Timeline for homotopy groups of an orbifold
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 4, 2015 at 12:54 | comment | added | Igor Rivin | The action on $\mathbb{H}^2$ gives the modular orbifold as quotient. The modular orbifold is a doubled triangle, with one vertex at infinity, so topologically a punctured sphere, so a disk. The action on $\mathbb{H}^3$ is that cross $\mathbb{R}.$ | |
| Nov 4, 2015 at 11:47 | comment | added | sphere | @IgorRivin Can you elaborate more, please ? How do you obtain that the quotient is homotopy equivalent to $\mathbb{D}^{3}$. BTW is $\mathbb{D}^{3}=\{(x,y,z,t)\in \mathbf{R}^{4}| x^{2}+y^{2}+z^{2}+t^{2}\leq 1 \}$ ? | |
| Nov 4, 2015 at 11:06 | history | edited | Igor Rivin | CC BY-SA 3.0 |
corrected mental bug.
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| Nov 4, 2015 at 11:05 | comment | added | Igor Rivin | @HJRW Oh, I see, the disagreement is about the cusp. You are absolutely right. Told you I was underslept :( | |
| Nov 4, 2015 at 11:01 | comment | added | Igor Rivin | @HJRW I generally count the orbifold points as part of the space, is this what the disagreement is about? | |
| Nov 4, 2015 at 6:41 | comment | added | HJRW | @IgorRivin, perhaps it's helpful to remember, which I know you do well, that $PSL(2,\mathbb{Z})$ is virtually free. This couldn't be the case if its action on the hyperbolic plane were cocompact. | |
| Nov 4, 2015 at 6:34 | comment | added | HJRW | @IgorRivin, the quotient of $\mathbb{H}^2$ by $PSL(2,\mathbb{Z})$ is the modular orbifold, which is a punctured 2-sphere with order-2 and -3 cone points. (This follows from the usual picture of the fundamental domain for the action on $\mathbb{H}^2$.) | |
| Nov 3, 2015 at 23:40 | comment | added | Daniel Litt | Perhaps the confusion is that the topological quotient is not $\mathbb{S}^2$, but rather $\mathbb{R}^2$ (it's the $j$-line...) | |
| Nov 3, 2015 at 19:36 | comment | added | Igor Rivin | @HJRW Actually, unless I am confused (possible, due to lack of sleep), the quotient of $\mathbb{H}^2$ by $SL(2, Z)$ is a sphere (topologically), and I think the quotient of $\mathbb{H}^3$ is topologically just that cross a line. No? | |
| Nov 3, 2015 at 19:31 | comment | added | HJRW | $S^2$? Don't you mean the 3-ball? | |
| Nov 3, 2015 at 19:28 | history | answered | Igor Rivin | CC BY-SA 3.0 |