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

The Infinitesimal CommutatorWhy derivations of PROP-models form a Lie algebra1. PROP-models and automorphisms2. Derivations as tangent vectors3. Addition comes from first-order multiplication4. The infinitesimal square5. The two kernels6. The area layer is the tangent space again7. The commutator lands in the area layer

 

The Infinitesimal Commutator

Why derivations of PROP-models form a Lie algebra

The aim of this note is to prove the following fact in a way that avoids the usual calculation.

Let $\mathsf{P}$$\mathsf P$ be a $k$$k$-linear PROP, and let $A$$A$ be a model of $\mathsf{P}$$\mathsf P$. Then the derivations of $A$$A$ are closed under the commutator bracket

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

Equivalently,

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

is a Lie subalgebra of ${\mathrm{End}}_k(A)$$\operatorname{End}_k(A)$.

For associative algebras this is familiar. One checks directly that if $D$$D$ and $E$$E$ satisfy the Leibniz rule, then so does $DE-ED$$DE-ED$. But that calculation is not the reason. It is only the shadow of a more structural fact:

$$\text{the commutator of derivations is the area term of a group commutator.}$$

More precisely, derivations are tangent vectors to an automorphism group functor. Addition of derivations comes from first-order multiplication of infinitesimal automorphisms. The bracket of derivations comes from the group commutator of two infinitesimal automorphisms placed in two independent infinitesimal directions.

The proof is then almost formal.

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1. PROP-models and automorphisms

Let $k$$k$ be a commutative base ring, and let $\mathsf{P}$$\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 identities, composition, tensor product, and the symmetric group actions.

Equivalently, $A$$A$ gives a strict symmetric monoidal $k$$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$,

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

Therefore

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

is a group-valued functor.

No representability is assumed. The argument only uses the functor of points.

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2. Derivations as tangent vectors

Consider the dual numbers

$$k[ϵ]/({ϵ}^2).$$

There is a quotient map

$$k[ϵ]/({ϵ}^2)\to k,ϵ\mapsto 0.$$

Applying $G_A$$G_A$ gives a group homomorphism

$$G_A(k[ϵ]/({ϵ}^2))\to G_A(k).$$

Define the tangent space at the identity by

$$\mathrm{Lie}(G_A)=\ker (G_A(k[ϵ]/({ϵ}^2))\to G_A(k)).$$

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

Since

$$A{\otimes }_{k}k[ϵ]/({ϵ}^2)\cong A\oplus ϵA,$$

such an automorphism is necessarily of the form

$$g=\mathrm{id}+ϵD$$

for some $k$$k$-linear endomorphism

$$D:A\to A.$$

Its inverse is

$$g^{-1}=\mathrm{id}-ϵD,$$

because

$${ϵ}^2=0.$$

Now impose the condition that $g$$g$ preserves the PROP-structure. For every operation $p:m\to n$$p:m\to n$, we require

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

over $k[ϵ]/({ϵ}^2)$$k[\epsilon]/(\epsilon^2)$.

Expand

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

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

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

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$ gives

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

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

So a derivation of $A$$A$ is a $k$$k$-linear endomorphism $D:A\to A$$D:A\to A$ such that, for every operation $p:m\to n$$p:m\to n$,

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

Equivalently,

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

Derivations are tangent vectors to the automorphism functor.

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3. Addition comes from first-order multiplication

Before proving closure under commutators, one should first see why derivations are closed under addition.

Let

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

Then

$$\mathrm{id}+ϵD,\mathrm{id}+ϵE$$

are elements of

$$G_A(k[ϵ]/({ϵ}^2)).$$

Since $G_A(k[ϵ]/({ϵ}^2))$$G_A(k[\epsilon]/(\epsilon^2))$ is a group, their product is again an infinitesimal automorphism:

$$(\mathrm{id}+ϵD)(\mathrm{id}+ϵE)\in G_A(k[ϵ]/({ϵ}^2)).$$

But

$$(\mathrm{id}+ϵD)(\mathrm{id}+ϵE)=\mathrm{id}+ϵ(D+E)+{ϵ}^{2}DE.$$

Since ${ϵ}^2=0$$\epsilon^2=0$, this becomes

$$\mathrm{id}+ϵ(D+E).$$

Therefore

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

Scalar closure is just as formal. For $a\in k$$a\in k$, the map

$$k[ϵ]/({ϵ}^2)\to k[ϵ]/({ϵ}^2),ϵ\mapsto aϵ$$

sends

$$\mathrm{id}+ϵD$$

to

$$\mathrm{id}+ϵ(aD).$$

Hence

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

So

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

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

At first order, group multiplication becomes addition.

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4. The infinitesimal square

To see the bracket, one needs two independent infinitesimal directions.

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

$${ϵ}_1{ϵ}_2\ne 0.$$

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

Now define

$$R_0=R/({ϵ}_1{ϵ}_2).$$

Equivalently,

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

There are quotient maps

$$R⟶R_0⟶k.$$

The first quotient kills the area term

$${ϵ}_1{ϵ}_2,$$

and the second quotient kills the two first-order directions

$${ϵ}_1,{ϵ}_2.$$

Geometrically, the arrows reverse:

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

The picture is:

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

The bracket lives in the difference between the square and its two axes.

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5. The two kernels

Apply $G_A$$G_A$ to

$$R\to R_0\to k.$$

We get group homomorphisms

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

Define

$$K_R=\ker (G_A(R)\to G_A(k)),$$

and

$$K_0=\ker (G_A(R_0)\to G_A(k)).$$

So $K_R$$K_R$ is the second-order infinitesimal unit neighbourhood, while $K_0$$K_0$ is its first-order quotient.

There is a natural homomorphism

$$K_R\to K_0.$$

The key point is that $K_0$$K_0$ is abelian.

Indeed, in $R_0$$R_0$ we have

$${ϵ}_1^2={ϵ}_2^2={ϵ}_1{ϵ}_2=0.$$

Thus every element of $K_0$$K_0$ has the form

$$\mathrm{id}+{ϵ}_{1}D+{ϵ}_{2}E$$

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

The product is

$$\begin{array}{rl}& (\mathrm{id}+{ϵ}_{1}D+{ϵ}_{2}E)(\mathrm{id}+{ϵ}_1{D}^{\prime }+{ϵ}_2{E}^{\prime })\\ & =\mathrm{id}+{ϵ}_1(D+D^{\prime })+{ϵ}_2(E+E^{\prime }),\end{array}$$

because all products of first-order terms vanish in $R_0$$R_0$.

Therefore $K_0$$K_0$ is an additive group:

$$K_0\cong {\mathrm{Der}}_{\mathsf{P}}(A){ϵ}_1\oplus {\mathrm{Der}}_{\mathsf{P}}(A){ϵ}_2.$$

In particular, $K_0$$K_0$ is abelian.

Hence the homomorphism

$$K_R\to K_0$$

factors through the abelianization of $K_R$$K_R$:

$$K_R\to K_R^{\mathrm{ab}}\to K_0.$$

Therefore every group commutator in $K_R$$K_R$ maps to the identity in $K_0$$K_0$.

Equivalently,

$$[K_R,K_R]\subseteq \ker (K_R\to K_0).$$

But

$$\ker (K_R\to K_0)=\ker (G_A(R)\to G_A(R_0)).$$

This is the area layer.

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6. The area layer is the tangent space again

The kernel

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

consists of automorphisms which become the identity after killing ${ϵ}_1{ϵ}_2$$\epsilon_1\epsilon_2$.

Therefore every element of this kernel has the form

$$\mathrm{id}+{ϵ}_1{ϵ}_{2}F$$

for some $k$$k$-linear endomorphism $F:A\to A$$F:A\to A$.

Because

$$({ϵ}_1{ϵ}_2{)}^2=0,$$

its inverse is

$$\mathrm{id}-{ϵ}_1{ϵ}_{2}F.$$

The condition that

$$\mathrm{id}+{ϵ}_1{ϵ}_{2}F$$

preserves the PROP-structure is precisely the first-order condition saying that $F$$F$ is a derivation.

Thus

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

Since

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

we may also write

$$\ker (G_A(R)\to G_A(R_0))\cong \mathrm{Lie}(G_A)\cdot {ϵ}_1{ϵ}_2.$$

This is the tangent space labelled by the area element.

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7. The commutator lands in the area layer

Take

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

and put

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

inside

$$K_R=\ker (G_A(R)\to G_A(k)).$$

Their group commutator

$$c=g_D{g}_E{g}_D^{-1}{g}_E^{-1}$$

also lies in $K_R$$K_R$.

Now reduce modulo ${ϵ}_1{ϵ}_2$$\epsilon_1\epsilon_2$. In

$$R_0=R/({ϵ}_1{ϵ}_2),$$

the group $K_0=\ker (G_A(R_0)\to G_A(k))$$K_0=\ker(G_A(R_0)\to G_A(k))$ is first-order and hence abelian. Therefore the image of the commutator $c$$c$ in $K_0$$K_0$ is the identity.

Thus

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

But this kernel is the area-labelled tangent space

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

Hence

$$c=\mathrm{id}+{ϵ}_1{ϵ}_{2}F$$

 

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

Easy to see that

$$F=DE-ED.$$

Since $F$$F$ is already known to be a derivation, we conclude that

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

This proves that derivations of a PROP-model are closed under commutators.