Definition of multiple sum

Question

Suppose we have an abelian group $(G,+)$. What is the formal definition of multiple sums such as $\sum_{i_1 \in A} \sum_{i_2 \in A_{i_1}} \cdots \sum_{i_n \in A_{i_{n-1}}}f(i_1,\ldots,i_n)$?

Thanks in advance!

Answer 1

Assuming all the sets in the subscripts are finite, you can give a recursive definition:

\[ \sum_{i \in A} f(i) = \begin{cases} 0 &\text{if }A = \emptyset \\ f(a) + \textstyle\sum_{i \in A \setminus \{a\}} f(i) &\text{if }a \in A \end{cases} \]

Now you can just apply the above definition repeatedly to eliminate each $\sum$ from left to right.

Two things need to be proven here: first, that any $\sum$ can be eliminated by application of these rules, and second, that the result does not depend on which $a$ is chosen in the second clause.

The first is proven by induction on the size of $A$. If $|A| = 0$ then it is empty, so the first clause applies and the $\sum$ is eliminated. If I can eliminate sums over sets of size $n$, then I can eliminate sums over sets of size $n + 1$ using the second clause. So every sum can be eliminated.

Now, clearly when $|A| = 0$ the result does not depend on which $a$ is chosen, because no $a$ is chosen, and moreover when $|A| = 1$ there is only one possible choice, so no ambiguity is introduced. Now suppose $|A| \ge 2$, so it contains two distinct elements $a \not= b$, say. It suffices to prove that $f(a) + \sum_{i \in A \setminus \{a\}} f(i) = f(b) + \sum_{i \in A \setminus \{b\}} f(i)$. \[\begin{align} f(a) + \sum_{i \in A \setminus \{a\}} f(i) &= f(a) + f(b) + \sum_{i \in A \setminus \{a,b\}} f(i) \\ &= f(b) + f(a) + \sum_{i \in A \setminus \{a,b\}} f(i) \\ &= f(b) + \sum_{i \in A \setminus \{b\}} f(i). \end{align}\]