Let $S_X$ be the barycentric subdivision operator of a topological space $X$ in singular theory. (The one standard algebraic topology texts such as Hatcher and Munkres) There is also a machinery so called barycentric subdivision in simplicial theory. The idea of constructions are the same, but they are different. I wonder if there is a reason not to distinguish them in terminologies.

So, my question is whether $S_X$ is just a machinery to prove Excision theorem. I skimmed my texts, but I clould not find $S_X$ after the proof of excision theorem.

  • $\begingroup$ I'm not an expert, but I'm pretty sure the answer is yes. I never saw the operator show up except in setting up the excision theorem. $\endgroup$ – Danu Oct 6 '16 at 10:20
  • $\begingroup$ This "operator" as seen in Hatcher's text is indeed a rather arbitrary thing. To prove the excision, you need to subdivide simplices into smaller simplices, and that "operator" gives you one canonical procedure. mathoverflow.net/questions/31035 $\endgroup$ – user144221 Oct 6 '16 at 11:38
  • $\begingroup$ I'm fond of the cubical approach in Sch\"on, R. , "Acyclic models and excision", Proc. Amer. Math. Soc. 59~(1) (1976) 167--168, as the "standard" subdivision of a cube into small cubes is very easy to picture, and can be exploited in other ways. $\endgroup$ – Ronnie Brown Oct 6 '16 at 20:26

Barycentric subdivision, or rather subdivision of simplicial sets, does indeed show up elsewhere in algebraic topology. See the answers from the link in Alejo's comment. Here are a couple more:

  • The subdivision functor has a right adjoint $\operatorname{Ex}$, which is used to construct the Kan fibrant replacement functor $\operatorname{Ex}^\infty$ in the model structure on simplicial sets.

  • "The second subdivision of a simplicial set is a simplicial complex." "The second subdivision of a category is a poset." See the second part of May's book-in-progess on finite spaces for an explanation of these statements.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.