The Free Smooth Ring Monad and Codifferential Category

Codifferential Categories: Definition, Axioms, and the Free Smooth Ring Monad

This is a learning note on codifferential categories.

The goal is to understand the definition first, including the axiom diagrams, and then to see why the free $C^{\mathrm{\infty }}$$C^\infty$-ring monad gives a central example.

The basic slogan is:

$$\boxed{{\displaystyle \text{A codifferential category is a categorical setting for function algebras and their differentials.}}}$$

More concretely, it is a category equipped with a monad $S$$S$ such that $SA$$S A$ behaves like an algebra of functions on $A$$A$, together with a natural operation

$$d_A:SA\to SA\otimes A.$$

This operation should be read as

$$f⟼df.$$

The axioms say that this abstract operation satisfies the ordinary rules of calculus: constants have zero derivative, products satisfy the Leibniz rule, linear functions have constant derivative, substitution satisfies the chain rule, and second derivatives commute.

---

The Ambient Category

A codifferential category begins with an additive symmetric monoidal category

$$(\mathcal{C},\otimes ,\mathsf{k},\sigma ).$$

Here:

• $\otimes$$\otimes$ is the tensor product;

• $\mathsf{k}$$\mathsf{k}$ is the monoidal unit;

• ${\sigma }_{A,B}:A\otimes B\to B\otimes A$$\sigma_{A,B}:A\otimes B\to B\otimes A$ is the symmetry;

• each hom-set $\mathcal{C}(A,B)$$\mathcal C(A,B)$ is a commutative monoid;

• composition and tensoring preserve addition.

In this context, “additive” does not necessarily mean abelian. It means that morphisms can be added and that zero morphisms exist.

This matters because the product rule has a sum in it:

$$d(fg)=fdg+gdf.$$

Categorically, that sum is a sum of morphisms.

---

Algebra Modalities

The next ingredient is an algebra modality.

An algebra modality is a monad

$$S:\mathcal{C}\to \mathcal{C}$$

with unit and multiplication

$${\eta }_A:A\to SA,$$

$${\mu }_A:SSA\to SA,$$

such that every object $SA$$S A$ carries a natural commutative monoid structure

$$m_A:SA\otimes SA\to SA,$$

$$u_A:\mathsf{k}\to SA.$$

The intended interpretation is:

$$\boxed{{\displaystyle SA=\text{the algebra of functions on }A.}}$$

Under this interpretation:

• $m_A$$m_A$ is multiplication of functions;

• $u_A$$u_A$ picks out the constant function $1$$1$;

• ${\eta }_A$$\eta_A$ inserts linear functions;

• ${\mu }_A$$\mu_A$ performs substitution of functions into functions.

The monad laws are:

$$\begin{array}{c}SSSA\stackrel{{\mu }_{SA}}{\to }SSA\stackrel{{\mu }_A}{\to }SA\end{array}=\begin{array}{c}SSSA\stackrel{S({\mu }_A)}{\to }SSA\stackrel{{\mu }_A}{\to }SA\end{array}$$

and

$$\begin{array}{c}SA\stackrel{{\eta }_{SA}}{\to }SSA\stackrel{{\mu }_A}{\to }SA\end{array}=1_{SA}=\begin{array}{c}SA\stackrel{S({\eta }_A)}{\to }SSA\stackrel{{\mu }_A}{\to }SA.\end{array}$$

The commutative monoid structure on $SA$$S A$ means that the following diagrams commute.

Associativity:

$$\begin{array}{c}SA\otimes SA\otimes SA\stackrel{m_A\otimes 1}{\to }SA\otimes SA\stackrel{m_A}{\to }SA\end{array}=\begin{array}{c}SA\otimes SA\otimes SA\stackrel{1\otimes m_A}{\to }SA\otimes SA\stackrel{m_A}{\to }SA.\end{array}$$

Unitality:

$$\begin{array}{c}\mathsf{k}\otimes SA\stackrel{u_A\otimes 1}{\to }SA\otimes SA\stackrel{m_A}{\to }SA\end{array}=\begin{array}{c}\mathsf{k}\otimes SA\stackrel{\cong }{\to }SA\end{array}$$

and

$$\begin{array}{c}SA\otimes \mathsf{k}\stackrel{1\otimes u_A}{\to }SA\otimes SA\stackrel{m_A}{\to }SA\end{array}=\begin{array}{c}SA\otimes \mathsf{k}\stackrel{\cong }{\to }SA.\end{array}$$

Commutativity:

$$\begin{array}{c}SA\otimes SA\stackrel{{\sigma }_{SA,SA}}{\to }SA\otimes SA\stackrel{m_A}{\to }SA\end{array}=\begin{array}{c}SA\otimes SA\stackrel{m_A}{\to }SA.\end{array}$$

Finally, the monad multiplication ${\mu }_A:SSA\to SA$$\mu_A:S S A\to S A$ must preserve the commutative monoid structure.

Multiplication preservation:

$$\begin{array}{c}SSA\otimes SSA\stackrel{m_{SA}}{\to }SSA\stackrel{{\mu }_A}{\to }SA\end{array}=\begin{array}{c}SSA\otimes SSA\stackrel{{\mu }_A\otimes {\mu }_A}{\to }SA\otimes SA\stackrel{m_A}{\to }SA.\end{array}$$

Unit preservation:

$$\begin{array}{c}\mathsf{k}\stackrel{u_{SA}}{\to }SSA\stackrel{{\mu }_A}{\to }SA\end{array}=\begin{array}{c}\mathsf{k}\stackrel{u_A}{\to }SA.\end{array}$$

So an algebra modality is not merely a monad. It is a monad whose free algebras $SA$$S A$ naturally behave like commutative algebras.

---

The Deriving Transformation

A deriving transformation on an algebra modality $S$$S$ is a natural transformation

$$d_A:SA\to SA\otimes A.$$

The intended meaning is:

$$\boxed{{\displaystyle d_A(f)=df.}}$$

In ordinary differential calculus, if $A={\mathbb{R}}^n$$A=\mathbb R^n$, then $SA$$S A$ should be some space of smooth functions on ${\mathbb{R}}^n$$\mathbb R^n$, and $SA\otimes A$$S A\otimes A$ should behave like the space of $1$$1$-forms.

For example, one should have

$$d(f)=\sum _{i=1}^n\frac{\partial f}{\partial x_i}\otimes e_i.$$

Naturality of $d$$d$ means that for every morphism $h:A\to B$$h:A\to B$, the following diagram commutes:

$$\begin{array}{c}SA\stackrel{Sh}{\to }SB\stackrel{d_B}{\to }SB\otimes B\end{array}=\begin{array}{c}SA\stackrel{d_A}{\to }SA\otimes A\stackrel{Sh\otimes h}{\to }SB\otimes B.\end{array}$$

This says that differentiation is compatible with change of variables along maps in the base category.

---

Definition of a Codifferential Category

A codifferential category is an additive symmetric monoidal category

$$(\mathcal{C},\otimes ,\mathsf{k},\sigma )$$

equipped with:

1. an algebra modality

$$(S,\mu ,\eta ,m,u);$$

2. a deriving transformation

$$d_A:SA\to SA\otimes A;$$

3. five axioms corresponding to the ordinary rules of differential calculus.

The five axioms are:

Axiom	Categorical name	Calculus meaning
$[d.1]$$[d.1]$	Constant rule	$d(c)=0$$d(c)=0$
$[d.2]$$[d.2]$	Product rule	$d(fg)=fdg+gdf$$d(fg)=f\,dg+g\,df$
$[d.3]$$[d.3]$	Linear rule	$d(x_i)=d{x}_i$$d(x_i)=dx_i$
$[d.4]$$[d.4]$	Chain rule	$d(g(f))=g^{\prime }(f)df$$d(g(f))=g'(f)\,df$
$[d.5]$$[d.5]$	Interchange rule	Mixed partials commute

Now we write these axioms as diagrams.

Throughout the diagrams, tensor associativity and unit isomorphisms are suppressed.

---

Axiom $[d.1]$$[d.1]$: Derivative of a Constant

The unit map

$$u_A:\mathsf{k}\to SA$$

selects the constant function $1$$1$.

The axiom says that constants have zero derivative:

$$d_A\circ u_A=0.$$

Diagrammatically:

$$\begin{array}{c}\mathsf{k}\stackrel{u_A}{\to }SA\stackrel{d_A}{\to }SA\otimes A\end{array}=\begin{array}{c}\mathsf{k}\stackrel{0}{\to }SA\otimes A.\end{array}$$

This is the categorical form of

$$d(c)=0.$$

---

Axiom $[d.2]$$[d.2]$: Leibniz Rule

The multiplication

$$m_A:SA\otimes SA\to SA$$

is multiplication of functions.

The derivative of a product can be computed in two ways.

First multiply, then differentiate:

$$\begin{array}{c}SA\otimes SA\stackrel{m_A}{\to }SA\stackrel{d_A}{\to }SA\otimes A.\end{array}$$

Second, differentiate the right factor and multiply:

$$\begin{array}{c}SA\otimes SA\stackrel{1\otimes d_A}{\to }SA\otimes SA\otimes A\stackrel{m_A\otimes 1}{\to }SA\otimes A.\end{array}$$

Third, differentiate the left factor and multiply:

$$\begin{array}{c}SA\otimes SA\stackrel{d_A\otimes 1}{\to }SA\otimes A\otimes SA\stackrel{1\otimes {\sigma }_{A,SA}}{\to }SA\otimes SA\otimes A\stackrel{m_A\otimes 1}{\to }SA\otimes A.\end{array}$$

The Leibniz rule says that the first composite is the sum of the second and third composites:

$$\begin{array}{c}SA\otimes SA\stackrel{m_A}{\to }SA\stackrel{d_A}{\to }SA\otimes A\end{array}=\begin{array}{c}SA\otimes SA\stackrel{1\otimes d_A}{\to }SA\otimes SA\otimes A\stackrel{m_A\otimes 1}{\to }SA\otimes A\end{array}+\begin{array}{c}SA\otimes SA\stackrel{d_A\otimes 1}{\to }SA\otimes A\otimes SA\stackrel{1\otimes {\sigma }_{A,SA}}{\to }SA\otimes SA\otimes A\stackrel{m_A\otimes 1}{\to }SA\otimes A.\end{array}$$

This is the categorical form of

$$d(fg)=fdg+gdf.$$

---

Axiom $[d.3]$$[d.3]$: Derivative of a Linear Function

The monad unit

$${\eta }_A:A\to SA$$

inserts linear functions into the algebra of functions.

The axiom says that the derivative of a linear function is the corresponding constant differential:

$$d_A\circ {\eta }_A=u_A\otimes 1_A.$$

Diagrammatically:

$$\begin{array}{c}A\stackrel{{\eta }_A}{\to }SA\stackrel{d_A}{\to }SA\otimes A\end{array}=\begin{array}{c}A\cong \mathsf{k}\otimes A\stackrel{u_A\otimes 1_A}{\to }SA\otimes A.\end{array}$$

In coordinates, this says:

$$d(x_i)=1\otimes e_i.$$

If $1\otimes e_i$$1\otimes e_i$ is written as $d{x}_i$$dx_i$, then this is simply

$$d(x_i)=d{x}_i.$$

---

Axiom $[d.4]$$[d.4]$: Chain Rule

The monad multiplication

$${\mu }_A:SSA\to SA$$

encodes substitution.

Intuitively, an element of $SSA$$S S A$ is a function whose variables are themselves functions on $A$$A$. Applying ${\mu }_A$$\mu_A$ substitutes those inner functions into the outer function.

The chain rule compares two ways of differentiating a substituted function.

First substitute, then differentiate:

$$\begin{array}{c}SSA\stackrel{{\mu }_A}{\to }SA\stackrel{d_A}{\to }SA\otimes A.\end{array}$$

Second, differentiate the outer function first:

$$\begin{array}{c}SSA\stackrel{d_{SA}}{\to }SSA\otimes SA.\end{array}$$

Then substitute the coefficient part and differentiate the inner function part:

$$\begin{array}{c}SSA\otimes SA\stackrel{{\mu }_A\otimes d_A}{\to }SA\otimes SA\otimes A.\end{array}$$

Finally multiply the two $SA$$S A$ factors:

$$\begin{array}{c}SA\otimes SA\otimes A\stackrel{m_A\otimes 1}{\to }SA\otimes A.\end{array}$$

The chain rule axiom is:

$$\begin{array}{c}SSA\stackrel{{\mu }_A}{\to }SA\stackrel{d_A}{\to }SA\otimes A\end{array}=\begin{array}{c}SSA\stackrel{d_{SA}}{\to }SSA\otimes SA\stackrel{{\mu }_A\otimes d_A}{\to }SA\otimes SA\otimes A\stackrel{m_A\otimes 1}{\to }SA\otimes A.\end{array}$$

This is the categorical form of the ordinary chain rule.

For smooth functions, if

$$g\in C^{\mathrm{\infty }}({\mathbb{R}}^m)$$

and

$$f_1,\dots ,f_m\in C^{\mathrm{\infty }}({\mathbb{R}}^n),$$

then

$$d(g(f_1,\dots ,f_m))=\sum _{j=1}^m\frac{\partial g}{\partial y_j}(f_1,\dots ,f_m)d{f}_j.$$

The monad multiplication $\mu$$\mu$ is substitution, and the chain rule says that differentiation is compatible with substitution.

---

Axiom $[d.5]$$[d.5]$: Interchange Rule

The interchange rule says that second derivatives are symmetric.

First differentiate once:

$$SA\stackrel{d_A}{\to }SA\otimes A.$$

Then differentiate the $SA$$S A$ component again:

$$SA\otimes A\stackrel{d_A\otimes 1}{\to }SA\otimes A\otimes A.$$

This gives a second derivative object. The interchange axiom says that swapping the two $A$$A$-directions gives the same result:

$$\begin{array}{c}SA\stackrel{d_A}{\to }SA\otimes A\stackrel{d_A\otimes 1}{\to }SA\otimes A\otimes A\end{array}=\begin{array}{c}SA\stackrel{d_A}{\to }SA\otimes A\stackrel{d_A\otimes 1}{\to }SA\otimes A\otimes A\stackrel{1\otimes {\sigma }_{A,A}}{\to }SA\otimes A\otimes A.\end{array}$$

This is the categorical form of

$$\frac{{\partial }^{2}f}{\partial x_i\partial x_j}=\frac{{\partial }^{2}f}{\partial x_j\partial x_i}.$$

Thus the deriving transformation behaves like a genuine differential operator, not merely an arbitrary derivation-like map.

---

Why Is It Called “Codifferential”?

The prefix “co” comes from the comparison with differential categories.

A differential category usually uses a coalgebra modality, often written as

$$!$$

and the deriving transformation has the shape

$$!A\otimes A\to !A.$$

A codifferential category uses an algebra modality $S$$S$, and the deriving transformation has the dual-looking shape

$$SA\to SA\otimes A.$$

So the conceptual shift is:

$$\begin{array}{c}\text{differential category}\\ \text{nonlinear maps and directional differentiation}\end{array}$$

versus

$$\begin{array}{c}\text{codifferential category}\\ \text{function algebras and differentials}\end{array}$$

The codifferential viewpoint is close to the algebraic geometry intuition:

$$\boxed{{\displaystyle \text{study a space through its algebra of functions.}}}$$

In this perspective, $SA$$S A$ is the function algebra, and $d_A:SA\to SA\otimes A$$d_A:S A\to S A\otimes A$ sends a function to its differential.

---

The Free Smooth Ring Monad

Now we come to the main example.

Let

$$\mathcal{C}=\mathbb{R}\text{-Vec}$$

with its usual tensor product

$${\otimes }_{\mathbb{R}}$$

and monoidal unit

$$\mathbb{R}.$$

There is a monad

$$S_{\mathrm{\infty }}:\mathbb{R}\text{-Vec}\to \mathbb{R}\text{-Vec}$$

called the free $C^{\mathrm{\infty }}$$C^\infty$-ring monad.

For finite-dimensional vector spaces,

$$S_{\mathrm{\infty }}({\mathbb{R}}^n)=C^{\mathrm{\infty }}({\mathbb{R}}^n).$$

Thus $S_{\mathrm{\infty }}({\mathbb{R}}^n)$$S_\infty(\mathbb R^n)$ is the vector space of smooth real-valued functions on ${\mathbb{R}}^n$$\mathbb R^n$.

The algebra structure

$$m_{{\mathbb{R}}^n}:S_{\mathrm{\infty }}({\mathbb{R}}^n)\otimes S_{\mathrm{\infty }}({\mathbb{R}}^n)\to S_{\mathrm{\infty }}({\mathbb{R}}^n)$$

is pointwise multiplication:

$$m_{{\mathbb{R}}^n}(f\otimes g)=fg.$$

The unit

$$u_{{\mathbb{R}}^n}:\mathbb{R}\to S_{\mathrm{\infty }}({\mathbb{R}}^n)$$

sends a scalar to the corresponding constant function.

The deriving transformation is the usual differential:

$$d_{{\mathbb{R}}^n}:C^{\mathrm{\infty }}({\mathbb{R}}^n)\to C^{\mathrm{\infty }}({\mathbb{R}}^n)\otimes {\mathbb{R}}^n$$

given by

$$d_{{\mathbb{R}}^n}(f)=\sum _{i=1}^n\frac{\partial f}{\partial x_i}\otimes e_i.$$

If one writes

$$d{x}_i=1\otimes e_i,$$

then this becomes the familiar formula

$$df=\sum _{i=1}^n\frac{\partial f}{\partial x_i}d{x}_i.$$

So the abstract deriving transformation is exactly ordinary differentiation in this example.

---

The Lawvere Theory Source of the Monad

The free smooth ring monad can be obtained from Lawvere theories.

There is a Lawvere theory of real vector spaces, denoted here by

$${\mathrm{Lin}}_{\mathbb{R}}.$$

There is also a Lawvere theory of smooth rings, denoted here by

$$\mathrm{Smooth}.$$

There is an inclusion

$${\mathrm{Lin}}_{\mathbb{R}}\hookrightarrow \mathrm{Smooth},$$

because every linear operation is a smooth operation.

Passing to models gives a forgetful functor

$$U:C^{\mathrm{\infty }}\text{-Ring}\to \mathbb{R}\text{-Vec}.$$

This functor forgets the smooth operations and remembers only the underlying real vector space.

It has a left adjoint

$$F:\mathbb{R}\text{-Vec}\to C^{\mathrm{\infty }}\text{-Ring}.$$

Therefore it gives a monad

$$S_{\mathrm{\infty }}=UF$$

on $\mathbb{R}\text{-Vec}$$\mathbb R\text{-Vec}$.

This is the free $C^{\mathrm{\infty }}$$C^\infty$-ring monad.

So the sequence of ideas is:

$$\begin{array}{c}{\mathrm{Lin}}_{\mathbb{R}}\hookrightarrow \mathrm{Smooth}\\ \Downarrow \\ C^{\mathrm{\infty }}\text{-Ring}\rightleftarrows \mathbb{R}\text{-Vec}\\ \Downarrow \\ S_{\mathrm{\infty }}=UF\end{array}$$

The codifferential structure is extra structure on this monad: the monad gives smooth functions and substitution, while the deriving transformation gives differentiation.

---

How the Axioms Look in the Smooth Example

For the free smooth ring monad, the five codifferential axioms become ordinary facts from calculus.

The constant rule says:

$$d(c)=0.$$

The product rule says:

$$d(fg)=fdg+gdf.$$

The linear rule says:

$$d(x_i)=d{x}_i.$$

The chain rule says:

$$d(g(f_1,\dots ,f_m))=\sum _{j=1}^m\frac{\partial g}{\partial y_j}(f_1,\dots ,f_m)d{f}_j.$$

The interchange rule says:

$$\frac{{\partial }^{2}f}{\partial x_i\partial x_j}=\frac{{\partial }^{2}f}{\partial x_j\partial x_i}.$$

Thus the abstract diagrams are not arbitrary. They are exactly the usual laws of smooth differential calculus, written in a form that makes sense in an additive symmetric monoidal category.

---

Comparison with the Polynomial Monad

It is useful to compare $S_{\mathrm{\infty }}$$S_\infty$ with the symmetric algebra monad.

The symmetric algebra monad

$$\mathrm{Sym}:\mathbb{R}\text{-Vec}\to \mathbb{R}\text{-Vec}$$

satisfies

$$\mathrm{Sym}({\mathbb{R}}^n)=\mathbb{R}[x_1,\dots ,x_n].$$

It is the free commutative algebra monad.

The free smooth ring monad satisfies

$$S_{\mathrm{\infty }}({\mathbb{R}}^n)=C^{\mathrm{\infty }}({\mathbb{R}}^n).$$

So one may think of $S_{\mathrm{\infty }}$$S_\infty$ as a smooth enhancement of $\mathrm{Sym}$$\mathrm{Sym}$:

$$\mathbb{R}[x_1,\dots ,x_n]\hookrightarrow C^{\mathrm{\infty }}({\mathbb{R}}^n).$$

The polynomial monad captures polynomial functions and polynomial substitution. The smooth monad captures smooth functions and smooth substitution.

Both admit a differential operation of the form

$$d(f)=\sum _i\frac{\partial f}{\partial x_i}\otimes e_i.$$

The smooth case is the one closest to ordinary calculus.