Proposition. Zero modes and the nearest-neighbor mismatch [fdip-000T]
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October 17, 2026

On the periodic grid, let $e$ be the number of even periods among $N_1,\ldots ,N_d$. The centered Dirac operator has exactly $2^e$ zero momentum blocks and \[\dim _{\mathbb C}\ker D=2^{d+e}.\] This is the kernel dimension for the explicitly chosen reducible module $S$, whose dimension is $2^d$. The unit-weight graph Laplacian \[L_{\mathrm {nn}}=\sum _{j=1}^d (2I_H-T_j-T_j^{-1})\] has only the zero momentum in its scalar kernel. Moreover $-D^2\ne I_S\otimes L_{\mathrm {nn}}$ on every permitted grid, including grids with all periods odd. These are finite consequences of the Fourier blocks; the centered-difference convention comes from [wingate2025lattice, Section 9.1, Equations (9.2), (9.6)-(9.7), pp. 19-20].

Proof.

October 17, 2026

The forward-difference factorization from Lemma [fdip-0014], applied in each coordinate, gives \[L_{\mathrm {nn}}=\sum _j(T_j-I_H)^*(T_j-I_H).\] For $N_j\ge 3$, each undirected nearest-neighbor edge occurs once in the corresponding squared-difference energy. On the character with momentum $p$, the eigenvalue is \[\begin {aligned} \ell (p)&=\sum _j(2-2\cos p_j)\\ &=4\sum _j\sin ^2(p_j/2). \end {aligned}\] The positive Dirac square has scalar block $s(p)=\sum _j\sin ^2p_j$. The double-angle identity gives \[\ell (p)-s(p)=4\sum _j\sin ^4(p_j/2).\] Choose $r_1=1$ and the other labels zero. Since $N_1\ge 3$, $\sin (\pi/N_1)>0$, so the difference is strictly positive on that block. This proves the operator mismatch.

Every summand of $s(p)$ is nonnegative. Hence $s(p)=0$ exactly when each $\sin p_j=0$. In the range $0\le r_j<N_j$, this means $r_j=0$, or $r_j=N_j/2$ when $N_j$ is even. There are exactly $2^e$ choices. On each such block $D(p)=0$ on all of $S$. Every other block is invertible, with two-sided inverse \[D(p)^{-1}=-D(p)/s(p),\] because $D(p)^2=-s(p)I_S$. Completeness of the Fourier basis therefore gives $\dim \ker D=2^e\dim S=2^{d+e}$. In contrast, $\ell (p)=0$ forces $r_j=0$ for every $j$; thus $\dim \ker (I_S\otimes L_{\mathrm {nn}})=2^d$.

For $d=1,N=4$, the alternating scalar field $a(x)=(-1)^x$ has momentum $p=\pi $. Its centered difference is zero, whereas $L_{\mathrm {nn}}a=4a$. Consequently every field $x\mapsto a(x)v$, $v\in S$, is killed by $D$. Together with the constant fields these give the four-dimensional kernel. Write $K_2=(2I_H-T^2-T^{-2})/4$, so $-D^2=I_S\otimes K_2$. The two scalar Laplacians have the following weighted graphs:

$\text{[math]}$

Left: the 4 unit-weight nearest-neighbor edges. Right: the two-step edges, with vertices rearranged to display the parity classes $\{0,2\}$ and $\{1,3\}$. At period $4$, $T^2=T^{-2}$, so $K_2=(I_H-T^2)/2$: the two shift contributions of weight $1/4$ combine into one undirected edge of weight $1/2$ for each opposite pair. The alternating field is constant on each right-hand component, explaining its zero energy there.

For $d=1,N=3$, momentum $p=2\pi/3$ instead gives \[s(p)=\tfrac 34,\qquad \ell (p)=3.\] Only $r=0$ is a zero momentum, yet the operators still differ. Thus extra exact zero momenta occur precisely when at least one period is even, whereas the square mismatch always occurs. The excluded case $N=2$ has $T=T^{-1}$, making its centered difference identically zero.

The finite kernel dimension above is not a physical species or chirality count. Friedan's Nielsen-Ninomiya theorem concerns free fermions on the infinite lattice $\mathbb Z^d$ for odd $d$, with continuous torus momenta. Its Hamiltonian is quadratic, phase invariant, translation invariant and local, with a Hermitian momentum symbol. The locality discussion reduces the Hamiltonian requirement to continuity; the internal-charge projections must be smooth. The zero set is finite and isolated, with specified relativistic Clifford-linear behavior near each zero. Under these hypotheses the total chiral index vanishes separately for each charge [friedan1982nielsen, Sections 1-2, pp. 481-484; Equations (6), (10)-(11)]. A finite sampled grid in arbitrary dimension with the chosen reducible module computes a different quantity. The odd-period example neither contradicts that theorem nor establishes a continuum limit. No Lorentzian model or factorization of the earlier weighted precision is supplied by this centered operator.