Timeline for Pontryagin square on $Y\times S^1$ where $Y$ is three-dimensional
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Oct 7, 2013 at 17:27 | comment | added | Yuji Tachikawa | Of course I shouldn't have trusted what's available on the internet written by anonymous persons! Thank you very much, Oscar, for helping me. (The original question which I erased out of shame was why we don't have $n^2=Sq^1 n=v_1 n=0$. ) | |
| Oct 7, 2013 at 17:16 | comment | added | Yuji Tachikawa | Thanks, I erased the question because I thought I should study this basic piece of algebraic topology first...Somehow there are many places on the internet where the cupping with $v_k$ equals $Sq^k$ even when it doesn't land in the top degree, like in Wikipedia: en.wikipedia.org/wiki/Stiefel-Whitney_class#Wu_classes or nLab: ncatlab.org/nlab/show/Wu+class . Doing the computation in $\mathbb{RP}^n$ it's clear it only works when the product lands in the top degree. Hmm ... | |
| Oct 7, 2013 at 17:14 | comment | added | Oscar Randal-Williams | No, cupping with $v_1$ only agrees with $Sq^1$ when it lands in the top degree, in this case 3 (this is the definition of $v_1$). | |
| Aug 21, 2013 at 17:45 | vote | accept | Yuji Tachikawa | ||
| Aug 21, 2013 at 16:33 | comment | added | Oscar Randal-Williams | $n^2n' +n (n')^2 = Sq^1(n n')$ and $Sq^1 : H^2(Y;\mathbb{Z}/2) \to H^3(Y;\mathbb{Z}/3)$ is given by cup product with the first Wu class $v_1$ (by definition), which is equal to the first Stiefel--Whitney class $w_1$, and vanishes as $Y$ is orientable. | |
| Aug 21, 2013 at 15:15 | comment | added | Yuji Tachikawa | Thanks, but $\mathfrak{P}(n\otimes x)=2n^3$ doesn't seem to satisfy $\mathfrak{P}((n+n')\otimes x)=\mathfrak{P}(n\otimes x)+\mathfrak{P}(n'\otimes x)$ as I don't see $n^2 n'+ n' n^2$ to vanish in general... | |
| Aug 21, 2013 at 14:55 | comment | added | Oscar Randal-Williams | I had been led astray by a strange formula. I think its right now (and fits with your example). | |
| Aug 21, 2013 at 14:54 | history | edited | Oscar Randal-Williams | CC BY-SA 3.0 |
added 373 characters in body
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| Aug 21, 2013 at 13:44 | history | edited | Oscar Randal-Williams | CC BY-SA 3.0 |
Gave correct argument
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| Aug 21, 2013 at 12:56 | comment | added | Oscar Randal-Williams | You probably didn't, and I probably did. Let me think about it. | |
| Aug 21, 2013 at 12:25 | comment | added | Yuji Tachikawa | Also, I'm a bit confused. Take $Y=\mathbb{RP}^3$, and consider a flat $SO(3)$ bundle over $Y\times S^1$, such that the holonomy $a$ around the generator of $\pi_1(\mathbb{RP}^3)$ and the holonomy $b$ around $S^1$ are given by $a=diag(-1,-1,1)$ and $b=diag(-1,1,-1)$ respectively. This bundle has $p_1=0$, and $w_2=\alpha^2+\alpha x$ where $\alpha$ is the generator of $H^1(\mathbb{RP}^3,\mathbb{Z}/2)$. Then $\mathfrak{P}(w_2)=\alpha^3 x$ according to your formula, but there's a general formula saying $p_1=\mathfrak{P}(w_2)$ mod 4, i.i.r.c. Where did I make a mistake? | |
| Aug 21, 2013 at 12:16 | comment | added | Yuji Tachikawa | Thanks; where can I have a look at the list of $H^i(K(\mathbb{Z}/2,1);\mathbb{Z}/4)$ and $H^i(K(\mathbb{Z},1);\mathbb{Z}/4)$? | |
| Aug 21, 2013 at 11:40 | history | answered | Oscar Randal-Williams | CC BY-SA 3.0 |