Leibnitz identity

Question

Is there any explanation (probably, categorial) why operators, satisfying Leibnitz rule are so omnipresent and natural in mathematics?

Answer 1

To expand on Mariano's comment, let $A$ be, say, an $\mathbb{R}$-algebra and $\phi : \mathbb{R} \to \text{Aut}(A)$ a one-parameter group of automorphisms of it. Suppose that we can make sense of the derivative of $\phi$. Then letting $D = d \phi_0(1) \in \text{End}(A)$ we find that differentiating the condition that $\phi$ is a family of automorphisms gives that $D$ satisfies the Leibniz identity. In other words, the Leibniz identity is the infinitesimal analogue of preserving multiplication. (For example, the standard derivative of a one-variable function is the derivation associated to the one-parameter group of translations.)

In fact $A$ need not be associative for this argument to go through. This makes the above a nice way to think about, for example, the Jacobi identity for a Lie bracket: it says precisely that $[x, -]$ is a derivation, and this is simply because it is the infinitesimal version of the adjoint action.

A more algebraic way to think about this condition is as follows. Suppose $f : A[\epsilon]/\epsilon^2 \to A[\epsilon]/\epsilon^2$ is an $\epsilon$-linear homomorphism, where $\epsilon$ is central. We think of this as an infinitesimal deformation of a homomorphism of $A$. Then $f$ is determined by what it does to $A$, and we can write

$$f(a) = f_0(a) + \epsilon f_1(a).$$

Then the condition that $f$ is linear gives that $f_0, f_1$ are linear, and the condition that $f$ preserves multiplication gives that

$$f(ab) = (f_0(a) + \epsilon f_1(a)))(f_0(b) + \epsilon f_1(b)) = f_0(ab) + \epsilon f_1(ab)$$

which gives that $f_0$ is a homomorphism and $f_1$ is an $f_0$-derivation. This is a fairly concrete way to think about the role of derivations in Hochschild cohomology, especially since it generalizes to infinitesimal deformations of the multiplication of $A$ as well.