Binomial formula for powers of a derivation

Statement

Suppose $R$ is a Non-associative ring (?) (i.e., a not necessarily associative ring -- this includes the associative ring, Lie ring, and other cases) and $d$ is a derivation of $R$. For any nonnegative integer $n$, if we denote by $d^{n}$ the composition of $d$ with itself $n$ times, with $d^{0}$ defined as the identity map.

We then have:

$\!d^{n}(x*y)=\sum _{i=0}^{n}{\binom {n}{i}}\left(d^{i}(x)*d^{n-i}(y)\right)$

Note that this in particular applies to the case of a Derivation of a Lie ring (?) and a Derivation of an associative ring (?).

Related facts

Applications

• Exponential of derivation is automorphism under suitable nilpotency assumptions

Proof

Proof idea

The idea is to use the Leibniz rule (which establishes the case $n=1$; $n=0$ is vacuous) and prove by induction. The proof is almost exactly like the proof of the binomial formula for $(x+y)^{n}$ for a commutative associative algebra.

Proof details: case $n=0$

In the case $n=0$, both sides are $x*y$, so the equality holds vacuously.

Proof details: case $n=1$

In this case, the left side is $d(x*y)$. The right side is:

$\!{\binom {1}{0}}(d^{0}(x)*d^{1}(y))+{\binom {1}{1}}(d^{1}(x)*d^{0}(y))=(x*dy)+(dx*y)$

By the Leibniz rule, the left side equals the right side.

Proof details: inductive step

Inductive hypothesis: $\!d^{n-1}(x*y)=\sum _{i=0}^{n-1}{\binom {n-1}{i}}d^{i}(x)*d^{n-1-i}(y)$

To prove: $\!d^{n}(x*y)=\sum _{i=0}^{n}{\binom {n}{i}}d^{i}(x)*d^{n-1-i}(y)$

Proof: We have by definition:

$\!d^{n}(x*y)=d\left(d^{n-1}(x*y)\right)$

By the inductive hypothesis, we can expand the inside and we get:

$\!d^{n}(x*y)=d\left\{\sum _{i=0}^{n-1}{\binom {n-1}{i}}d^{i}(x)*d^{n-1-i}(y)\right\}$

Since $d$ is additive, we can pull it inside the summation on the right side and get:

$\!d^{n}(x*y)=\sum _{i=0}^{n-1}{\binom {n-1}{i}}d\left\{d^{i}(x)*d^{n-1-i}(y)\right\}$

Now using the Leibniz rule on the inside, we get:

$\!d^{n}(x*y)=\sum _{i=0}^{n-1}{\binom {n-1}{i}}\left\{(d^{i+1}(x)*d^{n-(i+1)}(y))+(d^{i}(x)*d^{n-i}(y))\right\}$

We now rearrange the summation and obtain:

$\!d^{n}(x*y)=\sum _{i=0}^{n}\left({\binom {n-1}{i}}+{\binom {n-1}{i-1}}\right)(d^{i}(x)*d^{n-i}(y))$

We use Pascal's identity on the sum of binomial coefficients and obtain what we want:

$\!d^{n}(x*y)=\sum _{i=0}^{n}{\binom {n}{i}}d^{i}(x)*d^{n-1-i}(y)$

References

• Limits of abelian subgroups of finite p-groups by Jonathan Lazare Alperin and George Isaac Glauberman, Journal of Algebra, ISSN 00218693, Volume 203, Page 533 - 566(Year 1998): ^{Weblink for Elsevier copyMore info}, Proposition 2.5(a)