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Jul 1, 2010 at 5:01 comment added Dev Sinha 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.
Jul 1, 2010 at 0:43 comment added Greg Kuperberg 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.)
Jul 1, 2010 at 0:20 comment added Kevin Walker 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.
Jun 30, 2010 at 23:50 comment added Greg Kuperberg 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.
Jun 30, 2010 at 23:17 comment added Dev Sinha Nice - I had started thinking along the lines Greg suggests but this is what I was looking for (and should have thought of myself).
Jun 30, 2010 at 23:16 vote accept Dev Sinha
Jun 30, 2010 at 23:00 history answered Kevin Walker CC BY-SA 2.5