Theorem. Only the zero-momentum block survives the free Wilson term [fdip-009V]
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  • November 1, 2026

Take the Wilson operator at trivial links, with \(a>0\) and real \(r\ne 0\), on the periodic grid with \(N_j\ge 3\). Its kernel consists exactly of constant fields. In particular, \[\dim _{\mathbb C}\ker D_{W,a,1}=2^d q.\] For the continuous momentum symbol, a corner having \(k\) coordinates equal to \(\pi \) receives scalar Wilson term \(2rk/a\); it is nonzero for \(k\ge 1\).

Proof. 

  • November 1, 2026

The finite Fourier decomposition has momenta \(p_j=2\pi n_j/N_j\), with \(0\le n_j<N_j\). These are dimensionless lattice momenta; their physical representatives are \(p_j/a\). Put \[\begin {aligned} A_a(p)&=\frac {i}{a}\sum _j\Gamma _j\sin p_j,\\ M_a(p)&=\frac {r}{a}\sum _j(1-\cos p_j). \end {aligned}\] The Wilson block is \(A_a(p)+M_a(p)I\). Hermitian gamma matrices and their anticommutation relations give \[\begin {aligned} A_a(p)^*&=-A_a(p),\\ A_a(p)^2&=-a^{-2}\sum _j\sin ^2p_j\,I. \end {aligned}\] Since \(M_a(p)\) is real and scalar, the mixed terms cancel: \[\begin {aligned} &(A_a+M_aI)^*(A_a+M_aI)\\ &\qquad =\left (a^{-2}\sum _j\sin ^2p_j+M_a^2\right )I. \end {aligned}\] Thus a block has a nonzero kernel precisely when both \(\sum _j\sin ^2p_j=0\) and \(M_a(p)=0\). Because \(r\ne 0\) and each \(1-\cos p_j\ge 0\), the latter condition already forces every \(\cos p_j=1\). In the chosen momentum range this means \(p=0\). Conversely the entire block at \(p=0\) is zero. Its dimension is \(\dim (S\otimes F)=2^d q\), and the zero Fourier character is the constant field.

The same formula is defined for every real momentum modulo \(2\pi \). At \(p=\pi \epsilon \), with \(\epsilon \in \{0,1\}^d\), the sine part vanishes and \[M_a(\pi \epsilon )=\frac {2r}{a}\sum _j\epsilon _j.\] This proves the corner claim, whether or not a particular finite grid samples that corner exactly.

For \(q=1\), periods \((4,4)\) and \((4,6)\) both give kernel dimension \(4\); periods \((4,4,4)\) give 8. The same formula applies to odd periods: zero momentum is always present and every other block is invertible.

The scalar addition is the free Wilson denominator mechanism in [wingate2025lattice, Section 9.3, Equations (9.15)–(9.16), p. 21]. For \(r>0\), its value at each nonzero corner grows as \(1/a\). This is an exact free-symbol statement, not a continuum-limit construction or an interacting mass calculation. The cancellation above uses the scalar free block and proves no corresponding kernel formula for a general connection. The surviving block still contains the whole chosen reducible \(S\otimes F\).