3
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ 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 $\endgroup$ – ASKASK Feb 26 '17 at 9:43
  • 2
    $\begingroup$ @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. $\endgroup$ – Mariano Suárez-Álvarez Feb 26 '17 at 9:51
  • $\begingroup$ @MarianoSuárez-Álvarez good point. This is why I try not to math late at night. $\endgroup$ – ASKASK Feb 26 '17 at 9:52
  • 2
    $\begingroup$ $M_2$ 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 $1$. $\endgroup$ – Mariano Suárez-Álvarez Feb 26 '17 at 9:53
  • 1
    $\begingroup$ (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) $\endgroup$ – Mariano Suárez-Álvarez Feb 26 '17 at 10:14
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ IIRC this is actually included in Dietrich Burde's list of examples. Posting this anyway, because this example can be followed without any Lie algebra experience whatsoever. $\endgroup$ – Jyrki Lahtonen Feb 26 '17 at 12:09
  • $\begingroup$ This is a good idea, I think! $\endgroup$ – Dietrich Burde Feb 26 '17 at 12:45
5
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ Alphonse might be looking for an associative ring, though. $\endgroup$ – darij grinberg Feb 26 '17 at 15:43
  • $\begingroup$ @darijgrinberg $R$ is associative; $2$-step nilpotent Lie algebras do not only satisfy the Jacobi identity, but also the associative law. $\endgroup$ – Dietrich Burde Feb 26 '17 at 16:02
  • $\begingroup$ But is the pre-Lie structure associative as well? $\endgroup$ – darij grinberg Feb 26 '17 at 16:40
  • $\begingroup$ Yes, one can also find an associative pre-Lie structure for Heisenberg Lie algebras, given by matrix multiplication. $\endgroup$ – Dietrich Burde Feb 26 '17 at 16:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.