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I am trying to understand kind answer (somewhat generalized) by David Speyer on my previous question What is growth of ass. algebra with 3 generators and relation a1a2a3 + a2a3a1 +a3a1a2 - a1a3a2 - a2a1a3 -a3a2a1 ?

Consider free associative algebra $A$, denote $a_1, ... a_N$ to be its generators; take any homogeneous non-commutative polynomial $P(a_i)$. Consider $A/P$.

Let us choose any monomial $a_1^{k_1}a_2^{k_2}...a_N^{k_N}$ in $P$.

Question what are the conditions on $P$ such that monomials non-containing subsequence $a_1^{k_1}a_2^{k_2}...a_N^{k_N}$ will form a basis in $A/P$ ?

Remark: obviously such monomials will always span $A/P$, because we can rewrite relation $P=0$ in the form $a_1^{k_1}a_2^{k_2}...a_N^{k_N} = P - a_1^{k_1}a_2^{k_2}...a_N^{k_N}$ so using this relation we can always avoid $a_1^{k_1}a_2^{k_2}...a_N^{k_N}$.

*Example Positive * Consider relation $ab=ba$, so factor is commutative, then obviously monomials not-containing $ab$ i.e. monomials $b^pa^q$ form basis.

(Moreover it is true for $ab = q ba$, even for $q=0$).

*Example Negative * Consider $a^2 = ab$, the monomials not-containing $a^2$ are linear dependent: for example $aba= a b^2$.

*Example non-clear * It not clear for whether in the case of my original question i.e. $P = a_1a_2a_3 + a_2a_3a_1 +a_3a_1a_2 - a_1a_3a_2 - a_2a_1a_3 -a_3a_2a_1$, the construction works or not ?

I do not fully understand David's argument: " Let $u_1$ be the lexicographically first $u_i$ and, if that monomial occurs as $u_i$ more than once, choose the one so that $v_i$ is lexicographically earliest. Then the monomial $u_1(a_1a_2a_3)v_1$ only occurs in the term $u1Δv1$ and no others, so it can't cancel out. So the right hand of the sum contains the nonstandard term $u1(a_1a_2a_3)v_1$, a contradiction."

Because: Consider $u_1 = a_3$ - "lexigraphically biggest", take $v_2=a_3$ then $u_1(a_1a_2a_3 )= a_3a_1a_2a_3 = (a_3a_1a_2)a_3 = (a_3a_1a_2)v_2$ . So the $u_1(a_1a_2a_3 )$ is contained in $Δv1$ . This contradicts the claim "monomial only occurs in the term and no others". Am I wrong ?

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By the way, "obviously such monomials will always span $A/P$" is actually not clear at all, - you can go into circles when trying to eliminate the given monomial! –  Vladimir Dotsenko Dec 7 '11 at 11:02

2 Answers 2

I don't know specifically about this example, but you should look at George Bergman's 1979 paper "The Diamond Lemma for Ring Theory," published in Advances.

The question he asks is: given an algebra $A$ defined by generators $x_1, \dots, x_n$ and relations in the form $ W_i = f_i$, where each $W_i$ is a monomial in the $x_j$'s and each $f_i$ is a polynomial in the $x_j$'s, when does the set of words not containing the monomials $W_i$ give a basis for $A$?

The relevant result is Theorem 1.2, so there's not much to get through in order to understand what's going on. Bergman's paper is very clear, very readable, and Section 2 of the paper gives some interesting background on how he came to the results.

The Diamond Lemma handles much more general situations than the one you describe (one homogeneous relation) so perhaps some simplification is possible, but that is where you should start.

Edit: I would add that the answer depends on which monomial you choose from the polynomial $P$. For instance, in your "negative example" with relation $a^2 = ab$, you point out that $aba = ab^2$, so that monomials not containing $a^2$ are not linearly independent. On the other hand, monomials not containing $ab$ are linearly independent. So there is some subtlety here.

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Well, if someone can give somewhat more self-contained answer it would be great ! –  Alexander Chervov Dec 6 '11 at 10:33
I can send you a copy of the paper, but what is your email address? –  MTS Dec 6 '11 at 17:18

What are the conditions on $P$ such that monomials non-containing the given monomial $m$ from $P$ will form a basis in $A/P$?

In general, there might be some coincidences here (so I am not giving any necessary conditions), but the result which David Speyer was using relies on the following claim (sufficient conditions):

  • Let us call a total ordering of monomial admissible if for every four monomials $m\le n$, $p\le q$ we have $mp\le nq$.

  • Let us assume that we chose an admissible ordering, and that $m$ is the leading term of $P$ with respect to that ordering. We say that a monomial $m$ forms a self-overlap if there exist monomials $k,l$ of degree less than the degree of $m$ such that $mk=lm$. Let me describe a procedure of "resolving" self-overlaps. For each self-overlap (there can be many, check $m=a^4$), let us form the polynomial $Pk-lP$. Try to "reduce" it: if you see in it a monomial divisible by $m$, replace that occurrence of $m$ by $m-\frac{1}{c}P$, where $c$ is the coefficient of $m$ in $P$. After a finite number of steps there will be nothing to reduce (since the ordering is admissible and $m$ is the leading term of $P$). If the result is zero, we call that overlap resolvable, otherwise we call it non-resolvable.

  • Now, the answer to your question is that the monomials not divisible by $m$ form a basis if (1) there is an admissible ordering ordering for which $m$ is the leading monomial of $P$, and (2) all overlaps are resolvable. This is a particular case of Bergman's Diamond Lemma suggested in the other answer.

[In the case you were interested in, the reasonably understood lexicographic order singles out $a_1a_2a_3$, it does not form any self-overlaps, so there is nothing to check.]

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