Commutators (from physics background) are defined as $[X, Y]=XY-YX$. Do they satisfy the Jacobi identity for free?
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$\begingroup$This is even true in much greater generality. Algebra products $(x,y)\mapsto x\cdot y$ are called Lie-admissible, if$$ [x,y]:=x\cdot y-y\cdot x $$satisfies the Jacobi identity. There is a large literature in physics and mathematics on these Lie-admissible algebras. One prominent example are pre-Lie algebras. Here pre-Lie algebra means that we have$$ (x,y,z)=(y,x,z) $$for all $x,y,z$, where $(x,y,z):=x\cdot (y\cdot z)-(x\cdot y)\cdot z$ denotes the associator. If this product is even associative, then clearly$(x,y,z)=0=(y,x,z)$, so that the Jacobi identity holds for the commutator.
$\endgroup$ $\begingroup$Yes. Indeed, if $X$ $Y$, and $Z$ are square matrices of the same size, then
$[X, Y], Z] + [[Y, Z], X] + [[Z, X], Y] = [(XY - YX), Z] + [(YZ - ZY), X] + [(ZX - XZ), Y]$$= XYZ - YXZ - ZXY + ZYX + YZX - ZYX$$- XYZ + XZY + ZXY - XZY - YZX + YXZ = 0. \tag 1$
It is easy to see that this holds when $X$, $Y$, $Z$ are elements of any associative algebra; they needn't be matrices.
If the Jacobi identity isn't free, it certainly isn't expensive!
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