Proposition. Adjoints from chosen dualities [fdip-001P]
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October 20, 2026

Let \(R\) be a commutative unital ring and let \(M,N,P\) be unital \(R\)-modules with chosen perfect symmetric pairings \(b_M,b_N,b_P\) and duality maps \(\Phi _M,\Phi _N,\Phi _P\). For an ordinary \(R\)-linear map \(A:M\to N\), precomposition defines \[A^\vee :N^\vee \longrightarrow M^\vee ,\qquad A^\vee \ell =\ell \circ A.\] Its adjoint for chosen perfect pairings is \[A^\dagger =\Phi _M^{-1}A^\vee \Phi _N:N\longrightarrow M.\] Products denote composition, with the right factor first, as in composition and field multiplication. This is the unique linear map satisfying \[b_M(x,A^\dagger y)=b_N(Ax,y)\qquad (x\in M,\ y\in N).\]

For \(A,A':M\to N\), \(B:N\to P\) and \(r\in R\), \[\begin {aligned} (A+A')^\dagger &=A^\dagger +(A')^\dagger ,& (rA)^\dagger &=rA^\dagger ,\\ 0^\dagger &=0,&I_M^\dagger &=I_M,\\ (BA)^\dagger &=A^\dagger B^\dagger ,& (A^\dagger )^\dagger &=A. \end {aligned}\] If \(A\) is an isomorphism, then \(A^\dagger \) is an isomorphism and \[(A^{-1})^\dagger =(A^\dagger )^{-1}.\] An endomorphism \(L:M\to M\) is self-adjoint for this pairing when \(L^\dagger =L\). In particular, the inverse of an invertible self-adjoint endomorphism is self-adjoint.

Proof.

October 20, 2026

The composite \(\ell \circ A\) is linear. Evaluation at \(x\) gives \[\begin {aligned} A^\vee (\ell +k)(x)&=\ell (Ax)+k(Ax),\\ A^\vee (r\ell )(x)&=r\ell (Ax). \end {aligned}\] Hence \(A^\vee \) is linear, and so is the composite defining \(A^\dagger \). Also \((BA)^\vee =A^\vee B^\vee \) and \(I^\vee =I\), since both sides have the same value on every functional and argument. Because \(\Phi _M\Phi _M^{-1}=I\), \[b_M(x,A^\dagger y)=(A^\vee \Phi _Ny)(x)=b_N(Ax,y).\] If \(K:N\to M\) has the same identity, then \(\Phi _M(Ky)=\Phi _M(A^\dagger y)\) for every \(y\). Injectivity gives \(K=A^\dagger \); existence also used surjectivity of \(\Phi _M\).

Bilinearity shows that \(A^\dagger +(A')^\dagger \), \(rA^\dagger \), \(0\) and \(I_M\) represent the identities for \(A+A'\), \(rA\), \(0\) and \(I_M\), respectively. Uniqueness proves the 4 stated laws. For composition, apply the adjoint identity twice: \[\begin {aligned} b_M(x,A^\dagger B^\dagger z) &=b_N(Ax,B^\dagger z)\\ &=b_P(BAx,z). \end {aligned}\] This proves the product-reversal law. Symmetry gives \[\begin {aligned} b_N(y,Ax)&=b_N(Ax,y)\\ &=b_M(x,A^\dagger y)=b_M(A^\dagger y,x). \end {aligned}\] Thus \(A\) is the adjoint of \(A^\dagger \), proving involution.

Finally, adjointing \(AA^{-1}=I_N\) and \(A^{-1}A=I_M\) gives \[(A^{-1})^\dagger A^\dagger =I_N,\qquad A^\dagger (A^{-1})^\dagger =I_M.\] These are the two inverse identities. If \(A^\dagger =A\), they identify \((A^{-1})^\dagger \) with \(A^{-1}\). No order or division is used.

In particular, for \(d:M\to N\) and a self-adjoint \(W:N\to N\), the composite \(L=d^\dagger Wd:M\to M\) satisfies \[L^\dagger =d^\dagger W^\dagger d=L,\qquad b_M(f,Lg)=b_N(df,Wdg).\] The first equality uses product reversal and double adjunction; the second is the characterizing identity for \(d\). Taking \(g=f\) gives its energy identity, without any order assumption on \(R\).

For a finite function module with its counting form, \(\Phi ^{-1}=\Psi \) from Lemma [fdip-001F]; the formula is exactly \(\Psi A^\vee \Phi \) from Proposition [fdip-001H]. Hence it gives the same ordinary \(R\)-linear involution on endomorphisms. The proof above extends that direct construction to different carriers.

The Hom convention is [stacks2017stacks, Section 10.10], Tag 0581.

TauCeti's Module.dualEndAlgEquiv expresses reversed dual composition for reflexive modules. It concerns dual endomorphisms, before the chosen symmetric identifications used here. Mathlib's LinearMap.BilinForm.toDual constructs a duality from a nondegenerate form in the finite-dimensional field case. Here the dualities are explicitly assumed over \(R\); the argument above proves the resulting adjoint laws. A nontrivial coefficient involution and Hermitian forms require different scalar conventions.