27
votes

Accepted

### Torsion in the Atiyah–Hirzebruch spectral sequence of a classifying space

Of course, in any spectral sequence $E_{r+1}$ is a subquotient of $E_r$ (the kernel of $d_r$ divided by the image of $d_r$). And in general new torsion can appear in the sense of torsion elements in $...

20
votes

Accepted

### Sullivan conjecture for compact Lie groups

You were right to single out Lie groups as potentially interesting. In [Topology 5 (1966), 241-243], Brayton Gray showed that the homotopy group of maps $[BS^1, S^3]$ was uncountable. Indeed, he ...

17
votes

Accepted

### Classifying space BG and contractable space EG

The easiest way to construct an explicit contracting homotopy
is to observe that EG is the geometric realization of the nerve of the groupoid G//G,
which has G as its set of objects and exactly one ...

16
votes

Accepted

### quotient space of Eilenberg-MacLane space

Suppose a group $H$, not necessarily finite, acts on an Eilenberg-MacLane space $BN$. The homotopy quotient $BN/H$ (which agrees with the ordinary quotient if the action of $H$ is free) fits into a ...

16
votes

Accepted

### For which G is BLG weak homotopy equivalent to LBG?

[UPDATE: There were some mistakes in the first version. Here is a more careful account.]
I'll work everywhere with CGWH spaces, so I have a Cartesian closed category.
Note that $BLG$ is always path-...

14
votes

Accepted

### Infinite Grassmannian does not have the homotopy type of a finite-dimensional complex

We have $H^*(BO(k); \mathbb{Z}_2) \cong \mathbb{Z}_2[w_1, \dots, w_k]$ where $\deg w_i = i$. In particular, $H^n(BO(k); \mathbb{Z}_2) \neq 0$ for every $n$ as $w_1^n$ is a non-zero element. Therefore $...

14
votes

### If $G$ is a topological group that contains a torsion element, then the classifying space $BG$ is infinite-dimensional?

Here is sort of a canonical example.
Consider $GL(\mathbb H)$ the group of invertible operators on a Hilbert space. By Kuipers theorem it is contractible. But $GL(\mathbb H)$ acts freely and properly ...

13
votes

Accepted

### Equivariant classifying spaces from classifying spaces

Added Aug 2016: I've written this up, available at https://arxiv.org/abs/1608.02999
$\def\Hom{\mathrm{Hom}} \def\Map{\mathrm{Map}} \def\ad{\mathrm{ad}}$
I think this is true. I'll sketch a ...

13
votes

Accepted

### When does $BG \to BA$ loop to a homomorphism?

If $G$ is a compact connected topological group and $A$ is a locally compact abelian topological group, then for any map $f:BG\to BA$ the looped map $\Omega f:\Omega BG\to \Omega BA$ is homotopically ...

13
votes

Accepted

### About the cohomology of $BG^\delta$. Making a Lie group discrete

I will only attempt to answer your first question. The reason there is no contradiction is that it is not true for arbitrary spaces that $H^{\ast}(X;\mathbb Q) = H^{\ast}(X;\mathbb Z) \otimes \mathbb ...

12
votes

Accepted

### classifying space of orthogonal groups

$BO$ is the connected component of the zeroth space of a spectrum called the real K-theory spectrum. This spectrum represents a cohomology theory, namely real K-theory, and this means that $BO$ has ...

12
votes

Accepted

### Oriented Bordism Group and Un-Oriented Bordism Group of points $pt$

Unoriented cobordism: can be read off from the structure of the unoriented cobordism ring (calculated in Thom's thesis): $\Omega_6^O = (\mathbb Z/2)^3$, $\Omega_7^O = \mathbb Z/2$, $\Omega_8^O = (\...

11
votes

### quotient space of Eilenberg-MacLane space

$K(\pi,1)/G$ is not necessarily $K(\pi\times G,1)$ for example take $G=Z$, $K(Z,1)=S^1$. $S^1=R/(t_1=x\rightarrow x+1)$ the quotient of $S^1$ by the group $Z/n$ generated by the transformation induced ...

11
votes

Accepted

### Homotopy type of a specific discrete monoid

This space is contractible, and so all of its homotopy groups are trivial.
Define two elements in $M$ by:
$$
\begin{align*}
A(x) &=
\begin{cases} 2x &\text{if }x \leq 1/2\\1 &\text{if }x \...

11
votes

Accepted

### Map from a classifying space to a stack

You're almost there! The problem is that, as you've surmised, the group $\mathrm{Aut}(x)$ does not capture enough of the geometric structure of $G$. But that's easily solved:
For every $x\in X$ we ...

11
votes

Accepted

### Characteristic classes of non-linear sphere bundles

For many values of $n$, the answer to both questions is no. Since the fundamental groups of $BX$ and $B\mathrm{Diff}(S^n)$ are finite for $n\ge5$ (This uses that $\pi_0\mathrm{Diff}_\partial(D^n)$ is ...

11
votes

Accepted

### Trivial group cohomology induces trivial cohomology of subgroups

For any abelian group $A$ we have a canonical isomorphism $\bigwedge^2A\to H_2(A,\mathbb{Z})$, given by the (anti-symmetric) Pontrjagin product $H_1(A,\mathbb{Z})\times H_1(A,\mathbb{Z}) \to H_2(A,\...

10
votes

### Formality of classifying spaces

This is an old question. But sometimes old questions get answered!
Benson, Greenlees, Formality of cochains on BG
Here is the abstract:
Let $G$ be a compact Lie group with maximal torus $T$. If $|N_G(...

10
votes

### classifying space of orthogonal groups

$BO$ can be defined as the colimit over $(k,n)$ of Grassmanians $G_k(\Bbb R^n)$ of $k$-dimensional linear subspaces of $\Bbb R^n$ (the limit over $n$ is defined by standard inclusions $\Bbb R^n \...

10
votes

Accepted

### Stable homotopy groups of $QX$

The "Snaith splitting" gives the following spectrum level statement: for a pointed connected space $X$, there is a weak equivalence:
$$
\Sigma^\infty_+ (\Omega^\infty \Sigma^\infty X) \simeq \bigvee_{...

10
votes

Accepted

### Classifying space of a non-discrete group and relationship between group homology and topological homology of Lie groups

You may want to look at the classical paper of Jack Milnor, "On the homology of Lie groups made discrete." The Friedlander-Milnor conjecture states that the map $BG^\delta \to BG$ (where $G$ ...

10
votes

### Pullbacks of classifying spaces

This is not true. Note that $BG$ is only well-defined up to homotopy equivalence, so the only question that makes sense is when the square
$$\begin{array}{ccc} B\big(G \underset H\times H'\big) & \...

9
votes

### Homotopy Type of the Based Mapping Space $Map_*^{(k,l)}(\mathbb{C}P^2,BU(2))$

[This version has been updated in response to comments from the OP]
Recall that $B$ gives an endofunctor of the category of abelian topological groups. We can apply $B$ to the obvious map $\mathbb{Z}...

9
votes

Accepted

### Source request for $H^*(B\mathrm{TOP},\mathbb{Q}) \cong H^*(BO,\mathbb{Q})$

The canonical map $BPL\rightarrow BTOP$ is a rational homotopy equivalence by works of Thom, Novikov, Kirby-Siebenmann and others. In fact the homotopy fiber $TOP/PL$ is a $K(\mathbb{Z}/2,3)$. A nice ...

9
votes

Accepted

### classifying maps of Whitney sums of vector bundles

$BO(n)$ is the infinite-dimensional Grassmannian $Gr(n,\infty)$ of $n$-planes in ${\mathbf R}^\infty$. There is a natural direct sum operation
$$\oplus\colon Gr(n,\infty)\times Gr(m,\infty)\to Gr(n+m,\...

9
votes

Accepted

### Interesting properties in $...\to K(\mathbb{Z}_4,1) \overset{f}{\to} K(\mathbb{Z}_2,1)\overset{g}{\to}K(\mathbb{Z}_2,2) \to ...$

Represent $p$ by the identity map $id: \mathbb{Z}_2 \to \mathbb{Z}_2$.
Then $(p\cup p)(a,b) = p(a)p(b)$ is non-zero only on the 2-chain $(1,1)$. Namely, as a polynomial mod 2, $(p\cup p)(a,b) = ab$. ...

8
votes

### Classifying space as the geometric realization of the nerve of $G$ viewed as a small category

You should ignore simplicial objects at first, and just consider groupoids. In the following, you can let $G$ be a topological group such that $e\hookrightarrow G$ is a closed cofibration. All ...

8
votes

### Cohomology of $BE_8$ and $BSU(2)$

I believe Appendix 1. in ``Finite H-spaces and Lie Groups" by Frank Adams
shows that BE8 has 2,3 and 5-torsion. The letter from E8 at the end of this paper is also quite amusing:
....Be it therefore ...

8
votes

### Classification of fibrations for classifying spaces $B^2\mathbb{Z}_2$ and $BSO(3)$ or $BO(3)$

Firstly, yes, your examples are all correct. However, in example~(B) we just have $O(3)=\{\pm I\}\times SO(3)$ as groups, so your fibration is just the product of
$$BSO(3)\xrightarrow{1}BSO(3)\to 1$$...

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