Identity with repeatedly taking the commutator of a ring element
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.
ring-theory noncommutative-algebra
asked Nov 20 '18 at 6:07
MKeller
455
add a comment |
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.
ring-theory noncommutative-algebra
asked Nov 20 '18 at 6:07
MKeller
455
I haven't properly thought about your approach yet, but here is the "standard" trick for this sort of identity: Let $A$ be the ring. Let $L : A to A$ be the map sending each $x$ to $bx$, and let $R : A to A$ be the map sending each $x$ to $xb$. Then, the operators $L$ and $R$ are $mathbb{Z}$-linear and commute. But the left hand side of your identity is $sumlimits_{i=0}^k L^i R^{k-i} a$, whereas the right hand side is $sumlimits_{j=0}^k dbinom{k+1}{j+1} L^{k-j} left(R-Lright)^j a$. So it remains to ...
– darij grinberg
Nov 20 '18 at 6:19
... prove that $sumlimits_{i=0}^k L^i R^{k-i} = sumlimits_{j=0}^k dbinom{k+1}{j+1} L^{k-j} left(R-Lright)^j$. This should follow from binomial-style manipulations (treating $L$ and $R$ as two arbitrary commuting elements).
– darij grinberg
Nov 20 '18 at 6:19
Ah, yes, the identity $sumlimits_{i=0}^k x^i y^{k-i} = sumlimits_{j=0}^k dbinom{k+1}{j+1} x^{k-j} left(y-xright)^j$ holds for two arbitrary commuting elements $x$ and $y$. To prove it, it suffices to do so when $x$ and $y$ are two commuting indeterminates in a polynomial ring. Multiply both sides by $x-y$ (this is allowed, since $x-y$ is not a zero-divisor in a polynomial ring), so that the left hand side simplifies to $x^{k+1} - y^{k+1}$. Rewrite this using the binomial formula for $y^{k+1} = left(left(y-xright) + xright)^{k+1}$.
– darij grinberg
Nov 20 '18 at 6:22
As to your induction... You want to simplify $left(sum_{j=0}^{k-1} dbinom{k}{j} b^{left(k-1right)-j} a^{(j)} right) b$ so that it looks more like $sum_{j=0}^{k} dbinom{k+1}{j} b^{k-j} a^{(j)}$. So you want to commute the $b$ past the $a^{(j)}$. Of course, it doesn't just commute, but you have $a^{(j)} b = a^{(j+1)} + b a^{(j)}$. So your sum splits into two, with one sum getting its index shifted. I think you can finish it from here.
– darij grinberg
Nov 20 '18 at 6:28
add a comment |
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.
ring-theory noncommutative-algebra
asked Nov 20 '18 at 6:07
MKeller
455
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.
ring-theory noncommutative-algebra
ring-theory noncommutative-algebra
asked Nov 20 '18 at 6:07
MKeller
455
asked Nov 20 '18 at 6:07
MKeller
455
asked Nov 20 '18 at 6:07
MKeller
455
asked Nov 20 '18 at 6:07
MKeller
455
asked Nov 20 '18 at 6:07
MKeller
455
455
I haven't properly thought about your approach yet, but here is the "standard" trick for this sort of identity: Let $A$ be the ring. Let $L : A to A$ be the map sending each $x$ to $bx$, and let $R : A to A$ be the map sending each $x$ to $xb$. Then, the operators $L$ and $R$ are $mathbb{Z}$-linear and commute. But the left hand side of your identity is $sumlimits_{i=0}^k L^i R^{k-i} a$, whereas the right hand side is $sumlimits_{j=0}^k dbinom{k+1}{j+1} L^{k-j} left(R-Lright)^j a$. So it remains to ...
– darij grinberg
Nov 20 '18 at 6:19
... prove that $sumlimits_{i=0}^k L^i R^{k-i} = sumlimits_{j=0}^k dbinom{k+1}{j+1} L^{k-j} left(R-Lright)^j$. This should follow from binomial-style manipulations (treating $L$ and $R$ as two arbitrary commuting elements).
– darij grinberg
Nov 20 '18 at 6:19
Ah, yes, the identity $sumlimits_{i=0}^k x^i y^{k-i} = sumlimits_{j=0}^k dbinom{k+1}{j+1} x^{k-j} left(y-xright)^j$ holds for two arbitrary commuting elements $x$ and $y$. To prove it, it suffices to do so when $x$ and $y$ are two commuting indeterminates in a polynomial ring. Multiply both sides by $x-y$ (this is allowed, since $x-y$ is not a zero-divisor in a polynomial ring), so that the left hand side simplifies to $x^{k+1} - y^{k+1}$. Rewrite this using the binomial formula for $y^{k+1} = left(left(y-xright) + xright)^{k+1}$.
– darij grinberg
Nov 20 '18 at 6:22
As to your induction... You want to simplify $left(sum_{j=0}^{k-1} dbinom{k}{j} b^{left(k-1right)-j} a^{(j)} right) b$ so that it looks more like $sum_{j=0}^{k} dbinom{k+1}{j} b^{k-j} a^{(j)}$. So you want to commute the $b$ past the $a^{(j)}$. Of course, it doesn't just commute, but you have $a^{(j)} b = a^{(j+1)} + b a^{(j)}$. So your sum splits into two, with one sum getting its index shifted. I think you can finish it from here.
– darij grinberg
Nov 20 '18 at 6:28
add a comment |
I haven't properly thought about your approach yet, but here is the "standard" trick for this sort of identity: Let $A$ be the ring. Let $L : A to A$ be the map sending each $x$ to $bx$, and let $R : A to A$ be the map sending each $x$ to $xb$. Then, the operators $L$ and $R$ are $mathbb{Z}$-linear and commute. But the left hand side of your identity is $sumlimits_{i=0}^k L^i R^{k-i} a$, whereas the right hand side is $sumlimits_{j=0}^k dbinom{k+1}{j+1} L^{k-j} left(R-Lright)^j a$. So it remains to ...
– darij grinberg
Nov 20 '18 at 6:19
... prove that $sumlimits_{i=0}^k L^i R^{k-i} = sumlimits_{j=0}^k dbinom{k+1}{j+1} L^{k-j} left(R-Lright)^j$. This should follow from binomial-style manipulations (treating $L$ and $R$ as two arbitrary commuting elements).
– darij grinberg
Nov 20 '18 at 6:19
Ah, yes, the identity $sumlimits_{i=0}^k x^i y^{k-i} = sumlimits_{j=0}^k dbinom{k+1}{j+1} x^{k-j} left(y-xright)^j$ holds for two arbitrary commuting elements $x$ and $y$. To prove it, it suffices to do so when $x$ and $y$ are two commuting indeterminates in a polynomial ring. Multiply both sides by $x-y$ (this is allowed, since $x-y$ is not a zero-divisor in a polynomial ring), so that the left hand side simplifies to $x^{k+1} - y^{k+1}$. Rewrite this using the binomial formula for $y^{k+1} = left(left(y-xright) + xright)^{k+1}$.
– darij grinberg
Nov 20 '18 at 6:22
As to your induction... You want to simplify $left(sum_{j=0}^{k-1} dbinom{k}{j} b^{left(k-1right)-j} a^{(j)} right) b$ so that it looks more like $sum_{j=0}^{k} dbinom{k+1}{j} b^{k-j} a^{(j)}$. So you want to commute the $b$ past the $a^{(j)}$. Of course, it doesn't just commute, but you have $a^{(j)} b = a^{(j+1)} + b a^{(j)}$. So your sum splits into two, with one sum getting its index shifted. I think you can finish it from here.
– darij grinberg
Nov 20 '18 at 6:28
I haven't properly thought about your approach yet, but here is the "standard" trick for this sort of identity: Let $A$ be the ring. Let $L : A to A$ be the map sending each $x$ to $bx$, and let $R : A to A$ be the map sending each $x$ to $xb$. Then, the operators $L$ and $R$ are $mathbb{Z}$-linear and commute. But the left hand side of your identity is $sumlimits_{i=0}^k L^i R^{k-i} a$, whereas the right hand side is $sumlimits_{j=0}^k dbinom{k+1}{j+1} L^{k-j} left(R-Lright)^j a$. So it remains to ...
– darij grinberg
Nov 20 '18 at 6:19
I haven't properly thought about your approach yet, but here is the "standard" trick for this sort of identity: Let $A$ be the ring. Let $L : A to A$ be the map sending each $x$ to $bx$, and let $R : A to A$ be the map sending each $x$ to $xb$. Then, the operators $L$ and $R$ are $mathbb{Z}$-linear and commute. But the left hand side of your identity is $sumlimits_{i=0}^k L^i R^{k-i} a$, whereas the right hand side is $sumlimits_{j=0}^k dbinom{k+1}{j+1} L^{k-j} left(R-Lright)^j a$. So it remains to ...
– darij grinberg
Nov 20 '18 at 6:19
... prove that $sumlimits_{i=0}^k L^i R^{k-i} = sumlimits_{j=0}^k dbinom{k+1}{j+1} L^{k-j} left(R-Lright)^j$. This should follow from binomial-style manipulations (treating $L$ and $R$ as two arbitrary commuting elements).
– darij grinberg
Nov 20 '18 at 6:19
... prove that $sumlimits_{i=0}^k L^i R^{k-i} = sumlimits_{j=0}^k dbinom{k+1}{j+1} L^{k-j} left(R-Lright)^j$. This should follow from binomial-style manipulations (treating $L$ and $R$ as two arbitrary commuting elements).
– darij grinberg
Nov 20 '18 at 6:19
Ah, yes, the identity $sumlimits_{i=0}^k x^i y^{k-i} = sumlimits_{j=0}^k dbinom{k+1}{j+1} x^{k-j} left(y-xright)^j$ holds for two arbitrary commuting elements $x$ and $y$. To prove it, it suffices to do so when $x$ and $y$ are two commuting indeterminates in a polynomial ring. Multiply both sides by $x-y$ (this is allowed, since $x-y$ is not a zero-divisor in a polynomial ring), so that the left hand side simplifies to $x^{k+1} - y^{k+1}$. Rewrite this using the binomial formula for $y^{k+1} = left(left(y-xright) + xright)^{k+1}$.
– darij grinberg
Nov 20 '18 at 6:22
Ah, yes, the identity $sumlimits_{i=0}^k x^i y^{k-i} = sumlimits_{j=0}^k dbinom{k+1}{j+1} x^{k-j} left(y-xright)^j$ holds for two arbitrary commuting elements $x$ and $y$. To prove it, it suffices to do so when $x$ and $y$ are two commuting indeterminates in a polynomial ring. Multiply both sides by $x-y$ (this is allowed, since $x-y$ is not a zero-divisor in a polynomial ring), so that the left hand side simplifies to $x^{k+1} - y^{k+1}$. Rewrite this using the binomial formula for $y^{k+1} = left(left(y-xright) + xright)^{k+1}$.
– darij grinberg
Nov 20 '18 at 6:22
As to your induction... You want to simplify $left(sum_{j=0}^{k-1} dbinom{k}{j} b^{left(k-1right)-j} a^{(j)} right) b$ so that it looks more like $sum_{j=0}^{k} dbinom{k+1}{j} b^{k-j} a^{(j)}$. So you want to commute the $b$ past the $a^{(j)}$. Of course, it doesn't just commute, but you have $a^{(j)} b = a^{(j+1)} + b a^{(j)}$. So your sum splits into two, with one sum getting its index shifted. I think you can finish it from here.
– darij grinberg
Nov 20 '18 at 6:28
As to your induction... You want to simplify $left(sum_{j=0}^{k-1} dbinom{k}{j} b^{left(k-1right)-j} a^{(j)} right) b$ so that it looks more like $sum_{j=0}^{k} dbinom{k+1}{j} b^{k-j} a^{(j)}$. So you want to commute the $b$ past the $a^{(j)}$. Of course, it doesn't just commute, but you have $a^{(j)} b = a^{(j+1)} + b a^{(j)}$. So your sum splits into two, with one sum getting its index shifted. I think you can finish it from here.
– darij grinberg
Nov 20 '18 at 6:28
add a comment |
1 Answer
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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 $xin R$ and $yin R$, we define the
commutator $left[ x,yright] in R$ of $x$ and $y$ by $left[
x,yright] =xy-yx$. Define a sequence $left( a^{left( 0right)
},a^{left( 1right) },a^{left( 2right) },ldotsright) $ of elements
of $R$ recursively by setting
begin{align*}
a^{left( 0right) } & =aqquadtext{and}\
a^{left( kright) } & =left[ a^{left( k-1right) },bright]
qquadtext{for each }kgeq 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( jright) }
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}aunderbrace{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( jright) }
=underbrace{dbinom{0+1}{0+1}}_{=1}underbrace{b^{0-0}}_{=b^0
=1}underbrace{a^{left( 0right) }}_{=a}=a,
end{equation}
we obtain $sumlimits_{i=0}^0 b^i ab^{0-i}=sumlimits_{j=0}^0 dbinom{0+1}{j+1}
b^{0-j}a^{left( jright) }$. 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-1right) -i} & =sum_{j=0}^{K-1}
dbinom{left( K-1right) +1}{j+1}b^{left( K-1right) -j}a^{left(
jright) }\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{left( K-1right) -j}a^{left(
jright) }
label{darij1.pf.thm.1.2}
tag{2}
end{align}
(since $left( K-1right) +1=K$).
For every nonnegative integer $j$, we have
begin{align*}
a^{left( j+1right) } & =left[ a^{left( jright) },bright]
qquadleft( text{by the recursive definition of }left( a^{left(
0right) },a^{left( 1right) },a^{left( 2right) },ldotsright)
right) \
& =a^{left( jright) }b-ba^{left( jright) }
end{align*}
(by the definition of $left[ a^{left( jright) },bright] $) and thus
begin{equation}
a^{left( jright) }b=ba^{left( jright) }+a^{left( j+1right)
}.
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 aunderbrace{b^{K-i}
}_{substack{=b^{left( K-iright) -1}b\text{(since }K-igeq
1\text{(because }ileq K-1text{))}}}+b^{K}aunderbrace{b^{K-K}}_{=b^0
=1}=sum_{i=0}^{K-1}b^i ab^{left( K-iright) -1}b+b^{K}a.
end{equation}
In view of
begin{align*}
& sum_{i=0}^{K-1}b^i aunderbrace{b^{left( K-iright) -1}}
_{substack{=b^{left( K-1right) -i}\text{(since }left( K-iright)
-1=left( K-1right) -itext{)}}}b\
& =sum_{i=0}^{K-1}b^i ab^{left( K-1right) -i}b=left( sum_{i=0}
^{K-1}b^i ab^{left( K-1right) -i}right) b=left( sum_{j=0}
^{K-1}dbinom{K}{j+1}b^{left( K-1right) -j}a^{left( jright) }right)
b\
& qquadleft(
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-1right) -j}
underbrace{a^{left( jright) }b}_{substack{=ba^{left( jright)
}+a^{left( j+1right) }\text{(by eqref{darij1.pf.thm.1.3})}}}=sum
_{j=0}^{K-1}dbinom{K}{j+1}b^{left( K-1right) -j}left( ba^{left(
jright) }+a^{left( j+1right) }right) \
& =sum_{j=0}^{K-1}dbinom{K}{j+1}underbrace{b^{left( K-1right) -j}
b}_{substack{=b^{left( left( K-1right) -jright) +1}=b^{K-j}
\text{(since }left( left( K-1right) -jright) +1=K-jtext{)}
}}a^{left( jright) }+sum_{j=0}^{K-1}dbinom{K}{j+1}underbrace{b^{left(
K-1right) -j}}_{substack{=b^{K-left( j+1right) }\text{(since }left(
K-1right) -j=K-left( j+1right) text{)}}}a^{left( j+1right) }\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright) }
+underbrace{sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-left( j+1right)
}a^{left( j+1right) }}_{substack{=sum_{j=1}^{K}dbinom{K}{j}
b^{K-j}a^{left( jright) }\text{(here, we substituted }jtext{ for
}j+1text{ in the sum)}}}\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright) }+sum_{j=1}
^{K}dbinom{K}{j}b^{K-j}a^{left( jright) },
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( jright) }+sum_{j=1}
^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }+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(
jright) }\
& =sum_{j=0}^{K}left( dbinom{K}{j+1}+dbinom{K}{j}right) b^{K-j}
a^{left( jright) }\
& =underbrace{sum_{j=0}^{K}dbinom{K}{j+1}b^{K-j}a^{left( jright) }
}_{substack{=sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright)
}+dbinom{K}{K+1}b^{K-K}a^{left( Kright) }\text{(here, we have split off
the addend for }j=Ktext{ from the sum)}}}\
& qquad+underbrace{sum_{j=0}^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }
}_{substack{=sum_{j=1}^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }
+dbinom{K}{0}b^{K-0}a^{left( 0right) }\text{(here, we have split off
the addend for }j=0text{ from the sum)}}}\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright) }
+underbrace{dbinom{K}{K+1}}_{substack{=0\text{(by
eqref{darij1.pf.thm.1.8})}}}b^{K-K}a^{left( Kright) }\
& qquad+sum_{j=1}^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }
+underbrace{dbinom{K}{0}}_{=1}underbrace{b^{K-0}}_{=b^{K}}
underbrace{a^{left( 0right) }}_{=a}\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright) }+sum_{j=1}
^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }+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(
jright) }.
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.
answered Nov 26 '18 at 3:00
darij grinberg
10.2k33062
add a comment |
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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 $xin R$ and $yin R$, we define the
commutator $left[ x,yright] in R$ of $x$ and $y$ by $left[
x,yright] =xy-yx$. Define a sequence $left( a^{left( 0right)
},a^{left( 1right) },a^{left( 2right) },ldotsright) $ of elements
of $R$ recursively by setting
begin{align*}
a^{left( 0right) } & =aqquadtext{and}\
a^{left( kright) } & =left[ a^{left( k-1right) },bright]
qquadtext{for each }kgeq 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( jright) }
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}aunderbrace{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( jright) }
=underbrace{dbinom{0+1}{0+1}}_{=1}underbrace{b^{0-0}}_{=b^0
=1}underbrace{a^{left( 0right) }}_{=a}=a,
end{equation}
we obtain $sumlimits_{i=0}^0 b^i ab^{0-i}=sumlimits_{j=0}^0 dbinom{0+1}{j+1}
b^{0-j}a^{left( jright) }$. 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-1right) -i} & =sum_{j=0}^{K-1}
dbinom{left( K-1right) +1}{j+1}b^{left( K-1right) -j}a^{left(
jright) }\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{left( K-1right) -j}a^{left(
jright) }
label{darij1.pf.thm.1.2}
tag{2}
end{align}
(since $left( K-1right) +1=K$).
For every nonnegative integer $j$, we have
begin{align*}
a^{left( j+1right) } & =left[ a^{left( jright) },bright]
qquadleft( text{by the recursive definition of }left( a^{left(
0right) },a^{left( 1right) },a^{left( 2right) },ldotsright)
right) \
& =a^{left( jright) }b-ba^{left( jright) }
end{align*}
(by the definition of $left[ a^{left( jright) },bright] $) and thus
begin{equation}
a^{left( jright) }b=ba^{left( jright) }+a^{left( j+1right)
}.
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 aunderbrace{b^{K-i}
}_{substack{=b^{left( K-iright) -1}b\text{(since }K-igeq
1\text{(because }ileq K-1text{))}}}+b^{K}aunderbrace{b^{K-K}}_{=b^0
=1}=sum_{i=0}^{K-1}b^i ab^{left( K-iright) -1}b+b^{K}a.
end{equation}
In view of
begin{align*}
& sum_{i=0}^{K-1}b^i aunderbrace{b^{left( K-iright) -1}}
_{substack{=b^{left( K-1right) -i}\text{(since }left( K-iright)
-1=left( K-1right) -itext{)}}}b\
& =sum_{i=0}^{K-1}b^i ab^{left( K-1right) -i}b=left( sum_{i=0}
^{K-1}b^i ab^{left( K-1right) -i}right) b=left( sum_{j=0}
^{K-1}dbinom{K}{j+1}b^{left( K-1right) -j}a^{left( jright) }right)
b\
& qquadleft(
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-1right) -j}
underbrace{a^{left( jright) }b}_{substack{=ba^{left( jright)
}+a^{left( j+1right) }\text{(by eqref{darij1.pf.thm.1.3})}}}=sum
_{j=0}^{K-1}dbinom{K}{j+1}b^{left( K-1right) -j}left( ba^{left(
jright) }+a^{left( j+1right) }right) \
& =sum_{j=0}^{K-1}dbinom{K}{j+1}underbrace{b^{left( K-1right) -j}
b}_{substack{=b^{left( left( K-1right) -jright) +1}=b^{K-j}
\text{(since }left( left( K-1right) -jright) +1=K-jtext{)}
}}a^{left( jright) }+sum_{j=0}^{K-1}dbinom{K}{j+1}underbrace{b^{left(
K-1right) -j}}_{substack{=b^{K-left( j+1right) }\text{(since }left(
K-1right) -j=K-left( j+1right) text{)}}}a^{left( j+1right) }\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright) }
+underbrace{sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-left( j+1right)
}a^{left( j+1right) }}_{substack{=sum_{j=1}^{K}dbinom{K}{j}
b^{K-j}a^{left( jright) }\text{(here, we substituted }jtext{ for
}j+1text{ in the sum)}}}\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright) }+sum_{j=1}
^{K}dbinom{K}{j}b^{K-j}a^{left( jright) },
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( jright) }+sum_{j=1}
^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }+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(
jright) }\
& =sum_{j=0}^{K}left( dbinom{K}{j+1}+dbinom{K}{j}right) b^{K-j}
a^{left( jright) }\
& =underbrace{sum_{j=0}^{K}dbinom{K}{j+1}b^{K-j}a^{left( jright) }
}_{substack{=sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright)
}+dbinom{K}{K+1}b^{K-K}a^{left( Kright) }\text{(here, we have split off
the addend for }j=Ktext{ from the sum)}}}\
& qquad+underbrace{sum_{j=0}^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }
}_{substack{=sum_{j=1}^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }
+dbinom{K}{0}b^{K-0}a^{left( 0right) }\text{(here, we have split off
the addend for }j=0text{ from the sum)}}}\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright) }
+underbrace{dbinom{K}{K+1}}_{substack{=0\text{(by
eqref{darij1.pf.thm.1.8})}}}b^{K-K}a^{left( Kright) }\
& qquad+sum_{j=1}^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }
+underbrace{dbinom{K}{0}}_{=1}underbrace{b^{K-0}}_{=b^{K}}
underbrace{a^{left( 0right) }}_{=a}\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright) }+sum_{j=1}
^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }+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(
jright) }.
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.
answered Nov 26 '18 at 3:00
darij grinberg
10.2k33062
add a comment |
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 $xin R$ and $yin R$, we define the
commutator $left[ x,yright] in R$ of $x$ and $y$ by $left[
x,yright] =xy-yx$. Define a sequence $left( a^{left( 0right)
},a^{left( 1right) },a^{left( 2right) },ldotsright) $ of elements
of $R$ recursively by setting
begin{align*}
a^{left( 0right) } & =aqquadtext{and}\
a^{left( kright) } & =left[ a^{left( k-1right) },bright]
qquadtext{for each }kgeq 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( jright) }
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}aunderbrace{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( jright) }
=underbrace{dbinom{0+1}{0+1}}_{=1}underbrace{b^{0-0}}_{=b^0
=1}underbrace{a^{left( 0right) }}_{=a}=a,
end{equation}
we obtain $sumlimits_{i=0}^0 b^i ab^{0-i}=sumlimits_{j=0}^0 dbinom{0+1}{j+1}
b^{0-j}a^{left( jright) }$. 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-1right) -i} & =sum_{j=0}^{K-1}
dbinom{left( K-1right) +1}{j+1}b^{left( K-1right) -j}a^{left(
jright) }\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{left( K-1right) -j}a^{left(
jright) }
label{darij1.pf.thm.1.2}
tag{2}
end{align}
(since $left( K-1right) +1=K$).
For every nonnegative integer $j$, we have
begin{align*}
a^{left( j+1right) } & =left[ a^{left( jright) },bright]
qquadleft( text{by the recursive definition of }left( a^{left(
0right) },a^{left( 1right) },a^{left( 2right) },ldotsright)
right) \
& =a^{left( jright) }b-ba^{left( jright) }
end{align*}
(by the definition of $left[ a^{left( jright) },bright] $) and thus
begin{equation}
a^{left( jright) }b=ba^{left( jright) }+a^{left( j+1right)
}.
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 aunderbrace{b^{K-i}
}_{substack{=b^{left( K-iright) -1}b\text{(since }K-igeq
1\text{(because }ileq K-1text{))}}}+b^{K}aunderbrace{b^{K-K}}_{=b^0
=1}=sum_{i=0}^{K-1}b^i ab^{left( K-iright) -1}b+b^{K}a.
end{equation}
In view of
begin{align*}
& sum_{i=0}^{K-1}b^i aunderbrace{b^{left( K-iright) -1}}
_{substack{=b^{left( K-1right) -i}\text{(since }left( K-iright)
-1=left( K-1right) -itext{)}}}b\
& =sum_{i=0}^{K-1}b^i ab^{left( K-1right) -i}b=left( sum_{i=0}
^{K-1}b^i ab^{left( K-1right) -i}right) b=left( sum_{j=0}
^{K-1}dbinom{K}{j+1}b^{left( K-1right) -j}a^{left( jright) }right)
b\
& qquadleft(
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-1right) -j}
underbrace{a^{left( jright) }b}_{substack{=ba^{left( jright)
}+a^{left( j+1right) }\text{(by eqref{darij1.pf.thm.1.3})}}}=sum
_{j=0}^{K-1}dbinom{K}{j+1}b^{left( K-1right) -j}left( ba^{left(
jright) }+a^{left( j+1right) }right) \
& =sum_{j=0}^{K-1}dbinom{K}{j+1}underbrace{b^{left( K-1right) -j}
b}_{substack{=b^{left( left( K-1right) -jright) +1}=b^{K-j}
\text{(since }left( left( K-1right) -jright) +1=K-jtext{)}
}}a^{left( jright) }+sum_{j=0}^{K-1}dbinom{K}{j+1}underbrace{b^{left(
K-1right) -j}}_{substack{=b^{K-left( j+1right) }\text{(since }left(
K-1right) -j=K-left( j+1right) text{)}}}a^{left( j+1right) }\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright) }
+underbrace{sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-left( j+1right)
}a^{left( j+1right) }}_{substack{=sum_{j=1}^{K}dbinom{K}{j}
b^{K-j}a^{left( jright) }\text{(here, we substituted }jtext{ for
}j+1text{ in the sum)}}}\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright) }+sum_{j=1}
^{K}dbinom{K}{j}b^{K-j}a^{left( jright) },
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( jright) }+sum_{j=1}
^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }+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(
jright) }\
& =sum_{j=0}^{K}left( dbinom{K}{j+1}+dbinom{K}{j}right) b^{K-j}
a^{left( jright) }\
& =underbrace{sum_{j=0}^{K}dbinom{K}{j+1}b^{K-j}a^{left( jright) }
}_{substack{=sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright)
}+dbinom{K}{K+1}b^{K-K}a^{left( Kright) }\text{(here, we have split off
the addend for }j=Ktext{ from the sum)}}}\
& qquad+underbrace{sum_{j=0}^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }
}_{substack{=sum_{j=1}^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }
+dbinom{K}{0}b^{K-0}a^{left( 0right) }\text{(here, we have split off
the addend for }j=0text{ from the sum)}}}\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright) }
+underbrace{dbinom{K}{K+1}}_{substack{=0\text{(by
eqref{darij1.pf.thm.1.8})}}}b^{K-K}a^{left( Kright) }\
& qquad+sum_{j=1}^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }
+underbrace{dbinom{K}{0}}_{=1}underbrace{b^{K-0}}_{=b^{K}}
underbrace{a^{left( 0right) }}_{=a}\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright) }+sum_{j=1}
^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }+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(
jright) }.
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.
answered Nov 26 '18 at 3:00
darij grinberg
10.2k33062
add a comment |
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 $xin R$ and $yin R$, we define the
commutator $left[ x,yright] in R$ of $x$ and $y$ by $left[
x,yright] =xy-yx$. Define a sequence $left( a^{left( 0right)
},a^{left( 1right) },a^{left( 2right) },ldotsright) $ of elements
of $R$ recursively by setting
begin{align*}
a^{left( 0right) } & =aqquadtext{and}\
a^{left( kright) } & =left[ a^{left( k-1right) },bright]
qquadtext{for each }kgeq 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( jright) }
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}aunderbrace{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( jright) }
=underbrace{dbinom{0+1}{0+1}}_{=1}underbrace{b^{0-0}}_{=b^0
=1}underbrace{a^{left( 0right) }}_{=a}=a,
end{equation}
we obtain $sumlimits_{i=0}^0 b^i ab^{0-i}=sumlimits_{j=0}^0 dbinom{0+1}{j+1}
b^{0-j}a^{left( jright) }$. 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-1right) -i} & =sum_{j=0}^{K-1}
dbinom{left( K-1right) +1}{j+1}b^{left( K-1right) -j}a^{left(
jright) }\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{left( K-1right) -j}a^{left(
jright) }
label{darij1.pf.thm.1.2}
tag{2}
end{align}
(since $left( K-1right) +1=K$).
For every nonnegative integer $j$, we have
begin{align*}
a^{left( j+1right) } & =left[ a^{left( jright) },bright]
qquadleft( text{by the recursive definition of }left( a^{left(
0right) },a^{left( 1right) },a^{left( 2right) },ldotsright)
right) \
& =a^{left( jright) }b-ba^{left( jright) }
end{align*}
(by the definition of $left[ a^{left( jright) },bright] $) and thus
begin{equation}
a^{left( jright) }b=ba^{left( jright) }+a^{left( j+1right)
}.
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 aunderbrace{b^{K-i}
}_{substack{=b^{left( K-iright) -1}b\text{(since }K-igeq
1\text{(because }ileq K-1text{))}}}+b^{K}aunderbrace{b^{K-K}}_{=b^0
=1}=sum_{i=0}^{K-1}b^i ab^{left( K-iright) -1}b+b^{K}a.
end{equation}
In view of
begin{align*}
& sum_{i=0}^{K-1}b^i aunderbrace{b^{left( K-iright) -1}}
_{substack{=b^{left( K-1right) -i}\text{(since }left( K-iright)
-1=left( K-1right) -itext{)}}}b\
& =sum_{i=0}^{K-1}b^i ab^{left( K-1right) -i}b=left( sum_{i=0}
^{K-1}b^i ab^{left( K-1right) -i}right) b=left( sum_{j=0}
^{K-1}dbinom{K}{j+1}b^{left( K-1right) -j}a^{left( jright) }right)
b\
& qquadleft(
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-1right) -j}
underbrace{a^{left( jright) }b}_{substack{=ba^{left( jright)
}+a^{left( j+1right) }\text{(by eqref{darij1.pf.thm.1.3})}}}=sum
_{j=0}^{K-1}dbinom{K}{j+1}b^{left( K-1right) -j}left( ba^{left(
jright) }+a^{left( j+1right) }right) \
& =sum_{j=0}^{K-1}dbinom{K}{j+1}underbrace{b^{left( K-1right) -j}
b}_{substack{=b^{left( left( K-1right) -jright) +1}=b^{K-j}
\text{(since }left( left( K-1right) -jright) +1=K-jtext{)}
}}a^{left( jright) }+sum_{j=0}^{K-1}dbinom{K}{j+1}underbrace{b^{left(
K-1right) -j}}_{substack{=b^{K-left( j+1right) }\text{(since }left(
K-1right) -j=K-left( j+1right) text{)}}}a^{left( j+1right) }\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright) }
+underbrace{sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-left( j+1right)
}a^{left( j+1right) }}_{substack{=sum_{j=1}^{K}dbinom{K}{j}
b^{K-j}a^{left( jright) }\text{(here, we substituted }jtext{ for
}j+1text{ in the sum)}}}\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright) }+sum_{j=1}
^{K}dbinom{K}{j}b^{K-j}a^{left( jright) },
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( jright) }+sum_{j=1}
^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }+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(
jright) }\
& =sum_{j=0}^{K}left( dbinom{K}{j+1}+dbinom{K}{j}right) b^{K-j}
a^{left( jright) }\
& =underbrace{sum_{j=0}^{K}dbinom{K}{j+1}b^{K-j}a^{left( jright) }
}_{substack{=sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright)
}+dbinom{K}{K+1}b^{K-K}a^{left( Kright) }\text{(here, we have split off
the addend for }j=Ktext{ from the sum)}}}\
& qquad+underbrace{sum_{j=0}^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }
}_{substack{=sum_{j=1}^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }
+dbinom{K}{0}b^{K-0}a^{left( 0right) }\text{(here, we have split off
the addend for }j=0text{ from the sum)}}}\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright) }
+underbrace{dbinom{K}{K+1}}_{substack{=0\text{(by
eqref{darij1.pf.thm.1.8})}}}b^{K-K}a^{left( Kright) }\
& qquad+sum_{j=1}^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }
+underbrace{dbinom{K}{0}}_{=1}underbrace{b^{K-0}}_{=b^{K}}
underbrace{a^{left( 0right) }}_{=a}\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright) }+sum_{j=1}
^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }+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(
jright) }.
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.
answered Nov 26 '18 at 3:00
darij grinberg
10.2k33062
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 $xin R$ and $yin R$, we define the
commutator $left[ x,yright] in R$ of $x$ and $y$ by $left[
x,yright] =xy-yx$. Define a sequence $left( a^{left( 0right)
},a^{left( 1right) },a^{left( 2right) },ldotsright) $ of elements
of $R$ recursively by setting
begin{align*}
a^{left( 0right) } & =aqquadtext{and}\
a^{left( kright) } & =left[ a^{left( k-1right) },bright]
qquadtext{for each }kgeq 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( jright) }
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}aunderbrace{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( jright) }
=underbrace{dbinom{0+1}{0+1}}_{=1}underbrace{b^{0-0}}_{=b^0
=1}underbrace{a^{left( 0right) }}_{=a}=a,
end{equation}
we obtain $sumlimits_{i=0}^0 b^i ab^{0-i}=sumlimits_{j=0}^0 dbinom{0+1}{j+1}
b^{0-j}a^{left( jright) }$. 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-1right) -i} & =sum_{j=0}^{K-1}
dbinom{left( K-1right) +1}{j+1}b^{left( K-1right) -j}a^{left(
jright) }\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{left( K-1right) -j}a^{left(
jright) }
label{darij1.pf.thm.1.2}
tag{2}
end{align}
(since $left( K-1right) +1=K$).
For every nonnegative integer $j$, we have
begin{align*}
a^{left( j+1right) } & =left[ a^{left( jright) },bright]
qquadleft( text{by the recursive definition of }left( a^{left(
0right) },a^{left( 1right) },a^{left( 2right) },ldotsright)
right) \
& =a^{left( jright) }b-ba^{left( jright) }
end{align*}
(by the definition of $left[ a^{left( jright) },bright] $) and thus
begin{equation}
a^{left( jright) }b=ba^{left( jright) }+a^{left( j+1right)
}.
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 aunderbrace{b^{K-i}
}_{substack{=b^{left( K-iright) -1}b\text{(since }K-igeq
1\text{(because }ileq K-1text{))}}}+b^{K}aunderbrace{b^{K-K}}_{=b^0
=1}=sum_{i=0}^{K-1}b^i ab^{left( K-iright) -1}b+b^{K}a.
end{equation}
In view of
begin{align*}
& sum_{i=0}^{K-1}b^i aunderbrace{b^{left( K-iright) -1}}
_{substack{=b^{left( K-1right) -i}\text{(since }left( K-iright)
-1=left( K-1right) -itext{)}}}b\
& =sum_{i=0}^{K-1}b^i ab^{left( K-1right) -i}b=left( sum_{i=0}
^{K-1}b^i ab^{left( K-1right) -i}right) b=left( sum_{j=0}
^{K-1}dbinom{K}{j+1}b^{left( K-1right) -j}a^{left( jright) }right)
b\
& qquadleft(
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-1right) -j}
underbrace{a^{left( jright) }b}_{substack{=ba^{left( jright)
}+a^{left( j+1right) }\text{(by eqref{darij1.pf.thm.1.3})}}}=sum
_{j=0}^{K-1}dbinom{K}{j+1}b^{left( K-1right) -j}left( ba^{left(
jright) }+a^{left( j+1right) }right) \
& =sum_{j=0}^{K-1}dbinom{K}{j+1}underbrace{b^{left( K-1right) -j}
b}_{substack{=b^{left( left( K-1right) -jright) +1}=b^{K-j}
\text{(since }left( left( K-1right) -jright) +1=K-jtext{)}
}}a^{left( jright) }+sum_{j=0}^{K-1}dbinom{K}{j+1}underbrace{b^{left(
K-1right) -j}}_{substack{=b^{K-left( j+1right) }\text{(since }left(
K-1right) -j=K-left( j+1right) text{)}}}a^{left( j+1right) }\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright) }
+underbrace{sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-left( j+1right)
}a^{left( j+1right) }}_{substack{=sum_{j=1}^{K}dbinom{K}{j}
b^{K-j}a^{left( jright) }\text{(here, we substituted }jtext{ for
}j+1text{ in the sum)}}}\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright) }+sum_{j=1}
^{K}dbinom{K}{j}b^{K-j}a^{left( jright) },
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( jright) }+sum_{j=1}
^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }+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(
jright) }\
& =sum_{j=0}^{K}left( dbinom{K}{j+1}+dbinom{K}{j}right) b^{K-j}
a^{left( jright) }\
& =underbrace{sum_{j=0}^{K}dbinom{K}{j+1}b^{K-j}a^{left( jright) }
}_{substack{=sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright)
}+dbinom{K}{K+1}b^{K-K}a^{left( Kright) }\text{(here, we have split off
the addend for }j=Ktext{ from the sum)}}}\
& qquad+underbrace{sum_{j=0}^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }
}_{substack{=sum_{j=1}^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }
+dbinom{K}{0}b^{K-0}a^{left( 0right) }\text{(here, we have split off
the addend for }j=0text{ from the sum)}}}\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright) }
+underbrace{dbinom{K}{K+1}}_{substack{=0\text{(by
eqref{darij1.pf.thm.1.8})}}}b^{K-K}a^{left( Kright) }\
& qquad+sum_{j=1}^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }
+underbrace{dbinom{K}{0}}_{=1}underbrace{b^{K-0}}_{=b^{K}}
underbrace{a^{left( 0right) }}_{=a}\
& =sum_{j=0}^{K-1}dbinom{K}{j+1}b^{K-j}a^{left( jright) }+sum_{j=1}
^{K}dbinom{K}{j}b^{K-j}a^{left( jright) }+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(
jright) }.
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.
answered Nov 26 '18 at 3:00
darij grinberg
10.2k33062
edited yesterday
answered Nov 26 '18 at 3:00
darij grinberg
10.2k33062
answered Nov 26 '18 at 3:00
darij grinberg
10.2k33062
answered Nov 26 '18 at 3:00
darij grinberg
10.2k33062
10.2k33062
add a comment |
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I haven't properly thought about your approach yet, but here is the "standard" trick for this sort of identity: Let $A$ be the ring. Let $L : A to A$ be the map sending each $x$ to $bx$, and let $R : A to A$ be the map sending each $x$ to $xb$. Then, the operators $L$ and $R$ are $mathbb{Z}$-linear and commute. But the left hand side of your identity is $sumlimits_{i=0}^k L^i R^{k-i} a$, whereas the right hand side is $sumlimits_{j=0}^k dbinom{k+1}{j+1} L^{k-j} left(R-Lright)^j a$. So it remains to ...
– darij grinberg
Nov 20 '18 at 6:19
... prove that $sumlimits_{i=0}^k L^i R^{k-i} = sumlimits_{j=0}^k dbinom{k+1}{j+1} L^{k-j} left(R-Lright)^j$. This should follow from binomial-style manipulations (treating $L$ and $R$ as two arbitrary commuting elements).
– darij grinberg
Nov 20 '18 at 6:19
Ah, yes, the identity $sumlimits_{i=0}^k x^i y^{k-i} = sumlimits_{j=0}^k dbinom{k+1}{j+1} x^{k-j} left(y-xright)^j$ holds for two arbitrary commuting elements $x$ and $y$. To prove it, it suffices to do so when $x$ and $y$ are two commuting indeterminates in a polynomial ring. Multiply both sides by $x-y$ (this is allowed, since $x-y$ is not a zero-divisor in a polynomial ring), so that the left hand side simplifies to $x^{k+1} - y^{k+1}$. Rewrite this using the binomial formula for $y^{k+1} = left(left(y-xright) + xright)^{k+1}$.
– darij grinberg
Nov 20 '18 at 6:22
As to your induction... You want to simplify $left(sum_{j=0}^{k-1} dbinom{k}{j} b^{left(k-1right)-j} a^{(j)} right) b$ so that it looks more like $sum_{j=0}^{k} dbinom{k+1}{j} b^{k-j} a^{(j)}$. So you want to commute the $b$ past the $a^{(j)}$. Of course, it doesn't just commute, but you have $a^{(j)} b = a^{(j+1)} + b a^{(j)}$. So your sum splits into two, with one sum getting its index shifted. I think you can finish it from here.
– darij grinberg
Nov 20 '18 at 6:28