bio  website  homepages.abdn.ac.uk/… 

location  Aberdeen, United Kingdom  
age  
visits  member for  5 years, 1 month 
seen  15 mins ago  
stats  profile views  4,680 
Lecturer at University of Aberdeen, with interests in Algebraic and Differential Topology and their applications.
5h

revised 
bar construction and loop space cohomology
Added extra paragraph of explanation 
23h

comment 
cohomology ring of basepointpreserving maps on the 3sphere
You seem to have a lot of detailed questions in this general area. Do you have an adviser you could ask? 
1d

answered  Steenrod operations on cohomology of grassmannians 
1d

answered  bar construction and loop space cohomology 
1d

comment 
unordered configuration space over spheres and Euclidean spaces
Is it possible you meant to ask for relations between $B(\mathbb{R}^{n+1},k)$ and $B(S^n,k)$? 
Aug
25 
comment 
$G$CW complex structure of universal a $\mathcal{F}$space
Ah, OK. Maybe it would be better to ask a separate question about Bredon cohomology of joins. Before doing so, you could think about how this relates to the question mathoverflow.net/questions/211122/… given that $Y\ast Z\simeq \Sigma Y\wedge Z$. 
Aug
24 
comment 
$G$CW complex structure of certain Gspace
The join of discrete sets is naturally a simplicial complex, hence a CW complex. The $G$action on $G/H$ induces a diagonal $G$action on $X$, under which it becomes a $G$CW complex. 
Aug
21 
comment 
$G$CW complex structure of universal a $\mathcal{F}$space
The action of $G$ on your space $X$ is not free (its isotropy groups are all conjugates of $H$) so I don't see why the Bredon cohomology should reduce to the ordinary cohomology of the quotient. 
Aug
21 
answered  $G$CW complex structure of universal a $\mathcal{F}$space 
Aug
20 
comment 
When is a circle fibration a circle bundle?
The paper 131.220.77.52/lueck/data/… could be relevant here. The authors define primary and secondary obstructions for a fibration to be homotopy equivalent to a fibre bundle. These obstructions vanish here, I think, as $Wh(\mathbb{Z})=0$ and any homotopy equivalence of a circle is homotopic to a homeomorphism. But this doesn't quite answer your question (and anyway might be overkill). 
Aug
20 
comment 
Cohomology of $G_3(\mathbb{R}^5)$
Thank you for the reference, it told me everything I needed to know. By the way, there is a second author, Yuji Kodama. 
Aug
20 
accepted  Cohomology of $G_3(\mathbb{R}^5)$ 
Aug
19 
comment 
Can the KanThurston theorem be turned into some kind of equivalence between groups and spaces?
Groups up to isomorphism are the same thing as pathconnected aspherical complexes (i.e. $K(G,1)$'s) up to homotopy equivalence. This is much more elementary than the KanThurston theorem, however. 
Aug
16 
asked  Cohomology of $G_3(\mathbb{R}^5)$ 
Aug
10 
comment 
When is the Thom class the Poincare dual of the zero section?
If you represent cohomology classes by proper maps from manifolds (e.g. as in Quillen's "Elementary proofs of some results in cobordism theory using Steenrod operations") then there is a certain sense in which this is always true. 
Aug
3 
comment 
Manifold approximations to $BO(3)$
@user51223: In response to your first comment, because we can view $BO(2)$ as the total space of the Borel fibration $BSO(2)\times_{\mathbb{Z}/2} S^\infty$. 
Aug
2 
comment 
Manifold approximations to $BO(3)$
@user51223: I want something I can at least compute the integral cohomology of. Following Allen's suggestion, $G_3(\mathbb{R}^6)$ might be in this category (although I've not found an easy reference for that yet). 
Aug
2 
awarded  Yearling 
Jul
31 
comment 
Manifold approximations to $BO(3)$
@AllenHatcher: That's a good point. I guess I was hoping for something similar to the Dold manifolds, using some nice approximations to $BSO(3)$. Perhaps in general one has to look at finite Grassmannians, though. I may delete the question. 
Jul
31 
revised 
Manifold approximations to $BO(3)$
added closed condition 