Exponential of derivation is automorphism under suitable nilpotency assumptions

Statement

Weaker version: global powering and torsion assumptions

Suppose $R$ is a non-associative ring (i.e., a not necessarily associative ring) and $d$ is a derivation of $R$ satisfying the following:

$d$ is nilpotent.
The nilpotency of $d$ is at most one more than the powering threshold for $R$ . In other words, there exists a natural number $n$ such that $d^{n}=0$ and $R$ is powered for all primes strictly less than $n$ .
The binilpotency of $d$ is at most one more than the torsion-free threshold for $R$ . In other words, there exists a natural number $m$ such that $d^{i}(x)*d^{j}(y)=0$ for all $x,y\in R,i+j\geq m$ , and $R$ is $p$ -torsion-free for all primes $p<m$ .

Then, $d$ is an exponentiable derivation. Explicitly, if $d^{n}=0$ , then the function:

$\exp(d):=\operatorname {id} +{\frac {d}{1!}}+{\frac {d^{2}}{2!}}+\dots +{\frac {d^{n-1}}{(n-1)!}}$

is an automorphism of $R$ .

Stronger version: image powering and torsion assumptions

Suppose $R$ is a non-associative ring (i.e., a not necessarily associative ring) and $d$ is a derivation of $R$ . Suppose that $d$ satisfies the following three conditions:

$d$ is locally nilpotent.
$d$ is an infinitely powered endomorphism of $R$ , i.e., the powering threshold for $d$ is $\infty$ .
$d$ is an infinitely bi-torsion-free endomorphism of $R$ , i.e., the bi-torsion-free threshold for $d$ is $\infty$ .

Then, $d$ is an exponentiable derivation, i.e., the exponential of $d$ exists and is an automorphism of $R$ .

Note on conditions

Note that conditions (1) and (2) are necessary for us to even be able to talk of the exponential of a derivation. Condition (3) is what helps us prove that the exponential must be an automorphism. It's possible that there are weaker versions of (3) that still provide the same guarantee, so the above form sufficient but not necessary conditions for a derivation to be exponentiable.

Truncated exponential version

Further information: truncated exponential of derivation is automorphism under suitable nilpotency assumptions

Particular cases

Assumption on $R$	What can we say about condition (2)?	What can we say about condition (3)?	Overall conclusion
The additive group of $R$ is torsion-free	Needs to be checked as usual	Always true for nilpotent derivations	Conditions (1) and (2) suffice to show that a given nilpotent derivation is exponentiable.
The additive group of $R$ is uniquely divisible, i.e., $R$ is an algebra over the rationals	Always true for nilpotent derivation	Always true for nilpotent derivation	All nilpotent derivations are exponentiable. Neither condition need to be checked.
$R$ is an algebra over a field of characteristic $p$	Equivalent to checking that $d^{p}=0$	Equivalent to checking that $d^{i}(x)*d^{j}(y)=0$ for all $x,y\in R,i+j\geq p$ .
The additive group of $R$ is an abelian p-group, i.e., every element has order a power of $p$ .	Equivalent to checking that $d^{p}=0$	Equivalent to checking that $d^{i}(x)*d^{j}(y)=0$ for all $x,y\in R,i+j\geq p$ .

Related facts

Facts used

Proof

Proof with global powering and torsion assumptions

Given: $R$ is a non-associative ring (i.e., a not necessarily associative ring) and $d$ is a derivation of $R$ satisfying the following:

$d$ is nilpotent.
The nilpotency of $d$ is at most one more than the powering threshold for $R$ . In other words, there exists a natural number $n$ such that $d^{n}=0$ and $R$ is powered for all primes strictly less than $n$ .
The binilpotency of $d$ is at most one more than the torsion-free threshold for $R$ . In other words, there exists a natural number $m$ such that $d^{i}(x)*d^{j}(y)=0$ for all $x,y\in R,i+j\geq m$ , and $R$ is $p$ -torsion-free for all primes $p<m$ .

To prove: $d$ is an exponentiable derivation. Explicitly, if $d^{n}=0$ , then the function:

$\exp(d):=\operatorname {id} +{\frac {d}{1!}}+{\frac {d^{2}}{2!}}+\dots +{\frac {d^{n-1}}{(n-1)!}}$

is an automorphism of $R$ .

Proof: We will show that the following is a nilpotent derivation with divided powers:

$d^{(i)}:=\left\lbrace {\begin{array}{rl}{\frac {d^{i}}{i!}},&0\leq i<n\\0,&i\geq n\\\end{array}}\right.$

Once we have shown this, Fact (1) will complete the proof.

Note that once we have established that the above is a derivation, it is clearly nilpotent. So, we only need to check that it is a derivation.we need to check the divided power conditions and the Leibniz rule conditions. Let's check these.

Divided power conditions: We need to show that for all nonnegative integers $i,j$ ,

$d^{(i)}d^{(j)}={\binom {i+j}{i}}d^{(i+j)}$ .

Case on $i,j$	Left side	Right side
$i,j,i+j$ are all less than $n$	${\frac {d^{i}d^{j}}{i!j!}}$ simplifies to ${\frac {d^{i+j}}{(i+j)!}}$	${\frac {(i+j)!}{i!j!}}{\frac {d^{i+j}}{(i+j)!}}$ simplifies to ${\frac {d^{i+j}}{(i+j)!}}$ .
$i,j$ are both less than $n$ but $i+j\geq n$	${\frac {d^{i}d^{j}}{i!j!}}$ simplifies to ${\frac {d^{i+j}}{(i+j)!}}=0$ , where the last step is because $d^{n}=0$	0 by definition
One or both of $i,j$ is greater than or equal to $n$	A product of two terms, at least one of which is 0, hence it is 0.	0 by definition.

Leibniz rule conditions: We need to show that, for all nonnegative integers $k$ :

$d^{(k)}(x*y)=\sum _{i+j=k}d^{(i)}(x)*d^{(j)}(y)$

We make three cases:

Case on $k$	What we need to show in this case	Proof
$k<n$ , so $k!$ can be inverted	${\frac {d^{k}}{k!}}(xy)=\sum _{i+j=k}{\frac {1}{i!j!}}d^{i}(x)d^{j}(y)$	Fact (2) gives us that: $d^{k}(xy)=\sum _{i+j=k}{\frac {k!}{i!j!}}d^{i}(x)d^{j}(y)$ Since $k!$ is invertible, we can divide both sides by $k!$ and get the result.
$n\leq k<m$ , so it is true that $k!u=0\implies u=0$ (by the torsion-free condition)	$0=\sum _{i+j=k,\max\{i,j\}<n}{\frac {1}{i!j!}}d^{i}(x)*d^{j}(y)$ (Note that the other terms in the summation are zero anyway).	By Fact (2): $d^{k}(xy)=\sum _{i+j=k}{\frac {k!}{i!j!}}d^{i}(x)d^{j}(y)$ First, break up the summation on the right into the cases where $\max\{i,j\}\geq n$ and $\max\{i,j\}<n$ . We get: $d^{k}(xy)=\sum _{i+j=k,\max\{i,j\}\geq n}{\frac {k!}{i!j!}}d^{i}(x)d^{j}(y)+\sum _{i+j=k,\max\{i,j\}<n}{\frac {k!}{i!j!}}d^{i}(x)d^{j}(y)$ The left side and all summands for the first summation on the right side are zero by the assumption $d^{n}=0$ , so we get: $0=\sum _{i+j=k}{\frac {k!}{i!j!}}d^{i}(x)d^{j}(y)$ The right side now is $k!$ times the expression we want to prove is zero, and we therefore get the result by the torsion-free assumption.
Case $k\geq m$ , so $d^{i}(x)*d^{j}(y)=0$ whenever $i+j=k$ (by the binilpotency condition)	$0=\sum _{i+j=k,\max\{i,j\}<n}{\frac {1}{i!j!}}d^{i}(x)*d^{j}(y)$ (Note that the other terms in the summation are zero anyway).	The binilpotency condition tells us that all summands on the right side are zero, hence the summation is zero.