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About a year and a half ago, I was at a talk by Martin Hyland where he suggested that the Jacobi identity is to the associative law as the anticommutative law is to the commutative law. I think this was in the context of some kind of duality for operads, but I didn't understand at the time.

More recently, I've come to understand that the associative law and the Jacobi identity are essentially the same in the following sense: they both make self-action representations possible. Indeed, the associative law says $(x \cdot y) \cdot {-} = (x \cdot {-}) \circ (y \cdot {-})$ and the Jacobi identity says $[[x, y], {-}] = [[x, {-}], [y, {-}]]$

Question. Is there a way to make this precise in the language of universal algebra or (enriched) category theory, and are there other examples?

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    I am amused to discover that you had asked the [same question](http://mathoverflow.net/questions/21152) some years earlier.2014-02-09

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This isn't an answer to your question, but this analogy has always annoyed me because the action of a monoid (say) on itself by left multiplication is always faithful, but the action of a Lie algebra on itself by left bracket is generally not.

A better analogy is to think of the action of a group on itself by conjugation, which is not always faithful but on the other hand preserves group structure. Analogously, the left bracket is a derivation for itself (and this is also equivalent to the Jacobi identity). Indeed, differentiating the conjugation action of a Lie group $G$ exactly gets you the Lie bracket on $\mathfrak{g}$.

This vaguely suggests that you might want to look at the literature on quandles.

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    Interesting suggestion about the quandles, thanks. The self-distributivity axiom confirms my intuition that distributivity is what is needed to make a self-action into an endomorphism of the appropriate structure.2012-06-25