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Hot answers tagged classifying-spaces

13

What functor does $K(G, 1)$ represent for nonabelian $G$?

K(G,1) aka BG classifies G-bundles — i.e. G-coverings, if G is discrete. (Details can be found e.g. in May's Concise Course in Algebraic Topology.) Usual definition of Cech cohomology works for $H^1(X;G)$ even in non-abelian case (but it's just the usual cocycle definition of G-bundle). As for universal coefficient theorem, even if $H_1(X;\mathbb Z)$ is ...

9

Why is the cohomology of a $K(G,1)$ group cohomology?

Akhil, you're thinking of this the opposite of how I think group cohomology was discovered. The concept of group cohomology originally centered around the questions about the (co)homology of $K(\pi,1)$-spaces, by people like Hopf (he called them aspherical rather than $K(\pi,1)$ spaces, and Hopf preferred homology to cohomology at that point). I think the ...

5

Finite dimensional Eilenberg-Maclane spaces

There is never a $K(G, n)$ for $n \geq 2$ which is a finite complex (and $G$ nontrivial). In fact, a finite complex has finitely generated homotopy groups in all dimensions (by Serre's mod $\mathcal{C}$ theory applied to the universal cover). So one reduces to seeing that a $K(\mathbb{Z}, n)$ or a $K(\mathbb{Z}/p^k, n)$ cannot be a finite complex. In both ...

5

Group structure on Eilenberg-MacLane spaces

The standard classifying space functor $B$ from topological groups to topological spaces is product preserving, so it takes abelian topological groups to abelian topological groups. Start with an abelian group $G$ as a discrete topological group, so a $K(G,0)$. Apply the functor $B$ iteratively $n$ times to reach $B^nG$, which is an abelian topological ...

3

Group structure on Eilenberg-MacLane spaces

Well, an interval is a $K(\{e\},1)$, where $\{e\}$ is the trivial group, and an interval has no group structure that can make it a topological group (every continuous map has a fixed point). So there certainly are some $K(G,n)$'s out there that cannot be made into topological groups. Maybe if you refine the question there could be some sort of answer.

3

Why is the cohomology of a $K(G,1)$ group cohomology?

Well, you'll probably want a more conceptual proof, but one thing you can do is check they are computed by the same chain complex: for $K(G,1)$ take the simplicial construction of the classifying space $BG$ and compute its cohomology in the usual way for simplicial sets (using the dual to the complex of formal linear combinations of simplices); for the group ...

3

Why is the cohomology of a $K(G,1)$ group cohomology?

This is really a comment on ryan'sanswer: I have to disagree with Ryan. Group cohomology was in its early stages before Eilenberg and Maclane came along. There are awful and ugly formulations of just $H^1$ and $H^2$ that lead me to believe that they must have been formulated before E&M did their work. I am thinking of factor sets and cocycle conditions ...

2

Every principal $G$-bundle over a surface is trivial if $G$ is compact and simply connected: reference?

The homotopy theoretic proof is as follows: Let $E \longrightarrow \Sigma$ be a principal $G$-bundle over a surface $\Sigma$. Such a bundle is determined by a homotopy class $[f_E] \in [\Sigma, BG]$ by classifying space theory. Since $G$ is simply connected (and presumably connected), the classifying space $BG$ is $2$-connected (i.e. connected, simply ...

2

Classifying space of the reals

If we can construct a space, which we will call $EG$ such that our group acts transitively and freely, then we may set $BG=EG/G$. But if we have $\mathbb{R}$ act on $\mathbb{ER}=\mathbb{R}^2$ by $r∗(x,y)=(x+r,y)$, this will do the trick. Thus we get that $B\mathbb{R}=\mathbb{R}$. Their are other models as well.

2

Is there a classifying space for covering maps?

The answer to both your questions is yes, and Qiaochu gave the basic idea. The base space is $BS_n$ and the fiber is $ES_n$. You can make this concrete (very analogous to Grassmannians) by using the model $BS_n \equiv C_n(\mathbb R^\infty) / S_n$ and $ES_n = C_n(\mathbb R^\infty)$ where $C_n$ indicates the configuration space of $n$ labelled points in ...

1

Why is the group of covering transformations relative to the quotient map isomorphic to a subgroup of the Fundamental Group?

By the end of step 2, you know that $(\tilde{X},q)$ is the universal covering of $Y$, so there is a map $\Psi : \Pi_1(Y,y_0) \to CoveringMaps(\tilde{X},q)$ defined just like $\Phi$ : For any loop $\gamma : [0;1] \to Y$, $\Psi(\tilde{\gamma})(x_0)$ is the endpoint of the lift of $\gamma$ starting from $x_0$. Furthermore, $\Psi$ is a group isomorphism between ...

1

Homotopy-Fibre Sequence of Classifying Spaces

Assume you have the exact sequence $$1\rightarrow H\rightarrow G\rightarrow G/H\rightarrow 1$$Then it induces a fibration $$BH\rightarrow BG\rightarrow B(G/H)$$ as we imagine some large enough total space $EG$ whose quotient by $G$ is $BG$, by $H$ is $BH$, etc. Now assume we know $B(G/H)$ and $B(H)$ but do not know $BG$, then we need certain invariants to ...

1

How to correct a wrong proof about the Birman exact sequence?

See Andy Putman's answer to this math overflow question. This at least gives you good references. To answer your question about torsion, almost all values of $s,g$, and $r$ will have torsion in the mapping class group, and moduli space will only be a rational classifying space. Just build a symmetric looking surface with those values and isometries of that ...

1

Cohomology of $\Bbb CP^{\infty}=BU_1, BU_2,\dots$ : A reference request

Which class are you taking? This material is not easy. You should ask your professor to ask for a proof or some hints. There is a "simple" proof not using spectral sequences at here and is quite readable. Notice there is an obvious mistake in the proof. I hope David Speyer or someone else can give an answer on the spectral sequence part(which I do not ...

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