Theorem. Graph energy and the constant modes [fdip-0008]
AGENTDRAFTED
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.
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.
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.