Skip to main content
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
Results tagged with
Search options not deleted user 43326

Homotopy theory, homological algebra, algebraic treatments of manifolds.

17 votes

(co)homology of symmetric groups

The answer to the question 1) is "yes". The classifying space of the symmetric group is of finite type, so its integral homology is determined by its rational homology and $p$-local homology for all …
user43326's user avatar
  • 3,051
11 votes
Accepted

Why does the Steenrod algebra act faithfully on $H^\ast(BC_p)$?

As is commented by @Connor Malin, the action of the Steenrod algebra on $H^*B\mathbb{Z}/p$ is not faithful. Consider the case $p=2$. $Sq^3Sq^1$ acts trivially on $H^*(B \mathbb{Z}/2)$, since $Sq^{2n …
user43326's user avatar
  • 3,051
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 $H=\Sigma _3$ …
user43326's user avatar
  • 3,051
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 $d>deg(\alpha )$ …
user43326's user avatar
  • 3,051
8 votes
Accepted

Equivalence of the total spaces of two Serre fibrations with equivalent fibers

Note that $\pi _1(B)$ acts on $\pi _0(F)$. So the ending of the long exact sequence for the fibration $$\cdots\rightarrow \pi _1(B)\rightarrow \pi _0(F)\rightarrow \pi _0(E)$$ is not just an exact …
user43326's user avatar
  • 3,051
8 votes
Accepted

Atiyah Bott-Shapiro orientation Vs Anderson-Brown-Peterson Splitting

According to Michael Hopkins, Mark Hovey, Spin cobordism determines real K-theory, Mathematische Zeitschrift 210.1 (1992): 181-196, 4th page of the pdf file, the Atiyah-Bott-Shapiro Orientation is …
user43326's user avatar
  • 3,051
7 votes
Accepted

cohomology of iterated loop space on spheres

As is known from J.W. Milnor, J.C. Moore, "On the structure of Hopf algebras" Ann. of Math. (2), 81 : 2 (1965) pp. 211–264, the algebra structure of an underlying algebra of a Hopf algebra is quite li …
user43326's user avatar
  • 3,051
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 …
user43326's user avatar
  • 3,051
6 votes
Accepted

Reference for: $p$-primary component of $\pi^S_k$ is $\Bbb Z_p$ when $k=2l(p-1)-1$

This follows easily from Theorem 4.4.20 of Ravenel's book, Complex cobordism and stable homotopy groups of spheres". The elements in $Ext^i$ with $i>1$ have total degrees greater than or equal to $2p …
user43326's user avatar
  • 3,051
6 votes

Thom space, homotopy group and cohomology group

More generally, we have $$H^* Thom(BG,E) \cong H^*(BG) e(E)$$ as modules over the Steenrod algebra, with e(E) the Euler class of $E$. I guess the base space doesn't have to be a classifying space. …
user43326's user avatar
  • 3,051
6 votes
1 answer
322 views

Homology of the product of spaces with integer coefficients and the Massey products

Consider $H_*(X\wedge Y;Z)$, where $X=Y=BZ/2$ for concreteness' sake. If we write $e_i$ the generator of $H_i(BZ/2;Z/2)$., we see that the $E_2=E_{\infty}$ term of the Bockstein spectral sequence f …
user43326's user avatar
  • 3,051
6 votes
1 answer
3k views

Double coset formulas for Orthogonal groups [Solved]

According to Madsen-Brumfiel "Evaluation of the Transfer and the Universal Surgery Classes" Inventiones mathematicae 32 (1976): 133-170 Theorem 3.11, we can compute the composition $BO(1)^2\stackrel …
user43326's user avatar
  • 3,051
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 …
user43326's user avatar
  • 3,051
5 votes
0 answers
214 views

The action of Steenrod squares on the indeterminate in Bullett-Macdonald identities

The Bullet-Macdonald identity (c.f. On the Adem relations)is the following: $$P(s^2+st)P(t^2)=P(t^2+st)P(s^2).$$ where we have $P(s)=\Sigma _s s^iSq^i$. This is known to be equivalent to the Adem rela …
user43326's user avatar
  • 3,051
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 …
user43326's user avatar
  • 3,051

15 30 50 per page