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Questions tagged [classifying-spaces]

A classifying space $BG$ of a topological group $G$ is the quotient of a weakly contractible space $EG$ by a free action of $G$. When $G$ is a discrete group $BG$ has homotopy type of $K(G,1)$ and (co)homology groups of $BG$ coincide with group cohomology of $G$.

2 votes
1 answer
56 views

Suppose that $E_0, E_1 \rightarrow M$ are two $k$-dimensional vector bundles over a manifold $M$ classified by maps $\phi_0, \phi_1: M \rightarrow BGL(k)$. If $\phi_0, \phi_1$ are homotopic, then $E_0,...
user39598's user avatar
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0 votes
0 answers
17 views

I am interested in understanding nilpotent Lie groups; the simplest nontrivial one has central extension $$0 \to [N,N] \to N \to \mathbb{R}^m \to 0.$$ I have seen that such extensions (as Lie groups!) ...
Student's user avatar
  • 63
4 votes
1 answer
145 views

The following is a paragraph from Quillen's Appendix Q: On the group completion of a simplicial monoid. Let $M$ be a simplicial monoid and let $\text{Nerv}(M): (p, q) \mapsto (M_q)^p$ be the ...
okabe rintarou's user avatar
1 vote
0 answers
56 views

Given an exact sequence of topological / simplicial groups: \begin{align*} 1\longrightarrow K\longrightarrow G\longrightarrow H\longrightarrow1 \end{align*} we have a fibration of classifying spaces: \...
Junyeong Park's user avatar
0 votes
0 answers
75 views

Let $G$ be a (derived) group scheme. View it as a stack in some fixed topology and define $BG$, the classifying stack, to be the geometric realization of the simplicial diagram $$ \cdots \substack{\...
Chris Kuo's user avatar
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7 votes
2 answers
164 views

Consider the "determinant bundle" construction that takes a $n$-dimensional vector bundle $p\colon X \rightarrow B$ and produces a line bundle so that each fiber $X_b$ becomes the exterior ...
plougue's user avatar
  • 73
0 votes
0 answers
103 views

Let $G$ be a discrete simplicial group. Then its classifying space $(BG)$ can be realized from a simplicial set constructed via the bar construction. In particular, when $G$ is a discrete (ordinary) ...
Student's user avatar
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11 votes
0 answers
463 views

I'd like to solve or find a reference for Exercise 2 (a), Chapter 9 in the book Curvature and Characteristic Classes by Johan L. Dupont. https://mathscinet.ams.org/mathscinet/article?mr=500997 ...
Qing Lan's user avatar
3 votes
2 answers
193 views

Let $R$ be a commutative ring with unity, and let $G$ and $H$ be compact Lie groups with $BG$ and $BH$ as their respective classifying spaces. If there exists isomorphism $H^j(BH; R) \to H^j(BG; R)$ ...
ThePiv's user avatar
  • 63
3 votes
2 answers
250 views

I am trying to understand what the classifying space functor $G \mapsto BG$ does to group homomorphisms $f : H \to G$ for $H, G \in \textbf{TopGrp}$. Is there any suggested reference on the ...
Arnav Das's user avatar
  • 218
1 vote
0 answers
57 views

I am currently reading Topology of fibre bundles by Steenrod in an effort to understand Milnor's exotic sphere paper. In section 19 we are given the classifying theorem for universal bundles that ...
marta fernandez's user avatar
0 votes
0 answers
73 views

By way of example, suppose we have principal bundle of orthonormal frames on a manifold $M$ $$SO(n)\hookrightarrow PM\rightarrow M.$$ Does there exist a construction (I'm aiming at at least a ...
MRU's user avatar
  • 323
0 votes
0 answers
87 views

On the Wikipedia page it is stated that there is no algorithm that classifies manifolds up to homeomorphism because of the "word problem". And therefore one can deduce that classifying ...
Flynn Fehre's user avatar
-1 votes
1 answer
99 views

There is category of groups and there is classification theorem of finite simple groups. There is category of topological spaces and the topological spaces can be classified by topological invariants. ...
TomR's user avatar
  • 1,479
1 vote
0 answers
96 views

To show space $\bigcup_{n\geq 1} \frac{U_{2n}}{U_{n}\times U_{n}}$ has the same homotopy type as $BU = \bigcup_{k\geq 1}BU(k)$, where $BU(k)=\bigcup_{n\geq k} \frac{U_{n}}{U_{k}\times U_{n-k}}$ and $...
Rkb's user avatar
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