Timeline for When are uniform embeddings quasisymetric
Current License: CC BY-SA 4.0
9 events
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| Mar 29, 2022 at 13:08 | history | edited | Vitali Kapovitch | CC BY-SA 4.0 |
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| Mar 29, 2022 at 13:02 | comment | added | Carlos_Petterson | Cool or weakly quasi-symmetric (assuming $X$ is connected); ok cool, then that also covers both examples (bi-Lipschitz and the snowflake). Thanks Vitali really nice construction btw, it really helped :) | |
| Mar 29, 2022 at 13:01 | comment | added | Vitali Kapovitch | ok, yes, that would work too. | |
| Mar 29, 2022 at 13:01 | comment | added | Carlos_Petterson | Ah okay, so then I guess the "correct condition" is simply $f$ is quasi-symmetric and we cannot hope for much more? | |
| Mar 29, 2022 at 13:00 | comment | added | Vitali Kapovitch | @Carl_Petterson As I said I think you need $f$ to be bi-Lipschitz. I see no reason why any weaker modulus of continuity would suffice. For example should be possible to construct counterexamples when $f$ is bi-Hölder but not bi-Lipschitz. | |
| Mar 29, 2022 at 12:54 | comment | added | Carlos_Petterson | This is an extremely nice example. So in your opinion what would you expect that we need of the inverse $f^{-1}$ since your example shows that uniform continuity of $f^{-1}$ isn't enough? My intuition was based on the snowflaking $f:(X,d)\mapsto(X,d^{\alpha})$ ($\alpha\in (0,1)$) in which case $f$ has moduli $\alpha$ and $\cdot^{1/\alpha}$ . | |
| Mar 29, 2022 at 12:53 | history | edited | Vitali Kapovitch | CC BY-SA 4.0 |
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| Mar 29, 2022 at 12:51 | vote | accept | Carlos_Petterson | ||
| Mar 29, 2022 at 12:48 | history | answered | Vitali Kapovitch | CC BY-SA 4.0 |