Standard Lyndon bases of Lie algebras and enveloping algebras

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 19 May 2015

This is an excerpt of the paper Standard Lyndon bases of Lie algebras and enveloping algebras by Pierre Lalonde Arun Ram.

The first author's research was supported by FCAR and NSERC grants. The second author's research was partially supported by a National Science Foundation Postdoctoral Fellowship.

Standard bases

Order the words in $A^{*}$ by setting $u ⪯ w if {\begin{cases} ∣ u ∣ < ∣ w ∣ \\ or \\ ∣ u ∣ = ∣ v ∣ and u \geq v . \end{cases}$ This is a total order on words with the additional property that there are a finite number of words less than any given word.

Suppose that $J$ is a Lie ideal of $Lie (A)$ and that $I$ is the ideal in $ℚ [A^{*}]$ generated by $J .$ Let $𝔤 = Lie (A) / J and U 𝔤 = ℚ [A^{*}] / I .$ It follows from (1.6) and (1.8), that $U 𝔤$ is the enveloping algebra of $𝔤 .$ Define a Lyndon word to be Lie-standard with respect to $J$ if its bracketing $b [ℓ]$ cannot be written as a sum of bracketings of strictly smaller Lyndon words modulo the ideal $J$ of $Lie (A)$ with respect to the ordering $⪯ .$ Define a word $w$ to be standard with respect to $I$ if $w$ cannot be written as a sum of strictly smaller words modulo the ideal $I,$ again with respect to the ordering $⪯ .$ Make the following notations: $L$ is the set of Lyndon words, $S L$ is the set of Lie-standard Lyndon words, $S$ is the set of standard words. The standard words that we have defined are essentially a Gröbner basis. The following two theorems are the standard results from the Gröbner basis context.

The set of elements $b [ℓ]$ where $ℓ \in S L,$ is a basis of $𝔤 = Lie (A) / J .$

Proof.

The set of all $b [ℓ],$ where $ℓ \in L,$ spans $𝔤 .$ If $ℓ$ is not Lie-standard then $b [ℓ]$ can be written as a linear combination of bracketings of Lyndon words modulo $J$ which are smaller than $ℓ .$ If any of these words is not standard, express it as a sum of smaller words. Continue this process until all the words in the expansion are standard. The process must stop as the number of words smaller than any given word is finite. Thus the elements $b [ℓ],$ where $ℓ \in S L,$ span $𝔤 .$

We now show that the set of Lie-standard Lyndon words is linearly independent. Suppose that there was a nontrivial relation among them. Then this relation expresses the maximal word as a linear combination of lower words modulo $J,$ a contradiction to the standardness of the maximal word.

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The set of words in $S$ is a basis for $U 𝔤 = ℚ [A^{*}] / I .$

Proof.

The proof is exactly analogous to the proof of Theorem (2.1).

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We shall show that $S L = S \cap L,$ i.e. that the set of Lie-standard Lyndon words is the same as the set of standard Lyndon words (this is not a priori obvious).

$S \cap L \subseteq S L .$

Proof.

Let $m \in L .$ Suppose $m \notin S L .$ Then $b [m] = \sum_{\underset{n \in L}{n ≺ m}} a_{n} b [n] + x,$ for some $x \in J .$ Using (1.2) on each side, $m + \sum_{v ≺ m} b_{v} v = \sum_{n ≺ m} a_{n} (n + \sum_{w ≺ n} c_{w} w) + x,$ for some integers $b_{v}, a_{n}, c_{w} .$ Subtracting $\sum_{v ≺ m} b_{v} v$ from both sides, $m = \sum_{v ≺ m} d_{v} v + x$ for some integers $d_{v} .$ Since $x \in J \subseteq I,$ we have that $m \notin S .$

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Any factor of a standard word is a standard word.

Proof.

Suppose that $v$ is not standard so that we have $v = \sum_{m ≺ v} a_{m} m (mod I).$ Then $u v w = u (\sum_{m ≺ v} a_{m} m + x) w,$ where $x \in I .$ Since $I$ is an ideal $u x w \in I$ and since $u m w ≺ u v w$ for all $m,$ we have that $u v w$ is not standard.

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If $w \in S$ then $w$ has a unique factorization $w = ℓ_{1} \dots ℓ_{k}, ℓ_{i} \in S L, ℓ_{1} \geq \dots \geq ℓ_{k} .$

Let $ℓ = ℓ_{1} \dots ℓ_{k},$ $ℓ_{i} \in S L,$ and $ℓ_{1} \geq \dots \geq ℓ_{k} .$ Then $b [ℓ] = ℓ_{1} \dots ℓ_{k} + \sum_{m' ≺ l} b_{m'} m_{1}^{'} \dots m_{r}^{'} (mod I),$ where $m' = m_{1}^{'} \dots m_{r}^{'},$ $m_{i}^{'} \in S L$ for each $i,$ $m_{1}^{'} \geq \dots \geq m_{r}^{'},$ and $b_{m'} \in ℤ .$

Proof.

By (1.5) and the definition of the ordering $≺,$ $b [ℓ] = ℓ_{1} \dots ℓ_{k} + \sum_{m ≺ ℓ} a_{m} m .$ Expanding the sum in terms of standard words, $b [ℓ] = ℓ_{1} \dots ℓ_{k} + \sum_{\underset{m' ⪯ m ≺ ℓ}{m' \in S}} b_{m}^{'} m' (mod I).$ The result now follows from Corollary (2.5) since each $m'$ appearing in the sum has a unique factorization of the form $m' = m_{1}^{'} \dots m_{r}^{'},$ $m_{i}^{'} \in S L,$ $m_{1}^{'} \geq \dots \geq m_{r}^{'} .$

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Each of the following sets is a basis of $ℚ [A^{*}] / I .$ (B1) The set $S$ of standard words with respect to $I .$ (B2) The set of products $b [ℓ_{1}] \dots b [ℓ_{k}],$ where $ℓ_{i} \in S L$ and $ℓ_{1} \geq \dots \geq ℓ_{k} .$ (B3) The set of products $ℓ_{1} \dots ℓ_{k},$ where $ℓ_{i} \in S L$ and $ℓ_{1} \geq \dots \geq ℓ_{k} .$

Proof.

Statement (B1) is Theorem (2.2). (B2) is a basis by (2.1) and the Poincaré-Birkhoff-Witt theorem. Theorem (2.6) gives a triangular relation between the elements of the set (B3) and the elements of the set (B2) which proves that (B3) is a basis.

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(a) With notations as in Theorem (2.7), $(B1) = (B3).$ (b) $S L = S \cap L .$

Proof.

(a) Corollary (2.5) gives that $(B1) \subseteq (B3).$ Since these are both bases we must have $(B3) = (B1)$ (express the basis (B3) in terms of the basis (B1)).)

(b) Since $(B3) = (B1),$ $S L \subseteq S .$ Combining this with Lemma (2.3) we have that $S L = S \cap L .$

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The following proposition will help us to compute the standard Lyndon words by induction on the length of the words.

Let $ℓ$ be a standard Lyndon word. Then $ℓ$ is of the form $ℓ = ℓ_{1} ℓ_{2} \dots ℓ_{k} a,$ where (1) $ℓ_{i}$ are standard Lyndon words for all $1 \leq i \leq k,$ (2) $ℓ_{i}$ is a left factor of $ℓ_{i - 1}$ for all $i > 1,$ (3) $a \in A .$

Proof.

Let $m$ be the word $ℓ$ with the last letter removed. By (1.4) $m$ has a factorization $m = ℓ_{1} \dots ℓ_{k}$ into Lyndon words $ℓ_{i}$ such that $ℓ_{i} \geq ℓ_{i + 1}$ for all $1 \leq i \leq k - 1 .$ By Proposition (2.4), each of the factors $ℓ_{i}$ is standard since they are factors of the standard word $ℓ .$

It remains to prove that $ℓ_{i}$ is a left factor of $ℓ_{i - 1}$ for all $1 \leq i \leq k .$ Consider the following chain of inequalities. For $i > 1,$ $ℓ_{i} \leq ℓ_{i - 1} \leq ℓ_{1} < ℓ_{1} \dots ℓ_{k} a \leq ℓ_{i} \dots ℓ_{k} a,$ where the last inequality follows since the right hand side $ℓ_{i} \dots ℓ_{k} a$ is a right factor of the Lyndon word $ℓ_{1} \dots ℓ_{k} a .$ It follows easily from $ℓ_{i} \leq ℓ_{i - 1} < ℓ_{i} \dots ℓ_{k} a$ that $ℓ_{i}$ is a left factor of $ℓ_{i - 1}$ (consider these as words in a dictionary).

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