# Relaxing commutativity. For c1,c2 find q1,q2: (1) [c1,c2]=q1c2-q2c1 (2) [q1,q2]=0, (3)…

Consider some elements c1,c2 in some ring. Let me say that they are "relaxed commutative" if there exists two elements q1,q2, such that the following conditions hold:

(1) $[c_1,c_2]=c_1q_2-c_2q_1$

(2) $[q_1,q_2]=0$

(3) $[c_1,q_2]=[c_2,q_1]$

Question May be any two elements in any ring are always "relaxed commutative" or at least become such after embedding of the ring in some bigger ring ? (I am sure that NO, but how to prove it ?) . If $c_1, c_2$ are matrices say 2x2, 3x3 is this condition non-trivial, or any matrices are "relaxed non-commutative" ?

Examples (1) If [c1,c2]=0 then we can take q1=q2=0, so commutative elements are obviosly "relaxed commutative.

(2) If c1,c2 are representation of 2-dimensional Lie algebra i.e. [c1,c2]=k2c1-k1c2, for scalars k1,k2, then they are also "relaxed commutative" : q1=k1*Id; q2=k2*Id.

[EDIT one more question] Mark Sapir proposed 2x2 matrices c1,c2 for which even the equations (1),(3) do not have solutions q1,q2. This may rise the question whether the only matrices which c1,c2 which satisfy the conditions above comes from example 2 ? At least in 2x2 case ?

Motivation These relations appeares in my paper http://arxiv.org/abs/1203.5759 page 22 bottom, formula 64. Few days ago I have found another interpretation for them, so I feel they are quite natural in certain questions. What I want to understand how restrictive they are.

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The answers to the two parts of the first part of the question are "no" and "probably yes". Take the free associative algebra (over any ring, say, $\mathbb{Z}$) with free generators $x,y$. Then $x,y$ do not commute in your sense. The main obstacle is that $q_1,q_2$ in your definition must really commute and commuting elements in free rings have been described by Bergman. The description is very similar to the description of commuting elements in a free group/semigroup, it immediately implies that $q_1,q_2$ satisfying your conditions do not exist.

For every $c_1,c_2$, there probably exists a bigger ring, where $c_1,c_2$ "relaxly commute". Denote the initial ring by $C$. Your question is equivalent to the following one. Consider the ring $R$ given by the generators of $C$ plus $q_1,q_2$ and relations (1),(2),(3) plus all relations of $C$, and ask whether the natural homomorphism $\phi$ from $\langle c_1,c_2\rangle$ into $R$ is injective. I will prove it for the case when $C=\langle c_1, c_2\rangle$ is free, i.e. has no non-trivial relations (the general case should be similar, but more difficult). Notice that the relations (1), (2), (3) do not have any overlaps (use the order $c_1 < c_2 < q_1 < q_2$). Thus these relations form a (non-commutative) Groebner-Shirshov basis of $R$. Since every leading term in your relations contains $q_2$, no non-zero (non-commutative) polynomial in $c_1, c_2$ is equal to 0 in $R$, hence $\phi$ is injective. For general $C$, I would recommend reading this paper.

About the second part of the question. The answer is "no" (which also gives the "no" answer to the first part of the first part of the question). The matrices $\left(\begin{array}{ll} 1 & 2 \\\ 0 & 1\end{array}\right), \left(\begin{array}{ll} 1 & 0 \\\ 2 & 1\end{array}\right)$ do not relaxly commute in $M_2(\mathbb{C})$. In fact conditions (1) and (3) cannot be satisfied together for these matrices. It is easy to see: these conditions can be rewritten as a system of linear equations with entries of $q_1,q_2$ as unknowns, and this system is inconsistent.

Update 1. Here is the Maple program that proves inconsistency:

restart;with(linalg):

A:=matrix([[1,2],[0,1]]); B:=matrix([[1,0],[2,1]]):

C:=evalm(A&*B-B&*A):

P:=matrix([[a,b],[c,d]]): Q:=matrix([[x,y],[z,t]]):

W:=evalm(A&*P-B&*Q-C):

V:=evalm(A&*Q-Q&*A-B&*P+P&*B):

zz:=solve({W[1,1]=0, W[1,2]=0, W[2,1]=0,W[2,2]=0,V[1,1]=0,V[1,2]=0}):

subs(zz,evalm(V));

Output:

$$\left[ \begin {array}{cc} 0&0\\\ 16&0\end {array} \right]$$

Update 2. The matrices $\left(\begin{array}{lll} 1 & 1 & 0\\\ 0 & 1 & 0\\\ 0 & 0 & 1\end{array}\right), \left(\begin{array}{lll} 1 & 0 & 0\\\ 0 & 1 & 1\\\ 0 & 0 & 1\end{array}\right)$ relaxly commute. One can take $q_1=\left(\begin{array}{lll} 0 & 0 & -1\\\ 0 & 0 & 0\\\ 0 & 0 & 0\end{array}\right), q_2=\left(\begin{array}{lll} 0 & 0 & 0\\\ 0 & 0 & 0\\\ 0 & 0 & 0\end{array}\right)$.
The Lie algebra generated by these matrices (inside the Lie algebra of all $3\times 3$-matrices) is of dimension 4 spanned by the identity matrix and three matrix units $E_{i,j}$, $1\le i < j\le 3$. So it is the direct product of the Heisenberg Lie algebra and the 1-dimensional Lie algebra.