Timeline for Ideal of the free Lie algebra L(x,y) generated by x
Current License: CC BY-SA 4.0
17 events
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| Apr 3, 2020 at 18:00 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Mar 4, 2020 at 17:33 | comment | added | Sergei Ivanov | I published an answer on my question. | |
| Mar 4, 2020 at 17:31 | history | edited | Sergei Ivanov | CC BY-SA 4.0 |
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| Mar 4, 2020 at 17:30 | answer | added | Sergei Ivanov | timeline score: 1 | |
| Mar 4, 2020 at 17:16 | comment | added | Sergei Ivanov | Now I know the proof. Indeed, it is easy to check that $x,[x,y],[x,y,y],\dots$ freely generate the ideal $(x)$ as a Lie algebra using Theorem 2.9 of Section 2.4 of Reutenauer's book. And then it follows as it is written in the updated version of the question. Sorry for bothering you. | |
| Mar 4, 2020 at 17:05 | comment | added | Sergei Ivanov | By the way, I updated the question. Now I only need to prove that the ideal $(x)$ is freely generated as a Lie algebra by the elements $x,[x,y],[x,y,y],..$. | |
| Mar 4, 2020 at 17:03 | comment | added | Sergei Ivanov | No, it is not enough to have such a word. Even $V_1$ lives in all bi-degrees. If for a Lyndon word $w$ we denote by $[w]$ the corresponding element of the Lyndon basis, then $[x,[x^ny^m]] \in V_1.$ | |
| Mar 4, 2020 at 16:51 | comment | added | darij grinberg | The free Lie algebra is multigraded by $\left(\text{degree of }x, \text{degree of }y\right)$, so it suffices to prove that for any positive $n$ and $m$ there is a nonzero Lie bracket of multidegree $\left(n,m\right)$. For that, you just need a Lyndon word with $n$ $1$'s and $m$ $2$'s. For example, $11\cdots 1 22\cdots 2$ is such a word. | |
| Mar 4, 2020 at 16:46 | history | edited | Sergei Ivanov | CC BY-SA 4.0 |
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| Mar 4, 2020 at 15:47 | comment | added | Sergei Ivanov | I tried to use this explicit basis but I didn't succeed. | |
| Mar 4, 2020 at 15:42 | comment | added | darij grinberg | Ah, I see -- it bounds the number of $x$'s, not the total number of factors. Then it's a less trivial question than I expected. Still, any explicit basis of the free Lie algebra (e.g., using Lyndon words) should do the trick. | |
| Mar 4, 2020 at 15:37 | comment | added | Sergei Ivanov | No, $V_n$ lives in all homogeneous degrees for $n\geq 1$. For example, $V_1$ has the following element $[x,[x,y,...,y]].$ | |
| Mar 4, 2020 at 15:33 | history | edited | Sergei Ivanov | CC BY-SA 4.0 |
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| Mar 4, 2020 at 15:25 | review | Close votes | |||
| Mar 6, 2020 at 17:50 | |||||
| Mar 4, 2020 at 15:11 | comment | added | darij grinberg | Anyway, is $V = \operatorname{span}\left(x\right)$ here? Then you can just argue that $V_n$ lives in homogeneous degree $n$ and thus cannot cover the whole ideal. (Of course, you'd have to verify that the ideal has a nontrivial component in all positive degrees; but this is clear from the embedding of the free Lie algebra into the tensor algebra.) | |
| Mar 4, 2020 at 15:09 | comment | added | darij grinberg | Please don't use $\operatorname{id}$ for anything other than the identity map -- it's just too confusing. | |
| Mar 4, 2020 at 14:58 | history | asked | Sergei Ivanov | CC BY-SA 4.0 |