Derivation with divided powers

Definition

Usual notion: derivation with divided powers for all natural numbers

Suppose $R$ is a non-associative ring (i.e., a not necessarily associative ring). A derivation with divided powers for $R$ is the following data:

For every nonnegative integer $i$, an endomorphism $d^{(i)}$ of the additive structure of $R$, with $d^{(0)}$ equal to the identity map. We denote $d^{(1)}$ by $d$.

satisfying the following compatibility conditions:

Condition name	What it says
divided power conditions	For all nonnegative integers $i,j$, $d^{(i)}d^{(j)}={\binom {i+j}{i}}d^{(i+j)}$ where the multiplication on the left is composition of endomorphisms and the multiplication by ${\binom {i+j}{i}}$ on the right is just integer multiplication. Note that this also implies that $d^{(i)}$ commutes with $d^{(j)}$ for all $i,j$, and therefore commutativity need not be assumed explicitly.
Leibniz rule conditions (divided version of binomial formula for powers of a derivation)	For all nonnegative integers $n$, we have $d^{(n)}(x*y)=\sum _{i+j=n}d^{(i)}(x)*d^{(j)}(y)$ for all $x,y\in R$

where the multiplication on both sides is multiplication in the ring $R$.

The derivation in question is $d^{(1)}=d$, and this is indeed a derivation by the second condition applied to $n=1$.

Intuitively, $d^{(n)}$ is trying to be $d^{n}/n!$. Indeed, if $R$ is a $\mathbb {Q}$-algebra, then we can set $d^{(n)}=d^{n}/n!$ for all $n$, and this is the only feasible choice.

Derivation with divided powers up to a point

Suppose $R$ is a non-associative ring (i.e., a not necessarily associative ring) and $m$ is a natural number. A derivation with divided powers up to $m$ (which we may call a derivation with partial divided powers) for $R$ is the following data:

For every nonnegative integer $i\leq m$, an endomorphism $d^{(i)}$ of the additive structure of $R$, with $d^{(0)}$ equal to the identity map. We denote $d^{(1)}$ by $d$.

satisfying the following compatibility conditions:

Condition name	What it says
divided power conditions	For all nonnegative integers $i,j$ such that $i+j\leq m$, $d^{(i)}d^{(j)}={\binom {i+j}{i}}d^{(i+j)}$ where the multiplication on the left is composition of endomorphisms and the multiplication by ${\binom {i+j}{i}}$ on the right is just integer multiplication. Note that this also implies that $d^{(i)}$ commutes with $d^{(j)}$ for all $i,j$, and therefore commutativity need not be assumed explicitly.
Leibniz rule conditions (divided version of binomial formula for powers of a derivation)	For all nonnegative integers $n\leq m$, we have $d^{(n)}(x*y)=\sum _{i+j=n}d^{(i)}(x)*d^{(j)}(y)$ for all $x,y\in R$

where the multiplication on both sides is multiplication in the ring $R$.

The derivation in question is $d^{(1)}=d$, and this is indeed a derivation by the second condition applied to $n=1$.

Facts

• Exponential of nilpotent derivation with divided powers is automorphism

Related notions

• Derivation with divided Leibniz condition powers: Sequence satisfying the Leibniz rule conditions but not necessarily the divided power conditions.