The Infinitesimal Commutator: Why Derivations of PROP-Models Form a Lie Algebra

Infinitesimal Automorphisms and the Lie Algebra of PROP-Derivations1. PROP-models and automorphisms2. The dual numbers and the derivation condition3. Addition and scalar multiplication of derivations4. Usual Leibniz rules as examples5. The infinitesimal square6. The area layer is another copy of the tangent module7. The group commutator produces the bracket8. The theorem

 

Infinitesimal Automorphisms and the Lie Algebra of PROP-Derivations

The purpose of this note is to explain a structural fact:

$$\text{derivations of any PROP-model form a Lie algebra.}$$

For associative algebras, this says that if $D,E$$D,E$ are derivations, then

$$[D,E]=DE-ED$$

is again a derivation.

The same statement holds for Lie algebras, coalgebras, bialgebras, Frobenius algebras, Hopf-like structures, and more generally for models of any $k$$k$-linear PROP.

The reason is not a special Leibniz-rule computation. The reason is:

$$\text{a derivation is an infinitesimal automorphism.}$$

Addition of derivations comes from multiplying infinitesimal automorphisms over the dual numbers.
The commutator bracket comes from the group commutator over an infinitesimal square.

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1. PROP-models and automorphisms

Let $k$$k$ be a commutative ring, and let $\mathsf{P}$$\mathsf P$ be a one-coloured $k$$k$-linear PROP.

A model of $\mathsf{P}$$\mathsf P$ in $k$$k$-modules is a $k$$k$-module $A$$A$ together with structure maps

$$\rho (p):A^{\otimes m}\to A^{\otimes n}$$

for every operation

$$p:m\to n$$

in $\mathsf{P}$$\mathsf P$, compatible with composition, tensor product, identities, and symmetric group actions.

Equivalently, $A$$A$ determines a strict symmetric monoidal $k$$k$-linear functor

$$\rho :\mathsf{P}\to {\mathrm{End}}_A,$$

where

$${\mathrm{End}}_A(m,n)={\mathrm{Hom}}_k(A^{\otimes m},A^{\otimes n}).$$

For every commutative $k$$k$-algebra $R$$R$, put

$$A_R=A{\otimes }_{k}R.$$

The structure maps extend $R$$R$-linearly to

$${\rho }_R(p):A_R^{{\otimes }_{R}m}\to A_R^{{\otimes }_{R}n}.$$

Define the automorphism functor of the PROP-model $A$$A$ by

$$G_A(R)={\mathrm{Aut}}_{\mathsf{P},R}(A_R).$$

Thus an element

$$g\in G_A(R)$$

is an $R$$R$-linear automorphism

$$g:A_R\to A_R$$

such that, for every operation $p:m\to n$$p:m\to n$,

$$g^{\otimes n}\circ {\rho }_R(p)={\rho }_R(p)\circ g^{\otimes m}.$$

Hence

$$G_A:{\mathrm{CAlg}}_k\to \mathrm{Grp}$$

is a group-valued functor.

No representability is required. We only use the functor of points.

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2. The dual numbers and the derivation condition

Write

$$k[ϵ]=k[ϵ]/({ϵ}^2)$$

for the dual numbers.

The quotient map

$$k[ϵ]\to k,ϵ\mapsto 0$$

induces a group homomorphism

$$G_A(k[ϵ])\to G_A(k).$$

Define

$$\mathrm{Lie}(G_A)=\ker (G_A(k[ϵ])\to G_A(k)).$$

An element of this kernel is an infinitesimal automorphism of $A$$A$ reducing to the identity modulo $ϵ$$\epsilon$.

Since

$$A{\otimes }_{k}k[ϵ]\cong A\oplus ϵA,$$

every such infinitesimal automorphism is uniquely of the form

$$g=\mathrm{id}+ϵD$$

for a $k$$k$-linear endomorphism

$$D:A\to A.$$

Its inverse is

$$g^{-1}=\mathrm{id}-ϵD.$$

The condition that $g$$g$ preserves the PROP-structure is

$$g^{\otimes n}\circ \rho (p)=\rho (p)\circ g^{\otimes m}$$

over $k[ϵ]$$k[\epsilon]$, for every operation $p:m\to n$$p:m\to n$.

We suppress scalar extension from the notation.

Now

$$g^{\otimes r}=(\mathrm{id}+ϵD{)}^{\otimes r}.$$

Because ${ϵ}^2=0$$\epsilon^2=0$, at most one copy of $D$$D$ can appear. Hence

$$g^{\otimes r}=\mathrm{id}+ϵD^{(r)},$$

where

$$D^{(r)}=\sum _{i=1}^r{\mathrm{id}}^{\otimes (i-1)}\otimes D\otimes {\mathrm{id}}^{\otimes (r-i)}.$$

For $r=0$$r=0$, set

$$D^{(0)}=0.$$

Taking the coefficient of $ϵ$$\epsilon$ in

$$g^{\otimes n}\circ \rho (p)=\rho (p)\circ g^{\otimes m}$$

gives

$$D^{(n)}\circ \rho (p)=\rho (p)\circ D^{(m)}.$$

This is the general Leibniz rule for a PROP-model.

So we define

$${\mathrm{Der}}_{\mathsf{P}}(A)$$

to be the set of $k$$k$-linear endomorphisms $D:A\to A$$D:A\to A$ satisfying

$$D^{(n)}\circ \rho (p)=\rho (p)\circ D^{(m)}$$

for every operation $p:m\to n$$p:m\to n$ in $\mathsf{P}$$\mathsf P$.

Equivalently,

$${\mathrm{Der}}_{\mathsf{P}}(A)=\mathrm{Lie}(G_A).$$

Thus derivations are precisely tangent vectors at the identity of the automorphism functor.

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3. Addition and scalar multiplication of derivations

The dual numbers also explain why derivations form a $k$$k$-module.

Let

$$D,E\in {\mathrm{Der}}_{\mathsf{P}}(A).$$

Then

$$\mathrm{id}+ϵD,\mathrm{id}+ϵE$$

belong to

$$\ker (G_A(k[ϵ])\to G_A(k)).$$

Since this kernel is a subgroup of $G_A(k[ϵ])$$G_A(k[\epsilon])$, their product again lies in the kernel.

But

$$(\mathrm{id}+ϵD)(\mathrm{id}+ϵE)=\mathrm{id}+ϵ(D+E),$$

because ${ϵ}^2=0$$\epsilon^2=0$.

Therefore

$$D+E\in {\mathrm{Der}}_{\mathsf{P}}(A).$$

Similarly,

$$(\mathrm{id}+ϵD{)}^{-1}=\mathrm{id}-ϵD,$$

so

$$-D\in {\mathrm{Der}}_{\mathsf{P}}(A).$$

Thus ${\mathrm{Der}}_{\mathsf{P}}(A)$$\operatorname{Der}_{\mathsf P}(A)$ is an abelian group under addition.

Now let $a\in k$$a\in k$. The $k$$k$-algebra map

$$k[ϵ]\to k[ϵ],ϵ\mapsto aϵ$$

sends the infinitesimal automorphism

$$\mathrm{id}+ϵD$$

to

$$\mathrm{id}+ϵ(aD).$$

By functoriality of $G_A$$G_A$, this is again an infinitesimal automorphism. Hence

$$aD\in {\mathrm{Der}}_{\mathsf{P}}(A).$$

Therefore

$${\mathrm{Der}}_{\mathsf{P}}(A)\subseteq {\mathrm{End}}_k(A)$$

is a $k$$k$-submodule.

So the dual numbers already give the linear structure on derivations.

The remaining question is why this $k$$k$-module is closed under the commutator bracket.

For that, one needs two infinitesimal directions.

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4. Usual Leibniz rules as examples

Before proving closure under commutators, let us recover the familiar formulas.

For an associative algebra, the multiplication is an operation

$$\mu :2\to 1.$$

The corresponding structure map is

$$\mu :A\otimes A\to A.$$

The general equation becomes

$$D\circ \mu =\mu \circ (D\otimes \mathrm{id}+\mathrm{id}\otimes D).$$

In element notation,

$$D(xy)=D(x)y+xD(y).$$

For a Lie algebra, the bracket is an operation

$$[-,-]:2\to 1.$$

The formula becomes

$$D[x,y]=[Dx,y]+[x,Dy].$$

For a coalgebra, the comultiplication is an operation

$$\mathrm{\Delta }:1\to 2.$$

The corresponding formula is

$$(D\otimes \mathrm{id}+\mathrm{id}\otimes D)\circ \mathrm{\Delta }=\mathrm{\Delta }\circ D.$$

For a counit

$$\epsilon :A\to k,$$

which is an operation

$$1\to 0,$$

the formula says

$$\epsilon \circ D=0.$$

For a unit

$$\eta :k\to A,$$

which is an operation

$$0\to 1,$$

the formula says

$$D\circ \eta =0.$$

Thus the usual Leibniz rules are all instances of the same linearized PROP equation.

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5. The infinitesimal square

The dual numbers see one infinitesimal direction. They give the tangent module.

To see the bracket, we use two infinitesimal directions at once.

Let

$$R=k[{ϵ}_1,{ϵ}_2]/({ϵ}_1^2,{ϵ}_2^2).$$

As a $k$$k$-module,

$$R=k\oplus k{ϵ}_1\oplus k{ϵ}_2\oplus k{ϵ}_1{ϵ}_2.$$

The relations are

$${ϵ}_1^2=0,{ϵ}_2^2=0,$$

but the mixed term

$${ϵ}_1{ϵ}_2$$

is not killed.

Thus $R$$R$ remembers two first-order directions and their mixed second-order area term.

Now put

$$R_0=R/({ϵ}_1{ϵ}_2).$$

Equivalently,

$$R_0=k[{ϵ}_1,{ϵ}_2]/({ϵ}_1^2,{ϵ}_2^2,{ϵ}_1{ϵ}_2).$$

There is a sequence of quotient maps

$$R⟶R_0⟶k.$$

The first map kills the area term.
The second kills the two infinitesimal axes.

Contravariantly, the geometric picture is

$$\mathrm{Spec}k⟶\mathrm{Spec}{R}_0⟶\mathrm{Spec}R.$$

In words:

$$\text{origin}\subset \text{two infinitesimal axes}\subset \text{infinitesimal square}.$$

The bracket will be read from the area layer

$$\ker (G_A(R)\to G_A(R_0)).$$

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6. The area layer is another copy of the tangent module

Let

$$I=({ϵ}_1{ϵ}_2)\subset R.$$

Then

$$I^2=0,$$

and

$$0\to I\to R\to R_0\to 0$$

is a square-zero extension.

Consider the kernel

$$\ker (G_A(R)\to G_A(R_0)).$$

Its elements are PROP-automorphisms of $A_R$$A_R$ which become the identity after killing ${ϵ}_1{ϵ}_2$$\epsilon_1\epsilon_2$.

Every element of this kernel is uniquely of the form

$$\mathrm{id}+{ϵ}_1{ϵ}_{2}F$$

for some $k$$k$-linear endomorphism

$$F:A\to A.$$

Indeed, an $R$$R$-linear endomorphism of $A_R$$A_R$ which is zero modulo $R_0$$R_0$ has image in

$$A{\otimes }_k({ϵ}_1{ϵ}_2)k.$$

Since $({ϵ}_1{ϵ}_2)k$$(\epsilon_1\epsilon_2)k$ is generated by one square-zero element, such an endomorphism is determined by a $k$$k$-linear map $F:A\to A$$F:A\to A$.

The inverse is

$$(\mathrm{id}+{ϵ}_1{ϵ}_{2}F{)}^{-1}=\mathrm{id}-{ϵ}_1{ϵ}_{2}F.$$

The condition that

$$\mathrm{id}+{ϵ}_1{ϵ}_{2}F$$

preserves the PROP-structure is again the first-order condition

$$F^{(n)}\circ \rho (p)=\rho (p)\circ F^{(m)}$$

for every operation $p:m\to n$$p:m\to n$.

Thus

$$F\in {\mathrm{Der}}_{\mathsf{P}}(A).$$

Therefore

$$\ker (G_A(R)\to G_A(R_0))\cong {\mathrm{Der}}_{\mathsf{P}}(A)\cdot {ϵ}_1{ϵ}_2.$$

Using the chosen generator ${ϵ}_1{ϵ}_2$$\epsilon_1\epsilon_2$, one may identify this kernel with

$${\mathrm{Der}}_{\mathsf{P}}(A).$$

But conceptually it is better to keep the label:

$${\mathrm{Der}}_{\mathsf{P}}(A)\cdot {ϵ}_1{ϵ}_2.$$

It is the tangent module sitting in the area direction of the infinitesimal square.

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7. The group commutator produces the bracket

Let

$$D,E\in {\mathrm{Der}}_{\mathsf{P}}(A).$$

Place them in the two infinitesimal directions of $R$$R$:

$$g_D=\mathrm{id}+{ϵ}_{1}D,g_E=\mathrm{id}+{ϵ}_{2}E.$$

Since $D$$D$ and $E$$E$ are derivations, these are elements of

$$G_A(R).$$

Indeed, they are obtained from the dual-number infinitesimal automorphisms

$$\mathrm{id}+ϵD,\mathrm{id}+ϵE$$

by the $k$$k$-algebra maps

$$k[ϵ]\to R,ϵ\mapsto {ϵ}_1,$$

and

$$k[ϵ]\to R,ϵ\mapsto {ϵ}_2.$$

Since $G_A(R)$$G_A(R)$ is a group, the group commutator

$$[g_D,g_E]=g_D{g}_E{g}_D^{-1}{g}_E^{-1}$$

is again an element of

$$G_A(R).$$

Modulo the ideal $({ϵ}_1{ϵ}_2)$$(\epsilon_1\epsilon_2)$, the two infinitesimal directions no longer interact. Equivalently, in $R_0$$R_0$ one has

$${ϵ}_1{ϵ}_2=0.$$

Thus the two infinitesimal automorphisms commute modulo $R_0$$R_0$, and so

$$[g_D,g_E]$$

becomes the identity in

$$G_A(R_0).$$

Therefore

$$[g_D,g_E]\in \ker (G_A(R)\to G_A(R_0)).$$

By the previous section, this kernel is

$${\mathrm{Der}}_{\mathsf{P}}(A)\cdot {ϵ}_1{ϵ}_2.$$

Now compute the same commutator as an endomorphism of $A_R$$A_R$.

We have

$$g_D^{-1}=\mathrm{id}-{ϵ}_{1}D,g_E^{-1}=\mathrm{id}-{ϵ}_{2}E.$$

Hence

$$\begin{array}{rl}& =(\mathrm{id}+{ϵ}_{1}D)(\mathrm{id}+{ϵ}_{2}E)(\mathrm{id}-{ϵ}_{1}D)(\mathrm{id}-{ϵ}_{2}E)\\ & =\mathrm{id}+{ϵ}_1{ϵ}_2(DE-ED).\end{array}$$

The left-hand side lies in the area kernel

$${\mathrm{Der}}_{\mathsf{P}}(A)\cdot {ϵ}_1{ϵ}_2.$$

Therefore the coefficient of ${ϵ}_1{ϵ}_2$$\epsilon_1\epsilon_2$ must be a derivation:

$$DE-ED\in {\mathrm{Der}}_{\mathsf{P}}(A).$$

So derivations are closed under the commutator bracket.

The mechanism is simple:

$$\text{group commutator}⇝\text{area coefficient}⇝\text{Lie bracket}.$$

The commutator of two infinitesimal automorphisms has no linear part. Its first nontrivial term lies in the infinitesimal area direction. But the area direction is another copy of the tangent module. Hence the area coefficient is again a derivation.

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8. The theorem

Theorem.
Let $\mathsf{P}$$\mathsf P$ be a $k$$k$-linear PROP, and let $A$$A$ be a model of $\mathsf{P}$$\mathsf P$ in $k$$k$-modules. Then

$${\mathrm{Der}}_{\mathsf{P}}(A)$$

is a Lie subalgebra of

$${\mathrm{End}}_k(A)$$

under the commutator bracket

$$[D,E]=DE-ED.$$

In particular, derivations of any PROP-model form a Lie algebra.

Proof.
The dual-number argument shows that

$${\mathrm{Der}}_{\mathsf{P}}(A)$$

is a $k$$k$-submodule of ${\mathrm{End}}_k(A)$$\operatorname{End}_k(A)$.

The infinitesimal-square argument shows that, for any

$$D,E\in {\mathrm{Der}}_{\mathsf{P}}(A),$$

one has

$$DE-ED\in {\mathrm{Der}}_{\mathsf{P}}(A).$$

Thus ${\mathrm{Der}}_{\mathsf{P}}(A)$$\operatorname{Der}_{\mathsf P}(A)$ is closed under the commutator bracket.

Since the commutator bracket on ${\mathrm{End}}_k(A)$$\operatorname{End}_k(A)$ is $k$$k$-bilinear, alternating, and satisfies the Jacobi identity, the same is true on the submodule ${\mathrm{Der}}_{\mathsf{P}}(A)$$\operatorname{Der}_{\mathsf P}(A)$.

Therefore

$${\mathrm{Der}}_{\mathsf{P}}(A)$$

is a Lie algebra. $◻$$\square$