I'm referring to Cartan's "Elementary Theory of Analytic Functions of One or Several Complex Variables".

Recall that the order of a formal series $\sum_{n\ge0}a_nX^n$ is the smallest (if it exists) index $n\ge 0$ s.t. $a_n\ne0$; otherwise, the order is infinity, by convention.

Definition (p.11): A family $(S_i(X))_{i\in I}$ of series is said to be summable iff for any integer $k$, only a finite number of series of the family have order $< k$. In this case, the sum is defined to be the series $$S(X)=\sum_{i\in I}S_i(X)=\sum_{n\ge 0}(\sum_{i\in I}a_{n,i})X^n$$ which makes sense because the family is summable.

Two questions:

(Q1): the author claims (pp.11) that this "generalized addition is commutative and associative in a sense which the reader should specify".

I'm looking for a "sense", but I'm not sure at all what the author means. The main problem, at least for me, is that commutativity and associativity are properties defined for operations on some set, e.g. $+\colon \Bbb{R}\times \Bbb{R}\to \Bbb{R}$. But the generalized sum takes possibly infinitely many arguments. Thus, how should those properties look like in this generalized situation?

(Q2): For a formal series $S(X)=\sum_{n\ge0}a_nX^n$, define the formal derivative to be $S'(X)=\sum_{n\ge0}na_nX^{n-1}$. Clearly, we get a linear map $S\mapsto S'$. The author claims (pp.15) that a Leibniz-type rule holds, i.e. $(ST)'=S'T+ST'$, by saying that "it is sufficient to verify this formula in the particular case when $S,T$ are monomials, and it is clearly true then." Here, I'm not sure I understood the meaning of "clearly": I think it is a consequence of the fact that two power series are equal iff have the same coefficients of $X^n$ for every $n\ge0$. But for fixed $n\ge0$, we have to prove equality of products of polynomial, which follows from linearity and what we have already proved for monomials. Is this reasoning right or not?

Thank you in advance for your help.

  • 1
    $\begingroup$ For generalized commutativity, perhaps something like $\sum_{i \in I}S_i(X) = \sum_{i \in I}S_{\phi(i)}(X)$, where $\phi : I \to I$ is a bijection. $\endgroup$
    – user169852
    Aug 11 '19 at 23:32

(Q1): This seems to be dependent on interpretation, but for me, the most natural senses are the following:

Commutative: Given any permutation $\phi : I \to I,$ $$\sum_{i \in I} S_{\phi(i)}(X) = \sum_{i \in I} S_i(X)$$ Associative: Given any partition $\{I_j : j \in J\}$ of $I,$ $$\sum_{j \in J}\sum_{i \in I_j} S_i(X) = \sum_{i\in I} S_i(X)$$

(Q2): When Cartan said "clearly," he was talking about the fact that $$(X^n X^m)' = (X^n)'(X^m) + (X^n)(X^m)'$$ is clear. In order to reduce to this point, you use the fact that, as defined, the formal derivative is linear.

  • $\begingroup$ In case someone can't see how these generalize the usual properties of commutative and associative, consider $I = \{1, 2\}$ and $\phi(1) = 2, \phi(2) = 1$ for commutative, and consider $I = \{1,2,3\}, I_1 = \{1\}, I_2 = \{2,3\}$ for associative. $\endgroup$ Aug 11 '19 at 23:57

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