19
$\begingroup$

The classifying space of the nth symmetric group $S_n$ is well-known to be modeled by the space of subsets of $R^\infty$ of cardinality $n$. Various subgroups of $S_n$ have related models. For example, $B(S_i \times S_j)$ is modeled by subsets of $R^\infty$ of cardinality $i + j$ with $i$ points colored red and $j$ points colored blue. More fun: the wreath product $S_i \int S_j \subset S_{ij}$ has classifying space modeled by $ij$ points partitioned into $i$ sets of cardinality $j$ (but these sets are not "colored").

My question: is there a geometric model, preferably related to these, for classifying spaces of alternating groups? [Note: since any finite group is a subgroup of a symmetric group one wouldn't expect to find geometric models of arbitrary subgroups, but alternating groups seem special enough...]

Improve this question
$\endgroup$
4
  • 3
    $\begingroup$ For any subgroup $G\subset S_n$ you can proceed by: Take $E$ to be the universal cover of any classifying space of $S_n$. Then $E/G$ is a classifying space for $G$. This is implicit in the answers and in your examples. $\endgroup$
    – Bruce Westbury
    Commented Jul 1, 2010 at 3:10
  • $\begingroup$ Yes, but to get to more "natural" models requires at least a bit more than such a general construction (and presumably won't be possible for arbitrary subgroups). $\endgroup$
    – Dev Sinha
    Commented Jul 1, 2010 at 5:13
  • $\begingroup$ I don't understand the comment about "natural" models: the classifying space $BG$ is only defined up to homotopy and $BH=EG/H$ for $H<G$ is rather natural! $\endgroup$
    – Victor Protsak
    Commented Jul 30, 2010 at 7:31
  • $\begingroup$ Natural is perhaps the wrong word (because of the connotations of functoriality). What I mean is more "found in nature." For example you wouldn't want to understand the classifying space of Z/p \times Z/p as some kind of "configurations with ordering up to equivalence" - one model found in nature is the product of Lens spaces. $\endgroup$
    – Dev Sinha
    Commented Jul 30, 2010 at 8:15

2 Answers 2

Reset to default
22
$\begingroup$

$n$ linearly independent points in $R^\infty$ together with an orientation of the $n$-plane which they span.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Nice - I had started thinking along the lines Greg suggests but this is what I was looking for (and should have thought of myself). $\endgroup$
    – Dev Sinha
    Commented Jun 30, 2010 at 23:17
  • 2
    $\begingroup$ The two solutions are hardly different. Kevin is exactly describing a sign-ordering of the points, the only difference being that he removes linearly dependent sets of points. This deletion doesn't hurt anything because it has infinite codimension, but it also isn't necessary for the answer. $\endgroup$
    – Greg Kuperberg
    Commented Jun 30, 2010 at 23:50
  • $\begingroup$ My initial idea was the same as Greg's, but I didn't want to have to explain what a sign-ordering was. I also thought a sign-ordering might be exotic enough to disqualify the answer as simple. Also, note that the linear independence requirement arises naturally if we think of A_n as a subgroup SO(n). BSO(n) is oriented n-planes in R^\infty, and an (SO(n)/A_n)-bundle over this grassmanian gives the same answer as above. $\endgroup$
    – Kevin Walker
    Commented Jul 1, 2010 at 0:20
  • $\begingroup$ That's fair. The common theme in all of these constructions is to obtain one classifying space from another one by passing to a subgroup. (But strictly speaking you are using $A_n \subset SL(n)$. $A_n \subset SO(n)$ would give you orthogonal vectors rather than l.i. vectors.) $\endgroup$
    – Greg Kuperberg
    Commented Jul 1, 2010 at 0:43
  • $\begingroup$ The reason I like Kevin's answer is that it geometrically reflects that there is a pull-back square involving BA_n, BS_n, BSO(n) and BO(n). Of course, one sees this pullback algebraically by the definitions of A_n and SO(n)... And whether this is significantly different than the notion of sign ordering is a matter of taste, though it should be noted that orientations of linear subspaces is something which can engage a lay-person, bright high-school student, etc. $\endgroup$
    – Dev Sinha
    Commented Jul 1, 2010 at 5:01
12
$\begingroup$

Probably the right thing to do is to express the classifying space of $A_n$ as the non-trivial double cover of the classifying space of $S_n$. A point in the classifying space is then a set of $n$ points in $\mathbb{R}^\infty$ with a "sign ordering". A sign ordering is an equivalence class of orderings of the points, i.e., ways to number them from 1 to $n$, up to even permutations. I coined the term "sign ordering" by analogy with a cyclic ordering. But that name aside, the idea comes up all the time in various guises. For instance an orientation of a simplex is by definition a sign ordering of its vertices.

This is in the same vein as your other examples and you can of course do something similar with any subgroup $G \subseteq S_n$. You can always choose an ordering of the points up to relabeling by an element of $G$.

A bit more whimsically, you could call the configuration space of $n$ sign-ordered points in a manifold "the configuration space of $n$ fermions". Although a stricter model of the $n$ fermions is the local system or flat line bundle on $n$ unordered points, in which the holonomy negates the fiber when it induces an odd permutation of the points. This local system is similar to the sign-ordered space in the sense that the sign-ordered space is the associated principal bundle with structure group $C_2$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .