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Parenthesised tensor words of a fixed length are counted by the Catalan numbers

Statement

For $n \geq 1$ , let $a_{n}$ be the number of parenthesised tensor words on the letters $x_{1}, \dots, x_{n}$ with no inserted unit symbol. Then $a_{1} = 1$ and

$a_{n} = \sum_{k = 1}^{n - 1} a_{k} a_{n - k} (n \geq 2) .$

Equivalently, $a_{n} = C_{n - 1}$ , where the Catalan numbers are defined by $C_{0} = 1$ and

$C_{m} = \sum_{k = 0}^{m - 1} C_{k} C_{m - 1 - k} (m \geq 1) .$

Facts & Assumptions

Given: The recursive formation rule for parenthesised tensor words.

[L1]

A parenthesised tensor word is either one letter or a composite $(u \otimes v)$ built recursively, with the letters kept in order (Parenthesised tensor words and their evaluation functors).

Proof

technique · direct

1.1

For $n = 1$ there is only the word $x_{1}$ , so $a_{1} = 1$ .

givenL1

1.2

If $n \geq 2$ , every word has a unique outermost decomposition $(u \otimes v)$ , where $u$ uses the first $k$ letters and $v$ uses the remaining $n - k$ letters for a unique $k$ with $1 \leq k \leq n - 1$ . Conversely, every such pair $(u, v)$ produces a word on $n$ letters.

2.1

Therefore the words on $n$ letters are partitioned by the value of $k$ , and for fixed $k$ there are $a_{k} a_{n - k}$ choices. Summing over $k$ gives $a_{n} = \sum_{k = 1}^{n - 1} a_{k} a_{n - k}$ .

step 1.2algebra

3.1

The recurrence in step 2.1 is exactly the Catalan recurrence after the index shift $a_{n} = C_{n - 1}$ , and step 1.1 matches the initial value $C_{0} = 1$ . Hence $a_{n} = C_{n - 1}$ for all $n \geq 1$ .

step 1.1step 2.1algebra∎

Depends on

Parenthesised tensor words and their evaluation functors

Used by

The five bracketings of a four-fold tensor product with no inserted unit symbol Example

Dependency tree · two levels

2 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

P. Etingof, S. Gelaki, D. Nikshych, and V. Ostrik, Tensor Categories, Exercise 2.9.1 (standard reference, not scraped)