Timeline for Action of $\mathbb{Z}/3\mathbb{Z}$ on $P^{1}$
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jul 18, 2016 at 9:34 | history | edited | ThiKu | CC BY-SA 3.0 |
wikipedia link
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| Jul 18, 2016 at 8:53 | comment | added | ThiKu | You'd better think of $T$ as a map $P^1/G\to P^1$: all three points $z, \frac{1}{1-z}$ and $1-\frac{1}{z}$ are mapped to the same image. Of course one can compute that image, but I don't see a geometric interpretation of it. | |
| Jul 18, 2016 at 8:26 | vote | accept | Tensor_Product | ||
| Jul 18, 2016 at 8:26 | comment | added | Tensor_Product | SIr , I think there are gaps in my understanding. What I was thinking that each $x\in P^1$ goes to $orbit(x)$ under the map. Then how $T$ plays a role here. I understand that $T$ is $G-invariant$. What I know is that given a group action we have orbit space which is just set of $G-$ orbits and it is denoted by $X/G$ | |
| Jul 18, 2016 at 8:14 | history | edited | ThiKu | CC BY-SA 3.0 |
bracket
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| Jul 18, 2016 at 8:13 | comment | added | ThiKu | What is the point of the second question? Obviously $T$ is invariant under the $Z/3Z$-action, so it gives a well-defined map to $P^1/G$. | |
| Jul 18, 2016 at 8:10 | comment | added | Tensor_Product | Thank for taking time to answer .Can you give me some hint for second part of my question. | |
| Jul 18, 2016 at 8:03 | history | answered | ThiKu | CC BY-SA 3.0 |