On the Decomposition of the Laplacian on Metric Graphs

Abstract

We study the Laplacian on family preserving metric graphs. These are graphs that have a certain symmetry that, as we show, allows for a decomposition into a direct sum of one-dimensional operators whose properties are explicitly related to the structure of the graph. Such decompositions have been extremely useful in the study of Schrödinger operators on metric trees. We show that the tree structure is not essential, and moreover, obtain a direct and simple correspondence between such decompositions in the discrete and continuum case.

Appendix: Proof of Proposition 2.12

Proof of Proposition 2.12

We say that $n\in {\mathbb {N}}$ is “bad” if the number of edges of generation n is greater than the number of vertices of generation n and of the number of vertices of generation $n+1$. For such n, and $e=(i(e),t(e))$ such that $\mathrm{gen}(e)=n$, we add a vertex w, the edges $e_{1}=(i(e),w)$ and $e_{2}=(w,t(e))$ such that $l(e_{i})=\frac{l(e)}{2}$, and remove e from $E(\Gamma )$. In simple words, we divide the edge into two edges with equal length. Note that in the resulting graph, the numbers $n,\,n+1$ are not bad, because for every new edge e, there is a unique vertex $w_{e}$ which was added with e. In addition, none of the other generations became bad, because we did not remove any vertices from $V(\Gamma )$. Thus, if we follow this procedure for every bad n, the resulting graph will be locally balanced. Denote that graph by ${\overline{\Gamma }}$, and let ${\overline{\Delta }}=\Delta _{{\overline{\Gamma }}}$. Note that as measure spaces, there is no difference between $\Gamma$ and ${\overline{\Gamma }}$. The Kirchhoff boundary conditions assure us that the transformation $T{:}\,D(\Delta )\rightarrow D({\overline{\Delta }})$ defined by $T(\varphi )=\varphi$ is unitary, and maps $D(\Delta )$ bijectively onto $D({\overline{\Delta }}).$

It is left to prove that ${\overline{\Gamma }}$ is family preserving. Denote by $\overline{S_{n}}$ the n’th sphere in ${\overline{\Gamma }}$. Let $v,u\in \overline{S_{n}}$ such that v and u are forward neighbors. Consider the following cases:

• (i) $\overline{S_{n}}$ is not a part of $\Gamma$, meaning we added that sphere during the above process. In this case, v and u were added in the middle of the edges $(w_{1},z)$, $(w_{2},z)$ which means that $w_{1}$ and $w_{2}$ are forward neighbors in $\Gamma$. That implies that there exists a rooted graph automorphism $\tau {:}\,V(\Gamma )\rightarrow V(\Gamma )$ such that $\tau (w_{1})=\tau (w_{2})$, and $\forall k\ge 0\,\tau |_{S_{n+k}}= Id$. We define ${\overline{\tau }}{:}\,V({\overline{\Gamma }})\rightarrow V({\overline{\Gamma }})$ by ${\overline{\tau }}|_{V(\Gamma )}=\tau$, and if x was added in the middle of the edge $(v_{1},v_{2})$, ${\overline{\tau }}(x)$ will be the vertex that was added in the middle of $(\tau (v_{1}),\tau (v_{2}))$ (such vertex exists because of spherical symmetry). Now, we have that ${\overline{\tau }}(v)=u$, and $\forall k\ge 0\,{\overline{\tau }}|_{\overline{S_{n+k}}}= Id$, as required.

• (ii) $\overline{S_{n}}$ is a part of $\Gamma$. In that case, $\overline{S_{n+1}}$ is also a part of $\Gamma$, because an added vertex cannot connect vertices from the preceding sphere, so we can define ${\overline{\tau }}$ exactly as in case (i), and get the desired automorphism.

The proof for backward neighbors is exactly the same. $\square$