Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Homotopy theory, homological algebra, algebraic treatments of manifolds.
2
votes
Infinite loop spaces
If I am not mistaken, we get a counter-example to your last question easily.
First of all we have $\Omega Q\Sigma A=QA$ so we only need to find a map from
$QA$ to $QB$ that is not a $H$-map to get a …
4
votes
Accepted
The Thom map for the Brown-Peterson cohomology
Here is an "answer" which may be or not be good enough for your purpose, but which is easy to prove.
Let's start with Ravenel-Wilson-Yagita Theorem 1.20. Applied to Eilenberg-Maclane spaces, it impli …
5
votes
Accepted
Cohomology algebra of a fibration whose spectral sequence degenerates in the second term
The statement is false, here is a counterexample. First note that for a Lie Group $G$ and its closed subgroup $H$, we have a fibration $G/H\to BH\to BG$. $BG$ and $BH$ are not finite, but they are a …
8
votes
Accepted
What is $Sq^i(\alpha^j)$ for all $i$ and $j$?
The Cartan formula
$$Sq^i(xy)=\Sigma _{j+k=i}Sq^j(x)Sq^k(y)$$
together with the instability condition
$$Sq^d(\alpha)=\alpha ^2 \mbox{ if $d=deg(\alpha )$}, Sq^i(\alpha )=0 \mbox{ if $d>deg(\alpha )$ …
2
votes
Loop space generalization
[Edited after the comment below]
Well, I don't know exactly what you mean by "geometric interpretation", but
for topological group $G$ the unpointed mapping spaces $Map(BG,X)$ are examples of homotop …
10
votes
Accepted
Representation of finite groups in a compact Lie group
In "Maps from $B\pi$ into $X$" Quart. J. Math. Oxford Ser. (2) 39 (1988), no. 153, 117–127., Wojtkowiak proves that the natural map
$Rep(H,G)\rightarrow [BH,BG]$ is not surjective when $H=\Sigma _3$ …
4
votes
Isomorphism between the Burnside ring $A(G)$ and the zeroth $G$-equivariant stable homotopy ...
This is described in Chapter 5 of Lewis, May, Steinberger "Equivariant Stable homotopy theory" (with contribution by McClure). Basically the idea is that a generator of $A(G)$ is a $G$-set, so 0-dimen …
5
votes
Accepted
Equivalence of homotopy categories and model structure theory
You can find a "direct proof" in the paper by Curtis Simplicial Homotopy Theory (Advances in Mathematics, Volume 6, Issue 2, April 1971, Pages 107–209). The main ingredient is the use of barycent …
-1
votes
The Eilenberg-MacLane spectrum and retractions
If it were the case, then $K(Z,n)$ would be a retract of $\underline{ku}_n$ where $\underline{ku}_n$ is the $n$-th infinite loop space associated to the spectrum $ku$. However, $ku$ is what is called …
1
vote
Injectivity on rational homtopy implies surjectivity on rational cohomology for classifiying...
Write $kO$ for the connective $k$-theory, and $X$ for the connective delooping of $BTOP$. Then $H_*(BO;Q)$ and $H_*(BTOP,Q)$ are free commutative (in graded
sense) generated by $\pi _*(kO,Q)\cong \p …
3
votes
Accepted
Grading in Eilenberg-Moore spectral sequence
Different people use different notation on gradings, for example I would have called the bigrdading of $e_i$ $E_2^{1,2i}$. Supposing that this is not a typo, Quillen meant
by $k$ in $E_s^{j,k}$ the …
0
votes
Semidirect product of torus with cyclic group: representations/cohomology?
As to the cohomology, here is an answer, which is probably an "over-kill". Probably you can find an answer in much older literature like Borel's.
Since $BT$ has torsion-free homology,
COROLLARY 4.9 o …
7
votes
Accepted
What are the cohomology groups $H^d(BSO_\infty,Z)$ and $H^d(BO_\infty,Z)$?
The Theorem 1.5 and 1.6 you quote give the answer.
More precisely, for $SO$, in the range $d<6$, the only polynomial generators
are $p_1$ which has degree 4, $\delta(w_2)$ with degree 3 and $\delta(w …
5
votes
0
answers
326
views
The behaviour of the suspension homomorphism on $H_*(QX;Z/p)$ for odd $p$ (Reference request)
The mod $p$ homology of $QX=\Omega ^{\infty}\Sigma ^{\infty}X$ for connected $X$ was computed by Dyer-Lashof Homology of Iterated Loop Spaces, Amer. J. of Math., vol.84, No.1 pp 35-88. 1962 It follows …
4
votes
Accepted
spectral sequence with non-trivial action on coefficients
I guess here $\mathbb Z/2$-module means $\mathbb Z[\mathbb Z/2]$-module. So let's take $N=1$, $A=\mathbb Z$. $H_0(\mathbb Z/2,A)$ is the coinvariant
$A/\mathbb Z/2$ so it is $A$ in the case of the tr …