Periodicity and Even-Time Returns on the Lattice

Parity, Period, and Fourier Summation

A simple symmetric walk on a lattice alternates between two classes of sites: those whose coordinate sum is even and those whose coordinate sum is odd. This elementary parity fact has an important Fourier counterpart. The Fourier symbol takes the value \(1\) at one point of the torus and \(-1\) at the point shifted by \(\pi\) in every coordinate. Summing only the even powers keeps both values visible; replacing that sum too quickly by a geometric series can obscure the periodicity or mishandle an infinite return-probability sum.

Fix a dimension \(d\ge1\). On a probability space \((\Omega,\mathcal F,\mathbb P)\), let \((Y_n)_{n\ge1}\) be independent increments, each uniform on \(\{e_1,-e_1,\ldots,e_d,-e_d\}\), where \(e_j\) is the \(j\)th coordinate vector. Set \(X_0=0\) and \(X_n=\sum_{r=1}^nY_r\). This is a discrete-time Markov chain with state space \(S=\mathbb Z^d\). From every \(x\in S\), each of its \(2d\) nearest neighbors has transition probability \(1/(2d)\), and all other transition probabilities are zero. Thus every transition-kernel row sums to \(2d\cdot1/(2d)=1\). Write \(p_n=\mathbb P_0(X_n=0)\) for the return probability at time \(n\).

The period concerns possible return times, not whether return is certain, nor how large a return probability is. The first return time \(T_0^+=\inf\{n\ge1:X_n=0\}\) is also distinct from \(p_n\): the latter is the probability of being at the origin at a specified time and does not require that the walk avoided the origin earlier.

Proof. Each step changes the coordinate sum \(x_1+\cdots+x_d\) by \(1\) or \(-1\), so it changes the parity of that sum. Starting from any state, after an odd number of steps the walk is in the opposite parity class and cannot be back at its starting state. Hence all odd-time return probabilities are zero. For any integer \(m\ge1\), the path that alternates \(e_1,-e_1\) exactly \(m\) times returns after \(2m\) steps and has probability \((1/(2d))^{2m}>0\). Thus the possible positive return times are exactly the positive even integers. Their greatest common divisor is \(2\), proving the claim. \(\square\)

The Fourier Symbol and the Parity Shift

For \(\theta=(\theta_1,\ldots,\theta_d)\in[-\pi,\pi)^d\), define the Fourier symbol \(\phi(\theta)=\mathbb E[e^{\,i\theta\cdot Y_1}]\). The two steps in coordinate \(j\) contribute \(e^{i\theta_j}\) and \(e^{-i\theta_j}\), each with probability \(1/(2d)\), so \(\phi(\theta)=d^{-1}\sum_{j=1}^d\cos\theta_j\). In particular, \(-1\le\phi(\theta)\le1\).

Integrals over the torus below use normalized measure \(d\theta/(2\pi)^d\). Fourier inversion here follows directly from the identity that the normalized integral of \(e^{i\theta\cdot x}\) is \(1\) for \(x=0\) and \(0\) for every nonzero \(x\in\mathbb Z^d\). Indeed, for each fixed lattice point \(x\), that integral factors into one-dimensional integrals, and a nonzero coordinate contributes \((2\pi)^{-1}\int_{-\pi}^{\pi}e^{ik\theta}\,d\theta=0\) for a nonzero integer \(k\). Applying this identity to \(X_n\) and using independence of the increments gives the return formula.

Let \(\mathbf{\pi}=(\pi,\ldots,\pi)\), with coordinates interpreted modulo \(2\pi\) on the torus. Since \(\cos(\theta_j+\pi)=-\cos\theta_j\), the symbol satisfies \(\phi(\theta+\mathbf{\pi})=-\phi(\theta)\). Translation by \(\mathbf{\pi}\) preserves normalized torus measure. Thus the integral of \(\phi^n\) equals the integral of \((-\phi)^n\). For odd \(n\), this integral is its own negative and must be zero. This Fourier cancellation agrees with the pathwise parity proof; it does not mean that \(\phi(\theta)^n\) is zero at each \(\theta\).

Summing the Return Probabilities Without Losing Divergence

Define the return-probability sum, including the time-zero visit, by \(G=\sum_{n=0}^{\infty}p_n\), with the value \(+\infty\) allowed. Odd-time terms vanish, so \(G=\sum_{m=0}^{\infty}p_{2m}\). For a finite cutoff \(M\), Fourier inversion and linearity of a finite sum give \(\sum_{m=0}^{M}p_{2m}=\int\sum_{m=0}^{M}\phi(\theta)^{2m}\,d\theta/(2\pi)^d\). Since \(0\le\phi^2\le1\), these finite sums of even powers increase pointwise as \(M\) increases. The Monotone Convergence Theorem, a standard measure-theoretic result, therefore permits taking the increasing limit inside the integral, including when the result is infinite.

Proof. For each \(M\), the finite geometric-sum identity gives \(\sum_{m=0}^{M}\phi^{2m}=(1-\phi^{2M+2})/(1-\phi^2)\) wherever \(\phi^2\ne1\). If \(\phi^2=1\), the sum is \(M+1\). In either case, as \(M\) increases the sum increases to \(1/(1-\phi^2)\) when \(\phi^2<1\), and to \(+\infty\) when \(\phi^2=1\). Integrating the finite sums gives \(\sum_{m=0}^M p_{2m}\). Applying the Monotone Convergence Theorem to this nonnegative increasing sequence shows that its integral increases to the integral of the pointwise limit. The left side increases to \(G\), proving the formula in the extended sense. \(\square\)

This order of operations matters. Each even-power term is nonnegative, so the monotone limit retains the full value of the sum, including divergence. The formula describes the return-probability sum; it does not, by itself, say whether the integral is finite in a particular dimension. The behavior near points where \(|\phi|=1\) requires a separate integrability analysis.

A Common Fourier Pitfall

A tempting but incorrect step is to replace the even-power sum by \(1/(1-\phi)\). The geometric series for that expression sums powers \(\phi^n\), not just the even powers, and its pointwise behavior at \(\phi=-1\) does not represent the even series. For a concrete check, in dimension one at \(\theta=\pi\), \(\phi(\pi)=\cos\pi=-1\). The finite even-power sum is \(\sum_{m=0}^M(-1)^{2m}=M+1\), whereas \(1/(1-\phi(\pi))=1/2\). The correct even-power denominator is \(1-\phi^2\), with value \(+\infty\) at \(\phi^2=1\) in the limiting sum.

There is a related mistake in the odd-time Fourier calculation: \(\phi(\theta)^n\) is not pointwise zero for odd \(n\). For instance, in dimension one at \(\theta=\pi/3\), \(\phi(\theta)^3=(1/2)^3=1/8\). What vanishes is its integral, because shifting every torus coordinate by \(\pi\) changes its sign while preserving the measure. Keeping the distinction between pointwise values and integrated cancellation is essential when using Fourier formulas for periodic walks.