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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 ...


13

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 ...


9

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 ...


9

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 ...


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 ...


7

// 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 ...


7

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 ...


6

Let $G$ be a discrete group (i.e. a group equipped with the discrete topology), $BG$ its classifying space, and $EG$ the universal $G$-bundle. Then we have a fibration $$G \hookrightarrow EG \rightarrow BG,$$ and the long exact sequence of this fibration reads $$\cdots \to \pi_{n+1}(BG) \to \pi_n(G) \to \pi_n(EG) \to \pi_n(BG) \to \cdots \to \pi_0(G) \to ...


6

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 ...


6

The functor $B : \mathbf{TopGrp} \to \operatorname{Ho} \mathbf{Top}_*$ is not fully faithful. Indeed, it does not even preserve non-isomorphy: the multiplicative group $\mathbb{R}^\times$ and the discrete group $\mathbb{Z} / 2 \mathbb{Z}$ have the same classifying space (namely, $\mathbb{R P}^\infty$), so if the functor were fully faithful, ...


5

The correct and invariant statement is that the classifying space functor is an equivalence of $(\infty, 1)$-categories between grouplike $E_1$ spaces (this is a homotopically invariant version of "topological group") and pointed connected spaces (and every time I say "spaces" I mean "weak homotopy types"). I believe it's also true that every grouplike $E_1$ ...


5

For a very simple example, consider the poset $\{a,b,c,d,e,f\}$ with $a,b\leq c,d\leq e,f$. The classifying space of this poset is homeomorphic to $S^2$ (you can explicitly list out all the nondegenerate simplices in its nerve and draw a picture of them), which has plenty of higher homotopy groups. More generally, in fact, every simplicial complex is ...


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 ...


5

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 ...


5

I'll try to fill in some of the missing details, hopefully this will be enough. Let the unoriented Grassmanian be $X = \widetilde{\mathrm{Gr}}(k, \mathbb{R}^n) \cong SO(n) / (SO(k) \times SO(n-k))$. Assume $0 < k < n$ (otherwise there's not much to prove). There is thus a fiber bundle $SO(n) \to X$, with fiber $SO(k) \times SO(n-k)$. Since $SO(n)$ is ...


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

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

Let $M$ be a topological monoid. $M$ can be considered as a category internal to topological spaces and has a simplicial space $N_\bullet(M)$ as its nerve. (It's also called the internal nerve.) The geometric realization $BM=|N_\bullet(M)|$ of this simplicial space is the classifying space $BM$. (This includes the discrete case.) Unveiling the definitions, ...


4

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 ...


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, $B(S^1)^d\cong\mathbb(CP^\infty)^d$).


3

I think I have at least a sketch of an argument. Recall, every manifold admits a locally finite good open cover $\mathcal{U}$. Let us write $\mathcal{U}$ also for the poset category of nonempty finite intersections generated by elements of $\mathcal{U}$ ordered by inclusion. The nerve theorem in this case should apply and we deduce that ...


3

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 ...


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

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

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

The problem is that it is not true that the canonical map $i:\mathbb{Z}[[x]]\otimes\mathbb{Q}\to\mathbb{Q}[[x]]$ is an isomorphism. Any power series of the image of $i$ must have only finitely many different denominators on its coefficients, since an element of $\mathbb{Z}[[x]]\otimes\mathbb{Q}$ is a sum involving only finitely many elements of ...


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

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 ...



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