12,383 reputation
22669
bio website homepages.abdn.ac.uk/…
location Aberdeen, United Kingdom
age
visits member for 5 years, 1 month
seen 15 mins ago

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 base-point-preserving maps on the 3-sphere
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 G-space
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 Kan-Thurston theorem be turned into some kind of equivalence between groups and spaces?
Groups up to isomorphism are the same thing as path-connected aspherical complexes (i.e. $K(G,1)$'s) up to homotopy equivalence. This is much more elementary than the Kan-Thurston 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