Timeline for A question about groups of intermediate growth
Current License: CC BY-SA 3.0
8 events
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| Nov 11, 2011 at 23:41 | comment | added | kassabov | @valerio: I think most (if not all) groups of intermediate growth satisfy $\bar \zeta_k=1$. Our result with Igor is not strong enough to get the value of $\bar \zeta$ but my guess is that it is also 1. I feel that there are no groups (of intermediate growth) with $\bar \zeta >1$. | |
| Nov 11, 2011 at 23:34 | history | edited | kassabov | CC BY-SA 3.0 |
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| Nov 11, 2011 at 19:50 | comment | added | Valerio Capraro | @martin: do you think one use your corollary 1.3 to exibit at least a group of intermediate growth such that $\overline\zeta(G)=1$? | |
| Nov 11, 2011 at 10:02 | comment | added | Ben Green | When making my comment above I was addressing the question for "supremum" instead of "infimum". I feel one does need Gromov/Pansu to do that. | |
| Nov 10, 2011 at 22:46 | vote | accept | Valerio Capraro | ||
| Nov 10, 2011 at 22:45 | comment | added | Valerio Capraro | It seems good. This is nice, also because gives a direct proof in the case of polynomial growth, when all people with whom I talked was convinced of the necessity to use both Gromov and Pansu theorem. | |
| Nov 10, 2011 at 22:14 | comment | added | kassabov | I take the last comment back -- using submultiplicativity one can only show that $lim_k \sqrt[k]{\bar zeta_k(G)} =1 $ where $\bar \zeta_k(G) = \limsup_n \frac{B(kn+k)}{B(kn)}$. | |
| Nov 10, 2011 at 19:32 | history | answered | kassabov | CC BY-SA 3.0 |