Theorem. Four-cycles determine the fourth-power trace [fdip-00A7]
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  • November 3, 2026

Let \(G\) be a finite undirected simple graph, with no loops and no multiple edges, and assume \(G\) is triangle-free. Let each undirected edge \(e\) receive one orientation, from a source \(s(e)\) to a head \(h(e)\), and a transport \(U_e\in U(1)\). The reverse orientation carries \(U_e^{-1}\). For unit weights and vanishing mass, the connection background supplies the covariant Laplacian \(L_U\) on complex functions of the vertices. Its matrix in the standard basis is \[L_{h(e),s(e)}=-U_e,\qquad L_{s(e),h(e)}=-U_e^{-1},\qquad L_{vv}=\deg (v),\] and every other entry is zero. The same formula with every transport equal to \(1\) is written \(L_1\). Both operators are Hermitian, so the two traces below are real.

Write \(A\) for the off-diagonal part, so \(L_U=D-A\) with \(D\) the diagonal matrix of degrees. Thus \(A_{h(e),s(e)}=U_e\). For a 4-cycle \(c\), choose either orientation and let \(e^{i\varphi _c}\) be the product of the 4 transports along it. The opposite orientation gives the complex conjugate, so \(\cos \varphi _c\) does not depend on the choice.

The gauge-dependent part of the fourth-power trace is 8 times the sum of \(\cos \varphi _c-1\) over the undirected 4-cycles of \(G\), each taken once: \[\operatorname {Tr}L_U^4-\operatorname {Tr}L_1^4 =8\sum _c(\cos \varphi _c-1).\]

Proof. 

  • November 3, 2026

Expand the trace as a sum over quadruples of vertices. A factor \(L_{xy}\) is a diagonal degree when \(x=y\), and an off-diagonal transport when the two indices are the ends of an edge. The degree matrix does not depend on the connection. It is enough to show that every summand with fewer than 4 off-diagonal factors takes the same value at \(U\) and at the trivial connection, and then to evaluate the four-transport walks.

One off-diagonal factor cannot occur. The other three factors are diagonal, so they force all 4 vertices to agree, and \(A\) has zero diagonal. Three off-diagonal factors leave one stationary vertex and three steps. Those three steps would be a closed walk of length 3. No such walk exists: a simple graph has no loop that could close in one step, two vertices determine only one edge and an odd number of crossings of that edge ends at the other vertex, and three vertices of a closed walk would form a triangle. Thus every contribution with one or three off-diagonal factors vanishes for every connection.

Two off-diagonal factors are transports across one edge and back across the same edge. The other two factors are diagonal, so each equates its two indices. Rotating the 4 positions, the two moving steps are either consecutive or separated by a stay. In the consecutive case the walk leaves a vertex and returns on the next step. In the separated case it makes the same return after the stay. Either way the two steps are inverses, and a simple graph has only one edge between those vertices. The product of a transport with its inverse is \(1\), for every value in \(U(1)\). These summands therefore agree with the trivial connection.

The remaining sum is \(\operatorname {Tr}A_U^4-\operatorname {Tr}A_1^4\). Each of the 4 minus signs in \(L=D-A\) contributes a factor \(-1\), and 4 of them multiply to \(1\), so this is also the difference of the four-transport parts of \(\operatorname {Tr}L^4\). A summand is the product of the entries of \(A\) around a closed walk of length 4. If that walk is not a simple cycle, then it visits at most three vertices. The edges it actually uses form a connected subgraph of a triangle-free simple graph on those vertices, hence a tree. On a tree every closed walk has transport product \(1\): the first time the walk repeats a vertex, the two steps just taken are inverses, because any longer return would be a simple cycle, and deleting that immediate reversal multiplies the product by \(1\) and shortens the walk. By induction the product equals \(1\), which is also its value at the trivial connection.

A simple closed walk of length 4 is a 4-cycle. U(1) transports commute, so the product depends on the orientation but not on the starting vertex. The trace meets each undirected 4-cycle in 4 starting vertices and two orientations. One orientation contributes \(e^{i\varphi _c}\) at each start, and the other contributes the conjugate. The 8 products sum to \(8\cos \varphi _c\). At the trivial connection they sum to \(8\). Subtracting these contributions gives the displayed identity.

Now let both periods \(N_1\) and \(N_2\) be integers at least \(4\), and let the vertices be residue classes in \(\mathbb Z/N_1\mathbb Z\times \mathbb Z/N_2\mathbb Z\), with an undirected edge from each vertex to its 4 axis neighbors. These neighbors are distinct, so the graph is simple. It is also triangle-free. A closed walk of length 3 would need net displacement zero in each coordinate. Each coordinate that returns to its start uses an even number of steps, so the whole walk would have even length, except if some coordinate is displaced by a nonzero multiple of its period. Three steps cannot make a displacement of size at least \(4\). The trace identity therefore applies.

A 4-cycle on this torus is 4 axis steps through 4 distinct vertices whose vector sum is zero in the torus. Write \(r\), \(l\), \(u\) and \(d\) for the numbers of steps in the 4 axis directions. If both periods are at least \(5\), each net displacement has absolute value at most \(4\) and is divisible by a period, so both nets vanish. Then \(r=l\), \(u=d\) and \(r+u=2\). Two steps out and back along one axis visit at most three vertices. The remaining case is one step in each direction. Any order in which a direction is followed immediately by its opposite repeats a vertex. The only remaining orders are the two cyclic orders around a unit square. Thus every 4-cycle is a plaquette, and \[\operatorname {Tr}L_U^4-\operatorname {Tr}L_1^4 =-8\sum _p(1-\cos \varphi _p),\] the sum running once over plaquettes.

If instead a period equals \(4\), 4 steps along that axis close after one winding and visit 4 distinct vertices. There is one such undirected cycle for each position in the transverse direction. A net displacement of size \(4\) uses all 4 steps, so these windings do not mix the two axes. They are 4-cycles, and they are not plaquettes. The same identity includes them, each with its own cosine, on the same footing as the plaquettes. For a holonomy in \(U(1)\), the sum \[-8\sum _p(1-\cos \varphi _p)\] is the Wilson face sum at coupling \(8\). The Wilson action is that shape of face sum for a finite group, with each plaquette weighted by \(1-\operatorname {Re}\chi _\rho (U_p)/d\). The trace identity has the same face sum, up to sign and coupling, precisely when every 4-cycle is a plaquette. The finite-group hypothesis of that note remains in force. A period equal to \(4\) adds winding cycles that a sum over plaquettes does not contain.

[marcolli2013gaugenetworks, Theorem 28] computes a different trace. On a four-dimensional square lattice, with cutoff inversely proportional to the spacing and with \(f(x)=x^4\), the spectral action equals minus one fourth of the two orientation traces around each plaquette, together with a constant and the Higgs terms displayed there.

[marcolli2013gaugenetworks, Proposition 29] states the corresponding continuum functional. This note does not reproduce those calculations. The operator here is the scalar Laplacian of the connection background, not their lattice Dirac operator.

The determinant action is \(\kappa \log \det Q_U\), where \(Q_U=L_U+m^2I\) and \(m>0\). For positive matrices this is \(\kappa \operatorname {Tr}\log Q_U\), a spectral action with \(f=\log \) applied to the massive precision rather than to \(L_U\). On a square torus with both periods at least 6, the induced plaquette coefficient isolates a length-4 plaquette sum in the large-mass expansion of that logarithm, for a real scalar and the sign representation. The logarithm is not the fourth-power trace proved above.

The comparison of these spectral actions remains speculative. Nothing proved here identifies \(\operatorname {Tr}\log Q_U\) with \(\operatorname {Tr}L_U^4\), or either finite trace with a continuum limit. The continuum statement in [marcolli2013gaugenetworks, Proposition 29] is their computation, and it is not a step in the identity for 4-cycles.