# nLab derivation

### Context

#### Higher algebra

higher algebra

universal algebra

# Contents

## Disambiguation

For derivations in logic, see deduction.

## 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

$d\left(ab\right)=d\left(a\right)b+ad\left(b\right)\phantom{\rule{thinmathspace}{0ex}},$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

$d\left(ab\right)=d\left(a\right)\cdot b+a\cdot d\left(b\right)\phantom{\rule{thinmathspace}{0ex}},$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 graded derivation of degree $p$ on a graded algebra $A$ is a degree-$p$ graded-module homomorphism $d:A\to A$ such that

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

$d\left(ab\right)=d\left(a\right)ϵ\left(b\right)+ϵ\left(a\right)d\left(b\right).$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 ${\mathrm{Der}}_{k}\left(A,M\right)\subset {\mathrm{Hom}}_{k}\left(A,M\right)$ (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: $d\left(ab\right)=d\left(a\right)+ad\left(b\right)$.

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

$v:A\to N$v : A \to N

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

$\begin{array}{ccc}𝒪\left(n\right)\otimes {A}^{\otimes n}& \stackrel{}{\to }& A\\ {}^{\sum _{a+b=n-1}\mathrm{id}\otimes {\mathrm{id}}^{\otimes a}\otimes v\otimes {\mathrm{id}}^{\otimes b}}↓& & {↓}^{v}\\ {\oplus }_{a+b=n-1}𝒪\left(n\right)\otimes {A}^{\otimes a}\otimes N\otimes {A}^{\otimes b}& \to & N\end{array}\phantom{\rule{thinmathspace}{0ex}},$\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 $𝒪$-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 $\left(\infty ,1\right)$-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 $G\left(N\right):=N\oplus R$, equipped with the product operation

$\left({r}_{1},{n}_{1}\right)\cdot \left({n}_{2},{r}_{2}\right):=\left({r}_{1},{r}_{2},{n}_{1}{r}_{2}+{n}_{2}{r}_{1}\right)\phantom{\rule{thinmathspace}{0ex}}.$(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

$G\left(N\right)\to R$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 $R\to G\left(N\right)$ 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 ${\Omega }_{K}\left(R\right)$ 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

$\Omega :C\to {T}_{C}\phantom{\rule{thinmathspace}{0ex}}.$\Omega : C \to T_C \,.

An $\left(\infty ,1\right)$-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 $\Omega \left(R\right)\to N$ 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

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

where ${\Omega }^{1}\left(X\right)$ 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 ${C}^{\infty }\left(X\right)$ its algebra of smooth functions. Then the morphism

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

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

###### Proof

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

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

$f\left(x\right)=f\left(0\right)+\sum _{i}{x}_{i}{g}_{i}\left(x\right)$f(x) = f(0) + \sum_i x_i g_i(x)

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

$\delta \left(f\right)\left(0\right)=\sum _{i}\delta \left({x}_{i}\right)\frac{\partial f}{\partial {x}_{i}}\left(0\right)\phantom{\rule{thinmathspace}{0ex}}.$\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 $\left\{{x}_{i}\in {C}^{\infty }\left(X\right)\right\}$. Let ${v}_{\delta }\in T{ℝ}^{n}$ be the vector field

${v}_{\delta }==\sum _{i}\delta \left({x}_{i}\right)\frac{\partial }{\partial {x}_{i}}$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 $\mathrm{Vect}\left(X\right)\to \mathrm{Der}\left({C}^{\infty }\left(X\right)\right)$.

### Derivations of continuous functions

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

###### Proposition

The derivations $\delta :C\left(X\right)\to C\left(X\right)$ 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 $f\in C\left(X\right)$ vanishes at ${x}_{0}\in X$ also $\delta \left(f\right)$ vanishes att ${x}_{0}$, because we may write every function $g$ as $\left(g-g\left({x}_{0}\right)\right)+g\left({x}_{0}\right)$.

So let $f\in C\left(X\right)$ with $f\left({x}_{0}\right)=0$. 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

${g}_{2}:x↦\left\{\begin{array}{cc}f\left(x\right)/\sqrt{\mid f\left(x\right)\mid }& \mid f\left(x\right)\ne 0\\ 0& \mid f\left(x\right)=0\end{array}\phantom{\rule{thinmathspace}{0ex}}.$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

$\delta \left(f\right)\left({x}_{0}\right)=\delta \left({g}_{1}\right)\left({x}_{0}\right){g}_{2}\left({x}_{0}\right)+{g}_{1}\left({x}_{0}\right)\delta \left({g}_{2}\left({x}_{0}\right)\right)=0\phantom{\rule{thinmathspace}{0ex}}.$\delta(f)(x_0) = \delta(g_1)(x_0) g_2(x_0) + g_1(x_0) \delta(g_2(x_0)) = 0 \,.

## References

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

Revised on April 15, 2013 01:52:29 by Urs Schreiber (89.204.137.79)