Lemma. Counting adjoints on an indexed cycle [fdip-0014]
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  • October 20, 2026

Let \(R\) be a commutative unital ring. Take the cyclic shift on \(\mathcal F_R(\mathbb Z/n\mathbb Z)\), \(n\ge 1\), with its perfect counting duality \(b\). For the counting adjoint, write \(A^*=A^\dagger \). Then \[T^*=T^{-1},\qquad \Delta ^*=T^{-1}-I.\] Writing \(\nabla ^-=I-T^{-1}\) for the backward difference gives \[\Delta ^*\Delta =-\nabla ^-\Delta =2I-T-T^{-1}.\] Here \(2=1+1\); its invertibility is not required. For the real difference and kinetic signs, see [wingate2025lattice, Section 4, Equations (4.3) and (4.5), p. 3], with lattice spacing one. The finite periodic ring identities follow by reindexing, independently of the product rule.

Proof. 

  • October 20, 2026

The substitution \(j=i+1\) permutes the finite index set, so \[\begin {aligned} b(Tf,g)&=\sum _i f(i+1)g(i)\\ &=\sum _j f(j)g(j-1)=b(f,T^{-1}g). \end {aligned}\] Uniqueness of the counting adjoint gives \(T^*=T^{-1}\). Subtracting \(b(f,g)\) gives \(b(\Delta f,g)=b(f,(T^{-1}-I)g)\), so \(\Delta ^*=T^{-1}-I\). Multiplication gives \[(T^{-1}-I)(T-I)=2I-T-T^{-1}=-\nabla ^-\Delta .\] Consequently \(b(f,\Delta ^*\Delta f)=b(\Delta f,\Delta f)\). Over \(\mathbb R\) this counting form is the counting inner product, and the last expression is \(\sum _i(\Delta f(i))^2\ge 0\).

For the indexed cycle \(e_i:i\to i+1\), the edge-label isomorphism \(j(a)(i)=a(e_i)\) of the cyclic shift and edge-label identification preserves counting pairings: \(b_V(ja,jc)=\sum _i a(e_i)c(e_i)=b_E(a,c)\). Taking \(c=j^{-1}f\) shows that \(j^*=j^{-1}\), hence \(j^*j=I\). Thus \(\Delta =jd_R\) and \[\Delta ^*\Delta =d_R^\dagger j^*j d_R=d_R^\dagger d_R,\] the unit-weight graph Laplacian. At \(n=2\), the two oppositely oriented edges give \(b(f,d_R^\dagger d_Rf)=2(f(1)-f(0))^2\); they are not one simple edge. At \(n=1\), there is one loop, \(T=I\), and both incidence and kinetic maps vanish. An arbitrary incidence map still has distinct vertex and edge carriers; this factorization does not assert that \(\Delta \) is skew-adjoint or square-zero.