Skip to main content
13 events
when toggle format what by license comment
Sep 6, 2019 at 12:43 comment added Tim Campion @IJL Ah, thanks! I suppose this is fine, since the realization of a semisimplicial set is the same as the realization of the free simplicial set on it.
Sep 6, 2019 at 10:28 comment added IJL A delta complex is the geometric realization of a semi-simplicial set, not a simplicial set. If you want to make a simplicial set whose realization is the 2-sphere, you can do it with just one 0-simplex and one non-degenerate 2-simplex (i.e., identifying the whole boundary of the 2-simplex to the point). This isn't allowed in delta-complex land, where the cheapest way to make a 2-sphere is probably to identify the boundaries of two 2-simplices, making a 2-sphere with 3 0-cells, 3 1-cells and two 2-cells.
Sep 1, 2019 at 19:02 comment added Tim Campion @RyanBudney I've always been a little confused about the definition of a Delta-complex, but looking back, it seems to me that a Delta-complex is exactly the same thing as the geometric realization of a simplicial set! In any event, this is really cool, thanks!
Sep 1, 2019 at 15:46 comment added Ryan Budney If we interpret your question a little more broadly, it's well known that delta complexes give efficient (and very useful) descriptions for surfaces and "most" cusped hyperbolic 3-manifolds. There's some beautiful delta complex structures on $\mathbb CP^2$ and I imagine many of them fit into a natural family. Ben Burton and I found a delta complex structure on $\mathbb CP^2$ with just four $4$-dimensional simplices.
Sep 1, 2019 at 9:03 comment added Neil Strickland @skd The Postnikov tower for $\Omega^\infty K(n)$ is very thin and quite regular. The corresponding minimal Kan complex is a twisted form of $\prod_kK(\mathbb{Z}/p,2(p^n-1)k)$, and the first nontrivial $k$-invariant is the Milnor Bockstein operation $Q_n$. It seems plausible that there should be an explicit simplicial map realising $Q_n$, which might be describable in terms of the McClure-Smith operad. One might hope that this would involve congruences of binomial coefficients and so make contact with formal group theory.
Sep 1, 2019 at 4:51 comment added Tim Campion @skd I suppose the part that gets me is that you need to use a topologically enriched category. In order to "simplicialize" this, you need to take the singular complex of the topological direction, unless you've already got a simplicial structure on the topolgical part. And singular complexes are extravagantly large. So for instance I'm not sure how to get from this a simplicial model for $BU$, say, which doesn't have uncountably many cells. Hmm, I wonder if Snaith's theorem could help with this... And similarly Westerland's theorem for $K(n)$...
Sep 1, 2019 at 4:16 comment added skd @Tim, I don't know if the resulting simplicial model is "economical", but (and you probably already know this) one can often construct models for "moduli-theoretic" spaces (like the union ∐_n BU(n) and the zeroth space of Thom spectra) as geometric realizations of Kan complexes/quasicategories (groupoids of vector spaces and cobordism categories, in the above examples) given by the simplicial nerve of a topologically enriched category whose objects are the structures parametrized by the space, and whose morphisms impose the equivalence relation implicit in the moduli-theoretic problem.
Sep 1, 2019 at 4:16 comment added skd @NeilStrickland That sounds very interesting; could you elaborate a bit more?
Aug 31, 2019 at 23:43 comment added Neil Strickland There are some reasons to expect that there might be an efficient and explicit algebraic/combinatorial simplicial model for the space $\Omega^\infty K(n)$, but I do not believe that anyone has ever produced one. This is potentially a rather interesting problem.
Aug 31, 2019 at 22:11 comment added Tim Campion @DmitriPavlov Actually, that sounds like a great idea! I'll have to ponder what comes out, but it seems about as economical as one could like!
Aug 31, 2019 at 22:09 comment added Dmitri Pavlov Concerning the comment about CP^∞: why not just take Γ(Z[2]), where Γ:Ch→sAb is the Dold–Kan functor?
Aug 31, 2019 at 21:49 history edited Tim Campion CC BY-SA 4.0
added 227 characters in body
Aug 31, 2019 at 21:42 history asked Tim Campion CC BY-SA 4.0