# Tag Info

## Hot answers tagged classifying-spaces

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 ...

10

Recall that $[X,K(G,n)]=H^n(X;G)$. Hence $[K(\pi,n),K(\rho,n)]=H^n(K(\pi,n);\rho)$ — which (by Hurewicz theorem + universal coefficients) is exactly $\hom(\pi,\rho)$.

8

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 ...

6

Observe that any map $\mathbb{CP}^{n} \rightarrow \mathbb{CP}^{\infty}$ can be pushed down to the $2n$-skeleton of $\mathbb{CP}^{\infty}$, due to cellular approximation. This establishes a bijection between homotopy classes of maps $[\mathbb{CP}^{n}, \mathbb{CP}^{\infty}] \simeq [\mathbb{CP}^{n}, \mathbb{CP}^{m}]$ for all $m \geq n$. The left hand side ...

5

// This is essentially the same answer that Qiaochu Yuan deleted for some reason. Hence CW. By the Dold-Thom theorem $\pi_\bullet(\mathbb Z[X])=H_\bullet(X)$, where $\mathbb Z[X]$ is the free abelian group generated by $X$. Note that $\mathbb Z[X]$ is never connected ($\pi_0(\mathbb Z[X])=H_0(X)\supset\mathbb Z$) but it has reduced version: (fix a point ...

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

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 ...

4

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 ...

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 ...

3

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 ...

3

$K(G \rtimes H, 1)$ fits into a fibration sequence $$K(G, 1) \to K(G \rtimes H, 1) \to K(H, 1).$$ So for example one can access the homology and cohomology using the Serre spectral sequence. See this answer for some context.

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

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

$EG\times EH$ is a contractible space with free $G\times H$-action. The quotient by this action is $BG\times BH$. So $B(G\times H)\cong BG\times BH$ (in particular, $B(S^1)^d\cong\mathbb(CP^\infty)^d$).

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 ...

2

No. If unit octonions were $A_\infty$-space, $BS^7$ («$OP^\infty$») would be a space with cohomology ring $\mathbb Z[x]$, $\deg x=8$ (this follows from LHSS for Serre fibration $\text{pt}\to\Omega BS^7\cong S^7\to BS^7$) — which is impossible (see e.g. Corollary 4L.10 if Hatcher's «Algebraic topology»). P.S. In fact, there are no even just homotopy ...

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 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 The difference between these definitions boils down to two different descriptions of the category of simplices, and both yield the same$BG$. The first description of$BG$"thinks about" simplices as topological simplices, especially with language like "linear embedding". For the notation$\alpha:[m]\to [n]$in the second description, see simplex category. ... 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 ... 1 Turns out most models do produce a confidence or a score as well as a classification. I had trouble finding this in MATLAB but it does produce confidence values. This value is continuous and can be re-ordered to produce a lift chart 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 In general for a group$G$, we have$\Omega BG \simeq G$. Thus,$BG= K(H,n)$if and only if$G = K(H, n-1)$. That is to say, a necessary and sufficent condition is that$G\$ itself is an Eilenberg-Maclane space.

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