Spine representations for non-compact models of random geometry

Abstract

We provide a unified approach to the three main non-compact models of random geometry, namely the Brownian plane, the infinite-volume Brownian disk, and the Brownian half-plane. This approach allows us to investigate relations between these models, and in particular to prove that complements of hulls in the Brownian plane are infinite-volume Brownian disks.

Appendix: Some Laplace transforms

Recall the standard notation

$$\begin{aligned} \hbox {erfc}(x)=\frac{2}{\sqrt{\pi }}\int _x^\infty e^{-t^2}\,\text {d}t. \end{aligned}$$

Then the function $\chi _1$ defined for $x>0$ by

$$\begin{aligned} \chi _1(x)=\frac{1}{\sqrt{\pi }}x^{-1/2} - e^x\,\text {erfc}(\sqrt{x}) =\frac{1}{\sqrt{\pi }}\,e^x \int _{\sqrt{x}}^\infty \frac{1}{t^2}\,e^{-t^2}\,\text {d}t, \end{aligned}$$

(A.0)

satisfies, for every $\lambda >0$,

$$\begin{aligned} \int _0^\infty \text {d}x\,e^{-\lambda x}\,\chi _1(x)= (1+\sqrt{\lambda })^{-1}. \end{aligned}$$

(A.1)

This is easily verified via an integration by parts which gives for $\lambda >0$,

$$\begin{aligned} \int _0^\infty \text {erfc}(\sqrt{x})e^x\,e^{-\lambda x}\,\text {d}x = \frac{1}{\sqrt{\lambda }(1+\sqrt{\lambda })}. \end{aligned}$$

From the last two displays and an integration by parts, one checks that the function $\chi _2=\chi _1*\chi _1$, which satisfies

$$\begin{aligned} \int _0^\infty \text {d}x\,e^{-\lambda x}\,\chi _2(x)= (1+\sqrt{\lambda })^{-2}, \end{aligned}$$

(A.2)

is given for $x>0$ by

$$\begin{aligned} \chi _2(x)= e^x\text {erfc}(\sqrt{x})-2x\,\chi _1(x)= (2x+1)e^x\text {erfc}(\sqrt{x}) -\frac{2}{\sqrt{\pi }}x^{1/2}. \end{aligned}$$

Similar manipulations show that the function $\chi _3=\chi _1*\chi _1*\chi _1$ satisfying

$$\begin{aligned} \int _0^\infty \text {d}x\,e^{-\lambda x}\,\chi _3(x)= (1+\sqrt{\lambda })^{-3}. \end{aligned}$$

(A.3)

is given by

$$\begin{aligned} \chi _3(x)=\frac{2}{\sqrt{\pi }}(x^{3/2} + x^{1/2}) - 2x(x+\frac{3}{2})\,e^x\,\hbox {erfc}(\sqrt{x}). \end{aligned}$$

We observe that $\chi _1(x)>0$ for every $x>0$ (this is obvious from (A.0)) and thus we have also $\chi _3(x)>0$ for every $x>0$. Finally, we note that

$$\begin{aligned} \int _0^{\infty } \frac{1-e^{-\lambda x}}{x}\,\chi _3(x)\,\text {d}x&= \int _0^\lambda \text {d}\mu \int _0^\infty e^{-\mu x}\,\chi _3(x)\,\text {d}x\nonumber \\&= \int _0^\lambda \text {d}\mu (1+\sqrt{\mu })^{-3} = (1+\lambda ^{-1/2})^{-2}. \end{aligned}$$

(A.4)