Is there a $3$-commutative algebra?

Question

Let $k$ be a field of characteristic $0$. If $m\ge2$, I denote $P_m$ the standard polynomial in $m$ non-commutating indeterminates: $$P_m(X_1,\dotsc,X_m)=\sum_{\sigma\in\mathfrak S_m}\epsilon(\sigma)X_{\sigma(1)}\dotsm X_{\sigma(m)}.$$ We say that a $k$-algebra $A$ is $m$-commutative if $P_m$ is an identity over $A$, that is $$\forall a_1,\dotsc,a_m\in A,\qquad P_m(a_1,\dotsc,a_m)=0,$$ and $P_{m-1}$ is not. Remark that $2$-commutativity is just commutativity.

Amitsur–Levitzki's Theorem is that $M_n(k)$ is $(2n)$-commutative.

My question is whether there exist (interesting) algebras that are $m$-commutative for an odd $m$. For instance, what would be an example of a $3$-commutative algebra ?

Edit. As mentioned by user49822, $A$ may not be unital, otherwise $(2k+1)$-commutativity implies $(2k)$-commutativity by specifying $a_m=1$. Thus there remains the question of whether a non-unital algebra can be $3$-commutative.

Salvatore's answer gives an explicit, practical example of a $3$-commutative algebra. However it is a bit too narrow to my taste, because its ideal of polynomial identities contains (presumably : is generated by) the monomial $X_1X_2X_3$. This leads me to strengthen my question:

What is an algebra $A$ whose ideal of polynomial identities is $(P_3)$ ?

Answer 1

An easy example of 3-commutative algebra is $T_3(F)$, the algebra of all strictly upper triangular 3-by-3 matrices.

Another class of examples is given by the quandle rings $F[X]$, where $F$ is a field and $X$ a trivial quandle with more than one element.

Answer 2

This answer may be a bit anticlimactic, but you can take free associative algebra without $1$ (same as ideal of positive degree elements in tensor algebra) and look at its quotient by all products of degree 3 (or any other $n$). It's easy to see that obtained algebra will be strictly $n$-commutative in your sense (it's almost obvious: tensor algebra is strictly graded). Study of properties and representations of such nil-rings was quite popular few decades ago, but now mostly fell into obscurity.

Similarly you can take an ideal generated in free nonunital algebra by all substitutions of its elements into $n$-commutativity relation; quotient of tensor algebra by constructed ideal will be a free algebra in a variety defined by that relation. That varieties do not coincide, as witnessed by example above, so their free algebras would not be isomorphic as well.

Answer 3

I have just thought of another interesting example, which is a bit peculiar, in that the 3-commutativity property arises for a natural subspace of an algebra which itself does not satisfy any identity. Specifically, consider the algebra $D_1$ of polynomial differential operators on the line; it consists of linear combinations of elements $x^i\partial^j$, where $\partial x - x\partial =1$ holds. It is well known that this algebra does not satisfy any polynomial identity. However, let us consider the subspace (in fact, a Lie subalgebra) $W_1$ of this algebra consisting of all vector fields; it consists of linear combinations of elements $x^i\partial$. I claim that any three elements of this subspace 3-commute (in some papers, this is described by saying that 3-commutativity is a weak identity of $D_1$ with respect to its Lie subalgebra $W_1$). This follows from the fact that $$ f\partial g\partial h\partial=f\partial g (h\partial + h')\partial= f(gh\partial + g'h+gh')\partial^2+f(gh'\partial+gh''+g'h')\partial, $$ which can be written as $$ fgh\partial^3+(fg'h+2fgh')\partial^2+(fgh''+fg'h')\partial, $$ and this is clearly killed by total anti-symmetrization in $f,g,h$, and does not follow from any stronger identity.

Answer 4

This variety of algebras is well studied. In particular, the description of the underlying vector space of the free algebra in this variety follows immediately from

"A note on the T-ideal generated by $s_3(x_1,x_2,x_3)$" by G. D. James, Israel Journal of Mathematics, 29 (1978), 105–112.

The main result of that paper, when translated to free algebras, says that the degree $d$ component of the free algebra generated by $V$ is the following Schur functor of $V$:

$S^{2}(V)\oplus S^{1,1}(V)$ for $d=2$

$S^{4}(V)\oplus 2S^{3,1}(V)\oplus S^{2,2}(V)$ for $d=4$

$S^{d}(V)\oplus 2S^{d-1,1}(V)$ for $d=3$ or $d\ge 5$.

This description, in my opinion, is sufficiently pleasant to have good intuition of what these algebras are, present them by generators and relations, write down convenient bases etc.

Answer 5

Let $a,b,c$ be basis vectors over some commutative associative algebra $X$. Consider the algebra $A$ spanned by these basis vectors over $X$, with the multiplication rules $pq=q$ for all $p,q \in \{a,b,c\}$, and extend to all of $A$ by requiring that the multiplication is bilinear over $X$. Thus $(ua+vb+wc)(xa+yb+zc)=(u+v+w)(xa+yb+zc)$.

Write $S(xa+yb+zc)=x+y+z \in X$ for convenience, so that one can write $xy=S(x)y$ for all $x,y \in A$.

This gives an associative multiplication, since $(xy)z=S(x)yz=x(yz)$. Now since $ab - ba = b-a$ the algebra isn't commutative. But if I write $xyz+yzx+zxy-yxz-zyx-xzy$ we have $xyz=S(x)S(y)z=yxz$. Thus $P_3(x,y,z)=0$ for all $x,y,z \in A$.