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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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Quibble: the fact that you use $[ \cdot, \cdot ]$ for both the Lie bracket on $\mathfrak{g}$ and the commutator on $\text{End}(\mathfrak{g})$ is mildly confusing, since it looks like you could be taking the Lie bracket of $[x, z]$ and $[y, z]$ pointwise. – Qiaochu Yuan Jun 25 '12 at 8:33
I am amused to discover that you had asked the same question some years earlier. – Zhen Lin Feb 9 '14 at 19:35

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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The self-action of a monoid is faithful because the monoid operation has a unit element. I imagine there are semigroups whose self-action is not faithful though... – Zhen Lin Jun 25 '12 at 8:14
@Zhen: sure, for any set $S$ and any $s \in S$ there is a semigroup structure where every product is $s$. But my point is if one thinks of there being an "analogy functor" from groups to Lie algebras which sends associativity to the Jacobi identity then this "analogy functor" ought to send Cayley's theorem to Cayley's theorem. (My second point was misstated. What is true is that the Jacobi identity is equivalent to the statement that $y \mapsto [x, y]$ is always a derivation, which is the infinitesimal analogue of the statement that $h \mapsto ghg^{-1}$ is a group homomorphism.) – Qiaochu Yuan Jun 25 '12 at 8:17
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. – Zhen Lin Jun 25 '12 at 10:51

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