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Do you have an example of a non-commutative ring $R$ (possibly not unital) such that $$[[x,y],z]=0 \qquad \forall x,y,z \in R$$

where $[x,y]=xy-yx$ ?

Of course this is true if $R$ is commutative, since $[x,y]=0$ for any $x,y \in R$. I tried $R = M_2(\Bbb F_p)$, but it seems painful to check whether $[[x,y],z]=0$ holds or not. Maybe there is an easier idea? Thank you very much!

Alphonse
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    I would guess that it is probably not too difficult to find a Lie algebra with that property, so then by poincare birkhoff Witt, the universal enveloping algebra of such a Lie algebra must be what you are looking for – ASKASK Feb 26 '17 at 09:43
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    @ASKASK, it does not work like that. That the relation is satisfied in the Liee algebra certainly does not imply that all the elements of the enveloping algebra satisfy the identity. – Mariano Suárez-Álvarez Feb 26 '17 at 09:51
  • @MarianoSuárez-Álvarez good point. This is why I try not to math late at night. – ASKASK Feb 26 '17 at 09:52
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    $M_2$ does not work: it is not true that every element of the form $[x,y]$ is central. The center is one-dimensional and the set of elements which are commutators coincides with the set of the elements of trace zero, which has codimension $1$. – Mariano Suárez-Álvarez Feb 26 '17 at 09:53
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    (What is true in $M_2$ is that $[[x,y]^2,z]=0$ identically, so that the square of every commutator is central. This is called Hall's identity) – Mariano Suárez-Álvarez Feb 26 '17 at 10:14
  • Quantum mechanical operators for angular momentum, maybe? – Oscar Lanzi Feb 26 '17 at 11:35
  • @OscarLanzi : what do you mean? I don't these notions actually :-(. But I would be happy to learn about them. – Alphonse Feb 26 '17 at 12:23

3 Answers3

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Let $F$ be the free algbra generated by letters $a$ and $b$, endow it with its usual grading, and let $I$ be the ideal of $F$ generated by all elements of the form $[[x,y],z]$ with $x$, $y$ and $z$ in $F$. Since the iterated bracket is trilinear, it is enough to consider homogeneous elements $x$, $y$ and $z$ when generating $I$, and therefore $I$ is a homogeneous ideal. The ring $F/I$ satisfies the desired identity.

We want to show that $R$ is non trivial and non-commutative. If the three letters are homogeneous, the element $[[x,y],z]$ cannot be nonzero and have degree $\leq 2$. Indeed, that can only happen if one of $x$, $y$ or $z$ has degree zero, and then the iterated bracket is itself zero. It follows that $I$ does not contain $[a,b]$, so that $R$ is nontrivial and noncommutative.

Mariano Suárez-Álvarez
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The simplest example I can think of is the (non-unital) algebra of strictly upper triangular $3\times3$-matrices: $$ R=\left\{\left(\begin{array}{ccc}0&a&b\\0&0&c\\0&0&0\end{array}\right)\bigg\vert\ a,b,c\in\Bbb{R}\right\}. $$ All the commutators have $a=c=0$, but the $b$-component of $[x,y]$ is non-zero iff the $a$-component of $x$ and the $c$-component of $y$ are both non-zero. Therefore we also get $[[x,y],z]=0$ for all $x,y,z\in R$.

Unless I missed something we can actually make $R$ unital by allowing all the upper triangular matrices such that the diagonal entries are all equal. After all, adding a scalar multiple of $I_3$ is not going to change the commutators one bit.

Jyrki Lahtonen
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An example is the $(2n+1)$-dimensional Heisenberg Lie Algebra $R$, together with a non-commutative bilinear product $(x,y)\mapsto x\cdot y$ satisfying $$ x\cdot y-y\cdot y=[x,y]. $$ Such a bilinear product exists (in fact, several such products, also non-commutative ones, and also given by matrix multiplication) and is called a pre-Lie algebra structure on $R$. Of course we have $[[x,y],z]=0$, since the Heisenberg Lie algebra is $2$-step nilpotent as a Lie algebra, i.e., $[L,L]\subseteq Z(L)$. Here we consider $L$ as a matrix Lie algebra (taking a faithful linear representation if necessary).

Dietrich Burde
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