The Laplacian spectrum separates the pair

A cycle of $n$ nodes has Laplacian eigenvalues $2-2\cos(2\pi j/n)$. The spectrum of a disjoint union is the union of the spectra.

For the hexagon, $2-2\cos(2\pi j/6)$ for $j=0,\ldots,5$ gives $0,1,3,4,3,1$, that is $\{0,1,1,3,3,4\}$. For one triangle it gives $0,3,3$; for two, $\{0,0,3,3,3,3\}$. The spectra differ, so a network given spectral information can separate the pair that defeated colour refinement in I.14.B03.

The eigenvalue $0$ counts connected components. Since $\vec x^{\mathsf T}\mat L\vec x=\sum_{(u,v)\in E}(x_u-x_v)^2$ by (I.14.4), $\mat L\vec x=\vec 0$ exactly when $\vec x$ is constant across every edge, that is, constant on each component. There is one independent such vector per component: one for the hexagon, two for the triangles.

As positions the eigenvectors have two ambiguities. Each is defined only up to sign, so a network must give the same answer for $\vec v$ and $-\vec v$, or it depends on an arbitrary choice made by the eigensolver. And a repeated eigenvalue has no preferred basis: the two triangles’ eigenvalue $3$ has a four-dimensional eigenspace, and any rotation within it is as valid as any other. Positional features from the Laplacian therefore need architectures invariant to these choices, or they reintroduce exactly the dependence on arbitrary labels that equivariance was built to remove.