MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Take G to be a group. I care about discrete groups, but the answer in general would be welcome too. There are the various ways to construct the classifying space of G, bar construction, cellular construction if G is finitely presented, etc.

What I'm wondering about, is there a notion of a smooth classifying space? That is, when can a classifying space for a group be given a smooth structure?

share|cite|improve this question
up vote 11 down vote accepted

The answer to this does depend highly on the category in which you are prepared to work. If by "smooth structure" you mean "when is BG a finite dimensional manifold" then the answer is, as Andy says, "not many".

However if you are prepared to admit that there are more things that deserve the name "smooth" than just finite dimensional manifolds, then the answer ranges from "a few" to somewhere near "all".

To illustrate this with examples, the classifying space of ℤ is, of course, S1 whilst the classifying space of ℤ/2 is ℝℙ. Both are manifolds, but only the first is finite dimensional.

Here are some more details for the "somewhere near all". Take any topological model for BG. Then consider all continuous maps ℝ → BG. These correspond to G-bundles over ℝ. Amongst those will be certain bundles which deserve the name "smooth" bundles. By taking the corresponding curves, one determines a family of curves ℝ → BG which should be called "smooth". Using this one can define a Frölicher space structure on BG. (It is possible that you will get more smooth bundles than you bargained for this way. If that's a problem, you could work in the category of diffeological spaces but then you'd need to use all the ℝns).

In the middle, one can consider infinite dimensional manifolds. Then as your group is discrete it would be enough to ensure that you have a properly discontinuous action on an infinite sphere (there's a question somewhere around here about that being contractible). Some would say that your sphere "ought" to be the sphere in some Hilbert space. Failing that, if you have a faithful action on a Hilbert space (or more generally Banach space) with one or two topological conditions then you can quotient the general linear group by your group. Indeed, if your group is discrete then take the obvious action on ℓ2(G) (square summable sequences indexed by G).

A good example, but which is about as far from your situation for discrete groups as possible, is that of diffeomorphisms on a manifold. The classifying space of this group is the space of embeddings of that manifold in some suitable infinite dimensional space.

For more on the categories behind all this, see the nlab entries starting with generalised smooth spaces and the references therein. Also, anything by Kriegl, Michor, or Frolicher in the literature is worth a look.

share|cite|improve this answer

I'm not sure exactly what you mean here. One possible interpretation to your questions is "Which discrete groups have classifying spaces that are smooth manifolds"? For this, here are a few isolated facts.

1) One cheap necessary condition is that your group be torsion-free, as otherwise the group's cohomological dimension would be infinite.

2) If all you care about is that there is a classifying space that is a smooth manifold (not necessarily compact), then it is enough for the space to have a compact K(G,1) -- you could embed your K(G,1) in a high-dimensional R^n and then take a regular neighborhood.

3) A more useful thing is for your group to have a classifying space that is a compact manifold without boundary. You would then need your manifold to satisfy Poincare duality in an appropriate sense.

4) However, Poincare duality is not enough. Mike Davis has constructed Poincare duality groups that are not finitely presentable (and thus cannot have classifying spaces that are compact manifolds). If you require your group to be Poincare duality and finitely presentable, then I believe that it is open whether or not the group has a closed manifold classifying space.

A good survey on Poincare duality groups is Mike Davis's paper "Poincare Duality Groups", which is available on his webpage at

share|cite|improve this answer

This paper by Mostow is a great example of how a classifying space can be given a smooth structure, and how this smooth structure can be used to represent characteristic classes using differential forms on BG:

Mostow, Mark A.
The differentiable space structures of Milnor classifying spaces, simplicial complexes, and geometric realizations.
J. Differential Geom. 14 (1979), no. 2, 255--293.

share|cite|improve this answer

Since classifying spaces for compact Lie groups are constructed via direct limits of Grassmanians, you can really put smooth structures on them, where with a smooth structure I mean the structure of a manifold, modelled on a locally convex space (not necessarily finite-dimensional, not necessarily complete). This is due to the fact that "nice" direct limits of finite-dimensional manifolds admit such infinite-dimensional manifold structures (see for instance We also found no place where this is written up (although certainly well-known), so we spent some lines on this in (Lemma I.12).

share|cite|improve this answer
Nice to meet you on mathoverflow! – Ulrich Pennig May 7 '10 at 11:52
Yay, I drop a line just to join you two! – Peter Arndt May 7 '10 at 15:41

Well, perhaps not in the category you're interested in, but in algebraic geometry, the classifying space of any algebraic group (group scheme over a base) is "smooth" as an Artin Stack. The idea being that the notion of smoothness here is that it has a "nice" map from a smooth object. For a finite group, this just means that there's a smooth covering space, and then the classifying space is always just the stack obtained by quotienting out a single point by the group, and so it has a smooth cover, because a single point is smooth.

share|cite|improve this answer
For $G$ a Lie group, thinking of $BG$ as the smooth stack $*/G$ makes sense in the category of smooth manifolds as well as the algebraic category. – Jeffrey Giansiracusa Oct 4 '10 at 21:06

One general answer to this is to say: a generalized smooth thing is an oo-stack on some site of smooth test spaces like Diff or so. These are "smooth oo-groupoids" in a useful sense. There is some discussion aimed towards smooth classifying spaces from this perspective at motivation for sheaves, cohomology and higher stacks.

share|cite|improve this answer

I was directed here not for an ordinary BG or K(pi,1) but for K(A,n)

more details on that case?

share|cite|improve this answer
Where did you get sent here from? – jd.r Oct 4 '10 at 22:26
Hi Jim. I think you may be able to do it iteratively, taking K(A,n) = B^nA, and noting that (and this is conjecture on my part) for an abelian Frolicher group G there is a Frolicher abelian group structure on BG. @Josh, it was from the category theory mailing list, by Andrew Stacey. – David Roberts Oct 4 '10 at 23:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.