# Derivations

A map ${D}$ is a derivation if it satisfies the Leibniz rule: ${D(ab)=D(a)b+aD(b)}$. To make sense out of this, we need to be able to

• multiply arguments of ${D}$ together
• multiply values of ${D}$ by arguments of ${D}$

For example, if ${D\colon R\to M}$ where ${R}$ is a ring and ${M}$ is a two-sided module over ${R}$, then all of the above makes sense. In practice it often happens that ${M=R}$. In this case, the commutator (Lie bracket) of two derivations ${D_1,D_2}$ is defined as ${[D_1,D_2]=D_1\circ D_2-D_2\circ D_1}$ and turns out to be a derivation as well. If ${R}$ is also an algebra over a field ${K}$, then ${K}$-linearity of ${D}$ can be added to the requirements of being a derivation, but I am not really concerned about that.

What I am concerned about is that two of my favorite instances of the Leibniz rule are not explicitly covered by the ring-to-module derivations. Namely, for smooth functions ${\varphi\colon{\mathbb R}\rightarrow{\mathbb R}}$, ${F\colon{\mathbb R}\rightarrow{\mathbb R}^n}$ and ${G\colon{\mathbb R}\rightarrow{\mathbb R}^n}$ we have

$\displaystyle (\varphi F)' = \varphi' F + \varphi F' \quad \text{and} \quad (F\cdot G)' = F'\cdot G+F\cdot G' \ \ \ \ \ (1)$

Of course, ${{\mathbb R}^n}$ could be any ${{\mathbb R}}$-vector space ${V}$ with an inner product.

It seems that the most economical way to fit (1) into the algebraic concept of derivation is to equip the vector space ${{\mathbb R}\oplus V}$ with the product

$\displaystyle (\alpha,u)(\beta,v)= (\alpha\beta+u\cdot v, \alpha v+\beta u) \ \ \ \ \ (2)$

making it a commutative algebra over ${{\mathbb R}}$. Something tells me to put ${-u\cdot v}$ there, but I resist. Actually, I should have said “commutative nonassociative algebra”:

$\displaystyle \{(\alpha,u)(\beta,v)\}(\gamma,w) = (\alpha\beta+u\cdot v, \alpha v+\beta u) (\gamma,w) \\ \\ = (\alpha\beta\gamma+\gamma u\cdot v+ \alpha v\cdot w+\beta u\cdot w, \alpha\beta w + \gamma \alpha v+ \gamma \beta u +(u\cdot v) w)$

Everything looks nice, except for the last term ${(u\cdot v) w}$, which destroys associativity.

Now we can consider maps ${{\mathbb R}\rightarrow {\mathbb R}\oplus V}$, which are formal pairs of scalar functions and vector-valued functions. The derivative acts component-wise ${(\varphi,F)'=(\varphi',F')}$ and according to (1), it is indeed a derivation:

$\displaystyle \left\{(\varphi,F)(\psi,G)\right\}'= (\varphi,F)'(\psi,G)+(\varphi,F)(\psi,G)' \ \ \ \ \ (3)$

Both parts of (1) are included in (3) as special cases ${(\varphi,0)(0,F)}$ and ${(0,F)(0,G)}$.

If (2) has a name, I do not know it. Clifford algebras do a similar thing and are associative, but they are also larger. If I just want to say that (1) is a particular instance of a derivation on an algebra, (2) looks like the right algebra structure to use (maybe with ${-u\cdot v}$ if you insist). If ${V}$ has no inner product, the identity ${(\varphi F)' = \varphi' F + \varphi F'}$ can still be expressed via (2) using the trivial inner product ${u\cdot v=0}$.

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