# nLab derivation

This is about derivations on algebras. For derivations in logic, see deduction.

### Context

#### Higher algebra

higher algebra

universal algebra

# Contents

## Definitions

### Derivations on an algebra

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

$d : A \to A$

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

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

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

#### In a cancellative algebra

For $A$ a cancellative algebra, where for all $c \neq 0$, $c a = c b$ implies $a = b$ and $a c = b c$ implies $a = b$, the derivation satisfies the following identity: for any scalar $c$ in $k$, $d(c) = 0$. This follows from $k$-linearity, which says that for any scalar $c$ in $k$ and for any non-zero element $f$ in $A$, $d(c f) = c d(f)$ and $d(f c) = d(f) c$. The Leibniz rule states that $d(c f) = c d(f) + d(c) f$ and $d(f c) = d(f) c + f d(c)$. It follows that $d(c) f = 0$ and $f d(c) = 0$, and due to the cancellative property and the fact that $f$ is non-zero, it follows that $d(c) = 0$.

#### In a division algebra

For $A$ a division algebra, with left division $\backslash$ and right division $/$ the derivation satisfies the left and right quotient identities for $f$ in $A$ and non-zero $g$ in $A$:

$D(g\backslash f) = g\backslash D(f) - g\backslash (D(g) (g\backslash f)) \,,$

and

$D(f/g) = D(f)/g - ((f/g) D(g))/g \,,$

which reduces to the regular quotient rule? if the division algebra is associative and commutative:

$D\left(\frac{f}{g}\right) = \frac{D(f)}{g} - \frac{D(g) f}{g^2}$

### 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

$d : A \to N$

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

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

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.

A special case is where $A$ is a group algebra $k G$ of a group $G$, and $N$ is a left $G$-module, regarded as a bimodule where the right action is trivial. Here a derivation is also called a 1-cocycle, as used in group cohomology.

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

$d(a b) = d(a) b + (-1)^{p q} a d(b)$

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 $\epsilon: A \to B$, is a module homomorphism $d: A \to B$ such that

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

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

###### Proposition

Let $k Alg$ denote the category of commutative $k$-algebras, and $k Alg/k$ the slice, i.e., the category of commutative $k$-algebras $A$ equipped with an augmentation $\epsilon: A \to k$. Then the functor

$Der: (k Alg/k)^{op} \to k Mod$

which takes an augmented algebra to the module of augmented derivations $d: A \to k$ is represented by the augmented algebra $eval_0: k[x]/(x^2) \to k$.

###### Proof

Indeed, an algebra map $\delta: A \to k[x^2]/(x^2)$ which renders the triangle

$\array{ A & \stackrel{\delta}{\to} & k[x]/(x^2) \\ & \mathllap{\epsilon} \searrow & \downarrow \mathrlap{eval_0} \\ & & k }$

commutative may be uniquely written in the form $\delta(a) = \epsilon(a) + d(a)x$ for some $k$-module map $d: A \to k$, and then we have

$\array{ \epsilon(a b) + d(a b)x & = & \delta(a b) \\ & = & \delta(a)\delta(b) \\ & = & (\epsilon(a) + d(a)x)(\epsilon(b) + d(b)x) \\ & = & \epsilon(a)\epsilon(b) + (\epsilon(a)d(b) + d(a)\epsilon(b))x + d(a)d(b) x^2 \\ & = & \epsilon(a b) + (\epsilon(a)d(b) + d(a)\epsilon(b))x }$

and we conclude $d(a b) = \epsilon(a)d(b) + d(a)\epsilon(b)$, so that $d$ is an augmented derivation. Of course the calculation may be run in reverse, so that $\delta$ is an augmented algebra homomorphism precisely when $d$ is a derivation.

### Further variations

There are many further extensions, for examples derivations with values in an $A$-bimodule $M$ forming $Der_k(A,M) \subset Hom_k(A,M)$ (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 $A_\infty$-algebras and $L_\infty$-algebras. See also derivation on a group, which uses a modified Leibniz rule: $d(a b) = d(a) + a d(b)$.

### Derivations on algebras over a dg-operad

More generally, there is a notion of derivation for every kind of algebra over an operad over a dg-operad (at least).

###### Definition

Let $\mathcal{O}$ be a dg-operad (a chain complex-enriched operad). For $A$ an $\mathcal{O}$-algebra over an operad and $N$ a module over that algebra a derivation on $A$ with values in $N$ is a morphism

$v : A \to N$

in the underlying category of graded vector spaces, such that for each $n \gt 0$ we have a commuting diagram

$\array{ \mathcal{O}(n) \otimes A^{\otimes n} &\stackrel{}{\to}& A \\ {}^{\mathllap{\sum_{a+b=n-1} id \otimes id^{\otimes a} \otimes v \otimes id^{\otimes b}}}\downarrow && \downarrow^{\mathrlap{v}} \\ \oplus_{a+ b = n-1} \mathcal{O}(n) \otimes A^{\otimes a} \otimes N \otimes A^{\otimes b} &\to& N } \,,$

where the top horizontal morphism is that given by the $\mathcal{O}$-algebra structure of $A$ and the bottom that given by the $A$-module structure of $N$.

This appears as (Hinich, def. 7.2.1).

The theory of tangent complexes, Kähler differentials, etc. exists in this generality for derivations on algebras over an operad.

### Generalization to arbitrary $(\infty,1)$-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

$(r_1, n_1) \cdot (r_2, n_2) := (r_1 r_2, n_1 r_2 + n_2 r_1) \,.$

This comes with a natural morphism of rings

$G(N) \to R$

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 $p : Mod \to Ring$ 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 $Mod \to Ring$ 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

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

An $(\infty,1)$-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.

More algebraic examples:

• Let $A$ be any algebra over a ring. The constant function $D(a) = 0$ for all $a \in A$ is a derivation.

• Let $R[[x]]$ be a formal power series over a ring $R$. Then the formal derivative? is a derivation.

• A less obvious one: Let $R[X_{1},...,X_{n}]$ be a polynomial algebra over a commutative rig $R$. Define $L(P)=\underset{1 \le k \le n}{\sum}\frac{\partial P}{\partial X_{i}}X_{i}$. Then $L:R[X_{1},...,X_{n}] \rightarrow R[X_{1},...,X_{n}]$ is a derivation. Moreover, if we suppose that $R$ is a multiplicatively cancellable rig, then the homogeneous polynomials of degree $n$ are exactly the solutions of the differential equation $L(P)=n.P$ (see the page homogeneous polynomials).

### 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-forms yields a morphism

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

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 $C^\infty(X)$ its algebra of smooth functions. Then the morphism

$Vect(X) \to Der(C^\infty(X))$

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

###### Proof

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

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

$f(x) = f(0) + \sum_i x_i g_i(x)$

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

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

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

$v_\delta \coloneqq \sum_i \delta(x_i) \frac{\partial}{\partial x_i}$

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

### 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 trivial.

###### Proof

Observe that generally every derivation vanishes on the function 1 that is constant on $1 \in \mathbb{R}$. Therefore it is sufficient to show that if $f \in C(X)$ vanishes at $x_0 \in X$ also $\delta(f)$ vanishes at $x_0$, 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

$f = g_1 g_2$

with

$g_1 = \sqrt{|f|}$

and

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

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

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

Derivations on algebras over a dg-operad are discussed in section 7 of