Brownian-bridge perturbative expansion conjecture for TASEP cumulants

Let nn be a non-negative integer and let h0h_0 be an arbitrary sufficiently regular continuous function satisfying

h0(0)=h0(1)=0.h_0(0)=h_0(1)=0.

For independent standard Brownian bridges b1,,bnb_1,\ldots,b_n, define

fbb,n(s;h0)=bn(0+)bn(1)exp(smax[b1h0]sj=2nmax[bjbj1])b1,,bnexp(smax[b1h0]sj=2nmax[bjbj1])b1,,bn.f_{\mathrm{bb},n}(s;h_0)=-\frac{\left\langle b_n'(0^+)b_n'(1^-)\exp\left(-s\max[b_1-h_0]-s\sum_{j=2}^n\max[b_j-b_{j-1}]\right)\right\rangle_{b_1,\ldots,b_n}}{\left\langle\exp\left(-s\max[b_1-h_0]-s\sum_{j=2}^n\max[b_j-b_{j-1}]\right)\right\rangle_{b_1,\ldots,b_n}}.

Here max\max denotes the maximum over the bridge parameter, and define χ(v)=Li5/2(exp(v))/2π\chi(v)=-\operatorname{Li}_{5/2}(-\exp(v))/\sqrt{2\pi}; let ν(s)\nu(s) solve χ(ν(s))=s\chi'(\nu(s))=s. Brownian-bridge perturbative expansion conjecture.

fbb,n(s;h0)=s23+χ(ν(s))s+O(sn).f_{\mathrm{bb},n}(s;h_0)=\frac{s^2}{3}+\frac{\chi(\nu(s))}{s}+\mathcal{O}(s^n).

In particular, this perturbative expansion is independent of h0h_0 through order sn1s^{n-1}. The conjecture arises from exact Bethe-ansatz formulas and corresponding Brownian-bridge representations; direct probabilistic proofs were unknown in the source, while low-order cases had numerical support.

Sources & referencesView supporting material

Primary source

Kirone Mallick and Sylvain Prolhac, “Brownian bridges for late time asymptotics of KPZ fluctuations in finite volume”, arXiv:1805.03187 (2018).

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