Extracting an $\mathbb{N}$-filtered differential modality from a differential modality

TL;DR

This paper introduces the concept of an $ $-filtered differential modality, extending differential modalities with a degree notion, and shows how to derive such filtered modalities from existing ones, linking morphisms to polynomial maps.

Contribution

It establishes a method to obtain $ $-filtered differential modalities from standard differential modalities under mild conditions.

Findings

Every differential modality induces an $ $-filtered differential modality.

Morphisms in the filtered setting correspond to polynomial maps of bounded degree.

The $(n+1)$-th derivative of such morphisms vanishes.

Abstract

A differential modality is a comonad on an additive symmetric monoidal category $(\mathsf{C},\otimes,I)$, whose underlying functor we denote $!\colon\mathsf{C} \rightarrow \mathsf{C}$, together with some additional structure including a differential operator $\partial\colon!A \otimes A \rightarrow !A$. A morphism $f\colon !A \rightarrow B$ is interpreted as a smooth function from $A$ to $B$. The notion of an $\mathbb{N}$-filtered differential modality is a variant in which a notion of degree is present. Instead of a single functor $!\colon \mathsf{C} \rightarrow \mathsf{C}$, we ask for a family of functors $!_{\le n}\colon\mathsf{C} \rightarrow \mathsf{C}$ where $n \in \mathbb{N}$. Now, a morphism $f\colon !_{\le n} A \rightarrow B$ is interpreted as a smooth function from $A$ to $B$, with degree less than $n$ for some notion of degree. We prove that under mild conditions, every differential modality on an additive symmetric monoidal category with underlying functor $!\colon \mathsf{C} \rightarrow \mathsf{C}$ yields an $\mathbb{N}$-filtered differential modality with underlying functors $!_{\le n}\colon\mathsf{C} \rightarrow \mathsf{C}$. A morphism $f\colon !_{\le n}A \rightarrow B$ corresponds to a polynomial map of degree less than $n$ from $A$ to $B$, in the sense that the $(n+1)$-th derivative of $f$ is $0$.