Timeline for The action of Steenrod squares on the indeterminate in Bullett-Macdonald identities
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Sep 27, 2014 at 3:46 | comment | added | Jesse C. McKeown | If I had to come up with a geometric description of steenrod-acting-trivially-on-formal-variables, it would be in terms of the fiberwise cohomology of $X\times B(Z/pZ)$. edit-add: ... as a bundle over $B(Z/pZ)$. | |
| Sep 23, 2014 at 15:57 | comment | added | user43326 | Actually $F(X\times \vee _i S^i,HZ/2)$ should be irrelevant since $H^*(X\times \vee _i S^i,HZ/2)$ doesn't have the correct ring structure. | |
| Sep 23, 2014 at 9:39 | comment | added | user43326 | So would it be related to somehow the equivalence $F(X\times BZ/p,HZ/p)\cong F(X\times \vee _i S^i, HZ/2)$? (Their homotopy groups give $H^*(X)[[t]]$ with two different actions of Steenrod squares on $t$.) But I don't see how $F(X\times \vee _i S^i, HZ/2)$ gets into the picture. | |
| Sep 18, 2014 at 22:34 | comment | added | Jesse C. McKeown | I was puzzled by the two actions, as well! Smith's paper is nice (though, alas, there are typos as he feared...); the "nontrivial" indeterminates appear naturally from the interpretation of H*(X,p)[t] == H*(X \times K(p,1),p), while understanding the "total pth power" seems to want a different interpretation of H*(X)[t]; this comes up in B+M when they formally invert the fundamental class of K(p,1)... but, I don't know, so this is not an answer. Good luck! | |
| Sep 11, 2014 at 14:57 | history | asked | user43326 | CC BY-SA 3.0 |