# Is a random walk on an isoradial graph transient?

Is the random walk on an isoradial graph defined in Chelkak and Smirnov (2011) (in pages 3-4) transient?

Let us define the random walk on an isoradial graph $$\Gamma$$ starting from $$x$$ by, $$$$X_t = X_0 + \sum_{j=0}^{t-1} \xi_{X_j}^{(j)},$$$$ where $$X_0 = x$$ and the increments $$\xi_{u}^{(t)}$$ are independently and identically distributed for fixed vertex $$u \in \Gamma$$ and for all $$t \in \mathbb{N}$$. These are distributed according to, $$$$\mathbb{P}(\xi_u = u_k - u) = \frac{\tan{\theta_k}}{\sum_{s=1}^{n}\tan{\theta_s}},$$$$ for $$u_k$$ $$\sim$$ $$u$$ are adjacent vertices in $$\Gamma$$. The half-angle of the rhombus is denoted by $$\theta_k$$. These are uniformly bounded from $$0$$ and $$\pi/2$$.

I am hesitating whether $$(X_t)$$ is transient? For fixed $$u_0 \in \Gamma$$ and given the mesh size $$\delta > 0$$, the free Green's function is finite: $$G(u_0,u_0) = \frac{1}{2\pi}(\log \delta - \gamma_{Euler} -\log 2) < \infty$$. Does this imply that $$(X_t)$$ is transient? In the light of Pólya's theorem this feels counter-intuitive. Since increments of $$(X_t)$$ have zero expectations, that is $$\mathbb{E}(\Re \xi_u) = 0$$, $$\mathbb{E}(\Im \xi_u) = 0$$, and $$\Gamma$$ is embedded in $$\mathbb{C} \simeq \mathbb{R}^2$$, one may expect that $$(X_t)$$ is recurrent.