Malcev algebra

Remarks

With the Jacobian $J(x, y, z) = (xy)z + (yz)x + (zx)y$, the Malcev identity is equivalent, for anticommutative products, to $J(x, y, xz) = J(x, y, z)\, x$.

Example: every Lie algebra is a Malcev algebra, since $J = 0$ makes both sides of $J(x, y, xz) = J(x, y, z)\, x$ vanish. If $A$ is an alternative algebra, then $A$ with the commutator $[x, y] = xy - yx$ is a Malcev algebra.

Non-example: let $A$ have basis $x, y, z$ with $xy = y$, $yx = -y$, $yz = x$, $zy = -x$, and all other products of basis vectors zero. This product is anticommutative, and $J(x, y, z) = yz + xx + 0 = x$. In the form $J(u, v, uw) = J(u, v, w)\, u$, take $u = y$, $v = z$, $w = x$. The left side is $J(y, z, yx) = -J(y, z, y) = 0$, since $J$ is alternating. The right side is $J(y, z, x)\, y = J(x, y, z)\, y = xy = y \neq 0$. So the Malcev identity fails.