Timeline for Cobounded ⇒ cocompact?
Current License: CC BY-SA 2.5
32 events
| when toggle format | what | by | license | comment | |
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| Oct 29, 2010 at 4:09 | vote | accept | Anton Petrunin | ||
| Oct 27, 2010 at 23:00 | history | edited | Anton Petrunin | CC BY-SA 2.5 |
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| Oct 27, 2010 at 20:02 | answer | added | Anton Petrunin | timeline score: 4 | |
| Oct 23, 2010 at 21:03 | comment | added | user6976 | @Anton: MO does not send me replies to my comments, so perhaps it would be better if you copy them by email. The question about $R^d$ and $L$ is very nice. You may want to post a separate "follow up" question then. Also did you ask John Roe or Nigel Higson? I have a feeling that the original question should be known to specialists in functional analysis. Also Sergei Ivanov (who should be in Pennstate now) may know the answer. | |
| Oct 23, 2010 at 16:21 | history | edited | Anton Petrunin | CC BY-SA 2.5 |
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| Oct 23, 2010 at 16:20 | comment | added | Anton Petrunin | @Mark: Yes, sure. | |
| Oct 23, 2010 at 7:40 | comment | added | user6976 | @Anton: Is it possible that $L$ is finite dimensional, and $H$ is not while $diam(H/L)<1000$? I assumed that dim$(H)=\infty$. Or this $H$ and the $H$ above are different? Perhaps you meant $R^q/L$? | |
| Oct 23, 2010 at 4:28 | history | edited | Anton Petrunin | CC BY-SA 2.5 |
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| Oct 23, 2010 at 4:25 | comment | added | Anton Petrunin | @Mark: any positive integer | |
| Oct 23, 2010 at 4:22 | history | edited | Anton Petrunin | CC BY-SA 2.5 |
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| Oct 23, 2010 at 3:08 | comment | added | user6976 | @Anton: In your Comment, what is $q$? | |
| Oct 23, 2010 at 2:08 | history | edited | Anton Petrunin |
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| Oct 23, 2010 at 1:54 | history | edited | Anton Petrunin | CC BY-SA 2.5 |
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| Oct 22, 2010 at 19:32 | comment | added | Anton Petrunin | @Mark: I agree --- if true it should not be difficult... | |
| Oct 22, 2010 at 19:26 | comment | added | user6976 | @Anton: Translations are isometries (but not all isometries are translations). I was asking about a particular case of your question when $\Gamma\lt (H,+)$, that is $\Gamma$ is a subgroup of the additive group of $H$. I think this case should not be too difficult. | |
| Oct 22, 2010 at 19:14 | comment | added | Anton Petrunin | @Mark: No. Γ is a subgroup of isometries. But even if it is then I do not see a clear proof... | |
| Oct 22, 2010 at 18:51 | comment | added | user6976 | @Anton: Am I right and if $\Gamma$ is a subgroup of the additive group of $H$ acting by translation, then the answer to your question is "yes", because the quotient will always be a quotient of the Hilbert cube (provided it is bounded)? | |
| Oct 22, 2010 at 18:37 | comment | added | Anton Petrunin | @Mark: In the above comments people talk about different things and it creates a lot of misunderstandings. Simply start from scratch. | |
| Oct 22, 2010 at 16:47 | comment | added | Mariano Suárez-Álvarez | @Mark: how could possibly a transitive action (whose corresponding quotient has exactly one point!) be unbounded? In any case, the group $\Gamma$ in Anton's example is not the same as the one you used in your first comment. | |
| Oct 22, 2010 at 15:08 | comment | added | user6976 | @Anton: So you claim that Guntram's argument is wrong? Or you claim that there exists a transitive action with unbounded quotient? | |
| Oct 22, 2010 at 14:47 | history | edited | Anton Petrunin | CC BY-SA 2.5 |
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| Oct 22, 2010 at 14:45 | comment | added | Anton Petrunin | @Mark: There are a lot of misunderstanding above, but my example is correct. | |
| Oct 22, 2010 at 8:47 | comment | added | user6976 | @Anton: You should somehow reconcile your formula for the diameter with Guntam's fact. | |
| Oct 22, 2010 at 8:35 | comment | added | user6976 | @Guntam: Then the diameter should be 0? | |
| Oct 22, 2010 at 7:30 | comment | added | Guntram | @Mark: If the dimension of $H$ is at least 2, the group $\Gamma$ generated by vectors of integral norms is transitive: Given $f \in H, f\neq 0$, connect $v:=\frac{f}{||f||}$ to $-v$ via an arc $\phi\colon [0,1]\to H$ on the unit sphere and consider $w_t:=\phi(t)+f$. By the intermediate value theorem, $||w_{t_0}||$ will be integral for some $t_0$, so $f$ is the difference of two vectors of integral norm. | |
| Oct 22, 2010 at 3:33 | history | edited | Anton Petrunin | CC BY-SA 2.5 |
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| Oct 22, 2010 at 3:05 | comment | added | user6976 | @Anton: Since you have computed the diameter of the quotient, could you then give an example of two orbits arbitrary far apart? I thought of the following argument. Suppose that $f\in \ell_2$ has norm $M$. Subtract $[M]/M \cdot f$ (with integral norm $[M]$), get norm $<1$. Hence the diam. is at most 2. If I do not see something here, did you consider already all subgroups $\Gamma$ of the additive group of $H$ acting by translations? | |
| Oct 22, 2010 at 2:43 | history | edited | Anton Petrunin | CC BY-SA 2.5 |
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| Oct 22, 2010 at 1:55 | comment | added | Anton Petrunin | @Mark: (1) $diam=\infty$ (2) Γ is a group, what else can act? | |
| Oct 22, 2010 at 0:09 | comment | added | user6976 | @Anton: Of course you did not say what $\Gamma$ is. | |
| Oct 22, 2010 at 0:07 | comment | added | user6976 | Just a thought: Let $H=\ell_2$. Let $\Gamma$ be the subgroup of the additive group of $H$ generated by sequences with integral norms. Then $\Gamma$ acts on $H$ by translations. What is the diameter of $H/\Gamma$ and is the quotient compact? I do not have time to check it myself now. But it should not be that difficult. | |
| Oct 21, 2010 at 23:24 | history | asked | Anton Petrunin | CC BY-SA 2.5 |