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Let $A$ be a $K$-algebra ($K$ a field). Let $\delta:A\to A$ be a $K$-derivation. We say that these $K$-derivations form a Lie algebra by the commutator bracket.

But $[\delta_1,\delta_2]=\delta_1\delta_2-\delta_2\delta_1$, is this by composition or piecewise, or what? It feels very lazy for them to write 'by the commutator bracket' since a priori, the derivations were only defined as a vector space, so talking about the commutator seems ambiguous.

Since $A$ is a $K$-algebra both: $\delta_1(\delta_2(f))\in A$ and $\delta_1(f)\delta_2(f)\in A$, so I am confused which it is. If the algebra $A$ is commutative, which some people require in the definition of an algebra, then the latter always vanishes, so I would guess that's nonsense, which suggests to me that it is indeed composition. Am I right?

user462339
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    Yes it's composition – ziggurism Sep 13 '17 at 03:57
  • @ziggurism Of course this is subjective, but does calling this a commutator seem like a perversion? The commutator bracket is a way of turning a ring/algebra into a lie algebra, which implies here that the algebra of derivations has multiplication given by composition. Do you agree, or I am wrong? – user462339 Sep 13 '17 at 17:18
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    the $K$-linear endomorphisms of (the underlying vector space of) $A$ are naturally a ring/associative algebra under composition. Derivations live in this algebra. Commutators give this associative algebra a Lie algebra structure, yes. The derivations are a Lie subalgebra of the full associative algebra with commutator. In short, I agree with everything you said, and find no perversion. – ziggurism Sep 13 '17 at 17:52
  • Thank you very much. This makes me happy @ziggurism – user462339 Sep 13 '17 at 20:46

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Thanks to Ziggurism above.

$\text{End}_K(A)$ is a ring/associative algebra under composition, and as a result, it makes sense to talk about the commutator. Indeed, the commutator makes use of the multiplication of the algebra, and thus above it is indeed composition, rather than pointwise multiplication.

user462339
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