Timeline for A question on normal closures of elements in free groups.
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 24, 2013 at 10:38 | vote | accept | Alexey Kvashchuk | ||
| Mar 24, 2013 at 10:37 | vote | accept | Alexey Kvashchuk | ||
| Mar 24, 2013 at 10:38 | |||||
| Mar 24, 2013 at 10:14 | comment | added | Alexey Kvashchuk | Not sure how to edit comments, but $V^{-1} b^{-k} V$ in the above comment should be read as $V^{-1} \tilde{b}^{-k} V$ etc.... Anyhow, so now the question is that $b$ is root, and some powers of $p$ can be represented as $V^{-1} b^{-1} V\cdot U^{-1} b U $ form. For example, $p^m = V^{-1} b^{-1} V\cdot U^{-1} b U$ and $p^n = X^{-1} b^{-1} X\cdot Y^{-1} b Y$ for some $m, n \in \mathbb{Z}$ and $U, V, X, Y \in F$. What could be inferred about $p$ and $b$? | |
| Mar 24, 2013 at 9:57 | answer | added | HJRW | timeline score: 11 | |
| Mar 24, 2013 at 9:51 | comment | added | Alexey Kvashchuk | Thanks Henry. Let me go one step back and consider a case that led to an assumption of $b$ being root. If we drop that assumption we have that $b=\tilde{b}^k$ for some $k\geq 2$. Suppose that there is a proper power of $p$ that can be represented as $p^m= V^{-1} b^{-k}V\cdot U^{-1} b^{k}U$ for some $U$ and $V$. We then have that $$ p^m = \tilde{V}^{-k} \tilde{U}^{k} ,$$ where $\tilde{V} = V^{-1} b^{-1}V$,$\tilde{U} = U^{-1} b U$, and $|m|, k \geq 2$. Then by Schutzenberger-Lyndon we have that $\tilde{U}, \tilde{V}, p$ commute. So that $V = b^l U$ and $p=1$. | |
| Mar 24, 2013 at 8:31 | comment | added | HJRW | So, 1/2 is a universal lower bound for stable commutator length in free groups. Doesn't it follow that your $n_k\leq 2$? If so, then the possibilities are very restricted, by a theorem of Lyndon. I'll try to write more details later, if I have time. | |
| Mar 24, 2013 at 8:05 | history | asked | Alexey Kvashchuk | CC BY-SA 3.0 |