Timeline for Equivariant Homotopy
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Mar 9, 2015 at 11:03 | comment | added | PPR | @AllenKnutson, I see, thank you. So we can think of X as the "Klein Bottle", and if $j=0$, the range of the maps is $S^4$, if $j=1$ the range is a closed 4-dim disc embedded in $\mathbb{R}^5$, if $j=4$ I think we get a 1-dim line in $\mathbb{R}^5$. Does that make sense? Not sure about the other cases. At any rate, how do you then go about classifying those maps under regular homotopy? How do you relate $\pi_2(Y)\equiv\pi_{S^2}(Y)$ for some $Y$ to $\pi_{Klein}(Y)$? | |
| Mar 9, 2015 at 1:53 | comment | added | Allen Knutson | (1) Start with a cylinder including its boundary discs, so, $S^2$. Cutting out an open disc and $Z_2$-identifying its boundary circle is the same as gluing in a cross-cap; that's what you're doing at top and bottom of your cylindrical $S^2$. (2) An equivariant map $X\to Y$ induces a map $X/G\to Y/G$. (But not vice versa: consider $X=Y=Z_2$, with two equivariant maps that become the same in the quotient.) | |
| Mar 8, 2015 at 12:16 | comment | added | PPR | @WłodzimierzHolsztyński, yes, it is, I just wrote the parametrization explicitly to specify $f_1$. AllenKnutson, are you saying X/~ is the Klein-bottle? I don't see that.. I'm also unsure how you get from $G$-maps to quotient spaces. | |
| Mar 8, 2015 at 3:30 | comment | added | Allen Knutson | I guess you don't have group actions, just spaces with equivalence relations on them? In particular $X/\sim$ is $S^2$ with two cross-caps? | |
| Mar 8, 2015 at 3:09 | comment | added | Włodzimierz Holsztyński | Is $\ X := S^1\times[0;1]$ ? | |
| Mar 8, 2015 at 1:01 | comment | added | PPR | Perhaps equivariant homotopy is unrelated to this problem. Is there a way to relax the continuity condition on the group action? But still define a $G$-map $f:X\to S^4$ to be a map such that $f(f_1(g, (z,\theta))) = f_2(g, f(z, \theta))$. | |
| Mar 8, 2015 at 0:46 | comment | added | Allen Knutson | Now your $f_1$ isn't continuous. | |
| Mar 8, 2015 at 0:43 | comment | added | PPR | @AllenKnutson, thanks for your comment. It appears I made a mistake in the formulation of $f_1$ and so I have edited my question to fix it. I hope now it makes more sense that I have defined $f_1$ in three (or two if you will) lines. Since you have two circles at the boundaries of $X$, I wasn't sure if I could retract the cylinder into one circle or two, and if I retract it into two circles, what are the conditions between the two circles then, if any. | |
| Mar 8, 2015 at 0:41 | history | edited | PPR | CC BY-SA 3.0 |
Mistake in original formulation of $f_1$ fixed.
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| Mar 7, 2015 at 23:30 | comment | added | Allen Knutson | Since you can $G$-equivariantly retract the cylinder to a circle, I don't see the point in including the $z$ coordinate. (I also don't see why you've split up the definition of $f_1$ into three lines, instead of just $(-1,(z,\theta))\mapsto (z,-\theta) \forall z$.) | |
| Mar 7, 2015 at 22:21 | review | First posts | |||
| Mar 7, 2015 at 23:21 | |||||
| Mar 7, 2015 at 22:18 | history | asked | PPR | CC BY-SA 3.0 |