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Constant-term period of the triangle lattice potential

Example

In $R = Q [x^{\pm 1}, y^{\pm 1}]$ , take $W = x + y + x^{- 1} y^{- 1}$ . With $CT$ the coefficient of $x^{0} y^{0}$ , its formal classical period is $π_{W} (s) = \sum_{k \geq 0} CT (W^{k}) s^{k} = \sum_{d \geq 0} \frac{(3 d)!}{(d!)^{3}} s^{3 d} .$ This computation is only a Laurent constant-term identity. It makes no Gromov-Witten or mirror-equality claim, and the unmarked value $s = 1$ is not asserted to converge.

Facts & Assumptions

Given: The Laurent polynomial ring over $Q$ , the potential $W$ , and the formal variable $s$ .

[F1]

Intrinsic mirror algebra data and classical period: for a Laurent algebra with basis $x^{a} y^{b}$ and unit $1$ , the trace coefficient is the Laurent constant-term operator, so $π_{W} (s) = \sum_{k \geq 0} CT (W^{k}) s^{k}$ .

[A1]

The three summands of $W$ are $x$ , $y$ , and $x^{- 1} y^{- 1}$ , each occurring with coefficient $1$ .

Proof

1.1A1algebra

A term in the expansion of $W^{k}$ is determined by nonnegative integers $a, b, c$ counting the choices of $x$ , $y$ , and $x^{- 1} y^{- 1}$ , with $a + b + c = k$ . That term is $x^{a - c} y^{b - c}$ .

2.1A1step 1.1algebra

The term contributes to $CT (W^{k})$ exactly when $a - c = 0$ and $b - c = 0$ , hence exactly when $a = b = c = d$ . This forces $k = 3 d$ ; if $3 ∤ k$ , the constant term is zero.

3.1F1A1step 2.1algebra∎

For $k = 3 d$ , the multinomial theorem gives $(3 d)! / (d!)^{3}$ ordered choices with $d$ copies of each summand, and each choice contributes the constant monomial with coefficient $1$ . Thus $CT (W^{3 d}) = (3 d)! / (d!)^{3}$ , including $d = 0$ ; substituting these coefficients into [F1] gives the displayed period.

Remarks

The first nonconstant terms are $6 s^{3} + 90 s^{6} + 1680 s^{9}$ . They are finite coefficient checks of the formula; no geometric interpretation is being asserted here.

Depends on

Intrinsic mirror algebra data and classical period