Affine Geometry of Monoidal Functor Categories

 

From Free Enrichment to Monoidal DerivationsThe free layer: ordinary functor categoriesFinite-dimensional vector spaces as an affine-enriched categoryThe geometric layer: lax monoidal functorsOrdinary natural transformationsMonoidal natural transformationsPolynomial equations in coordinatesThe closed subscheme of monoidal natural transformationsCompositionStrong monoidal functorsWhy this is not ordinary enriched category theoryAutomorphisms in a Cartesian-enriched categoryReturning to affine schemesThe Lie algebra

 

From Free Enrichment to Monoidal Derivations

There are two layers in the story.

The first layer is formal:

$$\text{ordinary functor categories inherit enrichment.}$$

The second layer is geometric:

$$\text{monoidal natural transformations are cut out by polynomial equations.}$$

The point of this note is to separate these two layers.

The affine enrichment of ordinary representation categories such as group representations, quiver representations, associative algebra representations, and Lie algebra representations is essentially free. It comes from the general enriched functor category construction.

The genuinely geometric step appears when we pass from ordinary natural transformations to monoidal natural transformations. In finite-dimensional linear algebra, the latter are defined by polynomial equations, generally quadratic ones. This is why the natural base is not ${\mathrm{Vect}}_k$$\operatorname{Vect}_k$, but ${\mathrm{Aff}}_k$$\operatorname{Aff}_k$.

The final chain is:

$$\text{lax/strong monoidal functor}⟹\text{affine Hom-schemes}⟹\text{automorphism group scheme}⟹\text{Lie algebra of monoidal derivations}.$$

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The free layer: ordinary functor categories

Let $\mathcal{V}$$\mathcal V$ be a monoidal category with the relevant small products and equalizers. Let $\mathcal{A}$$\mathcal A$ be a $\mathcal{V}$$\mathcal V$-enriched category, and let $\mathcal{C}$$\mathcal C$ be an ordinary small category.

Then the ordinary functor category

$$[\mathcal{C},{\mathcal{A}}_0]$$

inherits a $\mathcal{V}$$\mathcal V$-enrichment, provided the required ends exist.

For two functors

$$F,G:\mathcal{C}\to {\mathcal{A}}_0,$$

the Hom-object is the end

$$\underset{―}{\mathrm{Nat}}(F,G)={\int }_{x\in \mathcal{C}}\mathcal{A}(Fx,Gx).$$

Equivalently,

$$\underset{―}{\mathrm{Nat}}(F,G)=\mathrm{Eq}(\prod _x\mathcal{A}(Fx,Gx)\rightrightarrows \prod _{a:x\to y}\mathcal{A}(Fx,Gy)).$$

The two arrows encode the two sides of naturality:

$${\eta }_x\mapsto G(a)\circ {\eta }_x$$

and

$${\eta }_x\mapsto {\eta }_y\circ F(a).$$

Thus the equalizer imposes the equations

$$G(a)\circ {\eta }_x={\eta }_y\circ F(a).$$

No closed structure on $\mathcal{V}$$\mathcal V$ is needed here. In particular, one should not assume that ${\mathrm{Aff}}_k$$\operatorname{Aff}_k$ is Cartesian closed. It is not. What is needed is the existence of the relevant products and equalizers.

Since

$${\mathrm{Aff}}_k\simeq ({\mathrm{CAlg}}_k{)}^{op}$$

and ${\mathrm{CAlg}}_k$$\operatorname{CAlg}_k$ is cocomplete, ${\mathrm{Aff}}_k$$\operatorname{Aff}_k$ has the limits needed for this construction.

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Finite-dimensional vector spaces as an affine-enriched category

The category ${\mathrm{Vect}}_k^{fd}$$\operatorname{Vect}^{fd}_k$ is enriched over ${\mathrm{Aff}}_k$$\operatorname{Aff}_k$ by

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

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

$${\underset{―}{\mathrm{Hom}}}_k(V,W)(R)={\mathrm{Hom}}_R(V{\otimes }_{k}R,W{\otimes }_{k}R).$$

Composition is induced by matrix multiplication, hence is a morphism of affine schemes:

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

Therefore, for every ordinary small category $\mathcal{C}$$\mathcal C$,

$$[\mathcal{C},{\mathrm{Vect}}_k^{fd}]$$

inherits an ${\mathrm{Aff}}_k$$\operatorname{Aff}_k$-enrichment:

$$\underset{―}{\mathrm{Nat}}(F,G)={\int }_{x\in \mathcal{C}}{\underset{―}{\mathrm{Hom}}}_k(Fx,Gx).$$

This explains, for free, the affine enrichment of many basic representation categories.

For a group $G$$G$,

$${\mathrm{Rep}}_k^{fd}(G)=[BG,{\mathrm{Vect}}_k^{fd}].$$

Equivalently,

$${\mathrm{Rep}}_k^{fd}(G)\simeq kG\text{-}{\mathrm{Mod}}^{fd}.$$

For a quiver $Q$$Q$, representations are functors from the path category $\mathsf{P}(Q)$$\mathsf P(Q)$:

$${\mathrm{Rep}}_k^{fd}(Q)=[\mathsf{P}(Q),{\mathrm{Vect}}_k^{fd}].$$

If $Q$$Q$ has finitely many vertices, this is equivalently

$${\mathrm{Rep}}_k^{fd}(Q)\simeq kQ\text{-}{\mathrm{Mod}}^{fd},$$

where $kQ$$kQ$ is the path algebra. For infinite quivers, one should either keep the path-category formulation or use the locally unital path algebra.

For a $k$$k$-algebra $A$$A$, finite-dimensional $A$$A$-modules also belong to this same layer. An $A$$A$-linear map

$$f:M\to N$$

is simply a $k$$k$-linear map satisfying

$$f(am)=af(m).$$

These are linear equations inside

$${\mathrm{Hom}}_k(M,N).$$

For a Lie algebra $\mathfrak{g}$$\mathfrak g$,

$${\mathrm{Rep}}_k^{fd}(\mathfrak{g})\simeq U(\mathfrak{g})\text{-}{\mathrm{Mod}}^{fd}.$$

Thus representations of groups, quivers, associative algebras, and Lie algebras all sit in the unary/module-theoretic layer. Their affine Hom-objects are obtained either from the general functor-category construction or from the elementary fact that module homomorphisms are cut out by linear equations.

This is not the main geometric point.

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The geometric layer: lax monoidal functors

Now let $\mathcal{C}$$\mathcal C$ be a small monoidal category. For notational simplicity, assume $\mathcal{C}$$\mathcal C$ is strict monoidal. In the non-strict case, one inserts the associativity and unit constraints.

Let

$$F,G:\mathcal{C}\to {\mathrm{Vect}}_k^{fd}$$

be two lax monoidal functors.

Thus $F$$F$ comes with structure maps

$${\mu }_{x,y}^F:F(x)\otimes F(y)\to F(x\otimes y),$$

and

$${\eta }^F:k\to F(\mathbf{1}).$$

Similarly, $G$$G$ comes with

$${\mu }_{x,y}^G:G(x)\otimes G(y)\to G(x\otimes y),$$

and

$${\eta }^G:k\to G(\mathbf{1}).$$

We are not constructing a moduli space of lax monoidal functors. The functors $F$$F$ and $G$$G$ are fixed.

The object we want is the affine scheme of monoidal natural transformations

$$F\Rightarrow G.$$

The construction proceeds in two steps:

$$\text{ordinary natural transformations}⇝\text{monoidal natural transformations}.$$

The first step is formal. The second step is geometric.

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Ordinary natural transformations

First ignore the monoidal structures.

An ordinary natural transformation

$$\alpha :F\Rightarrow G$$

is a family of linear maps

$${\alpha }_x:F(x)\to G(x)$$

such that, for every morphism

$$a:x\to y$$

in $\mathcal{C}$$\mathcal C$, one has

$$G(a)\circ {\alpha }_x={\alpha }_y\circ F(a).$$

Since all vector spaces involved are finite-dimensional, the possible maps ${\alpha }_x$$\alpha_x$ are represented by affine spaces

$${\underset{―}{\mathrm{Hom}}}_k(F(x),G(x)).$$

Hence ordinary natural transformations are represented by

$$\underset{―}{\mathrm{Nat}}(F,G)=\mathrm{Eq}(\prod _x{\underset{―}{\mathrm{Hom}}}_k(F(x),G(x))\rightrightarrows \prod _{a:x\to y}{\underset{―}{\mathrm{Hom}}}_k(F(x),G(y))).$$

The two arrows send a family $({\alpha }_x{)}_x$$(\alpha_x)_x$ to

$$(G(a)\circ {\alpha }_x{)}_a$$

and

$$({\alpha }_y\circ F(a){)}_a.$$

Thus this equalizer imposes precisely the ordinary naturality equations.

For every $k$$k$-algebra $R$$R$, its $R$$R$-points are ordinary $R$$R$-linear natural transformations

$$F_R\Rightarrow G_R,$$

where

$$F_R(x)=F(x){\otimes }_{k}R,G_R(x)=G(x){\otimes }_{k}R.$$

---

Monoidal natural transformations

Now we impose the condition that $\alpha$$\alpha$ is monoidal.

A natural transformation

$$\alpha :F\Rightarrow G$$

is monoidal if, for every pair $x,y\in \mathcal{C}$$x,y\in\mathcal C$, the square

$$\begin{array}{ccc}F(x)\otimes F(y)& \stackrel{{\mu }_{x,y}^F}{\to }& F(x\otimes y)\\ \downarrow {\alpha }_x\otimes {\alpha }_y& & \downarrow {\alpha }_{x\otimes y}\\ G(x)\otimes G(y)& \stackrel{{\mu }_{x,y}^G}{\to }& G(x\otimes y)\end{array}$$

commutes.

Equivalently,

$${\alpha }_{x\otimes y}\circ {\mu }_{x,y}^F={\mu }_{x,y}^G\circ ({\alpha }_x\otimes {\alpha }_y).$$

It must also satisfy the unit equation

$${\alpha }_{\mathbf{1}}\circ {\eta }^F={\eta }^G.$$

Thus monoidal natural transformations are ordinary natural transformations satisfying extra multiplicativity equations.

This is the crucial difference. Naturality is linear. Monoidality is generally quadratic.

---

Polynomial equations in coordinates

After choosing bases, this becomes completely concrete.

For each object $x\in \mathcal{C}$$x\in\mathcal C$, write the matrix entries of

$${\alpha }_x:F(x)\to G(x)$$

as variables

$$t_{ij}^x.$$

The ordinary naturality equation

$$G(a)\circ {\alpha }_x={\alpha }_y\circ F(a)$$

becomes a collection of linear polynomial equations in the variables $t_{ij}^x$$t^x_{ij}$ and $t_{ij}^y$$t^y_{ij}$.

The unit equation

$${\alpha }_{\mathbf{1}}\circ {\eta }^F={\eta }^G$$

is also linear.

The monoidal equation

$${\alpha }_{x\otimes y}\circ {\mu }_{x,y}^F={\mu }_{x,y}^G\circ ({\alpha }_x\otimes {\alpha }_y)$$

is different. The entries of

$${\alpha }_x\otimes {\alpha }_y$$

are products of entries of ${\alpha }_x$$\alpha_x$ and entries of ${\alpha }_y$$\alpha_y$. Therefore, in coordinates, the monoidal equation is generally quadratic.

Thus the coordinate ring of the Hom-object is obtained from the polynomial coordinate ring on all matrix entries of the ${\alpha }_x$$\alpha_x$ by quotienting by:

$$\text{naturality equations}+\text{unit equations}+\text{monoidal compatibility equations}.$$

The first two types are linear. The third type is generally quadratic.

This is the intuitive reason ${\mathrm{Aff}}_k$$\operatorname{Aff}_k$ is the correct base: ${\mathrm{Vect}}_k$$\operatorname{Vect}_k$ remembers linear structure, but ${\mathrm{Aff}}_k$$\operatorname{Aff}_k$ remembers polynomial equations.

---

The closed subscheme of monoidal natural transformations

Let

$$N=\underset{―}{\mathrm{Nat}}(F,G).$$

By definition of $N$$N$, there is a universal natural transformation over $N$$N$:

$${\alpha }^{\mathrm{univ}}:F_N\Rightarrow G_N.$$

For fixed $x,y\in \mathcal{C}$$x,y\in\mathcal C$, consider the two maps

$$F(x)\otimes F(y){\otimes }_k{\mathcal{O}}_N⟶G(x\otimes y){\otimes }_k{\mathcal{O}}_N$$

given by

$${\alpha }_{x\otimes y}^{\mathrm{univ}}\circ {\mu }_{x,y}^F$$

and

$${\mu }_{x,y}^G\circ ({\alpha }_x^{\mathrm{univ}}\otimes {\alpha }_y^{\mathrm{univ}}).$$

The condition that these two maps are equal is the vanishing of their difference.

Since all vector spaces involved are finite-dimensional, this difference is a section of the finite free ${\mathcal{O}}_N$$\mathcal O_N$-module

$${\mathrm{Hom}}_k(F(x)\otimes F(y),G(x\otimes y)){\otimes }_k{\mathcal{O}}_N.$$

The vanishing of such a section defines a closed subscheme of $N$$N$.

The unit condition

$${\alpha }_{\mathbf{1}}^{\mathrm{univ}}\circ {\eta }^F={\eta }^G$$

is closed in the same way.

Now impose these equations for all pairs $x,y\in \mathcal{C}$$x,y\in\mathcal C$. Since $\mathcal{C}$$\mathcal C$ is small, this is a set of equations. Their simultaneous vanishing defines a closed affine subscheme

$$\underset{―}{\mathrm{MonNat}}(F,G)\subseteq \underset{―}{\mathrm{Nat}}(F,G).$$

By construction,

$$\underset{―}{\mathrm{MonNat}}(F,G)(R)=\mathrm{MonNat}(F_R,G_R).$$

So $\underset{―}{\mathrm{MonNat}}(F,G)$$\underline{\operatorname{MonNat}}(F,G)$ represents $R$$R$-families of monoidal natural transformations.

This is the geometric step:

$$\boxed{{\displaystyle \text{monoidal natural transformations form a closed subscheme of ordinary natural transformations.}}}$$

The finite-dimensionality of the values of $F$$F$ and $G$$G$ is used here. It ensures that the relevant Hom-functors are represented by affine spaces of the expected form, and that the differences above are sections of finite free modules after base change.

---

Composition

Let

$$F\stackrel{\alpha }{\Rightarrow }G\stackrel{\beta }{\Rightarrow }H$$

be monoidal natural transformations.

Their composite is the ordinary composite natural transformation

$$\beta \circ \alpha :F\Rightarrow H,$$

defined pointwise by

$$(\beta \circ \alpha {)}_x={\beta }_x\circ {\alpha }_x.$$

We check that it is monoidal.

Compute:

$$(\beta \circ \alpha {)}_{x\otimes y}\circ {\mu }_{x,y}^F={\beta }_{x\otimes y}\circ {\alpha }_{x\otimes y}\circ {\mu }_{x,y}^F.$$

Since $\alpha$$\alpha$ is monoidal,

$${\alpha }_{x\otimes y}\circ {\mu }_{x,y}^F={\mu }_{x,y}^G\circ ({\alpha }_x\otimes {\alpha }_y).$$

Therefore

$$(\beta \circ \alpha {)}_{x\otimes y}\circ {\mu }_{x,y}^F={\beta }_{x\otimes y}\circ {\mu }_{x,y}^G\circ ({\alpha }_x\otimes {\alpha }_y).$$

Since $\beta$$\beta$ is monoidal,

$${\beta }_{x\otimes y}\circ {\mu }_{x,y}^G={\mu }_{x,y}^H\circ ({\beta }_x\otimes {\beta }_y).$$

Hence

$$(\beta \circ \alpha {)}_{x\otimes y}\circ {\mu }_{x,y}^F={\mu }_{x,y}^H\circ ({\beta }_x\otimes {\beta }_y)\circ ({\alpha }_x\otimes {\alpha }_y).$$

But

$$({\beta }_x\otimes {\beta }_y)\circ ({\alpha }_x\otimes {\alpha }_y)=({\beta }_x\circ {\alpha }_x)\otimes ({\beta }_y\circ {\alpha }_y).$$

Therefore

$$(\beta \circ \alpha {)}_{x\otimes y}\circ {\mu }_{x,y}^F={\mu }_{x,y}^H\circ ((\beta \circ \alpha {)}_x\otimes (\beta \circ \alpha {)}_y).$$

So $\beta \circ \alpha$$\beta\circ\alpha$ is monoidal.

On affine schemes, ordinary composition is induced by pointwise matrix multiplication. Hence it defines a regular morphism

$$\underset{―}{\mathrm{MonNat}}(G,H)\times \underset{―}{\mathrm{MonNat}}(F,G)⟶\underset{―}{\mathrm{MonNat}}(F,H).$$

The identity natural transformation is monoidal, so it gives the unit map

$$\mathrm{Spec}k⟶\underset{―}{\mathrm{MonNat}}(F,F).$$

Associativity and the unit laws follow from ordinary composition.

Therefore

$$\mathrm{LaxMon}(\mathcal{C},{\mathrm{Vect}}_k^{fd})$$

with monoidal natural transformations as morphisms is enriched over ${\mathrm{Aff}}_k$$\operatorname{Aff}_k$.

---

Strong monoidal functors

The strong case uses the same Hom-schemes.

A strong monoidal functor is a lax monoidal functor whose structure maps

$${\mu }_{x,y}^F:F(x)\otimes F(y)\to F(x\otimes y)$$

and

$${\eta }^F:k\to F(\mathbf{1})$$

are isomorphisms.

But once two strong monoidal functors $F,G$$F,G$ are fixed, the condition for

$$\alpha :F\Rightarrow G$$

to be a monoidal natural transformation is still

$${\alpha }_{x\otimes y}\circ {\mu }_{x,y}^F={\mu }_{x,y}^G\circ ({\alpha }_x\otimes {\alpha }_y),$$

and

$${\alpha }_{\mathbf{1}}\circ {\eta }^F={\eta }^G.$$

Thus the Hom-scheme between two strong monoidal functors is still

$$\underset{―}{\mathrm{MonNat}}(F,G).$$

So

$$\mathrm{StrMon}(\mathcal{C},{\mathrm{Vect}}_k^{fd})$$

is also enriched over ${\mathrm{Aff}}_k$$\operatorname{Aff}_k$.

The strong condition changes the objects. It does not change the Hom-equations between fixed objects.

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Why this is not ordinary enriched category theory

For ordinary functors,

$$\underset{―}{\mathrm{Nat}}(F,G)={\int }_x\underset{―}{\mathrm{Hom}}(Fx,Gx)$$

is formal.

For monoidal functors, the relevant Hom-object is

$$\underset{―}{\mathrm{MonNat}}(F,G)\subseteq \underset{―}{\mathrm{Nat}}(F,G),$$

cut out by multiplicative equations.

This is not a formal consequence of enrichment over an arbitrary base category. For example, monoidal natural transformations are generally not closed under addition, and therefore usually do not form a vector space.

The base ${\mathrm{Aff}}_k$$\operatorname{Aff}_k$ works because affine schemes encode equations, including nonlinear equations.

So the correct distinction is:

$$\boxed{{\displaystyle \text{ordinary natural transformations form an end;}}}$$

$$\boxed{{\displaystyle \text{monoidal natural transformations form a closed subscheme of that end.}}}$$

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Automorphisms in a Cartesian-enriched category

Let $\mathcal{V}$$\mathcal V$ be a Cartesian monoidal category with finite limits, and let $\mathcal{A}$$\mathcal A$ be a $\mathcal{V}$$\mathcal V$-enriched category.

For an object $X\in \mathcal{A}$$X\in\mathcal A$, set

$$E=\mathcal{A}(X,X).$$

Composition and identity make $E$$E$ a monoid object in $\mathcal{V}$$\mathcal V$:

$$m:E\times E\to E,$$

$$e:1\to E.$$

To define automorphisms, one should not simply say “the invertible elements of $E$$E$,” because existence of an inverse is not a finite-limit condition. Instead, include the inverse as part of the data.

Define

$$\mathrm{Aut}(X)=\mathrm{Eq}(E\times E\rightrightarrows E\times E),$$

where the two arrows are

$$(a,b)\mapsto (ab,ba)$$

and

$$(a,b)\mapsto (1,1).$$

Thus $\mathrm{Aut}(X)$$\operatorname{Aut}(X)$ internally represents pairs $(a,b)$$(a,b)$ satisfying

$$ab=1,ba=1.$$

Multiplication is

$$(a,b)(c,d)=(ac,db).$$

The unit is

$$(1,1),$$

and inversion is

$$(a,b)\mapsto (b,a).$$

These operations satisfy the group object axioms by the monoid axioms for $E$$E$.

Therefore:

$$\boxed{{\displaystyle \text{in a Cartesian-enriched category, endomorphisms form an internal monoid,}}}$$

and

$$\boxed{{\displaystyle \text{automorphisms form its internal group of units.}}}$$

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Returning to affine schemes

Now take

$$\mathcal{V}={\mathrm{Aff}}_k.$$

It is Cartesian monoidal under fiber product over $\mathrm{Spec}k$$\operatorname{Spec}k$, and it has finite limits.

Therefore, if $\mathcal{A}$$\mathcal A$ is enriched over ${\mathrm{Aff}}_k$$\operatorname{Aff}_k$, then every object $X\in \mathcal{A}$$X\in\mathcal A$ has an affine group scheme of automorphisms:

$$\underset{―}{\mathrm{Aut}}(X).$$

Apply this to

$$\mathcal{A}=\mathrm{LaxMon}(\mathcal{C},{\mathrm{Vect}}_k^{fd})$$

or

$$\mathcal{A}=\mathrm{StrMon}(\mathcal{C},{\mathrm{Vect}}_k^{fd}).$$

For a lax or strong monoidal functor $F$$F$, we obtain an affine group scheme

$${\underset{―}{\mathrm{Aut}}}_{\mathrm{Mon}}(F).$$

Its $R$$R$-points are the monoidal natural automorphisms of

$$F_R:\mathcal{C}\to {\mathrm{Vect}}_R^{fd}.$$

If $\mathcal{C}$$\mathcal C$ is finite, or if the defining equations are finitely generated in the relevant sense, this group scheme may be of finite type. Without such hypotheses, it is safer to call it an affine group scheme rather than an affine algebraic group.

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The Lie algebra

The Lie algebra of an affine group scheme $G$$G$ is defined using dual numbers:

$$\mathrm{Lie}(G)(R)=\ker (G(R[ϵ])\to G(R)),$$

where

$$R[ϵ]=R[t]/(t^2).$$

Hence

$$\mathrm{Lie}{\underset{―}{\mathrm{Aut}}}_{\mathrm{Mon}}(F)$$

consists of monoidal natural automorphisms of $F_{R[ϵ]}$$F_{R[\epsilon]}$ reducing to the identity modulo $ϵ$$\epsilon$.

Such an automorphism has the form

$${\alpha }_x={\mathrm{id}}_{F(x{)}_R}+ϵD_x.$$

Its inverse is

$${\mathrm{id}}_{F(x{)}_R}-ϵD_x.$$

The naturality condition for $\alpha$$\alpha$ linearizes to

$$F(a)D_x=D_{y}F(a)$$

for every morphism

$$a:x\to y$$

in $\mathcal{C}$$\mathcal C$.

Now linearize the monoidal condition

$${\alpha }_{x\otimes y}\circ {\mu }_{x,y}={\mu }_{x,y}\circ ({\alpha }_x\otimes {\alpha }_y).$$

The left-hand side is

$$(\mathrm{id}+ϵD_{x\otimes y})\circ {\mu }_{x,y}={\mu }_{x,y}+ϵD_{x\otimes y}\circ {\mu }_{x,y}.$$

The right-hand side is

$${\mu }_{x,y}\circ ((\mathrm{id}+ϵD_x)\otimes (\mathrm{id}+ϵD_y)).$$

Since ${ϵ}^2=0$$\epsilon^2=0$,

$$(\mathrm{id}+ϵD_x)\otimes (\mathrm{id}+ϵD_y)=\mathrm{id}+ϵ(D_x\otimes 1+1\otimes D_y).$$

Therefore the right-hand side is

$${\mu }_{x,y}+ϵ{\mu }_{x,y}\circ (D_x\otimes 1+1\otimes D_y).$$

Comparing the $ϵ$$\epsilon$-terms gives

$$D_{x\otimes y}\circ {\mu }_{x,y}={\mu }_{x,y}\circ (D_x\otimes 1+1\otimes D_y).$$

The unit equation linearizes to

$$D_{\mathbf{1}}\circ \eta =0.$$

Thus

$$\mathrm{Lie}{\underset{―}{\mathrm{Aut}}}_{\mathrm{Mon}}(F)$$

is the space of natural endomorphisms $D$$D$ satisfying

$$D_{x\otimes y}\circ {\mu }_{x,y}={\mu }_{x,y}\circ (D_x\otimes 1+1\otimes D_y),$$

and

$$D_{\mathbf{1}}\circ \eta =0.$$

These are the monoidal derivations of $F$$F$.

So

$$\boxed{{\displaystyle \mathrm{Lie}{\underset{―}{\mathrm{Aut}}}_{\mathrm{Mon}}(F)={\mathrm{Der}}_{\mathrm{Mon}}(F).}}$$

The bracket is the pointwise commutator:

$$[D,E{]}_x=D_x{E}_x-E_x{D}_x.$$

Closure under this bracket follows from the general fact that the Lie algebra of an affine group scheme is a Lie algebra.

If one wants to see the closure directly, it is just the cancellation of the mixed terms. If

$$D_{x\otimes y}\mu =\mu (D_x\otimes 1+1\otimes D_y)$$

and

$$E_{x\otimes y}\mu =\mu (E_x\otimes 1+1\otimes E_y),$$

then applying the two formulas twice gives

$$[D,E{]}_{x\otimes y}\mu =\mu ([D,E{]}_x\otimes 1+1\otimes [D,E{]}_y).$$

The unit condition is preserved because

$$D_{\mathbf{1}}\eta =0,E_{\mathbf{1}}\eta =0.$$

Thus monoidal derivations form a Lie algebra.

$$\boxed{{\displaystyle \text{lax/strong monoidal functor}\Rightarrow \text{affine Hom-schemes}\Rightarrow \text{automorphism group scheme}\Rightarrow \text{Lie algebra of monoidal derivations}.}}$$