# Tag Info

13

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 ...

12

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 ...

7

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 ...

5

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 ...

4

One can form $BM$ for any $A_\infty$-space and it's more or less the delooping ($A_\infty$-structure on a connected $H$-space $M$ is more or less the same thing as an equivalence $M\cong \Omega X$ for some $X$; the proof is more or less that $M\cong\Omega BM$). (AFAIR this can be generalized further but then there is a question of what properties do you ...

4

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

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 ...

3

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

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

As for $\mathcal Q$, over $[\Lambda]\in G(2,4)$ there lies $V/\Lambda$. Let $$\mathcal L\subset \mathbb G(1,3)\times\mathbb P^3\overset{\pi}{\longrightarrow}\mathbb G(1,3)$$ be the universal line. This means that $\pi^{-1}([\ell])=\ell$. Here, to give $\ell$ is to give $\mathbb P(\Lambda^\vee)$, exactly as to give $\mathbb P^3$ is to give $\mathbb ... 2 If we can construct a space, which we will call$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}$. Their are other models as well. 2 See Andy Putman's answer to this math overflow question. This at least gives you good references. To answer your question about torsion, almost all values of$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 ... 2 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 ... 2 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 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

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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