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darij grinberg
darij grinberg
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Here is a simple question regarding the standard Lyndon basis for the free Lie Algebra. Suppose I take two lyndon words $m$ and $n$ and their standard bracketings $B(m)$ and $B(n)$ as elements in the free lie algerbaLie algebra. Suppose further that $m < n$, so that $mn$ is a Lyndon word.

My question is when we express the bracket $[B(m),B(n)]$ in the Lyndon basis, is it of the form

$[B(m),B(n)] = B(mn) + \sum_{l>mn, |l| = |m|+|n|} a_{m,n}^{(l)} B(l)$

Here is a simple question regarding the standard Lyndon basis for the free Lie Algebra. Suppose I take two lyndon words $m$ and $n$ and their standard bracketings $B(m)$ and $B(n)$ as elements in the free lie algerba. Suppose further that $m < n$, so that $mn$ is a Lyndon word.

My question is when we express the bracket $[B(m),B(n)]$ in the Lyndon basis, is it of the form

$[B(m),B(n)] = B(mn) + \sum_{l>mn, |l| = |m|+|n|} a_{m,n}^{(l)} B(l)$

Here is a simple question regarding the standard Lyndon basis for the free Lie Algebra. Suppose I take two lyndon words $m$ and $n$ and their standard bracketings $B(m)$ and $B(n)$ as elements in the free Lie algebra. Suppose further that $m < n$, so that $mn$ is a Lyndon word.

My question is when we express the bracket $[B(m),B(n)]$ in the Lyndon basis, is it of the form

$[B(m),B(n)] = B(mn) + \sum_{l>mn, |l| = |m|+|n|} a_{m,n}^{(l)} B(l)$

put the coefficients $a_{m,n}^{(l)}$ in tensor form so that the "structure constants" problem appears.
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edit approved Jun 7, 2015 at 9:27
Duchamp Gérard H. E.
edit approved Jun 7, 2015 at 9:27
Duchamp Gérard H. E.
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Here is a simple question regarding the standard Lyndon basis for the free Lie Algebra. Suppose I take two lyndon words $m$ and $n$ and their standard bracketings $B(m)$ and $B(n)$ as elements in the free lie algerba. Suppose further that $m < n$, so that $mn$ is a Lyndon word.

My question is when we express the bracket $[B(m),B(n)]$ in the Lyndon basis, is it of the form

$[B(m),B(n)] = B(mn) + \sum_{l>mn, |l| = |m|+|n|} a_l B(l)$$[B(m),B(n)] = B(mn) + \sum_{l>mn, |l| = |m|+|n|} a_{m,n}^{(l)} B(l)$

Here is a simple question regarding the standard Lyndon basis for the free Lie Algebra. Suppose I take two lyndon words $m$ and $n$ and their standard bracketings $B(m)$ and $B(n)$ as elements in the free lie algerba. Suppose further that $m < n$, so that $mn$ is a Lyndon word.

My question is when we express the bracket $[B(m),B(n)]$ in the Lyndon basis, is it of the form

$[B(m),B(n)] = B(mn) + \sum_{l>mn, |l| = |m|+|n|} a_l B(l)$

Here is a simple question regarding the standard Lyndon basis for the free Lie Algebra. Suppose I take two lyndon words $m$ and $n$ and their standard bracketings $B(m)$ and $B(n)$ as elements in the free lie algerba. Suppose further that $m < n$, so that $mn$ is a Lyndon word.

My question is when we express the bracket $[B(m),B(n)]$ in the Lyndon basis, is it of the form

$[B(m),B(n)] = B(mn) + \sum_{l>mn, |l| = |m|+|n|} a_{m,n}^{(l)} B(l)$

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Chitrabhanu
Chitrabhanu
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Bracket of lyndon words?

Here is a simple question regarding the standard Lyndon basis for the free Lie Algebra. Suppose I take two lyndon words $m$ and $n$ and their standard bracketings $B(m)$ and $B(n)$ as elements in the free lie algerba. Suppose further that $m < n$, so that $mn$ is a Lyndon word.

My question is when we express the bracket $[B(m),B(n)]$ in the Lyndon basis, is it of the form

$[B(m),B(n)] = B(mn) + \sum_{l>mn, |l| = |m|+|n|} a_l B(l)$