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.