Theorem. The edge-source triple and inner fluctuations [fdip-0096]
✍️source

  • November 3, 2026

Let \(G\) be a finite oriented graph, with vertex set \(V\) and edge set \(E\). Write \(s(e)\) and \(h(e)\) for the source and head of an edge. No edge is a loop. Let the real Hilbert space have the counting inner product and the orthonormal basis indexed by \(V\) and by \(E\). Define \[(d_E\xi )(e)=\xi (h(e))-\xi (s(e)).\] Let \(D\) be the self-adjoint operator whose vertex-to-edge block is \(d_E\) and whose edge-to-vertex block is the adjoint \(d_E^{\dagger }\). A real function \(f\) on \(V\) acts by multiplication by \(f(v)\) on vertex \(v\) and by multiplication by \(f(s(e))\) on edge \(e\).

The operator norm of the commutator is read from the jumps on the edges that enter each vertex. For every vertex \(v\), \[\bigl |[D,f]\bigr | =\max _v\Bigl (\sum _{h(e)=v}\bigl (f(h(e))-f(s(e))\bigr )^{2}\Bigr )^{1/2}.\] If every vertex has at most one incoming edge, this norm is the maximum of \(|f(h(e))-f(s(e))|\) over the edges. On each connected component of the underlying undirected graph, the supremum \[\sup \bigl \{|f(j)-f(i)|:\bigl |[D,f]\bigr |\le 1\bigr \}\] equals the least number of edges joining the two vertices. Between distinct components that supremum is infinite. A directed cycle has one incoming edge at each vertex, so opposite vertices of a directed \(6\)-cycle are at distance \(3\).

Proof. 

  • November 3, 2026

Write \(\delta _e=f(h(e))-f(s(e))\). The edge component of \([D,f]\) applied to \((\xi ,\psi )\) is \(\delta _e\xi (h(e))\), and the vertex component is the negative adjoint of that edge map applied to \(\psi \). The operator norm is therefore the norm of the edge map \(C\xi (e)=\delta _e\xi (h(e))\). Its square length is \[\|C\xi \|^{2}=\sum _v\xi (v)^{2}\sum _{h(e)=v}\delta _e^{2}.\] The vertex basis is orthonormal, so the second sum is the eigenvalue at \(v\). The largest square root is the displayed maximum.

If each vertex has at most one incoming edge, the inner sum is a single square, and the norm is the largest absolute jump. The constraint then says that \(|f(h)-f(s)|\le 1\) on every edge. The graph-distance function from \(i\) changes by at most \(1\) along an edge, so it is allowed, and its value at \(j\) is the graph distance. Any allowed function changes by at most \(1\) along each edge, so it cannot exceed that distance. If no path joins \(i\) to \(j\), adding a constant on the component of \(j\) does not change the commutator, and the difference is unbounded.

The same norm is stricter where a vertex has two incoming edges. On the grid whose coordinates each run through \(0,1,2\), orient each edge in the positive direction of one axis. The distance from \((0,0)\) to \((1,1)\) is \(1+1/\sqrt {2}\). The Euclidean distance is \(\sqrt {2}\) and the graph distance is \(2\), and the commutator distance lies strictly between them.

Proof. 

  • November 3, 2026

Write \(x=f(1,0)-f(0,0)\), \(y=f(0,1)-f(0,0)\), \(p=f(1,1)-f(1,0)\) and \(q=f(1,1)-f(0,1)\). The vertex \((1,0)\) has only the incoming edge from \((0,0)\), and \((0,1)\) has only the incoming edge from \((0,0)\), so an allowed function satisfies \(x^{2}\le 1\) and \(y^{2}\le 1\). The vertex \((1,1)\) has those two incoming edges, so \(p^{2}+q^{2}\le 1\). The difference \(\delta =f(1,1)-f(0,0)\) equals both \(x+p\) and \(y+q\). Thus \((\delta -x)^{2}+(\delta -y)^{2}\le 1\) for some \(x,y\) in the interval from \(-1\) to \(1\). If \(\delta >1+1/\sqrt {2}\), then both \(\delta -x\) and \(\delta -y\) exceed \(1/\sqrt {2}\), and the sum of squares exceeds \(1\). Hence \(\delta \le 1+1/\sqrt {2}\).

The bound is attained by \(f(0,0)=0\), \(f(1,0)=f(0,1)=f(2,0)=1\), \(f(0,2)=1\), and the value \(1+1/\sqrt {2}\) at the other 4 vertices. The incoming square sums are \(1\) at \((1,0)\), \((0,1)\) and \((1,1)\), they are \(1/2\) at \((2,1)\) and \((1,2)\), and they vanish elsewhere. The norm is \(1\), and the difference of the endpoint values is \(1+1/\sqrt {2}\). It is larger than \(\sqrt {2}\) because \(1>1/\sqrt {2}\), and it is smaller than \(2\) because \(1/\sqrt {2}<1\).

Inner fluctuations fill the head slots of \(D\). For real functions \(a\) and \(b\), the edge component of \(a[D,b]\) applied to a vertex vector \(\xi \) is \[e\mapsto a(s(e))\bigl (b(h(e))-b(s(e))\bigr )\xi (h(e)).\] Suppose that at most one edge runs from any vertex to any other, and that the opposite orientation is absent. For an edge \(e\) from \(s\) to \(h\) and a real scalar \(c\), let \(a\) equal \(c\) at \(s\) and equal \(0\) elsewhere, and let \(b\) equal \(1\) at \(h\) and equal \(0\) elsewhere. The sum \(S=a[D,b]+b[D,a]\) is self-adjoint. Its edge component is \(c\,\xi (h)\) on \(e\) and zero on every other edge. Summing one such operator for each edge gives every real assignment \((c_e)\). Adding that sum to \(D\) replaces the edge formula by \[U_e\xi (h(e))-\xi (s(e)),\qquad U_e=1+c_e.\]

Proof. 

  • November 3, 2026

The displayed edge formula for \(a[D,b]\) is the commutator calculation already used for the norm: the terms in which \(b\) is read at the source cancel, and multiplication by \(a\) is evaluated at the source. With the two functions just chosen, \(a(s(e'))(b(h(e'))-b(s(e')))\) equals \(c\) on the edge from \(s\) to \(h\) and equals \(0\) on every other edge. The swapped product \(b(s(e'))(a(h(e'))-a(s(e')))\) is nonzero only on an edge from \(h\) to \(s\). No such edge is present.

It remains to check that \(b[D,a]\) is the adjoint of \(a[D,b]\). The edge block of \(a[D,b]\) multiplies the value at \(h\) by \(c\) and places it on \(e\). Its adjoint therefore multiplies the value on \(e\) by \(c\) and places it at \(h\). The vertex component of a commutator \(p[D,q]\), evaluated at a vertex \(v\), is \[-p(v)\sum _{h(e')=v}\bigl (q(h(e'))-q(s(e'))\bigr )\psi (e').\] For \(p=b\) and \(q=a\), the only vertex where \(b\) is nonzero is \(h\). The only incoming edge there whose source carries a nonzero value of \(a\) is \(e\), and on that edge \(a(h)-a(s)=-c\). The leading minus sign turns that term into \(c\,\psi (e)\) at \(h\), and the component is \(0\) at every other vertex. The two blocks agree. For the other pair, the vertex component of \(a[D,b]\) is supported at \(s\), and no edge entering \(s\) starts at \(h\), so that component is \(0\). The edge component of \(b[D,a]\) is \(0\) because the reverse edge is absent. Hence \(S\) is self-adjoint and its edge coefficient on \(e\) is \(c\).

The coefficients \(U_e\) constructed above are real. When \(U_e\ne 0\), \[U_e\xi (h(e))-\xi (s(e)) =U_e\bigl (\xi (h(e))-U_e^{-1}\xi (s(e))\bigr ).\] The second factor is the covariant difference of the scalar connection \(U_e^{-1}\). The two edge maps have the same kernel. This construction does not produce a matrix representation of a nonabelian group.

If every \(U_e\) equals \(\pm 1\), multiplication by \(U_e\) is an orthogonal map \(W\) of the edges, and \(W^{-1}=W\). Let \(d_1\) be the covariant difference and set \(d_2=Wd_1\). Then \(d_2^{\dagger }=d_1^{\dagger }W\). Conjugating the off-diagonal operator of \(d_2\) by the orthogonal map that is the identity on vertices and \(W\) on edges replaces the edge block \(d_2\) by \(Wd_2=d_1\) and replaces \(d_2^{\dagger }\) by \(d_1^{\dagger }\). The two self-adjoint operators are orthogonally equivalent.