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 $...
- 53.1k
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 ...
- 10.5k
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 ...
- 35.5k
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 ...
- 114k
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-...
- 54.5k
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 $...
- 18.5k
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 ...
- 7,185
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 ...
- 26.5k
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 ...
- 34.8k
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 ...
- 11.8k
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 ...
- 114k
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 = (\...
- 6,716
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 ...
- 3,644
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:
$$
\begin{align*}
A(x) &=
\begin{cases} 2x &\text{if }x \leq 1/2\\1 &\text{if }x \...
- 50.3k
11
votes
Accepted
Map from a classifying space to a stack
You're almost there! The problem is that, as you've surmised, the group $\mathrm{Aut}(x)$ does not capture enough of the geometric structure of $G$. But that's easily solved:
For every $x\in X$ we ...
- 16.1k
11
votes
Accepted
Characteristic classes of non-linear sphere bundles
For many values of $n$, the answer to both questions is no. Since the fundamental groups of $BX$ and $B\mathrm{Diff}(S^n)$ are finite for $n\ge5$ (This uses that $\pi_0\mathrm{Diff}_\partial(D^n)$ is ...
- 2,954
11
votes
Accepted
Trivial group cohomology induces trivial cohomology of subgroups
For any abelian group $A$ we have a canonical isomorphism $\bigwedge^2A\to H_2(A,\mathbb{Z})$, given by the (anti-symmetric) Pontrjagin product $H_1(A,\mathbb{Z})\times H_1(A,\mathbb{Z}) \to H_2(A,\...
- 9,361
10
votes
Formality of classifying spaces
This is an old question. But sometimes old questions get answered!
Benson, Greenlees, Formality of cochains on BG
Here is the abstract:
Let $G$ be a compact Lie group with maximal torus $T$. If $|N_G(...
- 5,827
10
votes
classifying space of orthogonal groups
$BO$ can be defined as the colimit over $(k,n)$ of Grassmanians $G_k(\Bbb R^n)$ of $k$-dimensional linear subspaces of $\Bbb R^n$ (the limit over $n$ is defined by standard inclusions $\Bbb R^n \...
- 18.5k
10
votes
Accepted
Stable homotopy groups of $QX$
The "Snaith splitting" gives the following spectrum level statement: for a pointed connected space $X$, there is a weak equivalence:
$$
\Sigma^\infty_+ (\Omega^\infty \Sigma^\infty X) \simeq \bigvee_{...
- 50.3k
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 $G$ ...
- 9,361
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
$$\begin{array}{ccc} B\big(G \underset H\times H'\big) & \...
- 26.3k
9
votes
Homotopy Type of the Based Mapping Space $Map_*^{(k,l)}(\mathbb{C}P^2,BU(2))$
[This version has been updated in response to comments from the OP]
Recall that $B$ gives an endofunctor of the category of abelian topological groups. We can apply $B$ to the obvious map $\mathbb{Z}...
- 54.5k
9
votes
Accepted
Source request for $H^*(B\mathrm{TOP},\mathbb{Q}) \cong H^*(BO,\mathbb{Q})$
The canonical map $BPL\rightarrow BTOP$ is a rational homotopy equivalence by works of Thom, Novikov, Kirby-Siebenmann and others. In fact the homotopy fiber $TOP/PL$ is a $K(\mathbb{Z}/2,3)$. A nice ...
- 9,732
9
votes
Accepted
classifying maps of Whitney sums of vector bundles
$BO(n)$ is the infinite-dimensional Grassmannian $Gr(n,\infty)$ of $n$-planes in ${\mathbf R}^\infty$. There is a natural direct sum operation
$$\oplus\colon Gr(n,\infty)\times Gr(m,\infty)\to Gr(n+m,\...
- 10.2k
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, $(p\cup p)(a,b) = ab$. ...
- 4,024
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 ...
- 33.1k
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 ...
- 5,241
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
$$BSO(3)\xrightarrow{1}BSO(3)\to 1$$...
- 54.5k
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