# Homotopy type of the geometric realization of a poset

Consider a set of $$n$$ elements $$S=\lbrace 1,\dots,n\rbrace$$ and $$\mathcal{P}(S)$$ to be the power set of $$S$$, which is a well-defined poset with respect to the inclusions. Now consider $$\emptyset\neq T\varsubsetneq S$$ and define $$A=\lbrace \emptyset\neq U\in\mathcal{P}(S)\mid T\nsubseteq U\rbrace$$. I want to compute the homotopy type of the geometric realization $$\left|A\right|$$.

My intuition tells me that $$\left|A\right|\simeq\mathbb{S}^{\left|T\right|-2}$$, however I don't know if this is right and how to prove it.

I know some facts such that if the poset has a maximal element or a minimal element then the geometric realization is contractible. Also, for some small cases I am able to do it by hand, however I don't know how to do it in general and whether there is some ''general'' procedure to study this problem.

• @SamHopkins Thanks!! You can add this as an answer if you want me to acept it. Jun 6 at 21:07
• Thanks! That's an interesting example. Jun 6 at 23:16
• How is the geometric realization of a poset defined? Is it the realization of its order complex (p6 of "Poset Topology: Tools and Applications" arxiv.org/pdf/math/0602226.pdf)? Jun 14 at 0:10
• @nasosev: yes, that's right. Jun 14 at 13:04
• @SamHopkins thanks. I was wondering, a topological space is itself a poset. So is there a relationship between the homotopy types of a space, and the realisation of the order complex of its poset of opens ? Jun 15 at 16:19

First recall that geometric realisation of posets preserves products: the projections $$P\xleftarrow{p}P\times Q\xrightarrow{q}Q$$ give a map $$(|p|,|q|)\colon |P\times Q|\to|P|\times|Q|$$, and it is a standard fact that this is a homeomorphism.

Next, if $$f_0,f_1\colon P\to Q$$ are morphisms of posets and $$f_0(x)\leq f_1(x)$$ for all $$x$$ then we can define a poset morphism $$h\colon \{0,1\}\times P\to Q$$ by $$h(i,x)=f_i(x)$$, and this gives a map $$|h|\colon [0,1]\times|P|\to|Q|$$ which is a homotopy between $$|f_0|$$ and $$|f_1|$$.

Now put $$A=\{U\subseteq S\;:\;U\neq\emptyset, U\not\supseteq T\}$$ as in the question. We can define $$f_0,f_1,f_2\colon A\to A$$ by $$f_0(U)=U$$ and $$f_1(U)=U\cup T^c$$ and $$f_2(U)=T^c$$. Then $$f_0\leq f_1\geq f_2$$, so the identity is homotopic to the constant map $$|f_2|$$, so $$|A|$$ is contractible.

Now suppose instead we consider the poset $$B=\{U\subseteq S\;:\;T\cap U\neq\emptyset,U\not\supseteq T\}\subset A$$, and the poset $$C=\{U\;:\;\emptyset\subset U\subset T\}$$. We have an inclusion $$i\colon C\to B$$ and a retraction $$r\colon B\to C$$ given by $$r(U)=T\cap U$$. These have $$ri=1$$ and $$ir\leq 1$$ so $$|i|$$ and $$|r|$$ are mutually inverse homotopy equivalences. Alternatively, we can say that $$B\simeq C\times D$$, where $$D$$ is the poset of all subsets of $$T^c$$. This gives $$|B|\simeq|C|\times|D|$$, where $$|D|$$ is homeomorphic to $$[0,1]^{|T^c|}$$ and in particular is contractible.

Now put $$M=\{m:T\to\mathbb{R}\;:\;\sum_{t\in T}m(t)=0\}$$ (which is a vector space of dimension $$|T|-1$$ with an obvious inner product). For $$U\in C$$ define $$f(U)=\chi_U-|U||T|^{-1}\in M$$. Extend this linearly over simplices to get a map $$f\colon |C|\to M$$. One can check that this is nowhere zero, so we can define $$f_1(x)=f(x)/\|f(x)\|$$, and this gives a homeomorphism $$|C|\to S(M)\simeq S^{|T|-2}$$.

• Thanks, I didn't realize that the above proof was wrong. This means that I have to work harder to get confortable with those tools. Jun 8 at 10:17

Once more, with feeling. Thanks to comments from Tyler Lawson and Neil Strickland.

Let me use $$B_n$$ for the finite Boolean lattice of subsets of $$[n] := \{1,2,\ldots,n\}$$ (your $$\mathcal{P}(S)$$), and let $$k=|T|$$. Then your $$A=\{U\subseteq[n]\colon U\neq \varnothing, T \not \subseteq U\}=((B_k\setminus \{\hat{1}\}) \times B_{n-k}) \setminus \{\hat{0}\}$$, where $$\hat{0}$$ means the minimum of a poset, and $$\hat{1}$$ the maximum. To understand the homotopy type of this product, we need a result of Quillen (stated as Theorem 5.1(b) in the paper "Canonical homeomorphisms of posets" by Walker cited below; see also the discussion in Section 5.1 of the notes of Wachs), which says that $$|P\times Q \setminus \{\hat{0}\}| \simeq |P\setminus \{\hat{0}\}| \ast |Q\setminus\{\hat{0}\}|,$$ for posets $$P,Q$$ which have minimums, where $$\ast$$ is the join of topological spaces. So in our case $$|A| \simeq |B_k \setminus \{\hat{0},\hat{1}\}| \ast |B_{n-k}\setminus \{\hat{0}\}|.$$ But $$B_{n-k}\setminus \{\hat{0}\}$$ is contractible (it has a maximum), so $$|A|$$ is also contractible.

Note that if we considered the poset $$B=\{U\subseteq [n]\colon U \cap T \neq \varnothing, T \not\subseteq U\}$$ as in the answer of Neil Strickland, then we have that $$B = (B_k \setminus \{\hat{0},\hat{1}\}) \times B_{n-k}$$. [This is the poset that the first version of my answer was implicitly about.] Here we need a slightly different fact about homotopy types of products, namely that $$|P\times Q| \simeq |P| \times |Q|,$$ where $$\times$$ means product of topological spaces (see Theorem 5.1(a) in the paper of Walker, or again Section 5.1 of the notes of Wachs). This means that $$|B| \simeq |B_k \setminus \{\hat{0},\hat{1}\}| \times |B_{n-k}|,$$ and since $$B_{n-k}$$ is contractible, this time we actually can say $$|B| \simeq |B_k \setminus \{\hat{0},\hat{1}\}| \simeq \mathbb{S}^{k-2}$$.

Quillen, Daniel, Homotopy properties of the poset of nontrivial p-subgroups of a group, Adv. Math. 28, 101-128 (1978). ZBL0388.55007.

Wachs, Michelle L., Poset topology: tools and applications, Miller, Ezra (ed.) et al., Geometric combinatorics. Providence, RI: American Mathematical Society (AMS); Princeton, NJ: Institute for Advanced Studies (ISBN 978-0-8218-3736-8/hbk). IAS/Park City Mathematics Series 13, 497-615 (2007). ZBL1135.06001.

Walker, James W., Canonical homeomorphisms of posets, Eur. J. Comb. 9, No. 2, 97-107 (1988). ZBL0661.06006.

• I'm not able to see the equality of posets - it seems to suggest that the subsets of interest have non-empty intersection with T, rather than just being non-empty? Jun 7 at 16:36
• @TylerLawson: Thanks for your comment. I've corrected the answer (or at least attempted to!). Jun 7 at 17:08
• @NeilStrickland: argh, of course you are right. I've corrected again. Jun 8 at 13:20