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.

Any help or references about this problem will be appreciated.

  • 1
    $\begingroup$ @SamHopkins Thanks!! You can add this as an answer if you want me to acept it. $\endgroup$
    – Marcos
    Jun 6 at 21:07
  • $\begingroup$ Thanks! That's an interesting example. $\endgroup$
    – Dry Bones
    Jun 6 at 23:16
  • $\begingroup$ 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)? $\endgroup$
    – nasosev
    Jun 14 at 0:10
  • $\begingroup$ @nasosev: yes, that's right. $\endgroup$ Jun 14 at 13:04
  • $\begingroup$ @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 ? $\endgroup$
    – nasosev
    Jun 15 at 16:19

2 Answers 2


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 $|C|\to S(M)\simeq S^{|T|-2}$.

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    $\begingroup$ 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. $\endgroup$
    – Marcos
    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.

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    $\begingroup$ 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? $\endgroup$ Jun 7 at 16:36
  • 2
    $\begingroup$ @TylerLawson: Thanks for your comment. I've corrected the answer (or at least attempted to!). $\endgroup$ Jun 7 at 17:08
  • $\begingroup$ @NeilStrickland: argh, of course you are right. I've corrected again. $\endgroup$ Jun 8 at 13:20

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