection is identity, so $\pi_n(X,x_0)$ is a subset of $X_n$. I don't know what is Connes' circle, but minimal Kan complexes of a given homotopy type are unique up to isomorphism, so if it is not isomorphic to the usual definition of $\mathbf{B}(\mathbb{Z})$, it can't be minimal Kan. – Valery Isaev Jul 7 '19 at 7:58