Formal sums over endomorphisms

Let $𝑉$ be a vector space over $𝕂$ and consider formal sums over the endomorphism ring $E n d 𝑉$ , denoted $(E n d 𝑉) {𝑧}$ .¹ We define the following operations: #m/def/fcalc

Operations

Summation

The sum of a family ${𝑋 𝑖 (𝑧)} 𝑖 \in 𝐼$ in $(E n d 𝑉) {𝑧}$ with $𝑋 𝑖 (𝑧) = \sum 𝑛 \in 𝕂 𝑥 𝑖 (𝑛) 𝑧 𝑛$ exists iff the ${𝑥 𝑖 (𝑛)} 𝑖 \in 𝐼$ are summable for all $𝑛 \in 𝕂$ , and is given by

\sum 𝑖 \in 𝐼 𝑋 𝑖 (𝑧) = \sum 𝑛 \in 𝕂 (\sum 𝑖 \in 𝐼 𝑥 𝑖 (𝑛)) 𝑧 𝑛

Multiplication

The product of a finite list $(𝑋 𝑖 (𝑧)) 𝑟 𝑖 = 1$ in $(E n d 𝑉) {𝑧}$ exists iff for every $𝑛 \in 𝕂$ , the set

{𝑟 \prod 𝑖 = 1 𝑥 𝑖 (𝑛 𝑖) : 𝑟 \sum 𝑖 = 1 𝑛 𝑖 = 𝑛 \land {𝑛 𝑖} 𝑟 𝑖 = 1 \subseteq 𝕂}

is summable and is defined as

𝑟 \prod 𝑖 = 1 𝑋 𝑖 (𝑧) = \sum 𝑛 \in 𝕂 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ \sum 𝑛 1 + \dots + 𝑛 𝑟 = 𝑛 𝑛 1, \dots, 𝑛 𝑟 \in 𝕂 𝑟 \prod 𝑖 = 1 𝑥 𝑖 (𝑛 𝑖) ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 𝑧 𝑛

Importantly, partitioning a product into existent subproducts and taking the product of those will give the same result, but the converse doesn't hold: Multiplication of formal sums fails to be associative, instead satisfying partial associativity.

Counterexamples

Consider the Formal delta $𝛿 (𝑧) \in 𝕂 [[𝑧, 𝑧 - 1]]$ . Then naïve manipulation would suggest

𝛿 (𝑧) = ((\infty \sum 𝑛 = 0 𝑧 𝑛) (1 - 𝑧)) 𝛿 (𝑧)! = (\infty \sum 𝑛 = 0 𝑧 𝑛) ((1 - 𝑧) 𝛿 (𝑧)) = 0

On the other hand, this triple product exists but contains a nonexistant subproduct

(\infty \sum 𝑛 = 0 𝑧 𝑛) (\infty \sum 𝑛 = 0 𝑧 - 𝑛) 0 = 0

Limits of multivariable formal sums

Let

𝑋 (𝑧 1, 𝑧 2) = \sum 𝑚, 𝑛 \in 𝕂 𝑥 (𝑚, 𝑛) 𝑧 𝑚 1 𝑧 𝑛 2 \in (E n d 𝑉) {𝑧 1, 𝑧 2}

Then $l i m 𝑧 1 \to 𝑧 2 𝑋 (𝑧 1, 𝑧 2)$ exists iff for every $𝑛 \in 𝕂$ the family ${𝑥 (𝑚, 𝑛 - 𝑚)} 𝑚 \in 𝕂$ is summable, and is given by

l i m 𝑧 1 \to 𝑧 2 (\sum 𝑚, 𝑛 \in 𝕂 𝑥 (𝑚, 𝑛) 𝑧 𝑚 1 𝑧 𝑛 2) = \sum 𝑛 \in 𝕂 (\sum 𝑚 \in 𝕂 𝑥 (𝑚, 𝑛 - 𝑚)) 𝑧 𝑛 2