Finite-Dimensional PROP Models Enriched over Aff_k: From monoidal naturality to derivations

PROP Models and Affine Hom-SchemesHomomorphisms, automorphisms, and derivations of finite-dimensional algebraic structures1. Finite-dimensional vector spaces already have affine Hom-spaces2. PROPs and their models3. Morphisms are monoidal natural transformations4. Why the naturality equations are polynomial5. The Hom-scheme of two PROP models6. Enrichment over affine schemes7. Examples8. Endomorphism monoid schemes9. Automorphism group schemes10. Derivations from automorphisms11. The Lie algebra functor12. Recovering the usual Leibniz rulesAssociative algebrasLie algebrasCoalgebrasThe point

 

PROP Models and Affine Hom-Schemes

Homomorphisms, automorphisms, and derivations of finite-dimensional algebraic structures

Let $k$$k$ be a field.

I want to record a small observation about finite-dimensional algebraic structures. The observation is simple, but it reorganizes several familiar notions in a useful way.

The usual starting point is to say that a homomorphism is a map preserving the operations. For example,

$$F(xy)=F(x)F(y)$$

for associative algebras, or

$$F([x,y])=[F(x),F(y)]$$

for Lie algebras.

But these formulas are not isolated accidents. They are instances of one naturality equation.

Similarly, derivations are often introduced by writing down a Leibniz rule. But for self-valued derivations, it is more conceptual to start from automorphisms: a derivation should be an infinitesimal automorphism. The Leibniz rule is then obtained by linearizing the condition of preserving the structure.

The rough slogan is:

$$\boxed{{\displaystyle \text{homomorphisms are natural transformations, and derivations are infinitesimal automorphisms.}}}$$

The affine geometry enters because these naturality equations are polynomial equations.

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1. Finite-dimensional vector spaces already have affine Hom-spaces

Before adding algebraic structure, start with finite-dimensional vector spaces.

Let $V,W$$V,W$ be finite-dimensional $k$$k$-vector spaces. The set

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

is not only a set, nor only a vector space. It is the set of $k$$k$-points of an affine space.

More precisely, define

$${\underset{―}{\mathrm{Hom}}}_k(V,W)=\mathrm{Spec}\mathrm{Sym}({\mathrm{Hom}}_k(V,W{)}^{\vee }).$$

This affine scheme represents the functor

$$R⟼{\mathrm{Hom}}_R(V{\otimes }_{k}R,W{\otimes }_{k}R)$$

on commutative $k$$k$-algebras. In other words,

$${\underset{―}{\mathrm{Hom}}}_k(V,W)(R)={\mathrm{Hom}}_R(V_R,W_R),$$

where

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

The finite-dimensionality assumption matters here. It ensures that the Hom functor is represented by an ordinary affine space.

Composition of linear maps is matrix multiplication. Hence composition is polynomial in coordinates, and gives a morphism of affine schemes

$${\underset{―}{\mathrm{Hom}}}_k(W,U){\times }_k{\underset{―}{\mathrm{Hom}}}_k(V,W)\to {\underset{―}{\mathrm{Hom}}}_k(V,U).$$

The identity map ${\mathrm{id}}_V$$\operatorname{id}_V$ is a $k$$k$-point of

$${\underset{―}{\mathrm{Hom}}}_k(V,V).$$

Equivalently, it is a morphism

$$\mathrm{Spec}k\to {\underset{―}{\mathrm{Hom}}}_k(V,V).$$

Since the monoidal unit of $({\mathbf{Aff}}_k,{\times }_k)$$(\mathbf{Aff}_k,\times_k)$ is $\mathrm{Spec}k$$\operatorname{Spec}k$, this is exactly the unit map required for enrichment.

Thus one may regard

$$\boxed{{\displaystyle {\mathbf{FinVect}}_k\text{ as enriched over }{\mathbf{Aff}}_k.}}$$

This is the ground-level affine geometry. Algebraic structures will be obtained by imposing polynomial equations inside these affine Hom-spaces.

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2. PROPs and their models

A convenient language for many algebraic structures is the language of PROPs.

I will use the following convention. A one-coloured $k$$k$-linear PROP $\mathsf{P}$$\mathsf P$ is a strict symmetric monoidal $k$$k$-linear category whose objects are

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

and whose tensor product on objects is addition:

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

A morphism

$$p:m\to n$$

in $\mathsf{P}$$\mathsf P$ should be thought of as an operation with $m$$m$ inputs and $n$$n$ outputs.

A model of $\mathsf{P}$$\mathsf P$ in finite-dimensional vector spaces is a strict symmetric monoidal functor

$$A:\mathsf{P}\to {\mathbf{FinVect}}_k.$$

If

$$A(1)=V,$$

then strict monoidality gives

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

A PROP operation

$$p:m\to n$$

is interpreted as a linear map

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

This notation covers many familiar operations:

$$\text{multiplication: }2\to 1,$$

$$\text{comultiplication: }1\to 2,$$

$$\text{unit: }0\to 1,$$

$$\text{counit: }1\to 0.$$

So Lie algebras, associative algebras, commutative algebras, coalgebras, bialgebras, and Frobenius algebras can all be treated in the same format.

The benefit of the PROP language is that the phrase “preserve the structure” becomes a naturality condition.

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3. Morphisms are monoidal natural transformations

Let

$$A,B:\mathsf{P}\to {\mathbf{FinVect}}_k$$

be two finite-dimensional $\mathsf{P}$$\mathsf P$-models. Write

$$A(1)=V,B(1)=W.$$

A morphism of models is a monoidal natural transformation

$$\eta :A\Rightarrow B.$$

Because $\eta$$\eta$ is monoidal, it is determined by its component at $1$$1$:

$${\eta }_1=f:V\to W.$$

The component at $n$$n$ is then

$${\eta }_n=f^{\otimes n}:V^{\otimes n}\to W^{\otimes n}.$$

Now take a PROP operation

$$p:m\to n.$$

Naturality of $\eta$$\eta$ gives the commutative square

$$\begin{array}{ccc}A(m)& \stackrel{A(p)}{\to }& A(n)\\ \downarrow {\eta }_m& & \downarrow {\eta }_n\\ B(m)& \stackrel{B(p)}{\to }& B(n).\end{array}$$

Since

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

and

$$B(m)=W^{\otimes m},B(n)=W^{\otimes n},$$

this square says precisely

$$B(p)\circ f^{\otimes m}=f^{\otimes n}\circ A(p).$$

This is the basic equation.

All the usual homomorphism conditions are instances of it.

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

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

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

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

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

$${\mathrm{\Delta }}_B\circ f=(f\otimes f)\circ {\mathrm{\Delta }}_A.$$

So the familiar formulas are not separate definitions. They are the same naturality equation seen through different PROPs.

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4. Why the naturality equations are polynomial

Now let us spell out why the equation

$$B(p)\circ f^{\otimes m}=f^{\otimes n}\circ A(p)$$

is a polynomial condition on the matrix entries of $f$$f$.

Choose bases

$$v_1,\dots ,v_r$$

of $V$$V$, and

$$w_1,\dots ,w_s$$

of $W$$W$.

Write

$$f(v_i)=\sum _{\alpha =1}^s{x}_{\alpha i}{w}_{\alpha }.$$

The scalars $x_{\alpha i}$$x_{\alpha i}$ are coordinate functions on the affine space

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

Fix a PROP operation

$$p:m\to n.$$

The two models give linear maps

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

and

$$B(p):W^{\otimes m}\to W^{\otimes n}.$$

Use multi-index notation. For

$$I=(i_1,\dots ,i_m),$$

write

$$v_I=v_{i_1}\otimes \cdots \otimes v_{i_m}.$$

For

$$J=(j_1,\dots ,j_n),$$

write

$$v_J=v_{j_1}\otimes \cdots \otimes v_{j_n}.$$

Similarly, for

$$K=({\alpha }_1,\dots ,{\alpha }_m),$$

write

$$w_K=w_{{\alpha }_1}\otimes \cdots \otimes w_{{\alpha }_m},$$

and for

$$L=({\beta }_1,\dots ,{\beta }_n),$$

write

$$w_L=w_{{\beta }_1}\otimes \cdots \otimes w_{{\beta }_n}.$$

Write the structure maps in coordinates:

$$A(p)(v_I)=\sum _J{A}_I^J{v}_J,$$

and

$$B(p)(w_K)=\sum _L{B}_K^L{w}_L.$$

We compute the two sides of the naturality equation on a basis tensor $v_I$$v_I$.

First,

$$f^{\otimes m}(v_I)=f(v_{i_1})\otimes \cdots \otimes f(v_{i_m}).$$

Since

$$f(v_{i_a})=\sum _{{\alpha }_a}{x}_{{\alpha }_a{i}_a}{w}_{{\alpha }_a},$$

we get

$$f^{\otimes m}(v_I)=\sum _K\left(\prod _{a=1}^m{x}_{{\alpha }_a{i}_a}\right)w_K.$$

Applying $B(p)$$B(p)$ gives

$$B(p)f^{\otimes m}(v_I)=\sum _K\left(\prod _{a=1}^m{x}_{{\alpha }_a{i}_a}\right)B(p)(w_K).$$

Using

$$B(p)(w_K)=\sum _L{B}_K^L{w}_L,$$

we obtain

$$B(p)f^{\otimes m}(v_I)=\sum _L\left[\sum _K{B}_K^L\prod _{a=1}^m{x}_{{\alpha }_a{i}_a}\right]w_L.$$

So the coefficient of $w_L$$w_L$ on the left-hand side is

$$\sum _K{B}_K^L\prod _{a=1}^m{x}_{{\alpha }_a{i}_a}.$$

This is a polynomial of degree $m$$m$ in the matrix entries of $f$$f$.

Now compute the right-hand side.

We have

$$A(p)(v_I)=\sum _J{A}_I^J{v}_J.$$

Applying $f^{\otimes n}$$f^{\otimes n}$ gives

$$f^{\otimes n}A(p)(v_I)=\sum _J{A}_I^J{f}^{\otimes n}(v_J).$$

But

$$f^{\otimes n}(v_J)=\sum _L\left(\prod _{b=1}^n{x}_{{\beta }_b{j}_b}\right)w_L.$$

Therefore

$$f^{\otimes n}A(p)(v_I)=\sum _L\left[\sum _J{A}_I^J\prod _{b=1}^n{x}_{{\beta }_b{j}_b}\right]w_L.$$

So the coefficient of $w_L$$w_L$ on the right-hand side is

$$\sum _J{A}_I^J\prod _{b=1}^n{x}_{{\beta }_b{j}_b}.$$

This is a polynomial of degree $n$$n$ in the matrix entries of $f$$f$.

Therefore the naturality equation is equivalent to the collection of polynomial equations

$$\sum _K{B}_K^L\prod _{a=1}^m{x}_{{\alpha }_a{i}_a}-\sum _J{A}_I^J\prod _{b=1}^n{x}_{{\beta }_b{j}_b}=0$$

for all input multi-indices $I$$I$ and output multi-indices $L$$L$.

The convention is that an empty product is $1$$1$. Thus the same formula also covers operations of type $0\to n$$0\to n$, such as units, and operations of type $m\to 0$$m\to 0$, such as counits.

This is the concrete reason why morphisms of finite-dimensional PROP-models are cut out by polynomial equations.

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5. The Hom-scheme of two PROP models

The affine space

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

parametrizes all underlying linear maps $f:V\to W$$f:V\to W$.

The equations

$$B(p)\circ f^{\otimes m}=f^{\otimes n}\circ A(p)$$

cut out precisely those linear maps which define monoidal natural transformations

$$A\Rightarrow B.$$

Thus we obtain a closed affine subscheme

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

Equivalently, this affine scheme represents the functor

$$R⟼{\mathrm{Hom}}_{\mathsf{P}{\text{-Mod}}_R}(A_R,B_R),$$

where

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

The ordinary Hom-set is recovered by taking $k$$k$-points:

$${\mathrm{Hom}}_{\mathsf{P}{\text{-Mod}}_k}(A,B)={\underset{―}{\mathrm{Hom}}}_{\mathsf{P}}(A,B)(k).$$

Thus the Hom-set is only the visible $k$$k$-point layer of a more geometric object.

One can package this as follows. For every operation $p:m\to n$$p:m\to n$, define a regular map

$${\mathrm{\Phi }}_p:{\underset{―}{\mathrm{Hom}}}_k(V,W)\to {\underset{―}{\mathrm{Hom}}}_k(V^{\otimes m},W^{\otimes n})$$

by

$${\mathrm{\Phi }}_p(f)=B(p)\circ f^{\otimes m}-f^{\otimes n}\circ A(p).$$

Then

$${\underset{―}{\mathrm{Hom}}}_{\mathsf{P}}(A,B)=\bigcap _p{\mathrm{\Phi }}_p^{-1}(0).$$

If the PROP is generated by specified operations, it is enough to impose these equations for those generators. Naturality for composites and tensor products follows formally from functoriality and monoidality.

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6. Enrichment over affine schemes

Let $A,B,C$$A,B,C$ be finite-dimensional $\mathsf{P}$$\mathsf P$-models.

Composition of underlying linear maps is polynomial, since it is matrix multiplication:

$$(g\circ f{)}_{ij}=\sum _r{g}_{ir}{f}_{rj}.$$

Moreover, the composite of two monoidal natural transformations is again a monoidal natural transformation.

Therefore composition gives a morphism of affine schemes

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

The identity morphism of $A$$A$ is the identity natural transformation

$${\mathrm{id}}_A:A\Rightarrow A.$$

It gives a $k$$k$-point

$$\mathrm{Spec}k\to {\underset{―}{\mathrm{Hom}}}_{\mathsf{P}}(A,A).$$

Since the monoidal unit of

$$({\mathbf{Aff}}_k,{\times }_k)$$

is

$$\mathrm{Spec}k,$$

this is the unit map required for enrichment.

Hence

$$\mathsf{P}{\text{-Mod}}_k^{\mathrm{fd}}$$

is enriched over

$${\mathbf{Aff}}_k.$$

The ordinary category is recovered by taking $k$$k$-points of each Hom-scheme.

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7. Examples

For Lie algebras, the main operation is the bracket

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

of type $2\to 1$$2\to 1$. A linear map

$$F:\mathfrak{g}\to \mathfrak{h}$$

is a Lie algebra homomorphism if

$$F([x,y{]}_{\mathfrak{g}})=[F(x),F(y){]}_{\mathfrak{h}}.$$

This is exactly the naturality equation

$$B(p)\circ F^{\otimes 2}=F\circ A(p).$$

In coordinates, if

$$[e_i,e_j{]}_{\mathfrak{g}}=\sum _r{c}_{ij}^r{e}_r,$$

and

$$[f_{\alpha },f_{\beta }{]}_{\mathfrak{h}}=\sum _{\gamma }{d}_{\alpha \beta }^{\gamma }{f}_{\gamma },$$

while

$$F(e_i)=\sum _{\alpha }{x}_{\alpha i}{f}_{\alpha },$$

then the homomorphism condition becomes

$$\sum _r{c}_{ij}^r{x}_{\gamma r}=\sum _{\alpha ,\beta }{d}_{\alpha \beta }^{\gamma }{x}_{\alpha i}{x}_{\beta j}.$$

So the Lie algebra Hom-scheme is cut out by quadratic equations.

For associative algebras, the multiplication

$$m:A\otimes A\to A$$

also has type $2\to 1$$2\to 1$. A homomorphism satisfies

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

If the algebras are unital, the unit operation has type $0\to 1$$0\to 1$, and the condition

$$F(1_A)=1_B$$

is another naturality equation.

For coalgebras, the comultiplication

$$\mathrm{\Delta }:C\to C\otimes C$$

has type $1\to 2$$1\to 2$. A coalgebra homomorphism satisfies

$${\mathrm{\Delta }}_D\circ F=(F\otimes F)\circ {\mathrm{\Delta }}_C.$$

Again, this is the same PROP naturality equation.

Thus multiplication, brackets, units, counits, and comultiplications are all handled by the same mechanism.

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8. Endomorphism monoid schemes

For a finite-dimensional $\mathsf{P}$$\mathsf P$-model $A$$A$, define

$${\underset{―}{\mathrm{End}}}_{\mathsf{P}}(A)={\underset{―}{\mathrm{Hom}}}_{\mathsf{P}}(A,A).$$

Since endomorphisms compose, this Hom-scheme carries a multiplication morphism

$${\underset{―}{\mathrm{End}}}_{\mathsf{P}}(A){\times }_k{\underset{―}{\mathrm{End}}}_{\mathsf{P}}(A)\to {\underset{―}{\mathrm{End}}}_{\mathsf{P}}(A).$$

The identity endomorphism gives a unit point

$$\mathrm{Spec}k\to {\underset{―}{\mathrm{End}}}_{\mathsf{P}}(A).$$

Thus

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

is an affine monoid scheme.

On coordinate rings, this means that

$$\mathcal{O}({\underset{―}{\mathrm{End}}}_{\mathsf{P}}(A))$$

is a commutative bialgebra. The comultiplication is dual to composition.

This applies equally to Lie algebras, associative algebras, coalgebras, and other finite-dimensional PROP-models.

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9. Automorphism group schemes

Let $A$$A$ be a finite-dimensional $\mathsf{P}$$\mathsf P$-model with underlying vector space

$$A(1)=V.$$

An endomorphism of $A$$A$ is an automorphism exactly when its underlying linear map

$$V\to V$$

is invertible.

Therefore

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

is the open subscheme of

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

where the determinant is invertible:

$${\underset{―}{\mathrm{Aut}}}_{\mathsf{P}}(A)=D(\det )\subseteq {\underset{―}{\mathrm{End}}}_{\mathsf{P}}(A).$$

Equivalently,

$${\underset{―}{\mathrm{Aut}}}_{\mathsf{P}}(A)={\underset{―}{\mathrm{End}}}_{\mathsf{P}}(A)\cap GL(V).$$

Thus

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

is an affine group scheme.

If

$$E=\mathcal{O}({\underset{―}{\mathrm{End}}}_{\mathsf{P}}(A)),$$

then

$$\mathcal{O}({\underset{―}{\mathrm{Aut}}}_{\mathsf{P}}(A))=E[{\det }^{-1}].$$

In a classical reduced setting over an algebraically closed field, one often calls this an affine algebraic group. Scheme-theoretically, the safer phrase is affine group scheme.

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10. Derivations from automorphisms

The word “derivation” is often introduced by writing down a Leibniz rule.

For an associative algebra, one says that a linear map $D:A\to A$$D:A\to A$ is a derivation if

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

For a Lie algebra, one says that $D:\mathfrak{g}\to \mathfrak{g}$$D:\mathfrak g\to\mathfrak g$ is a derivation if

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

These formulas are correct, but they are not the primitive idea.

For self-valued derivations, the more conceptual definition is

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

In words:

$$\boxed{{\displaystyle \text{derivations are infinitesimal automorphisms.}}}$$

This definition explains why the Leibniz rule has the form it does.

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11. The Lie algebra functor

Let $G$$G$ be an affine group scheme over $k$$k$.

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

$$R[\epsilon ]=R[X]/(X^2).$$

Thus

$$R[\epsilon ]=R\oplus R\epsilon ,{\epsilon }^2=0.$$

There is a projection

$$R[\epsilon ]\to R,a+\epsilon b\mapsto a.$$

The Lie algebra functor of $G$$G$ is

$$\mathrm{Lie}(G)(R)=\ker (G(R[\epsilon ])\to G(R)).$$

So $\mathrm{Lie}(G)(R)$$\operatorname{Lie}(G)(R)$ consists of those infinitesimal $R[\epsilon ]$$R[\varepsilon]$-points of $G$$G$ which reduce to the identity over $R$$R$.

For example,

$$\mathrm{Lie}(G{L}_n)(R)=M_n(R),$$

because every element of the kernel has the form

$$I+\epsilon A,A\in M_n(R),$$

and

$$(I+\epsilon A{)}^{-1}=I-\epsilon A.$$

Now take

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

Then an element of

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

is an automorphism of $A_R$$A_R$ over $R[\epsilon ]$$R[\varepsilon]$ which reduces to the identity modulo $\epsilon$$\varepsilon$.

Such an automorphism has the form

$$\mathrm{id}+\epsilon D,$$

where

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

It is automatically invertible as a linear map, with inverse

$$\mathrm{id}-\epsilon D.$$

The only remaining condition is that it preserves the PROP structure.

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12. Recovering the usual Leibniz rules

Let us now look carefully at what the general linearized equation says in familiar examples.

The general situation is this. We have an operation

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

An infinitesimal automorphism of the underlying vector space has the form

$$\mathrm{id}+\epsilon D,{\epsilon }^2=0.$$

To say that this infinitesimal automorphism preserves the operation $\theta$$\theta$ means

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

Expanding this equation to first order in $\epsilon$$\varepsilon$ gives

$$(\sum _{j=1}^n{1}^{\otimes (j-1)}\otimes D\otimes 1^{\otimes (n-j)})\circ \theta =\theta \circ (\sum _{i=1}^m{1}^{\otimes (i-1)}\otimes D\otimes 1^{\otimes (m-i)}).$$

This is the general derivation condition associated to the operation $\theta$$\theta$.

The usual Leibniz rules are obtained by applying this formula to particular operations.

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Associative algebras

Let $A$$A$ be an associative algebra with multiplication

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

Here the operation has type

$$2\to 1.$$

An infinitesimal automorphism of the underlying vector space is

$$\mathrm{id}+\epsilon D.$$

The condition that it preserves multiplication is

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

Evaluate both sides on a pure tensor $x\otimes y$$x\otimes y$.

The left-hand side is

$$(\mathrm{id}+\epsilon D)(xy)=xy+\epsilon D(xy).$$

The right-hand side is

$$\mu ((x+\epsilon D(x))\otimes (y+\epsilon D(y))).$$

Using bilinearity of multiplication, this becomes

$$(x+\epsilon D(x))(y+\epsilon D(y))=xy+\epsilon D(x)y+\epsilon xD(y)+{\epsilon }^{2}D(x)D(y).$$

Since

$${\epsilon }^2=0,$$

the last term vanishes. Hence the right-hand side is

$$xy+\epsilon (D(x)y+xD(y)).$$

Therefore multiplication is preserved to first order if and only if

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

Comparing the coefficients of $\epsilon$$\varepsilon$, we obtain

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

This is the usual Leibniz rule for associative algebras.

If the algebra is unital, there is also a unit operation

$$u:k\to A,1\mapsto 1_A.$$

Preserving the unit means

$$(\mathrm{id}+\epsilon D)\circ u=u.$$

Evaluating at $1\in k$$1\in k$, this says

$$1_A+\epsilon D(1_A)=1_A.$$

Thus

$$D(1_A)=0.$$

So for unital associative algebras, the infinitesimal automorphisms are precisely the linear maps $D:A\to A$$D:A\to A$ satisfying

$$D(xy)=D(x)y+xD(y),D(1_A)=0.$$

This is exactly the usual notion of a derivation.

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Lie algebras

Let $\mathfrak{g}$$\mathfrak g$ be a Lie algebra with bracket

$$[-,-]:\mathfrak{g}\otimes \mathfrak{g}\to \mathfrak{g}.$$

Again, this operation has type

$$2\to 1.$$

An infinitesimal automorphism of the underlying vector space is

$$\mathrm{id}+\epsilon D.$$

The condition that it preserves the bracket is

$$(\mathrm{id}+\epsilon D)([x,y])=[x+\epsilon D(x),y+\epsilon D(y)].$$

The left-hand side is

$$[x,y]+\epsilon D([x,y]).$$

The right-hand side is

$$[x+\epsilon D(x),y+\epsilon D(y)].$$

Using bilinearity of the Lie bracket, this expands to

$$[x,y]+\epsilon [D(x),y]+\epsilon [x,D(y)]+{\epsilon }^2[D(x),D(y)].$$

Since

$${\epsilon }^2=0,$$

the last term vanishes. Therefore the right-hand side is

$$[x,y]+\epsilon ([D(x),y]+[x,D(y)]).$$

Thus bracket preservation to first order is equivalent to

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

Comparing coefficients of $\epsilon$$\varepsilon$, we get

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

This is the usual derivation condition for Lie algebras.

So the Lie algebra Leibniz rule is not an independent mystery. It is simply the first-order expansion of the equation

$$F([x,y])=[F(x),F(y)]$$

near

$$F=\mathrm{id}.$$

Equivalently,

$$\mathrm{Der}(\mathfrak{g})=\mathrm{Lie}({\underset{―}{\mathrm{Aut}}}_{\mathrm{Lie}}(\mathfrak{g})).$$

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Coalgebras

Now let $C$$C$ be a coalgebra with comultiplication

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

This operation has type

$$1\to 2.$$

An infinitesimal automorphism of the underlying vector space is again

$$\mathrm{id}+\epsilon D.$$

The condition that it preserves comultiplication is

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

We compute both sides.

Using Sweedler notation,

$$\mathrm{\Delta }(c)=\sum c_{(1)}\otimes c_{(2)}.$$

The left-hand side is

$$((\mathrm{id}+\epsilon D)\otimes (\mathrm{id}+\epsilon D))\mathrm{\Delta }(c).$$

So

$$((\mathrm{id}+\epsilon D)\otimes (\mathrm{id}+\epsilon D))(\sum c_{(1)}\otimes c_{(2)})$$

equals

$$\sum (c_{(1)}+\epsilon D(c_{(1)}))\otimes (c_{(2)}+\epsilon D(c_{(2)})).$$

Expanding this gives

$$\sum c_{(1)}\otimes c_{(2)}+\epsilon \sum D(c_{(1)})\otimes c_{(2)}+\epsilon \sum c_{(1)}\otimes D(c_{(2)})+{\epsilon }^2\sum D(c_{(1)})\otimes D(c_{(2)}).$$

Since

$${\epsilon }^2=0,$$

the last term vanishes. Thus the left-hand side is

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

The right-hand side is

$$\mathrm{\Delta }((\mathrm{id}+\epsilon D)(c))=\mathrm{\Delta }(c+\epsilon D(c)).$$

By linearity of $\mathrm{\Delta }$$\Delta$, this is

$$\mathrm{\Delta }(c)+\epsilon \mathrm{\Delta }(D(c)).$$

Therefore comultiplication is preserved to first order if and only if

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

Comparing coefficients of $\epsilon$$\varepsilon$, we get

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

This is the usual coderivation condition.

If the coalgebra is counital, there is also a counit operation

$$ϵ:C\to k.$$

This has type

$$1\to 0.$$

Preserving the counit means

$$ϵ\circ (\mathrm{id}+\epsilon D)=ϵ.$$

Thus, for every $c\in C$$c\in C$,

$$ϵ(c+\epsilon D(c))=ϵ(c).$$

Equivalently,

$$ϵ(c)+\epsilon ϵ(D(c))=ϵ(c).$$

Hence

$$ϵ(D(c))=0$$

for all $c$$c$, or

$$ϵ\circ D=0.$$

So for a counital coalgebra, an infinitesimal automorphism is a linear map $D:C\to C$$D:C\to C$ satisfying

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

and

$$ϵ\circ D=0.$$

This is exactly the usual notion of a coderivation compatible with the counit.

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The point

The three examples have the same source.

For an operation

$$\theta :V^{\otimes m}\to V^{\otimes n},$$

an infinitesimal automorphism

$$\mathrm{id}+\epsilon D$$

must satisfy

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

The first-order part of this equation says that $D$$D$ acts on every output slot of $\theta$$\theta$ in the same way as it acts on every input slot of $\theta$$\theta$.

For multiplication and Lie brackets, there is one output and two inputs, so one obtains

$$D(\text{operation}(x,y))=\text{operation}(D(x),y)+\text{operation}(x,D(y)).$$

For comultiplication, there is one input and two outputs, so one obtains

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

Thus the Leibniz rule is not the root definition. It is the coordinate expression of a more structural statement:

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