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14

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 $\hom(\pi,\rho)$.


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 ...


8

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 ...


8

There is never a $K(G, n)$ for $n \geq 2$ which is a finite complex (and $G$ nontrivial). In fact, a finite simply connected 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 ...


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

The title and the body seem to be asking very different questions, so I'll answer them separately. Title question: Not quite. While it is true that every generalized cohomology theory is represented by a spectrum and conversely that every spectrum represents a generalized cohomology theory, maps between spectra are richer than maps between generalized ...


5

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 ...


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 are Chern classes defined on K-theory groups with values in cyclic homology and Hochschild homology, which are the exact analogue of characteristic classes for general algebras —in fact, these can be defined using those. You can find this explained in Loday's book on cyclic homology, in Karoubi's book on the same subject, in Rosenberg's introduction ...


4

First, let me make the weaker claim that $\mathbb{HP}^{\infty}$ is not naturally an H-space. The natural H-space structures on $\mathbb{RP}^{\infty}$ resp. $\mathbb{CP}^{\infty}$ come from the fact that they classify isomorphism classes of real resp. complex line bundles, which naturally have group structures given by taking the tensor product. This no ...


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

$K(G \rtimes H, 1)$ 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

In this answer I'll assume that $\mathbb{Z}_p$ means $\mathbb{Z}/p\mathbb{Z}$. For 2), more generally, let $G$ be a finite group and let $q$ be a prime not dividing $|G|$. Then $H^n(BG, \mathbb{F}_q)$ vanishes (for $n \ge 1$). One of many ways to see this is that under the given hypotheses $\mathbb{F}_q[G]$ is semisimple and so the category of ...


3

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 ...


3

The notion of "vector bundles up to stable equivalence" isn't defined for a general topological group $G$. Stable equivalence for vector bundles is defined in terms of taking direct sums with trivial bundles, but there is no analogous notion for a general topological group. So one side of the question is ill-defined.


3

Real resp. complex $n$-dimensional vector bundles are classified by homotopy classes of maps $M \to BO(n)$ resp. $M \to BU(n)$. Stable real resp. complex vector bundles are classified by homotopy classes (not stable homotopy classes) of maps $M \to BO$ resp. $M \to BU$, where $O$ is the stable orthogonal group and $U$ is the stable unitary group. This ...


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

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

$EG\times EH$ 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

I just consider the case $h=3$, but the argument is completely general. From the classification of closed surfaces, we know that $N_3$ is the connected sum of three projective spaces, so that $N_3$ be the quotient of an hexagon $P$ by identifying its sides as indicated by the following figure: Now let $U, V \subset N_3$ be subspaces illustrated by the ...


2

Given a discrete group $G$, any two $BG$'s are homotopy equivalent to each other. Also, we can simply construct $EG$ to be the universal cover of a $BG$. Also, every $EG$ is constructed in this manner. Also, since $BG$ is a CW complex, so is $EG$. So no, there is not one specific $EG$, because there is not one specific $BG$. However, the $BG$'s do all vary ...


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

No. A Poincaré duality space must in particular have vanishing cohomology above some degree, but a nontrivial finite cyclic group has nonvanishing cohomology in arbitrarily high degrees.



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