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 answers only not deleted user 10366
15 votes
Accepted

What is an unstable dual-Steenrod comodule?

Normally people think about Steenrod comodules as graded $\mathbb{Z}/2$-modules equipped with a graded coaction $\psi\colon M_*\to M_*[\xi_1,\xi_2,\dotsc]$. However, it is equivalent to consider ungr …
Neil Strickland's user avatar
7 votes
Accepted

Adem-Wu relations from Bullett-Macdonald identities

Larry Smith is not really using complex integration. Instead, he is using the residue map, which can be defined algebraically by the rule $$ \text{res}\left(\sum_{k=-N}^\infty a_k z^k dz\right)=a_{- …
Neil Strickland's user avatar
16 votes

$Sq^1$ cohomology of spaces

I remember that when I wrote my thesis I was unable to find references for some quite basic facts about this that everyone knew. It is quite possible that there were references that I did not succeed …
Neil Strickland's user avatar
9 votes

Steenrod squares as power operations vs. as cohomology operations

Here are some things that one can say for more general ring spectra. I choose to work with even periodic spectra, where $\pi_1(E)=0$ and $\pi_2(E)$ contains an invertible element, so that one can gen …
Neil Strickland's user avatar