Timeline for Coarsely trivial Borel cross section for $G\to G/N$
Current License: CC BY-SA 3.0
7 events
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| Mar 3, 2015 at 9:24 | comment | added | Dave Witte Morris | Hannes, I think the discrete Heisenberg group (my 2nd example) is a counterexample for your modified question. Since $\sigma(s)$ must differ from $s$ by an element of $N$, which is in the center, we have $\sigma(a)^{-n} \sigma(b)^{-n} \sigma(a)^n \sigma(b)^n = a^{-n} b^{-n} a^n b^n$, so essentially the same calculation, but with $\sigma(a)$ and $\sigma(b)$ in the place of $a$ and $b$, shows $$ z^{n^2} \sigma(1 \cdot N) = \sigma(1 \cdot N) \cdot \prod_{i=1}^{4n} \alpha(s_i, x_i) .$$ So $\alpha(s_i, x_i)$ cannot be bounded. | |
| Mar 3, 2015 at 8:43 | comment | added | Hannes Thiel | Do you know if there is also a counterexample to the modified question of the comment above? | |
| Mar 3, 2015 at 8:34 | comment | added | Hannes Thiel | Dear Dave, thank you for your answer. The reason I was thinking that semidirect products work is that I thought the cocycle records how far $\sigma(xy)$ is from $\sigma(x)\sigma(y)$ for $x,y\in G/N$. I understand now that the cocycle is in fact not recording this. Let us therefore consider $\alpha\colon G/N\times G/N\to N$ defined by the formula $\sigma(xy)\alpha(x,y)=\sigma(x)\sigma(y)$. Then, I should have asked my question for $\alpha$: Given a compact subset $K$ of $G/N$, can $\sigma$ be chosen s.t. $\alpha(K\times G/N)$ is pre-compact. Then the answer is 'yes' for semidirect products. | |
| Mar 3, 2015 at 8:26 | vote | accept | Hannes Thiel | ||
| Mar 2, 2015 at 21:07 | history | edited | Dave Witte Morris | CC BY-SA 3.0 |
Added a counterexample that is a semidirect product.
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| Mar 2, 2015 at 19:24 | history | edited | Dave Witte Morris | CC BY-SA 3.0 |
Clarifed that the answer is "not always", not "never".
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| Mar 2, 2015 at 19:16 | history | answered | Dave Witte Morris | CC BY-SA 3.0 |