9

The classifying space of a category $\scr{C}$ is obtained by taking its nerve $N\scr{C}$, which is the simplicial set defined by $$ N\mathscr{C}_n:= \mathrm{Fun}([n],\mathscr{C}) $$ and the classifying space is defined as $$ B\mathscr{C}:= |N\mathscr{C}| $$ the geometric realization of the nerve. The only concrete examples I have every played with are the classifying spaces of groups, $BG$. But these all end up being $K(G,1)$'s.

Question: What is an explicit example of a category $\mathscr{C}$ so that its classifying space has nontrivial higher homotopy groups.

I know that such things should exist; it is my understanding that Quillen's Q-construction takes a category $M$ and outputs a category $QM$ whose classifying space is the K-theory $K(M)$.

Thanks!

Eric Wofsey
  • 342,377
CWcx
  • 1,391
  • 3
    Eric already mentioned that you can get every homotopy from the nerve of a poset. Another, more surprising result of that kind is that you can get every connected homotopy type as the nerve of a monoid (thought of as a category with a single object)! That's a theorem of Dusa McDuff's. – Omar Antolín-Camarena Mar 04 '16 at 06:28

1 Answers1

16

For a very simple example, consider the poset $\{a,b,c,d,e,f\}$ with $a,b\leq c,d\leq e,f$. The classifying space of this poset is homeomorphic to $S^2$ (you can explicitly list out all the nondegenerate simplices in its nerve and draw a picture of them), which has plenty of higher homotopy groups. More generally, in fact, every simplicial complex is homeomorphic to the classifying space of a poset, namely its poset of faces (indeed, the nerve of the poset of faces is just the barycentric subdivision of the simplicial complex you started with). Since every space is weak homotopy equivalent to a simplicial complex, this means that every (weak) homotopy type can be the realized as the classifying space of a category.

Eric Wofsey
  • 342,377
  • 1
    That's so cool! So the poset above then comes from writing down $S^2$ as the simplicial complex glued from 3 2-simplices? For a general simplicial complex $X$, how do you define the ordering on the faces? – CWcx Mar 03 '16 at 02:44
  • 1
    Actually, the poset I gave is not the face poset of any simplicial complex; any face poset of a triangulation of $S^2$ would be more complicated. The face poset of a simplicial complex is just the set of all of its simplices, ordered by inclusion. – Eric Wofsey Mar 03 '16 at 02:46
  • ahh ok. Thanks! – CWcx Mar 03 '16 at 02:47
  • 4
    To see how I came up with that 6-element poset, you can note that if you take a poset $P$ and add two new elements to it which are greater than every element of $P$, you get a new poset $Q=P\cup{x,y}$ and the classifying space of $Q$ is the suspension of the classifying space of $P$ (each of the new points you've added gives you a cone over the classifying space of $P$). Doing this twice starting from the discrete poset ${a,b}$ gives you a $6$-element poset whose classifying space is the double suspension of a pair of points, i.e. $S^2$. – Eric Wofsey Mar 03 '16 at 02:49
  • 1
    It's kind of astounding that we can get every weak homotopy type from a poset. Do you know if this related to the fact that the algebraic K-theory spectrum of the category of finite sets is the sphere spectrum? – CWcx Mar 03 '16 at 04:28
  • I don't really see any particular connection, no. – Eric Wofsey Mar 03 '16 at 04:53