Indexed Summation does not Change under Permutation

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\mathbb A$ be one of the standard number systems $\N, \Z, \Q, \R, \C$.

Let $a$ and $b$ be integers.

Let $\closedint a b$ be the integer interval between $a$ and $b$.

Let $f: \closedint a b \to \mathbb A$ be a mapping.

Let $\sigma: \closedint a b \to \closedint a b$ be a permutation.


Then we have an equality of indexed summations:

$\displaystyle \sum_{i \mathop = a}^b \map f i = \sum_{i \mathop = a}^b \map f {\map \sigma i}$


Outline of Proof

Using Indexed Summation over Translated Interval, we reduce to the case $a = 1$.

Because Symmetric Group on n Letters is Generated by Standard Cycle and Adjacent Transposition, it suffices to do the cases where:

$\map \sigma i = i$ for all $i \in \closedint 1 b$ except for two consecutive integers
$\sigma$ is the standard cycle $\begin {bmatrix} 1 & \ldots & b \end {bmatrix}$


Proof

Let $a > b$.

Then by definition of indexed summation, both sides are $0$.

Let $a \le b$.

Reduction to $a=1$

By Indexed Summation over Translated Interval:

$\displaystyle \sum_{i \mathop = a}^b \map f i = \sum_{j \mathop = 1}^{b - a + 1} \map f {j - a + 1}$
$\displaystyle \sum_{i \mathop = a}^b \map f {\map \sigma i} = \sum_{j \mathop = 1}^{b - a + 1} \map f {\map \sigma {j - a + 1} }$

Let $n = b - a + 1$.

Because $a \le b$, we have $n \ge 1$.

By Translation of Integer Interval is Bijection, the mapping $T : \closedint 1 n \to \closedint a b$ defined by:

$\map T i = i + a - 1$

is a bijection.

We have:

$\displaystyle \sum_{i \mathop = a}^b \map f i = \sum_{j \mathop = 1}^n \map f {\map T j}$
$\displaystyle \sum_{i \mathop = a}^b \map f {\map \sigma i} = \sum_{j \mathop = 1}^n \map f {\map \sigma {\map T j} }$

By Inverse of Bijection is Bijection and Composite of Bijections is Bijection, the composition $T^{-1} \circ \sigma \circ T$ is a permutation of $\closedint 1 n$.

Suppose we have settled the case $a = 1$.

Then:

$\displaystyle \sum_{j \mathop = 1}^n \map f {\map T j} = \sum_{j \mathop = 1}^n \map {\paren {f \circ T} \circ \paren {T^{-1} \circ \sigma \circ T} } j$

By Composition of Mappings is Associative, this equals:

$\displaystyle \sum_{j \mathop = 1}^n \paren {f \circ \sigma \circ T} j$

Thus:

$\displaystyle \sum_{i \mathop = a}^b \map f i = \sum_{i \mathop = a}^b \map f {\map \sigma i}$

$\Box$


It remains to investigate the case $a = 1$.


The case $a = 1$

It remains to show that

$\displaystyle \sum_{i \mathop = 1}^n \map f i = \sum_{i \mathop = 1}^n \map f {\map \sigma i}$

for any permutation $\sigma$ of $\closedint 1 n$.

Let $S_n$ be the symmetric group on $n$ letters, that is, the group of permutations of $\closedint 1 n$.

Let $U \subset S_n$ be the subset of all permutations $\sigma$ satisfying

$\displaystyle \sum_{i \mathop = 1}^n \map f i = \sum_{i \mathop = 1}^n \map f {\map \sigma i}$

for all mappings $f: \closedint 1 n \to \mathbb A$.

$\Box$


Summation-preserving permutations form subgroup

We prove that $U$ is a subgroup of $S_n$, using Two-Step Subgroup Test.

We have to show that $U$ is non-empty.

By Identity Mapping is Permutation the identity mapping $I$ on $\closedint 1 n$ is a permutation.

By Identity Mapping is Right Identity, $f \circ I = f$.

Thus:

$\displaystyle \sum_{i \mathop = 1}^n \map f i = \sum_{i \mathop = 1}^n \map f {\map I i}$

Thus $I \in U$.

Let $\sigma, \tau \in U$.

We have to show that $\sigma \circ \tau$ is in $U$.

We have:

\(\displaystyle \sum_{i \mathop = 1}^n \map {f \circ \paren {\sigma \circ \tau} } i\) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \map {\paren {\paren {f \circ \sigma} \circ \tau} } i\) Composition of Mappings is Associative
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \map {\paren {f \circ \sigma} } i\) $\tau \in U$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \map f i\) $\sigma \in U$

Thus $\sigma \circ \tau \in U$.

We show that $\sigma^{-1} \in U$.

We have:

\(\displaystyle \sum_{i \mathop = 1}^n \map {\paren {f \circ \sigma^{-1} } } i\) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \map {\paren {\paren {f \circ \sigma^{-1} } \circ \sigma} } i\) $\sigma \in U$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \map {\paren {f \circ \paren {\sigma^{-1} \circ \sigma} } } i\) Composition of Mappings is Associative
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \map {\paren {f \circ I} } i\) Identity Mapping is Identity Element in Group of Permutations
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \map f i\) $I \in U$

Thus $\sigma^{-1} \in U$.

Thus $U$ is a subgroup of $S_n$.

$\Box$


Last adjacent transposition preserves summations

If $n = 1$, then by First Symmetric Group is Trivial Group and Subgroups of Trivial Group, $U = S_n$.

Let $n \ge 2$.

By Symmetric Group on n Letters is Generated by Standard Cycle and Adjacent Transposition, it suffices to show that $U$ contains an adjacent transposition and the standard cycle.

Let $\sigma \in S_n$ be the adjacent transposition $\begin {bmatrix} n - 1 & n \end {bmatrix}$.

We have:

\(\displaystyle \sum_{i \mathop = 1}^n \map f {\map \sigma i}\) \(=\) \(\displaystyle \sum_{i \mathop = 1}^{n - 1} \map f {\map \sigma i} + \map f {\map \sigma n}\) Definition of Indexed Summation
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^{n - 2} \map f {\map \sigma i} + \map f {\map \sigma {n - 1} } + \map f {\map \sigma n}\) Definition of Indexed Summation
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^{n - 2} \map f i + \map f n + \map f {n - 1}\) Definition of Adjacent Transposition
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^{n - 2} \map f i + \map f {n - 1} + \map f n\) Arithmetic Addition is Commutative
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^{n - 1} \map f i + \map f n\) Definition of Indexed Summation
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \map f i\) Definition of Indexed Summation

Thus $\begin {bmatrix} n - 1 & n \end {bmatrix} \in U$.

$\Box$


Standard cycle preserves summations

Let $\sigma$ be the standard cycle $\begin {bmatrix} 1 & \ldots & n \end {bmatrix}$.

We have:

\(\displaystyle \sum_{i \mathop = 1}^n \map f {\map \sigma i}\) \(=\) \(\displaystyle \sum_{i \mathop = 1}^{n - 1} \map f {\map \sigma i} + \map f {\map \sigma n}\) Definition of Indexed Summation, $n \ge 1$
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^{n - 1} \map f {i + 1} + \map f 1\) Definition of Standard Cycle
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 2}^n \map f i + \map f 1\) Indexed Summation over Translated Interval
\(\displaystyle \) \(=\) \(\displaystyle f(1) + \sum_{i \mathop = 2}^n \map f i\) Arithmetic Addition is Commutative
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \map f i\) Indexed Summation without First Term

Thus $\begin {bmatrix} 1 & \ldots & n \end {bmatrix} \in U$.

$\blacksquare$


Also see


Generalizations