Identity with repeatedly taking the commutator of a ring element

Question

This is taken from Jacobson's Basic Algebra 2e, it's 2.1.5

If $a$ and $b$ are elements of a ring, define $a^{(0)} =a, a^{(1)} = [a,b] = ab-ba$ and $a^{(k)}=[a^{(k-1)},b]$ Prove the following formula: $$\sum_{i=0}^k b^i a b^{k-i} = \sum_{j=0}^k {k+1 \choose j+1} b^{k-j}a^{(j)}$$

So, I want to use induction to prove this and have verified it for k=1 and k=2. I've worked on the left hand side and gotten $$\sum_{i=0}^k b^i a b^{k-i} =(\sum_{i=0}^{k-1} b^i a b^{(k-1)-i} )b + b^k a $$ The next part would be to make $$ \sum_{j=0}^{k-1} {k \choose j} b^{(k-1)-j}a^{(j)}$$ appear on the right hand side so that I can apply the inductive hypothesis. The only thing I can think of would be to use ${n+1 \choose k}={n \choose k } + {n \choose k-1}$. Maybe I've made an error but I believe this gives $$\sum_{j=0}^k {k+1 \choose j+1} b^{k-j}a^{(j)}=b(\sum_{j=0}^{k-1} {k \choose j+1} b^{(k-1)-j}a^{(j)}) + \sum_{j=0}^k {k \choose j}b^{k-j}a^{(j)} + {k \choose k+1}b^0 a^{(k)}$$ While the last term is $0$, because the extra $b$ appears on opposite sides of the first term, I can't easily equate them and cancel. So I think I'm barking up the wrong tree trying to manipulate the right hand side of the formula in this way.

My question is two fold: how to prove this identity, and what does this composition of the commutator $a^{(j)}$ represent? If it eventually hits 0 is that still some kind of measure for how near $a$ and $b$ are to commuting? If anybody has seen this identity before and it has some usefulness beyond the exercise of proving it, I would also love to hear that.

Answer 1

I outlined two solutions in the comments above; let me expand one of them (the inductive one) into full detail in order to have this question answered. Be warned: This is going to be a long computation with no twists or surprises.

Theorem 1. Let $a$ and $b$ be two elements of an (associative, unital, noncommutative) ring $R$. For any $x\in R$ and $y\in R$, we define the commutator $\left[ x,y\right] \in R$ of $x$ and $y$ by $\left[ x,y\right] =xy-yx$. Define a sequence $\left( a^{\left( 0\right) },a^{\left( 1\right) },a^{\left( 2\right) },\ldots\right) $ of elements of $R$ recursively by setting \begin{align*} a^{\left( 0\right) } & =a\qquad\text{and}\\ a^{\left( k\right) } & =\left[ a^{\left( k-1\right) },b\right] \qquad\text{for each }k\geq 1. \end{align*} Then, \begin{equation} \sum_{i=0}^{k} b^i ab^{k-i} = \sum_{j=0}^{k}\dbinom{k+1}{j+1}b^{k-j}a^{\left( j\right) } \label{darij1.eq.thm.1.claim} \tag{1} \end{equation} for each nonnegative integer $k$.

Proof of Theorem 1. We shall prove \eqref{darij1.eq.thm.1.claim} by induction on $k$:

Induction base: Comparing \begin{equation} \sum_{i=0}^0 b^i ab^{0-i}=\underbrace{b^0 }_{=1}a\underbrace{b^{0-0} }_{=b^0 =1}=a \end{equation} with \begin{equation} \sum_{j=0}^0 \dbinom{0+1}{j+1}b^{0-j}a^{\left( j\right) } =\underbrace{\dbinom{0+1}{0+1}}_{=1}\underbrace{b^{0-0}}_{=b^0 =1}\underbrace{a^{\left( 0\right) }}_{=a}=a, \end{equation} we obtain $\sum\limits_{i=0}^0 b^i ab^{0-i}=\sum\limits_{j=0}^0 \dbinom{0+1}{j+1} b^{0-j}a^{\left( j\right) }$. In other words, \eqref{darij1.eq.thm.1.claim} holds for $k=0$. This completes the induction base.

Induction step: Let $K$ be a positive integer. Assume that \eqref{darij1.eq.thm.1.claim} holds for $k=K-1$. We must prove that \eqref{darij1.eq.thm.1.claim} holds for $k=K$.

We have assumed that \eqref{darij1.eq.thm.1.claim} holds for $k=K-1$. In other words, \begin{align} \sum_{i=0}^{K-1}b^i ab^{\left( K-1\right) -i} & =\sum_{j=0}^{K-1} \dbinom{\left( K-1\right) +1}{j+1}b^{\left( K-1\right) -j}a^{\left( j\right) }\\ & =\sum_{j=0}^{K-1}\dbinom{K}{j+1}b^{\left( K-1\right) -j}a^{\left( j\right) } \label{darij1.pf.thm.1.2} \tag{2} \end{align} (since $\left( K-1\right) +1=K$).

For every nonnegative integer $j$, we have \begin{align*} a^{\left( j+1\right) } & =\left[ a^{\left( j\right) },b\right] \qquad\left( \text{by the recursive definition of }\left( a^{\left( 0\right) },a^{\left( 1\right) },a^{\left( 2\right) },\ldots\right) \right) \\ & =a^{\left( j\right) }b-ba^{\left( j\right) } \end{align*} (by the definition of $\left[ a^{\left( j\right) },b\right] $) and thus \begin{equation} a^{\left( j\right) }b=ba^{\left( j\right) }+a^{\left( j+1\right) }. \label{darij1.pf.thm.1.3} \tag{3} \end{equation}

Now, we can split off the addend for $i=K$ from the sum $\sum_{i=0}^{K} b^i ab^{K-i}$. We thus obtain \begin{equation} \sum_{i=0}^{K}b^i ab^{K-i}=\sum_{i=0}^{K-1}b^i a\underbrace{b^{K-i} }_{\substack{=b^{\left( K-i\right) -1}b\\\text{(since }K-i\geq 1\\\text{(because }i\leq K-1\text{))}}}+b^{K}a\underbrace{b^{K-K}}_{=b^0 =1}=\sum_{i=0}^{K-1}b^i ab^{\left( K-i\right) -1}b+b^{K}a. \end{equation} In view of \begin{align*} & \sum_{i=0}^{K-1}b^i a\underbrace{b^{\left( K-i\right) -1}} _{\substack{=b^{\left( K-1\right) -i}\\\text{(since }\left( K-i\right) -1=\left( K-1\right) -i\text{)}}}b\\ & =\sum_{i=0}^{K-1}b^i ab^{\left( K-1\right) -i}b=\left( \sum_{i=0} ^{K-1}b^i ab^{\left( K-1\right) -i}\right) b=\left( \sum_{j=0} ^{K-1}\dbinom{K}{j+1}b^{\left( K-1\right) -j}a^{\left( j\right) }\right) b\\ & \qquad\left( \begin{array} [c]{c} \text{this follows by multiplying both sides of}\\ \text{the equality \eqref{darij1.pf.thm.1.2} by }b \end{array} \right) \\ & =\sum_{j=0}^{K-1}\dbinom{K}{j+1}b^{\left( K-1\right) -j} \underbrace{a^{\left( j\right) }b}_{\substack{=ba^{\left( j\right) }+a^{\left( j+1\right) }\\\text{(by \eqref{darij1.pf.thm.1.3})}}}=\sum _{j=0}^{K-1}\dbinom{K}{j+1}b^{\left( K-1\right) -j}\left( ba^{\left( j\right) }+a^{\left( j+1\right) }\right) \\ & =\sum_{j=0}^{K-1}\dbinom{K}{j+1}\underbrace{b^{\left( K-1\right) -j} b}_{\substack{=b^{\left( \left( K-1\right) -j\right) +1}=b^{K-j} \\\text{(since }\left( \left( K-1\right) -j\right) +1=K-j\text{)} }}a^{\left( j\right) }+\sum_{j=0}^{K-1}\dbinom{K}{j+1}\underbrace{b^{\left( K-1\right) -j}}_{\substack{=b^{K-\left( j+1\right) }\\\text{(since }\left( K-1\right) -j=K-\left( j+1\right) \text{)}}}a^{\left( j+1\right) }\\ & =\sum_{j=0}^{K-1}\dbinom{K}{j+1}b^{K-j}a^{\left( j\right) } +\underbrace{\sum_{j=0}^{K-1}\dbinom{K}{j+1}b^{K-\left( j+1\right) }a^{\left( j+1\right) }}_{\substack{=\sum_{j=1}^{K}\dbinom{K}{j} b^{K-j}a^{\left( j\right) }\\\text{(here, we substituted }j\text{ for }j+1\text{ in the sum)}}}\\ & =\sum_{j=0}^{K-1}\dbinom{K}{j+1}b^{K-j}a^{\left( j\right) }+\sum_{j=1} ^{K}\dbinom{K}{j}b^{K-j}a^{\left( j\right) }, \end{align*} this rewrites as \begin{align} & \sum_{i=0}^{K}b^i ab^{K-i}\nonumber\\ & =\sum_{j=0}^{K-1}\dbinom{K}{j+1}b^{K-j}a^{\left( j\right) }+\sum_{j=1} ^{K}\dbinom{K}{j}b^{K-j}a^{\left( j\right) }+b^{K} a. \label{darij1.pf.thm.1.5} \tag{4} \end{align}

On the other hand, each nonnegative integer $j$ satisfies \begin{equation} \dbinom{K+1}{j+1}=\dbinom{K}{j+1}+\dbinom{K}{j} \label{darij1.pf.thm.1.7} \tag{5} \end{equation} (by the recurrence relation of the binomial coefficients). Also, the nonnegative integers $K$ and $K+1$ satisfy $K+1 > K$; thus, \begin{equation} \dbinom{K}{K+1}=0 \label{darij1.pf.thm.1.8} \tag{6} \end{equation} (because any two nonnegative integers $n$ and $k$ satisfying $k>n$ must satisfy $\dbinom{n}{k}=0$).

Now, \begin{align*} & \sum_{j=0}^{K}\underbrace{\dbinom{K+1}{j+1}}_{\substack{=\dbinom{K} {j+1}+\dbinom{K}{j}\\\text{(by \eqref{darij1.pf.thm.1.7})}}}b^{K-j}a^{\left( j\right) }\\ & =\sum_{j=0}^{K}\left( \dbinom{K}{j+1}+\dbinom{K}{j}\right) b^{K-j} a^{\left( j\right) }\\ & =\underbrace{\sum_{j=0}^{K}\dbinom{K}{j+1}b^{K-j}a^{\left( j\right) } }_{\substack{=\sum_{j=0}^{K-1}\dbinom{K}{j+1}b^{K-j}a^{\left( j\right) }+\dbinom{K}{K+1}b^{K-K}a^{\left( K\right) }\\\text{(here, we have split off the addend for }j=K\text{ from the sum)}}}\\ & \qquad+\underbrace{\sum_{j=0}^{K}\dbinom{K}{j}b^{K-j}a^{\left( j\right) } }_{\substack{=\sum_{j=1}^{K}\dbinom{K}{j}b^{K-j}a^{\left( j\right) } +\dbinom{K}{0}b^{K-0}a^{\left( 0\right) }\\\text{(here, we have split off the addend for }j=0\text{ from the sum)}}}\\ & =\sum_{j=0}^{K-1}\dbinom{K}{j+1}b^{K-j}a^{\left( j\right) } +\underbrace{\dbinom{K}{K+1}}_{\substack{=0\\\text{(by \eqref{darij1.pf.thm.1.8})}}}b^{K-K}a^{\left( K\right) }\\ & \qquad+\sum_{j=1}^{K}\dbinom{K}{j}b^{K-j}a^{\left( j\right) } +\underbrace{\dbinom{K}{0}}_{=1}\underbrace{b^{K-0}}_{=b^{K}} \underbrace{a^{\left( 0\right) }}_{=a}\\ & =\sum_{j=0}^{K-1}\dbinom{K}{j+1}b^{K-j}a^{\left( j\right) }+\sum_{j=1} ^{K}\dbinom{K}{j}b^{K-j}a^{\left( j\right) }+b^{K}a. \end{align*} Comparing this with \eqref{darij1.pf.thm.1.5}, we obtain \begin{equation} \sum_{i=0}^{K}b^i ab^{K-i}=\sum_{j=0}^{K}\dbinom{K+1}{j+1}b^{K-j}a^{\left( j\right) }. \end{equation} In other words, \eqref{darij1.eq.thm.1.claim} holds for $k=K$. This completes the induction step. Thus, \eqref{darij1.eq.thm.1.claim} is proven by induction. Hence, Theorem 1 follows. $\blacksquare$

Remark. Theorem 1 also holds if $R$ is a nonunital ring, provided that we interpret all the expressions appearing in \eqref{darij1.eq.thm.1.claim} appropriately. (For example, a product of the form "$b^0 a$" has to be interpreted as $a$ even though its sub-expression "$b^0$" is not defined.) The proof we gave above still applies to this situation.