Lemma. Adjoints preserve the chosen parity [fdip-001L]
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  • October 20, 2026

Let \(R\) be a commutative unital ring and \(V=V_0\sqcup V_1\) a finite partition. Equip \(H=\mathcal F_R(V)\) with its counting pairing \(b\), counting adjoint constructed in Proposition [fdip-001H], and parity grading \(A_0,A_1\) on \(\operatorname {End}_R(H)\). Then \(p_i^\dagger =p_i\) for the support projections, and \[a\in A_k\quad \Longrightarrow \quad a^\dagger \in A_k.\] Every multiplication operator \(M_f\), for \(f\in H\), is even and satisfies \(M_f^\dagger =M_f\).

Proof. 

  • October 20, 2026

For \(u,v\in H\), restriction of the counting sum gives \[b(p_iu,v)=\sum _{x\in V_i}u(x)v(x)=b(u,p_iv),\] so uniqueness of the adjoint gives \(p_i^\dagger =p_i\). The support summands are orthogonal: if \(u\in H_0\) and \(v\in H_1\), every term of \(b(u,v)\) is zero.

For \(a\in A_k\), evaluation on each support summand gives \[a=\sum _{i=0}^1p_{i+k}ap_i.\] Taking the ordinary adjoint, including its linearity and product reversal, yields \[a^\dagger =\sum _{i=0}^1p_i a^\dagger p_{i+k}.\] Each summand maps \(H_{i+k}\) into \(H_i\) and vanishes on the other support summand. Its degree is \(-k=k\) in \(\mathbb Z/2\mathbb Z\), proving the claim.

Multiplication by a field preserves support, so \(M_f\in A_0\). Commutativity of \(R\) gives \[b(M_fu,v)=\sum _x f(x)u(x)v(x)=b(u,M_fv).\] Uniqueness therefore gives \(M_f^\dagger =M_f\).

The degree statement depends on this orthogonal support decomposition and this symmetric counting pairing. It is not a law for an unspecified grading and involution. The involution on coefficients is the identity, including for complex coefficients; no Hermitian or positive conclusion is inferred. The graded composition convention is [stacks2017stacks, Section 22.25], Tag 09L1; the compatibility with the chosen pairing is proved above.