nLab
derivation

Definitions

Derivations on an algebra

For $A$ an algebra (over some ring $k$), a derivation on $A$ is a $k$-linear morphism

$$d:A\to A$$d : A \to A

such that for all $a,b\in A$ we have

$$d(ab)=d(a)b+ad(b),$$d(a b) = d(a) b + a d (b) \,,

This identity is called the Leibniz rule; compare it to the product rule in ordinary calculus (first written down by Gottfried Leibniz?).

Derivations with values in a bimodule

For $A$ an algebra (over some ring $k$) and $N$ a bimodule over $A$, a derivation of $A$ with values in $N$ is a $k$-linear morphism

$$d:A\to N$$d : A \to N

such that for all $a,b\in A$ we have

$$d(ab)=d(a)\cdot b+a\cdot d(b),$$d(a b) = d(a) \cdot b + a \cdot d (b) \,,

where on the dot on the right-hand side denotes the right (first term) and left (second term) action of $A$ on the bimodule $N$.

The previous definition is a special case of this one, where the bimodule is $N=A$, the algebra itself with its canonical left and right action on itself.

Graded derivations

A graded derivation of degree $p$ on a graded algebra $A$ is a degree-$p$ graded-module homomorphism $d:A\to A$ such that

$$d(ab)=d(a)b+(-1{)}^{\mathrm{pq}}ad(b)$$d(a b) = d(a) b + (-1)^{pq} a d(b)

whenever $a$ is homogeneous of degree $q$. (By default, the grade is usually $1$, or sometimes $-1$.)

Augmented derivations

An augmented derivation on an algebra $A$, augmented by an algebra homomorphism $ϵ:A\to B$, is a module homomorphism $d:A\to B$ such that

$$d(ab)=d(a)ϵ(b)+ϵ(a)d(b).$$d(a b) = d(a) \epsilon(b) + \epsilon(a) d(b) .

If you think about it, you should be able to figure out the definition of an augmented graded derivation.

Further variations

There are many further extensions, for examples derivations with values in an $A$-bimodule $M$ forming ${\mathrm{Der}}_k(A,M)\subset {\mathrm{Hom}}_k(A,M)$ (see also double derivation), skew-derivations in ring theory (with a twist in the Leibniz rule given by an endomorphism of a ring) and the dual notion of a coderivation of a coalgebra. The latter plays role in Koszul-dual definitions of $A_{\mathrm{\infty }}$-algebras and $L_{\mathrm{\infty }}$-algebras. See also derivation on a group, which uses a modified Leibniz rule: $d(ab)=d(a)+ad(b)$.

Generalization to arbitrary $(\mathrm{\infty }\mathrm{,1})$-categories

Another equivalent reformulation of the notion of derivations turns out to be useful for the vertical categorification of the concept:

for $N$ an $R$-module, there is the nilpotent extension ring $G(N):=N\oplus R$, equipped with the product operation

$$(r_1,n_1)\cdot (n_2,r_2):=(r_1,r_2,n_1{r}_2+n_2{r}_1).$$(r_1, n_1) \cdot (n_2, r_2) := (r_1, r_2, n_1 r_2 + n_2 r_1) \,.

This comes with a natural morphism of rings

$$G(N)\to R$$G(N) \to R

given by sending the elements of $N$ to 0. One sees that a derivation on $R$ with values in $N$ is precisely a ring homomorphism $R\to G(N)$ that is a section of this morphism.

In terms of the bifibration $p:\mathrm{Mod}\to \mathrm{Ring}$ of modules over rings, this is the same as a morphism from the module of Kähler differentials ${\Omega }_K(R)$ to $N$ in the fiber of $p$ over $R$.

While this is a trivial restatement of the universal property of Kähler differentials, it is this perspective that vastly generalizes:

we may replace $\mathrm{Mod}\to \mathrm{Ring}$ by the tangent (∞,1)-category projection $p:T_C\to C$ of any (∞,1)-category $C$. The functor that assigns Kähler differentials is then replaced by a left adjoint section of this projection

$$\Omega :C\to T_C.$$\Omega : C \to T_C \,.

An $(\mathrm{\infty }\mathrm{,1})$-derivation on an object $R$ with coefficients in an object $N$ in the fiber of $T_C$ over $R$ is then defined to be morphism $\Omega (R)\to N$ in that fiber.

More discussion of this is at deformation theory.

Examples

Derivations on an algebra

• Let $A$ consist of the smooth real-valued functions on an interval in the real line. Then differentiation is a derivation; this is the motivating example.
• Let $A$ consist of the holomorphic functions? on a region in the complex plane. Then differentiation is a derivation again.
• Let $A$ consist of the meromorphic functions? on a region in the complex plane. Then differentiation is still a derivation.
• Let $A$ consist of the smooth functions on a manifold (or generalized smooth space) $X$. Then any tangent vector field on $X$ defines a derivation on $A$; indeed, this serves as one definition of tangent vector field.
• Let $A$ consist of the germs of differentiable functions near a point $p$ in a smooth space $X$. Then any tangent vector at $a$ on $X$ defines a derivation on $A$ augmented by evaluation at $a$; again, this serves to define tangent vectors.
• Let $A$ consist of the smooth differential forms on a smooth space $X$. Then exterior differentiation is a graded derivation (of degree $1$).
• In any of the above examples containing the adjective ‘smooth’, replace it with $C^k$ and augment $A$ by the inclusion of $C^k$ into $C^{k-1}$. Then we have an augmented derivation.

There should be some more clearly algebraic examples (other than obvious things like restricting the above to polynomials), but I don't know how to state them.

Derivations with values in a bimodule

The standard example of a derivation not on an algebra, but with values in a bimodule is a restriction of the above case of the exterior differential acting on the deRham algebra of differential forms. Restricting this to 0-fomrs yields a morphism

$$d:C^{\mathrm{\infty }}(X)\to {\Omega }^1(X)$$d : C^\infty(X) \to \Omega^1(X)

where ${\Omega }^1(X)$ is the space of 1-forms on $X$, regarded as a bimodule over the algebra of functions in the obvious way.

A variation of this example is given by the Kähler differentials. These provide a universal derivation in some sense.

Derivations of smooth functions

Proposition Let $X$ be a smooth manifold and $C^{\mathrm{\infty }}(X)$ its algebra of smooth functions. Then the morphism

$$\mathrm{Vect}(X)\to \mathrm{Der}(C^{\mathrm{\infty }}(X))$$Vect(X) \to Der(C^\infty(X))

that sends a vector field $v$ to the derivation $v(-):C^{\mathrm{\infty }}(X)\to C^{\mathrm{\infty }}(X)$ is a bijection.

Proof This is true because $C^{\mathrm{\infty }}(X)$ satisfies the Hadamard lemma.

Since every smooth manifold is locally isomorphic ${ℝ}^n$, it suffices to consider this case. By the Hadamard lemma every function $f\in C^{\mathrm{\infty }}({ℝ}^n)$ may be written as

$$f(x)=f(0)+\sum _i{x}_i{g}_i(x)$$f(x) = f(0) + \sum_i x_i g_i(x)

for smooth $\{g_i\in C^{\mathrm{\infty }}(X)\}$ with $g_i(0)=\frac{\partial f}{\partial x_i}(0)$. Since any derivation $\delta :C^{\mathrm{\infty }}(X)\to C^{\mathrm{\infty }}(X)$ satisfiesthe the Leibniz rule, it follows that

$$\delta (f)(0)=\sum _i\delta (x_i)\frac{\partial f}{\partial x_i}(0).$$\delta(f)(0) = \sum_i \delta(x_i) \frac{\partial f}{\partial x_i}(0) \,.

Similarly, by translation, at all other points. Therefore $\delta$ is already fixed by its action of the coordinate functions $\{x_i\in C^{\mathrm{\infty }}(X)\}$. Let $v_{\delta }\in T{ℝ}^n$ be the vector field

$$v_{\delta }==\sum _i\delta (x_i)\frac{\partial }{\partial x_i}$$v_\delta = = \sum_i \delta(x_i) \frac{\partial}{\partial x_i}

then it follows that $\delta$ is the derivation coming from $v_{\delta }$ under $\mathrm{Vect}(X)\to \mathrm{Der}(C^{\mathrm{\infty }}(X))$.

Derivations of continuous functions

Let now $X$ be a topological manifold and $C(X)$ the algebra of continuous real-valued functions on $X$.

Proposition The derivations $\delta :C(X)\to C(X)$ are all tivial.

Proof Observe that generally every derivation vanishes on the function 1 that is constant on $1\in ℝ$. Therefore it is sufficient to show that if $f\in C(X)$ vanishes at $x_0\in X$ also $\delta (f)$ vanishes att $x_0$, because we may write every function $g$ as $(g-g(x_0))+g(x_0)$.

So let $f\in C(X)$ with $f(x_0)=0$. Then we may write $f$ as a product

$$f=g_1{g}_2$$f = g_1 g_2

with

$$g_1=\sqrt{\mid f\mid }$$g_1 = \sqrt{|f|}

and

$$g_2:x\mapsto \{\begin{array}{cc}f(x)/\sqrt{\mid f(x)\mid }& \mid f(x)\ne 0\\ 0& \mid f(x)=0\end{array}.$$g_2 : x \mapsto \left\{ \array{ f(x)/\sqrt{|f(x)|} & | f(x) \neq 0 \\ 0 & | f(x) = 0 } \right. \,.

Notice that indeed both functions are continuous. (But even if $X$ is a smooth manifold and $f$ a smooth function, $g_1$ will in general not be smooth.)

But also both functions vanish at $x_0$. This implies that

$$\delta (f)(x_0)=\delta (g_1)(x_0)g_2(x_0)+g_1(x_0)\delta (g_2(x_0))=0.$$\delta(f)(x_0) = \delta(g_1)(x_0) g_2(x_0) + g_1(x_0) \delta(g_2(x_0)) = 0 \,.