Derivations as Infinitesimal Automorphisms of PROP-Models: From Affine Hom-Schemes to Internal Lie Algebras

Automorphisms, Derivations, and the Affine Geometry of PROP-ModelsPROP-modelsHom-schemesAutomorphism functorsThe Lie functorThe usual Leibniz rulesInfinite dimensionThe Lie bracket without a Leibniz calculationAdditive closure

 

Automorphisms, Derivations, and the Affine Geometry of PROP-Models

A derivation is often introduced by a formula.

For an associative algebra,

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

For a Lie algebra,

$$D([x,y])=[D(x),y]+[x,D(y)].$$

For a coalgebra,

$$\mathrm{\Delta }D=(D\otimes 1+1\otimes D)\mathrm{\Delta }.$$

These formulas look different, but they express one idea:

A derivation is an infinitesimal automorphism.

PROP language makes this uniform.

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PROP-models

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

Its objects are natural numbers

$$0,1,2,\dots ,$$

with tensor product given by addition:

$$m\otimes n=m+n.$$

A morphism

$$\theta :n\to m$$

is an operation with $n$$n$ inputs and $m$$m$ outputs.

A finite-dimensional model of $\mathsf{P}$$\mathsf P$ is a strict symmetric monoidal $k$$k$-linear functor

$$A:\mathsf{P}\to {\mathrm{Vect}}_k^{\mathrm{fd}}.$$

Write

$$V=A(1).$$

Then

$$A(n)=V^{\otimes n}.$$

Thus every operation

$$\theta :n\to m$$

is interpreted as a linear map

$$A(\theta ):V^{\otimes n}\to V^{\otimes m}.$$

A morphism between models

$$A,B:\mathsf{P}\to {\mathrm{Vect}}_k^{\mathrm{fd}}$$

is a monoidal natural transformation. Since everything is generated by the object $1$$1$, such a transformation is determined by a single linear map

$$T:V=A(1)\to W=B(1).$$

Naturality for an operation $\theta :n\to m$$\theta:n\to m$ says

$$B(\theta )\circ T^{\otimes n}=T^{\otimes m}\circ A(\theta ).$$

This one equation contains all the usual homomorphism laws.

For multiplication $\mu :2\to 1$$\mu:2\to 1$, it gives

$$T(xy)=T(x)T(y).$$

For a Lie bracket $[-,-]:2\to 1$$[-,-]:2\to 1$, it gives

$$T([x,y])=[T(x),T(y)].$$

For a comultiplication $\mathrm{\Delta }:1\to 2$$\Delta:1\to 2$, it gives

$$\mathrm{\Delta }T=(T\otimes T)\mathrm{\Delta }.$$

So the familiar preservation rules are just naturality.

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Hom-schemes

Assume $V$$V$ and $W$$W$ are finite-dimensional.

The vector space

$${\mathrm{Hom}}_k(V,W)$$

is an affine space. A linear map $T:V\to W$$T:V\to W$ is described by its matrix entries.

For each operation $\theta :n\to m$$\theta:n\to m$, the condition

$$B(\theta )\circ T^{\otimes n}=T^{\otimes m}\circ A(\theta )$$

is polynomial in the matrix entries of $T$$T$.

Therefore the set of $\mathsf{P}$$\mathsf P$-model morphisms is represented by a closed affine subscheme

$${\mathrm{Hom}}_{\mathsf{P}}(A,B)\subseteq {\mathrm{Hom}}_k(V,W).$$

Equivalently, for each commutative $k$$k$-algebra $R$$R$,

$${\mathrm{Hom}}_{\mathsf{P}}(A,B)(R)$$

is the set of $R$$R$-linear maps

$$T:V_R\to W_R$$

satisfying

$$B_R(\theta )\circ T^{\otimes n}=T^{\otimes m}\circ A_R(\theta )$$

for all operations $\theta :n\to m$$\theta:n\to m$, where

$$V_R=V{\otimes }_{k}R,W_R=W{\otimes }_{k}R.$$

Composition is ordinary composition of linear maps, hence polynomial in matrix coordinates. Therefore composition defines morphisms of affine schemes

$${\mathrm{Hom}}_{\mathsf{P}}(B,C)\times {\mathrm{Hom}}_{\mathsf{P}}(A,B)\to {\mathrm{Hom}}_{\mathsf{P}}(A,C).$$

Thus finite-dimensional $\mathsf{P}$$\mathsf P$-models form a category enriched over affine $k$$k$-schemes.

The usual category is recovered by taking $k$$k$-points.

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Automorphism functors

For a model $A$$A$, define its automorphism functor by

$${\mathrm{Aut}}_{\mathsf{P}}(A)(R)={\mathrm{Aut}}_{{\mathsf{P}}_R}(A_R).$$

This is a group-valued functor on commutative $k$$k$-algebras.

If $A$$A$ is finite-dimensional, then

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

is represented by an affine group scheme: it is the open subfunctor of

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

where the determinant of the underlying linear map is invertible.

For example, if $A$$A$ is a finite-dimensional Lie algebra $\mathfrak{g}$$\mathfrak g$, then

$${\mathrm{Aut}}_{\mathrm{Lie}}(\mathfrak{g})$$

is an affine algebraic group. Its defining equations are exactly the polynomial equations

$$T([x,y])=[T(x),T(y)].$$

But representability is not essential. Even if $A$$A$ is infinite-dimensional and the automorphism functor is not represented by an affine scheme, the group functor still exists.

This distinction matters:

Finite-dimensionality gives geometry.

The automorphism group functor exists without it.

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The Lie functor

Let

$$G:k\text{-}\mathbf{Alg}\to \mathbf{Grp}$$

be any group functor.

Define its Lie functor by

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

This definition only uses the dual numbers. It does not require $G$$G$ to be representable.

Now take

$$G={\mathrm{Aut}}_{\mathsf{P}}(A).$$

An element of

$$\mathrm{Lie}(G)(R)$$

is an automorphism of $A_{R[ϵ]}$$A_{R[\epsilon]}$ reducing to the identity modulo $ϵ$$\epsilon$.

Such an automorphism has the form

$$\mathrm{id}+ϵD$$

for a unique $R$$R$-linear map

$$D:V_R\to V_R.$$

It is automatically invertible, with inverse

$$\mathrm{id}-ϵD.$$

The only remaining condition is that it preserves the $\mathsf{P}$$\mathsf P$-structure.

For an operation

$$\theta :n\to m,$$

structure preservation says

$$(\mathrm{id}+ϵD{)}^{\otimes m}\circ A_R(\theta )=A_R(\theta )\circ (\mathrm{id}+ϵD{)}^{\otimes n}.$$

Define

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

as an endomorphism of $V_R^{\otimes r}$$V_R^{\otimes r}$.

Since ${ϵ}^2=0$$\epsilon^2=0$,

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

Comparing coefficients of $ϵ$$\epsilon$ gives

$$D^{(m)}\circ A_R(\theta )=A_R(\theta )\circ D^{(n)}.$$

This is the general derivation rule.

So define

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

to be the set of $R$$R$-linear maps

$$D:V_R\to V_R$$

satisfying

$$D^{(m)}{A}_R(\theta )=A_R(\theta )D^{(n)}$$

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

We have proved

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

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

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The usual Leibniz rules

For an associative algebra, the operation is

$$\mu :V\otimes V\to V.$$

Here $n=2$$n=2$ and $m=1$$m=1$. The general derivation equation becomes

$$D\mu =\mu (D\otimes 1+1\otimes D),$$

that is,

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

For a Lie algebra, the operation is

$$[-,-]:V\otimes V\to V.$$

The same formula gives

$$D([x,y])=[D(x),y]+[x,D(y)].$$

For a coalgebra, the operation is

$$\mathrm{\Delta }:V\to V\otimes V.$$

Here $n=1$$n=1$ and $m=2$$m=2$, so the formula becomes

$$(D\otimes 1+1\otimes D)\mathrm{\Delta }=\mathrm{\Delta }D.$$

This is the coderivation rule.

The formulas differ only because the arities of the operations differ.

The source is the same:

$$\text{derivation}=\text{first-order automorphism}.$$

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Infinite dimension

The Hom-scheme construction used finite-dimensionality. If $V$$V$ is infinite-dimensional, the functor

$$R\mapsto V{\otimes }_{k}R$$

is generally not represented by an ordinary affine scheme.

So one should not expect

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

to be an affine group scheme in general.

But the group functor

$$R\mapsto {\mathrm{Aut}}_{{\mathsf{P}}_R}(A_R)$$

still makes sense.

The Lie functor also still makes sense:

$$\mathrm{Lie}({\mathrm{Aut}}_{\mathsf{P}}(A))(R)=\ker ({\mathrm{Aut}}_{\mathsf{P}}(A)(R[ϵ])\to {\mathrm{Aut}}_{\mathsf{P}}(A)(R)).$$

The same dual-number calculation still gives

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

So representability is not the foundation of the definition of derivation.

The correct foundation is the automorphism group functor.

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The Lie bracket without a Leibniz calculation

It remains to explain why derivations form a Lie algebra.

The standard proof expands the Leibniz rule and checks that

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

again satisfies it.

For general PROP-models, that is the wrong proof. It is a brute-force shadow of a cleaner group-theoretic argument.

Let

$$G={\mathrm{Aut}}_{\mathsf{P}}(A).$$

We already know

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

Let

$$B=R[{ϵ}_1,{ϵ}_2]/({ϵ}_1^2,{ϵ}_2^2),\delta ={ϵ}_1{ϵ}_2,$$

and

$$B_0=B/(\delta ).$$

The key point is the area-layer identification:

$$\ker (G(B)\to G(B_0))\cong \delta \mathrm{Lie}(G)(R).$$

For the automorphism functor, this says explicitly:

every element of the kernel is uniquely of the form

$$1+{ϵ}_1{ϵ}_{2}F$$

with

$$F\in \mathrm{Lie}(G)(R)={\mathrm{Der}}_{\mathsf{P}}(A)(R).$$

This is just the dual-number tangent calculation, but with the square-zero parameter

$$\delta ={ϵ}_1{ϵ}_2.$$

Now take

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

Then

$$g_D=1+{ϵ}_{1}D,g_E=1+{ϵ}_{2}E$$

are elements of $G(B)$$G(B)$.

Consider their group commutator

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

In $B_0$$B_0$, the product ${ϵ}_1{ϵ}_2$$\epsilon_1\epsilon_2$ is zero. Hence

$$(1+{ϵ}_{1}D)(1+{ϵ}_{2}E)=1+{ϵ}_{1}D+{ϵ}_{2}E=(1+{ϵ}_{2}E)(1+{ϵ}_{1}D).$$

Therefore the image of $c$$c$ in $G(B_0)$$G(B_0)$ is the identity.

So

$$c\in \ker (G(B)\to G(B_0)).$$

By the area-layer identification, there is a unique

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

such that

$$c=1+{ϵ}_1{ϵ}_{2}F.$$

Thus, before calculating any coefficient, we already know that the area coefficient is a derivation.

It remains only to identify it.

Inside the underlying endomorphism algebra,

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

Therefore

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

Hence

$$F=DE-ED.$$

But $F$$F$ was already known to be a derivation. Therefore

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

is again a derivation.

This proves closure under commutators without expanding the derivation rule.

Additive closure

The same infinitesimal viewpoint also explains why derivations are closed under addition.

Let

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

Equivalently,

$$1+ϵD,1+ϵE$$

are elements of

$${\mathrm{Aut}}_{\mathsf{P}}(A)(R[ϵ]/({ϵ}^2))$$

which reduce to the identity modulo $ϵ$$\epsilon$.

Since automorphisms form a group, their product is again an automorphism:

$$(1+ϵD)(1+ϵE)\in {\mathrm{Aut}}_{\mathsf{P}}(A)(R[ϵ]/({ϵ}^2)).$$

But because

$${ϵ}^2=0,$$

we have

$$(1+ϵD)(1+ϵE)=1+ϵ(D+E).$$

This element still reduces to the identity modulo $ϵ$$\epsilon$. Hence

$$1+ϵ(D+E)\in \mathrm{Lie}({\mathrm{Aut}}_{\mathsf{P}}(A))(R).$$

Using the identification

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

we obtain

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

Similarly,

$$(1+ϵD{)}^{-1}=1-ϵD,$$

so

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

The zero derivation corresponds to the identity automorphism

$$1+ϵ0=\mathrm{id}.$$

Thus derivations are closed under addition, additive inverses, and zero.

Scalar multiplication is equally formal. For $a\in R$$a\in R$, the map

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

sends

$$1+ϵD$$

to

$$1+ϵ(aD).$$

Therefore

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

So

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

is an $R$$R$-submodule of

$${\mathrm{End}}_R(V_R).$$

Again, no Leibniz formula needs to be expanded. Additive closure is forced by multiplication of first-order infinitesimal automorphisms.

The Jacobi identity follows because this bracket is the ordinary commutator bracket inside

$${\mathrm{End}}_R(V_R).$$

Thus

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

is a Lie algebra.

The derivation functor

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

is not merely a functor

$$k\text{-}\mathbf{Alg}\to k\text{-}\mathbf{LieAlg}.$$

It carries a stronger structure: it is an internal Lie algebra over the internal ring

$$\mathcal{O}={\mathbb{A}}_k^1$$

in the functor category

$$\mathrm{Fun}(k\text{-}\mathbf{Alg},\mathbf{Set}).$$

Equivalently, the pair

$$(\mathcal{O},\mathcal{D})$$

is a model of the many-sorted algebraic theory of a commutative ring together with a Lie algebra over it, with the ring sort fixed to

$$\mathcal{O}={\mathbb{A}}_k^1.$$