Timeline for Number of connective orbit types of torus actions
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 13, 2023 at 14:16 | comment | added | Mehmet Onat | @HenrikRüping What is the meaning of coprime $a,b$, as in the picture I added above? If they are not coprime, then $\{(at,bt)\mid t\in S^1\}$ is not a subgroup? | |
| Dec 13, 2023 at 14:14 | history | edited | Mehmet Onat | CC BY-SA 4.0 |
added 208 characters in body
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| Dec 13, 2023 at 13:31 | comment | added | Mehmet Onat | @HenrikRüping Thank you for your time and patient reply. | |
| Dec 13, 2023 at 13:04 | comment | added | HenrikRüping | Well maybe we should have a closer look at the example of $T^2$. Then any closed subgroup of $T^2$ is either trivial, the whole of $T^2$ or of the form $\{(at,bt)\mid t\in S^1\}$ for some coprime $a,b\in \mathbb{Z}$. They are all nonconjugate and they are all "of the type" $S^1\times \{1\}$. The map that assigns to a conjugacy class of a closed subgroup its type thus sends infinitely many different conjugacy classes to the same type. | |
| Dec 13, 2023 at 11:36 | comment | added | Mehmet Onat | @HenrikRüping For any $x \in X$, $G_x=\mathbb{S}^{1}\times \mathbb{S}^{1}\times \cdots \times \mathbb{S}^{1}\times \mathbb{Z}_{n_{1}}\times \mathbb{Z}_{n_{2}}\times \cdots \times \mathbb{Z}_{n_{k}}$? | |
| Dec 13, 2023 at 11:30 | comment | added | Mehmet Onat | @HenrikRüping Since $T$ is abelian, the conjugacy class $[(G_x)_0]$ consists of only $(G_x)_0$. So $(G_x)_0$ equals to $\mathbb{S}^{1}\times \mathbb{S}^{1}\times \cdots \times \mathbb{S}^{1}\times \left\{1\right\} \times \left\{ 1\right\} \times \cdots \times \left\{ 1\right\}$? | |
| Dec 13, 2023 at 11:01 | comment | added | HenrikRüping | No there are infinitely many non-cconjugate subgroups of the same type. The crucial point here is the difference between inner and non-inner automorphism. For example, for any two embeddings $f_1,f_2:S^1\to T^2$ there is an automorphism $\sigma$ of $T^2$ such that $f_2=\sigma\circ f_1$. However there need not be an inner automorphsm. | |
| Dec 13, 2023 at 10:28 | history | asked | Mehmet Onat | CC BY-SA 4.0 |