Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am looking for a list of classifying spaces $BG$ of groups $G$ (discrete and/or topological) along with associated covers $EG$; there does not seem to be such cataloging on the web. Or if not a list, just some further fundamental examples. For instance, here are the ones I have off the top of my head:

$B\mathbb{Z}_n=L_n^\infty$ with cover $S^\infty$ ($B\mathbb{Z}_2=\mathbb{R}P^\infty$)

$B\mathbb{Z}=S^1$ with cover $\mathbb{R}$

$BS^1=\mathbb{C}P^\infty$ with cover $S^\infty$

$B(F_2)=S^1\vee S^1$ with cover $\mathcal{T}$ (infinite fractal tree)

$BO(n)=BGL_n(\mathbb{R})=G_n(\mathbb{R}^\infty)$ with cover $V_n(\mathbb{R}^\infty)$

$B\mathbb{R}=\lbrace pt.\rbrace$ with cover $\mathbb{R}$

$B\langle a_1,b_1,\ldots,a_g,b_g\;|\;\prod_{i=1}^g[a_i,b_i]\rangle=M_g$ with cover $\mathcal{H}$ (hyperbolic plane tiled by $4g$-sided polygon)

And of course, $B(G_1\times G_2)=BG_1\times BG_2$, so I do not care that much about ''decomposable'' groups.

**The "associated cover" is the [weakly] contractible total space.

[Edit] I should make the comment that $BG$ will be different from $BG_\delta$, where $G_\delta$ denotes the topological group with discrete topology. For instance, the homology of $B\mathbb{R}_\delta$ has uncountable rank in all degrees (learned from a comment of Thurston).

share|improve this question
4  
What do you mean by "associated cover"? For example, $\mathbb{CP}^\infty$ is already simply connected and perhaps the use of the word "cover" is a bit misleading. What you are looking for is the total space $EG$ associated to $G$. Towards that end, apart from Milnor's original construction it is worth looking at Segal's construction of $EG$ as well. It's very nice and in particular implies that $EG$ (in a suitable model) is an abelian group if $G$ is abelian! –  Somnath Basu Feb 23 '11 at 9:03
    
Dave Benson and Steve Smith wrote a book on the clasifying spaces of the sporadic simple groups. –  Steve D Dec 31 '11 at 19:36
    
Why isn't this Community Wiki? –  Ryan Budney Dec 31 '11 at 20:13

9 Answers 9

For the symmetric group $\Sigma_n$, you can take \begin{align*} E\Sigma_n &= \{\text{injective functions } \{1,\dotsc,n\}\to\mathbb{R}^\infty \} \\\\ B\Sigma_n &= \{\text{subsets of size $n$ in } \mathbb{R}^\infty \} \end{align*}

Now let $G_n$ be the group of braids on $n$ strings, and let $H_n$ be the subgroup of pure braids. We have \begin{align*} BH_n &= \{\text{injective functions } \{1,\dotsc,n\}\to\mathbb{R}^2 \} \\\\ BG_n &= \{\text{subsets of size $n$ in } \mathbb{R}^2 \} \end{align*} These spaces have trivial homotopy groups $\pi_{k}(X)$ for $k\\geq 2$, so $$ EH_n=EG_n= \text{ universal cover of } BH_n = \text{ universal cover of } EH_n. $$ I think I see a proof that this space is homeomorphic to $\mathbb{R}^{2n}$, but I don't know if that is in the literature.

share|improve this answer
1  
By writing the elements of $BH_n$ as $n$-tuples of distinct points of $R^2$, so in particular, elements of $\mathbb{R}^{2n}$, you can view it as the complement of a subspace arrangement. &c. –  Mariano Suárez-Alvarez Feb 23 '11 at 18:24
3  
To add to Mariano's comment, there is a nice story for a kind of generalized braid group associated to a Weyl group (in the same way that Artin's braid group is associated to the symmetric group). Again, the classifying space can be represented as a quotient of a hyperplane complement. See papers of Brieskorn-Saito and Deligne from the early 70s. –  Chris Brav Feb 24 '11 at 7:47
    
Similarly to Neil's $B\Sigma_n$ model, you can view $BA_n$ as the collection of all subsets of $\mathbb R^\infty$ which span an affine $(n-1)$-dimensional subspace, and where the subspace is equipped with an orientation. This is a model Dev Sinha mentions in some of his talks. –  Ryan Budney Mar 6 '12 at 23:02

If $G$ is linear topological group, then a model for $EG$ may be taken to be an infinite Stiefel manifold. More precisely, if there is an faithful representation of $G$ into $GL_n(\mathbb{C})$, that gives a free action of $G$ on the space $V_n$ of $n$-frames in $\mathbb{C}^\infty$. $V_n$ is contractible, so $BG$ may be taken to be the quotient $BG = V_n / G$. If it's helpful to think of it this way, this is a fibre bundle over $BGL_n(\mathbb{C}) = G_n(\mathbb{C}^\infty)$, with fibre $GL_n(\mathbb{C}) / G$.

Another favorite example comes from spaces of embeddings: if $M$ is a compact manifold without boundary, then it is a consequence of Whitney's embedding theorem that $Emb(M, \mathbb{R}^\infty)$ is contractible. The group $G=Diff(M)$ of diffeomorphisms of $M$ acts freely on $Emb(M,\mathbb{R}^\infty)$ by precomposing an embedding with a diffeomorphism. Therefore a model for $BG$ is the quotient $Emb(M, \mathbb{R}^\infty) / Diff(M)$, which is often thought of as the space of subspaces of $\mathbb{R}^\infty$ diffeomorphic to $M$. One can combine this idea with the previous idea for subgroups of diffeomorphism groups.

Lastly, if there is a homomorphism $G \to H$ which is a homotopy equivalence, then of course there is a homotopy equivalence $BG \to BH$. So in the previous examples, one can for instance replace $GL_n(\mathbb{C})$ with $U(n)$, and $Diff^{+}(\Sigma)$ with the mapping class group $\Gamma(\Sigma) = \pi_0(Diff^{+}(\Sigma))$ for closed surfaces $\Sigma$.

share|improve this answer

If $P \to X$ is a $G$-principal bundle, then the space $map_G (P;EG)$ is contractible. Let $map_P(X;BG)$ be the space of all maps $f$ with $f^{\ast} EG \cong P$. There is an obvious map $map_G (P;EG) \to map_P (X;BG)$, a universal bundle for the gauge group $Aut(P)$ (automorphism group of $P$).

share|improve this answer

As Craig Westerland points out, the space $V_n$ of full-rank $\infty\times n$ complex matrices (aka the infinite Stiefel manifold) works for any subgroup of $GL_n({\Bbb C})$. That applies to all the examples listed in the question, except the free group -- but of course, it's nice to have alternative descriptions, such as the ones given above, where it's easier to get a handle on $BG$. (A compendium of these things would be nice to have.) Here are a few more:

(1) For $G$ upper triangular matrices in $GL_n({\Bbb C})$, $EG=V_n$ gives $BG=Fl(1,\ldots,n;{\Bbb C}^\infty)$.

(2) For $G=({\Bbb C}^\ast)^n$, you can take $EG=(V_1)^n$ and get $BG=({\Bbb CP}^\infty)^n$. Or you could take $EG=V_n$ and get $BG$ as the space of $n$-dimensional subspaces of ${\Bbb C}^\infty$, together with a splitting into lines. The latter has the advantage of coming with an obvious map to the space from (1), realizing the homotopy equivalence explicitly. (I mention this to point out that it can be useful to have different choices available, even for "decomposable" groups.)

(3) For $G=Sp(2n)$ (the compact symplectic group), take $EG$ to be "full-rank" $\infty\times n$ quaternionic matrices, and get $BG = Gr({\Bbb H^n},{\Bbb H}^\infty)$. (It has the same cell structure as the complex Grassmannian, but with cells in dimensions $4k$ instead of $2k$.)

(4) For $G=Sp_{2n}({\Bbb C})$, you can take $BG=Sp_{\infty}/(Sp_{2n}\times Sp_{\infty})$ (interpreted as a suitable limit).

share|improve this answer

Let $G$ be the group of all invertible operators on a Hilbert space $H$ that are of the form $1+K$, $K$ compact. Then the space of all invertible operators $GL(H)$ is a model for $EG$ and $BG$ is the identity component of the space of all units of the Calkin algebra $B(H)/K(H)$ (bounded modulo compact operators). By the way, $BG \simeq BU$, $G \simeq U$.

share|improve this answer
    
You assume H infinite dimensional, right? –  Jan Weidner Mar 7 '12 at 8:43
    
Yes, $H$ is infinite dimensional. –  Johannes Ebert Mar 7 '12 at 19:40

If $G$ is a topological group, let $\Omega(G)$ be the group of loops in $G$ based at $e$, and $PG$ the space of paths in $G$ starting from $e$. Then $\Omega(G)$ acts on $PG$ freely and $PG$ is contractible, so $G$ is a classifying space for $\Omega(G)$.

share|improve this answer

There are some finite-dimensional variants of Craig's example.

In the world of diffeomorphism groups of manifolds, Smale's theorem is that the group of diffeomorphisms of the 2-disc which restrict to the identity on the boundary circle, this is a contractible space. Denote this group by $Diff(D^2)$. There are fibre-bundles:

$$ Diff(D^2 \text{ fix } P_n) \to Diff(D^2) \to C_n(D^2)$$

where $C_n(D^2)$ is the configuration space of $n$ distinct points in the interior of $D^2$, and $P_n \subset D^2$ is an $n$-point subspace of $D^2$. This is how one sees the configuration space of $n$ distinct points in the disc as the classifying space of the pure braid group, which is $\pi_0 Diff(D^2 \text{ fix } P_n)$. Similarly,

$$ Diff(D^2, P_n) \to Diff(D^2) \to C_n(D^2)/\Sigma_n$$

is a fibre bundle, giving the unordered configuration space as the classifying space of the full braid group.

There are similar results in dimension $3$, where instead of Smale's theorem, you use Hatcher's, that $Diff(D^3)$ is contractible. So you can do similar things for configuration spaces, but in this case $Diff(D^3, P_n)$ is not homotopy-discrete if $n >1$. You can also make similar constructions for knots. Given a smooth embedding $f : [-1,1] \to D^3$ which agrees with the standard inclusion $x \longmapsto (x,0,0)$ on the boundary, you get fibre-bundles

$$Diff(D^3 \text{ fix } f([-1,1])) \to Diff(D^3) \to \mathcal{K}_{3,1}(f)$$

here $\mathcal K_{3,1}$ is the space of all knots in $D^3$, and $\mathcal K_{3,1}(f)$ is the path-component of $f$ in $\mathcal K_{3,1}$.

In this case the fiber is a homotopy-discrete group, and the groups can be fairly elaborate. In general, they're all finitely covered by groups that are products of pure braid groups, but the covers can be arbitrarily large. These groups have a somewhat similar feel to subgroups of the braid group that all preserve a common reduction curve system (in the sense of Thurston's classification of surface automorphisms).

share|improve this answer

You might see Milnor's Construction of Universal Bundles, I think.

share|improve this answer

Extending Johannes' answer a little further: Let $A$ be a separable unital $C^*$-algebra and denote by $\mathbb{K}$ the compact operators on a separable infinite dimensional Hilbert space. Let $M(A \otimes \mathbb{K})$ be the multiplier algebra of the stabilization $A \otimes \mathbb{K}$ of $A$.

By a theorem of Mingo the unitary group $U(M(A \otimes \mathbb{K}))$ of this multiplier algebra is contractible when equipped with the norm topology. In this topology it is also a Banach Lie group with Lie algebra the real Banach space of skew-adjoint operators in $M(A \otimes \mathbb{K})$. Let $e_1$ be a rank 1 projection in $\mathbb{K}$ and let $p = 1 \otimes e_1$. Then we have $$ p\; M(A \otimes \mathbb{K})\; p \cong p\;(A \otimes \mathbb{K})\;p \cong A $$ and $U(pM(A \otimes \mathbb{K})p)$ is a Banach Lie subgroup of $U(M(A \otimes \mathbb{K}))$. Using the exponential map (which exists and is not that bad for Banach Lie groups), we get that $U(M(A \otimes \mathbb{K})) / U(pM(A \otimes \mathbb{K})p)$ allows a local section. Therefore $$ U(M(A \otimes \mathbb{K})) \to U(M(A \otimes \mathbb{K})) / U(pM(A \otimes \mathbb{K})p) $$ is a model for $EG \to BG$ for $G = U(A)$, the unitary group of the $C^*$-algebra.

share|improve this answer

Your Answer

 
discard

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.