Indexed Summation does not Change under Permutation

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

Let $a$ and $b$ be integers.

Let $\left[{a \,.\,.\, b}\right]$ be the integer interval between $a$ and $b$.

Let $f : \left[{a \,.\,.\, b}\right] \to \mathbb A$ be a mapping.

Let $\sigma : \left[{a \,.\,.\, b}\right] \to \left[{a \,.\,.\, b}\right]$ be a permutation.

Then we have an equality of indexed summations:


 * $\displaystyle \sum_{i \mathop = a}^b f(i) = \sum_{i \mathop = a}^b f(\sigma(i))$

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 Adjacent Transpositions, it suffices to do the case where $\sigma(i) = i$ for all $i \in \left[{1 \,.\,.\, b}\right]$ except for two consecutive integers.

To do this, we use Indexed Summation over Adjacent Intervals to cut out the two terms that need to be switched.

Proof
Let $a > b$.

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

Let $a \leq b$.

Reduction to $a=1$
By Indexed Summation over Translated Interval:
 * $\displaystyle \sum_{i \mathop = a}^b f(i) = \sum_{j \mathop = 1}^{b-a+1} f(j-a+1)$
 * $\displaystyle \sum_{i \mathop = a}^b f(\sigma(i)) = \sum_{j \mathop = 1}^{b-a+1} f(\sigma(j-a+1))$

Let $n=b-a+1$.

Because $a\leq b$, we have $n\geq 1$.

By Translation of Integer Interval is Bijection, the mapping $T : \left[{1 \,.\,.\, n}\right] \to \left[{a \,.\,.\, b}\right]$ defined by:
 * $T(i) = i+a-1$

is a bijection.

We have:
 * $\displaystyle \sum_{i \mathop = a}^b f(i) = \sum_{j \mathop = 1}^{n} f(T(j))$
 * $\displaystyle \sum_{i \mathop = a}^b f(\sigma(i)) = \sum_{j \mathop = 1}^{n} f(\sigma(T(j)))$

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

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

Then $\displaystyle \sum_{j \mathop = 1}^{n} f(T(j)) = \sum_{j \mathop = 1}^{n} (f\circ T) \circ (T^{-1} \circ \sigma \circ T) (j)$.

By Composition of Mappings is Associative, this equals $\displaystyle \sum_{j \mathop = 1}^{n} (f \circ \sigma \circ T) (j)$.

Thus $\displaystyle \sum_{i \mathop = a}^b f(i) =\sum_{i \mathop = a}^b f(\sigma(i))$.

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

The case $a=1$
It remains to show that
 * $\displaystyle \sum_{i \mathop = 1}^n f(i) = \sum_{i \mathop = 1}^n f(\sigma(i))$

for any permutation $\sigma$ of $\left[{1 \,.\,.\, n}\right]$.

Let $S_n$ be the symmetric group on $n$ letters, that is, the group of permutations of $\left[{1 \,.\,.\, n}\right]$.

Let $U \subset S_n$ be the subset of all permutations $\sigma$ satisfying
 * $\displaystyle \sum_{i \mathop = 1}^n f(i) = \sum_{i \mathop = 1}^n f(\sigma(i))$

for all mappings $f : \left[{1 \,.\,.\, n}\right] \to \mathbb A$.

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 nonempty.

By Identity Mapping is Permutation the identity mapping $\operatorname{id}$ on $\left[{1 \,.\,.\, n}\right]$ is a permutation.

By Identity Mapping is Right Identity, $f\circ \operatorname{id} = f$.

Thus:
 * $\displaystyle \sum_{i \mathop = 1}^n f(i) = \sum_{i \mathop = 1}^n f(\operatorname{id}(i))$

Thus $\operatorname{id} \in U$.

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

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

We have

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

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

We have

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

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

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

Let $n\geq 2$.

By Symmetric Group on n Letters is Generated by Adjacent Transpositions, it suffices to show that $U$ contains all adjacent transpositions.

Surgery
Let $\sigma = \begin{bmatrix} k & k+1 \end{bmatrix}$ be an adjacent transposition in $S_n$, with $1\leq k \leq n-1$.

We have:

Thus $\sigma \in U$.

Also see

 * Change of Variables of Indexed Summation