Pairings and discrete differences [fdip-0016]
✍️sourceAGENTDRAFTED
- October 20, 2026
Pairings and discrete differences [fdip-0016]
✍️sourceAGENTDRAFTED
- October 20, 2026
How can an edge difference return to a vertex field, and when does its energy detect a constant field? The finite field modules and counting duality give a concrete starting point. Chosen perfect symmetric pairings extend the adjoint construction to different modules over a commutative ring. Applied to incidence, they give weighted energy and the cyclic kinetic factorization.
The component-kernel theorem identifies the additional role of positive real weights; a characteristic-two example shows why perfection alone is insufficient. An invertible precision then gives a unique response to each source, before any probability interpretation.
Definition 1. Chosen perfect symmetric pairings [fdip-001O]AGENTDRAFTED
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Definition 1. Chosen perfect symmetric pairings [fdip-001O]AGENTDRAFTED
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The perfect counting duality represents every linear functional by a unique field. To make the same construction on another carrier, let \(R\) be a commutative unital ring and \(M\) a unital \(R\)-module. Write \(M^\vee =\operatorname {Hom}_R(M,R)\) for its dual module. Every linear map here has coefficient map \(\operatorname {id}_R\), as in Remark [ca-001G].
A chosen perfect symmetric pairing on \(M\) is a
bilinear form \(b_M:M\times M\to R\) such that
\(b_M(x,y)=b_M(y,x)\) and the linear map
\[\Phi _M:M\longrightarrow M^\vee ,\qquad
\Phi _M(y)(x)=b_M(x,y)\]
is an isomorphism. Thus each functional \(\ell \)
has exactly one representative \(\Phi _M^{-1}\ell \), characterized by
\(b_M(x,\Phi _M^{-1}\ell )=\ell (x)\) for every \(x\). Bilinearity makes
\(\Phi _M\) linear. Symmetry identifies the other currying map
\(y\mapsto (x\mapsto b_M(y,x))\) with \(\Phi _M\), so both are
isomorphisms. This agrees with the two-sided perfect-pairing convention
in Mathlib's LinearMap.IsPerfPair. The ordinary Hom convention
is [stacks2017stacks, Section 10.10], Tag 0581.
Injectivity alone says that \(b_M(x,y)=0\) for every \(x\) forces \(y=0\); symmetry gives the analogous assertion in the other slot. It does not give every functional a representative; see a zero-radical pairing that is not perfect.
No finiteness, basis or projectivity hypothesis is imposed on \(M\); the specified duality is the assumption. The counting form on every finite function module satisfies it by the explicit inverse in Lemma [fdip-001F]. The zero module has its unique perfect pairing. If \(R\) is the zero ring, every unital module is zero since \(x=1x=0x=0\), so this case is included. Perfection does not assert positivity or determine a bilinear form from its diagonal values.
Example 2. A zero-radical pairing that is not perfect [fdip-001S]AGENTDRAFTED
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Example 2. A zero-radical pairing that is not perfect [fdip-001S]AGENTDRAFTED
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On the free rank-one \(\mathbb Z\)-module \(M=\mathbb Z\), the symmetric bilinear form \(b(x,y)=2xy\) separates vectors in both slots but is not a perfect pairing. Its currying map is \[\Phi :\mathbb Z\longrightarrow \operatorname {Hom}_{\mathbb Z} (\mathbb Z,\mathbb Z),\qquad \Phi (y)(x)=2xy.\]
If \(b(x,y)=0\) for every integer \(x\), testing against \(x=1\) gives \(2y=0\), hence \(y=0\). Symmetry gives the same conclusion in the other slot, so both radicals are zero and \(\Phi \) is injective. Every integer-linear functional \(\ell \) on \(\mathbb Z\) satisfies \(\ell (x)=x\ell (1)\), so it is determined by its value at \(1\). But \(\Phi (y)(1)=2y\) is even for every \(y\). Thus \(\Phi \) misses the linear functional \(x\mapsto x\), whose value at \(1\) is \(1\), and is not surjective. Freeness of \(M\) does not repair this failure of the chosen form.
The Hom convention is [stacks2017stacks, Section 10.10], Tag 0581; the calculation above proves the failure for this chosen form. For the perfect counting form and its explicit inverse, see Lemma [fdip-001F].
Proposition 3. Adjoints from chosen dualities [fdip-001P]AGENTDRAFTED
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Proposition 3. Adjoints from chosen dualities [fdip-001P]AGENTDRAFTED
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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.
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Proof.
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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.
Definition 4. Weighted graph differentials and adjoints [fdip-0007]AGENTDRAFTED
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Definition 4. Weighted graph differentials and adjoints [fdip-0007]AGENTDRAFTED
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Let \(R\) be a commutative unital ring, \(V,E\) finite sets and \(s,h:E\to V\) endpoint maps. The ring-valued incidence map is \(d_R:H_V\to H_E\), where \(H_V=\mathcal F_R(V)\), \(H_E=\mathcal F_R(E)\) and \(d_Rf(e)=f(h(e))-f(s(e))\). Equip both modules with the symmetric counting forms \[b_V(f,g)=\sum _v f(v)g(v),\qquad b_E(a,c)=\sum _e a(e)c(e).\] Their perfect counting dualities give the counting adjoint \(d_R^\dagger :H_E\to H_V\) by Proposition 3. It is characterized by \(b_E(d_Rf,a)=b_V(f,d_R^\dagger a)\).
The return map can be read directly from the endpoints: \[(d_R^\dagger a)(v) =\sum _{h(e)=v}a(e)-\sum _{s(e)=v}a(e).\]
Proof.
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Proof.
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Expanding and regrouping finite sums gives \[\begin {aligned} b_E(d_Rf,a) &=\sum _e\bigl (f(h(e))-f(s(e))\bigr )a(e)\\ &=\sum _v f(v)\left (\sum _{h(e)=v}a(e)-\sum _{s(e)=v}a(e)\right ). \end {aligned}\] This is the characterizing identity, so uniqueness identifies the map. Each loop occurs once in both sums and cancels; parallel edges remain separate summands. If \(V\) is empty, existence of \(s:E\to V\) forces \(E\) empty; the formula gives the unique map of zero modules.
For arbitrary weights \(w:E\to R\), let \(W:H_E\to H_E\) multiply pointwise by \(w\): \((Wa)(e)=w(e)a(e)\). The weighted graph Laplacian is \[L=d_R^\dagger Wd_R:H_V\longrightarrow H_V.\] Commutativity makes \(W\) self-adjoint, since \(b_E(Wa,c)=\sum _e w(e)a(e)c(e)=b_E(a,Wc)\). The weighted edge form \[b_{E,w}(a,c)=b_E(a,Wc)=\sum _e w(e)a(e)c(e)\] is symmetric bilinear. It is perfect if and only if every \(w(e)\) is a unit, meaning an invertible element of \(R\).
Proof.
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Proof.
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Its duality map is \(\Phi _EW\). Since \(\Phi _E\) is an isomorphism, \(\Phi _EW\) is an isomorphism exactly when \(W\) is. If every weight is a unit, multiplication by \(w(e)^{-1}\) gives both inverse identities pointwise. Conversely, if \(WJ=I\), evaluate \(WJ\delta _e=\delta _e\) at \(e\): \[w(e)(J\delta _e)(e)=1.\] Commutativity supplies the other inverse identity for this scalar, so \(w(e)\) is a unit. For \(E=\varnothing \), the unit condition is vacuous and \(W\) is the unique isomorphism of the zero module. In the zero ring, \(0=1\) is itself a unit, consistent with the same proof.
Even when some weights are nonunits, the represented return map \(d_R^\dagger W\) exists and satisfies \[b_{E,w}(d_Rf,a)=b_V(f,d_R^\dagger Wa).\] Its existence does not make the weighted edge form perfect. Thus it does not by itself give a two-sided adjoint involution for that form. All these ring identities are direct deductions from the chosen counting dualities. Their positive-real incidence convention is [spielman2009effective, Section 2.1, p. 4].
Orientation reversal leaves \(L\) unchanged. Choose which edges to reverse, and put \(\varepsilon (e)=-1\) on those edges and \(1\) on the others, interpreted in \(R\). The sign map \(Ja(e)=\varepsilon (e)a(e)\) satisfies \(J^\dagger =J\), \(J^2=I\), \(JW=WJ\), and the reversed incidence map is \(d'_R=Jd_R\). Therefore \[(d'_R)^\dagger Wd'_R=d_R^\dagger JWJd_R=d_R^\dagger Wd_R.\] The corresponding edge field is \(a'=Ja\), and \(b_{E,w}(Ja,Jc)=b_{E,w}(a,c)\). The proof includes characteristic two, loops and parallel edges. Endpoints and weights are otherwise fixed.
For the real incidence adapter, start with the incidence map \(D:\mathbb R^V\to \mathbb R^E\) and positive weights \(w(e)>0\). The counting forms are the counting inner products, and write \(D^*=d_{\mathbb R}^\dagger \). The edge form becomes the weighted inner product \[\langle a,c\rangle _{E,w}=\sum _e w(e)a(e)c(e).\] It is symmetric bilinear, and its self-pairing is a finite sum of nonnegative terms. Strictly positive weights make that sum zero only at the zero edge field. Keeping the vertex pairing counting, the weighted adjoint is \(D_w^\dagger =D^*W\), since \[\langle Df,a\rangle _{E,w} =\langle Df,Wa\rangle _E=\langle f,D^*Wa\rangle _V.\] Hence \(L=D^*WD=D_w^\dagger D\); \(D^*D\) is the unit-weight case.
For the coordinate recovery, use the ordered-value isomorphisms \(c_V,c_E\) and incidence matrix \(B=c_EDc_V^{-1}\). The endpoint return formula gives \(c_VD^*c_E^{-1}=B^T\). The multiplication map \(W\) has diagonal array \(\operatorname {diag}(w(e))\); writing the same letter for that array yields \([D_w^\dagger ]=B^TW\) and \([L]=B^TWB\). For a single edge \(0\to 1\) of weight \(c>0\), the two adjoints send an edge value \(a\) to \((-a,a)\) and \((-ca,ca)\), respectively. Under orientation reversal the sign array \(S\) gives \(B'=SB\), \(S^T=S\), \(S^2=I\), \(SWS=W\), and \((B')^TWB'=B^TWB\). A loop has a zero incidence row before and after reversal.
Mathlib's LinearMap.toMatrix_adjoint identifies adjoint
matrices with conjugate transposes in orthonormal bases. Over the reals
this is transpose. The coordinate edge basis is orthonormal for counting,
but not generally for the weighted pairing; the return-map calculation
above accounts for the extra \(W\).
Example 5. The weighted four-cycle Laplacian [fdip-004B]AGENTDRAFTED
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Example 5. The weighted four-cycle Laplacian [fdip-004B]AGENTDRAFTED
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Use the alternating four-cycle, with vertices \(0,1,2,3\), edges \(e_i:i\to i+1\), and positive real weights \((1,t,1,t)\). For a field \(f=(a,b,c,d)\), the incidence values are \(Df=(b-a,c-b,d-c,a-d)\). The counting adjoint sends an edge field to \[D^*(u_0,u_1,u_2,u_3) =(u_3-u_0,u_0-u_1,u_1-u_2,u_2-u_3).\] First multiply by \(W_t=\operatorname {diag}(1,t,1,t)\), then apply \(D^*\). The resulting \(L_t=D^*W_tD\) has coordinate matrix \[L_t=\begin {pmatrix} 1+t&-1&0&-t\\ -1&1+t&-t&0\\ 0&-t&1+t&-1\\ -t&0&-1&1+t \end {pmatrix}.\]
For example, its first coordinate is \(t(a-d)-(b-a)=(1+t)a-b-td\); the other coordinates follow from the same return formula. Thus \[L_t\delta _0=(1+t,-1,0,-t),\qquad L_t\mathbf 1=0,\] and the counting energy is \[\langle f,L_tf\rangle =(b-a)^2+t(c-b)^2+(d-c)^2+t(a-d)^2.\] The counting adjoint \(D^*\) does not depend on \(t\). The weighted return map \(D^*W_t\), and hence \(L_t\), do. When \(t=1\) the two return maps coincide; otherwise the factors \(t\) must be retained.
This calculation substitutes the four-cycle data into Definition 4; its incidence convention follows [spielman2009effective, Section 2.1, p. 4]. The cyclic energy factorization gives the corresponding operator identity.
Lemma 6. Counting adjoints on an indexed cycle [fdip-0014]AGENTDRAFTED
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Lemma 6. Counting adjoints on an indexed cycle [fdip-0014]AGENTDRAFTED
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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.
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Proof.
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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.
Theorem 7. Graph energy and the constant modes [fdip-0008]AGENTDRAFTED
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Theorem 7. Graph energy and the constant modes [fdip-0008]AGENTDRAFTED
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Let \(R\) be a commutative unital ring, \(V,E,s,h\) finite incidence data and \(w:E\to R\) arbitrary weights. With the counting forms and weighted graph Laplacian \(L=d_R^\dagger Wd_R\), one has \[L^\dagger =L,\qquad b_V(f,Lg)=\sum _e w(e)(d_Rf)(e)(d_Rg)(e).\] The kernel of \(d_R\) consists exactly of fields constant on each connected component of the underlying undirected graph, and \(\ker d_R\subseteq \ker L\). Two vertices are in the same component when a finite path joins them, with edge orientation ignored.
Proof.
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Proof.
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The weight map is self-adjoint. The adjoint laws give \(L^\dagger =d_R^\dagger W^\dagger d_R=L\), and the defining identity gives \[\begin {aligned} b_V(f,Lg)&=b_E(d_Rf,Wd_Rg)\\ &=\sum _e w(e)(d_Rf)(e)(d_Rg)(e). \end {aligned}\] In particular, \(b_V(f,Lf)=\sum _e w(e)(d_Rf)(e)^2\). Each edge occurs once and no factor \(1/2\) is used. The equality \(d_Rf=0\) says that endpoint values agree along every edge, hence along every path. Conversely, a componentwise constant field has zero difference on each edge. This identifies \(\ker d_R\) with the function module on the components. Such a field also satisfies \(Lf=d_R^\dagger W0=0\).
The energy is unchanged by edge-orientation reversal: each reversed edge negates both factors in its summand. Equivalently, the orientation-invariant return composite gives the same Laplacian.
Perfection and unit weights do not give the reverse kernel inclusion. Over \(\mathbb F_2\), take two vertices \(0,1\) and two distinct edges, both oriented \(0\to 1\), both of weight \(1\). The counting pairings are perfect and \(W=I\) is invertible. Nevertheless, \[\begin {aligned} d_R(a,b)&=(b-a,b-a),\\ d_R^\dagger (u_1,u_2)&=(-u_1-u_2,u_1+u_2),\\ L(a,b)&=(0,0). \end {aligned}\] For \(\delta _1=(0,1)\), one has \(d_R\delta _1=(1,1)\ne 0\), but its energy is \(1+1=0\). Hence \(\ker L=\mathbb F_2^2\) strictly contains \(\ker d_R=\{(a,a):a\in \mathbb F_2\}\). This failure comes from the absence of real positivity, not from nonunit weights or nonperfect forms.
Now specialize to \(R=\mathbb R\) and \(w(e)>0\). Write \(D=d_{\mathbb R}\). The Laplacian is symmetric and positive semidefinite, and \[\ker L=\ker D,\qquad \dim \ker L=k,\] where \(k\) is the number of connected components. The energy convention and connected-graph argument appear in [spielman2009effective, Section 2.1, pp. 4-5]; the ring identity above and the disconnected, loop and parallel-edge extensions are proved directly here.
Proof.
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Proof.
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The energy identity specializes to \[\langle f,Lf\rangle _V=\sum _e w(e)(Df)(e)^2\ge 0.\] If \(Lf=0\), this finite sum of nonnegative real terms is zero. Every summand is therefore zero; strict positivity of \(w(e)\) gives \(Df(e)=0\). The reverse inclusion was proved over \(R\). Indicators of the nonempty components span the componentwise constant fields. In a linear relation between them, evaluation at a vertex of any one component makes its coefficient zero. Thus they are a basis, proving the dimension statement. An action defined as half this quadratic form uses a separate convention.
If \(V\) is nonempty and connected, the kernel is the one-dimensional space spanned by the constant field \(\mathbf 1_V\). Its positive unit mode is \[|V|^{-1/2}\mathbf 1_V.\] An isolated vertex contributes one component and one kernel basis vector. If \(V\) is empty, the maps \(s,h:E\to V\) force \(E\) to be empty too: both the component count and kernel dimension are zero. Loops contribute zero energy, while distinct parallel edges each contribute their own positive summand.
For the alternating four-cycle, put \(u=1+t\). Direct multiplication gives \[L_t=\begin {pmatrix} u&-1&0&-t\\ -1&u&-t&0\\ 0&-t&u&-1\\ -t&0&-1&u \end {pmatrix}.\] For \(f=(a,b,c,d)\), the energy is \[\begin {aligned} f^TL_tf={}&(b-a)^2+t(c-b)^2\\ &+(d-c)^2+t(a-d)^2. \end {aligned}\] Every row of \(L_t\) sums to zero, and for \(t>0\) zero energy forces all 4 values equal. The comparison with the positive action identifies how this edge energy combines with a mass term.
The hypotheses have concrete consequences. With vertices \(0,1,2,3\) and just the unit-weight edges \(0\to 1\) and \(2\to 3\), the matrix has two diagonal blocks \(\bigl (\begin {smallmatrix}1&-1\\-1&1\end {smallmatrix}\bigr )\). Its kernel is \(\{(a,a,b,b):a,b\in \mathbb R\}\); connectivity is needed for a single constant mode.
On two vertices with one edge \(0\to 1\) of weight zero, the underlying graph is connected but \(L=0\), so its kernel is all of \(\mathbb R^2\). For nonnegative weights the same proof identifies the kernel using only the positive-weight edges. A zero weight also makes the proposed edge pairing degenerate. If instead this edge has weight \(-1\), then \[f^TLf=-(f(1)-f(0))^2,\] which is negative at \(f=(0,1)\). Positivity fails, and the edge form is not an inner product. This example is negative semidefinite; two disjoint edges with weights \(1\) and \(-1\) give an indefinite form, since fields supported on the separate edges have energies of opposite signs.
For unweighted simple graphs, Mathlib's
SimpleGraph.posSemidef_lapMatrix proves positivity of its
Laplacian. That model has no separate weighted edge set; its energy
formula sums both adjacency directions and divides by two. The
edge-once sum above also handles loops and parallel edges, and is proved
directly here.
Lemma 8. Inverse response for an invertible precision [fdip-001Q]AGENTDRAFTED
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Lemma 8. Inverse response for an invertible precision [fdip-001Q]AGENTDRAFTED
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Let \(M\) be a unital module over a commutative unital ring \(R\), with a chosen perfect symmetric pairing \(b\). Let \(L:M\to M\) be self-adjoint, and let \(\mu \in R\). The endomorphism \[Q=L+\mu I_M:M\longrightarrow M\] is self-adjoint. Assume additionally that \(Q\) is an isomorphism, and put \(G=Q^{-1}\). Then \(G^\dagger =G\), and for every source \(j\in M\) the equation \[Qf=j\] has the unique solution \(f=Gj\).
This is a direct consequence of the adjoint and inverse laws and the two-sided inverse convention of [stacks2017stacks, Definition 4.2.4], Tag 0017.
Proof.
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Proof.
- October 20, 2026
Linearity of the adjoint and \(I^\dagger =I\) give \(Q^\dagger =L^\dagger +\mu I=Q\). The inverse-adjoint law gives \(G^\dagger =(Q^\dagger )^{-1}=G\). Since \(QG=I\), \(Q(Gj)=j\). Conversely, if \(Qf=j\), then \(GQ=I\) gives \(f=Gj\).
For \(M=\mathcal F_R(V)\) on a finite set with its perfect counting duality, define the point kernel only after this inverse map exists: \[G(v,u)=(G\delta _u)(v).\] The indicator expansion and linearity give \[(Gf)(v)=\sum _{u\in V}G(v,u)f(u).\] Moreover, \[\begin {aligned} G(v,u)&=b(\delta _v,G\delta _u)\\ &=b(G\delta _v,\delta _u)=G(u,v). \end {aligned}\] Thus the symmetric array presents the already defined inverse. On the zero module the unique endomorphism is the identity and is invertible; for an empty carrier the kernel and these sums are empty.
A unit mass coefficient does not ensure invertibility over a ring. Over \(\mathbb Z\), take one unit-weight edge \(0\to 1\), the Laplacian \(L(a,b)=(a-b,b-a)\), and \(\mu =1\). Then \[Q(a,b)=(2a-b,2b-a).\] If \(Q(a,b)=(1,0)\), the second equation gives \(a=2b\), and the first gives \(3b=1\), impossible in \(\mathbb Z\). Thus \(Q\) is not onto, even though the mass and edge weight are units. The real massive Green operator acquires invertibility from positive energy and finite dimension. No positivity, spectrum or probability law is asserted over an arbitrary \(R\).