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Genus expansion is not a nonperturbative completion

Physical scope

Logical scope of a formal or asymptotic perturbative genus expansion in a fixed background; it makes no claim that a particular string expansion converges or diverges.

Statement

A formal genus expansion records coefficients weighted by the topology convention of String Coupling Genus Weight. Its formal existence alone proves neither convergence of its numerical sum nor existence or uniqueness of an exact nonperturbative theory, instanton sectors, or a string-field completion. These are separate constructions and status claims. This remark does not say that every string genus series diverges.

Statement references

String Coupling Genus Weight

Facts & Assumptions

Given: A fixed background and normalization in which complex coefficients $a_{0}, a_{1}, \dots$ are recorded against the closed topology powers as the formal Laurent series $F (g_{s}) = \sum_{g \geq 0} a_{g} g_{s}^{2 g - 2} = g_{s}^{- 2} A (g_{s}^{2})$ , where $A (x) = \sum_{g \geq 0} a_{g} x^{g} \in C [⁣ [x] ⁣]$ . The notation $C [⁣ [x] ⁣]$ means coefficient sequences with coefficientwise addition and Cauchy multiplication; $g_{s}$ is an indeterminate here. These algebraic operations assign no value at nonzero numerical $g_{s}$ . A fixed external normalization simply multiplies this series by its declared factor.

[F1]

String Coupling Genus Weight supplies the closed topology power and explicitly does not assert convergence or a complete measure.

Proof

1.1givenF1algebra

The formal series $A (x)$ is coefficient data with algebraic addition and multiplication. Substitution $x = g_{s}^{2}$ and multiplication by $g_{s}^{- 2}$ give exactly the powers $2 g - 2$ of [F1], but assign no value at any nonzero numerical $g_{s}$ . Therefore convergence cannot follow from the formal-series datum alone.

2.1givenstep 1.1algebra

Even an asymptotic expansion does not select a unique exact function: for $g_{s} > 0$ , let $f (g_{s}) = 0$ and $h (g_{s}) = e^{- 1 / g_{s}}$ . These are distinct. Put $x = 1 / g_{s}$ ; for every integer $N \geq 0$ , the positive exponential series gives $e^{x} \geq x^{N + 1} / (N + 1)!$ , hence $0 \leq g_{s}^{- N} e^{- 1 / g_{s}} = x^{N} e^{- x} \leq (N + 1)! / x \to 0$ . Thus $h (g_{s}) = O (g_{s}^{N})$ for every $N$ , so $f$ and $h$ have the same all-zero power asymptotic expansion. Coefficient data alone therefore does not determine a unique nonperturbative completion. This abstract calculation asserts nothing about a particular string model.

3.1F1step 1.1step 2.1algebra∎

Consequently convergence, nonperturbative sectors, and uniqueness require additional analytic or physical input; they are not logical consequences of the topology weights.

Remarks

Tong describes the genus expansion as formal/asymptotic. Model-specific instanton, matrix-model, and string-field constructions belong to their own interfaces and are not premises here.

Genus expansion is not a nonperturbative completion

Statement

Statement references

Facts & Assumptions

Proof

Remarks

Depends on

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