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Anybody knows a semi-simplicial model for $K(Z,2)$ having finite number of simplexes in any dimension? With some regular description? I have heard about big activity on triangulating $CP^n$ but this does not look providing stable regular answer.

(update) Many thanks for comments.

My motivation for the question is following As a current work lemma i have got such a model, hopefully (not all written yet). And i am wondering how much is known in such a surface level question.

The model is a semi-simplicial set. As k simplexes we take the set of all circular permutations of k+1 elements which are the same as oriented necklaces with k+1 beads colored by different ordered colors 0,...,k. The face map is deletion i-th colored bead. This is K(Z,2) and a sort of tame semi-simplicial model for Connes' cyclic simplex. My current simple proof is that this space classifies circle bundles. The related way of proof is like in comment of André Henriques. Semi-simplicial set of permutations you can see as a contractible semi-cyclic set and have to say words about trivial role of degenerations here.

Interestingly that here only restricting the complex up to dimension 2 you will got a deformed sphere (a couple of dunce-huts glued by "boundary circle"), but further restrictions up to dimension 2n will not provide models of $CP^n$.

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    $\begingroup$ This question can probably be reformulated in terms of finding a cyclic set (a functor from Connes' cyclic category to the category of sets), whose geometric realisation is an $ES^1$. Freeness is easily enforced: the (let's say unique) 0-simplex $p$ should have its two degeneracies be distinct $s_0p\not =s_1p$. So all that remains to do is to make the simplicial set (=underlying simplicial set of the cyclic set) contractible. $\endgroup$
    – André Henriques
    Commented Apr 29, 2015 at 10:42
  • $\begingroup$ Thank you André! Yes in fact as a lemma i have got a simple tame model for Connes' cyclic simplex. The speculation is different but sort of related to you plan. I was wondering how much is known. $\endgroup$
    – Nikolai Mnev
    Commented Apr 29, 2015 at 12:01
  • $\begingroup$ What do all those "semi-" mean? $\endgroup$
    – Fernando Muro
    Commented Sep 2, 2019 at 14:49
  • $\begingroup$ @FernandoMuro the post is outdated. For combinatorics of triangulations the natural category is semi-simplicial sets or $\Delta$-sets -simplicial sets without degeneracies but here everywhere degeneracies can be added canonically. So the “semi” is unnecessary $\endgroup$
    – Nikolai Mnev
    Commented Sep 2, 2019 at 16:05

2 Answers 2

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I have some idea, but I'm not sure it works well.

First I give a triangulation to $CP^1$: let's say that $CP^1$ is $C$ with a point $\infty$ added. Then we can fix 6 vertices, $0,1,-1,i,-i,\infty$ and divide $CP^1$ in 6 vertices, 12 edges and 8 triangles. Each complex number belongs to one of the 8 triangles according to the sign of its real part, the sign of its imaginary part and whether its modulus is greater or lower than 1.

So there are 26 types of points in $CP^1$: 3 choices for the sign (or zero) for the imaginary part, 3 choices for the sign (or zero) for the real part, 3 choices for the sign of the logarithm of the modulus would yield 27, but it is not possible to have 0 real, 0 imaginary and 1 modulus (we say that $\infty$ has modulus greater than 1, but its imaginary and real parts are 0). Note that a complex number (not $\infty$) has only 25 types: the ones different from the $\infty$ type.

More generally, every homogeneous $n+1$-tuple $[z_0,\dots, z_n]$ in $CP^n$ has exactly one representative whose first nonzero coordinate is 1. The following coordinates are just complex numbers, so each of them is in one of the 25 types stated before. So we associate to a point in $CP^n$ the following data: 1) The position of the first nonzero coordinate, which is some integer $0\leq k\leq n+1$; 2) the $n-k$-tuple of the types of complex numbers in the preferred representative.

Points with the same data are in the same internal part of the same simplex, and the dimension of a simplex is given by the number of < and > appearing in point 2 (I have no time now to explain in detail, but I hope it's quite clear).

For example, vertices are points of the form $[0,0,0,1,i,0,-i,0,1,-1,0]$, that is, $n+1$-tuplesmade just of $0,1,-1,i,-i$ with the first nonzero coordinate equal to 1.

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    $\begingroup$ The attaching map for the top cell in $\mathbb{C}P^2$ is the Hopf map $\eta\colon S^3\to S^2$. Thus, if you have a simplicial model of $\mathbb{C}P^2$, then it must encode a simplicial representative of $\eta$, which is necessarily quite complicated. This makes me skeptical about the above answer, but I have not worked through the details. $\endgroup$
    – Neil Strickland
    Commented Apr 29, 2015 at 14:01
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    $\begingroup$ In case it might be of some interest: hopf.math.purdue.edu/Sauvageot/simpl-hopf-model.pdf $\endgroup$
    – Fernando Muro
    Commented Apr 30, 2015 at 7:40
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    $\begingroup$ Thank you Fernando, very interesting. Actually Hopf bundle has pretty canonical semi-simplicial triangulation having 2 2-simplices in base and 6 3-simplices in total space. $\endgroup$
    – Nikolai Mnev
    Commented Apr 30, 2015 at 14:07
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Here is my model https://arxiv.org/abs/1908.04029 called in the paper $\pmb SC$. The homotopy issue only mentioned and postponed to more general later writings but it is exactly what was mentioned by @AndréHenriques — factor of symmetric cross-simplicial group $\pmb S$ (which is contractible) by free right acton of Connes’ cyclic group

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