aa r X i v : . [ m a t h . C O ] N ov Another Proof of Oscar Rojo’s Theorems
Hao Chen a,b , J¨urgen Jost b, a Ecole Polytechnique, Palaiseau, France b Max Planck Institute for Mathematics in the Sciences, Leipzig, Germany
Abstract
We present here another proof of Oscar Rojo’s theorems about the spectrum ofgraph Laplacian on certain balanced trees, by taking advantage of the symme-try properties of the trees in question, and looking into the eigenfunctions ofLaplacian.
Keywords:
Tree, Graph Laplacian, Graph spectrum
1. Introduction
Oscar Rojo has proved, first for balanced binary trees [1], then extended tobalanced trees such that vertices at the same level l are of the same degree d ( l )[2], that Theorem 1.
The spectrum of the graph Laplacian on such trees, is the unionof the eigenvalues of T j , ≤ j ≤ k , where T k is a tridiagonal k × k matrix T k = p d (2) − · · · p d (2) − d (2) . . . ... . . . . . . . . . ... . . . d ( k − p d ( k )0 · · · p d ( k ) d ( k ) (1) and T j with j < k is the j × j principal submatrix of T k . The multiplicity ofeach eigenvalue of T j as an eigenvalue of the Laplacian matrix is at least thedifference of population between level j et level j + 1 . Please note that the levels are numerated from the leaves’ level in OscarRojo’s papers, i.e. the leaves are at level 1 and the root at level k.He then found similar results [3] for a tree obtained by identifying the rootsof two balanced trees. We call all the trees studied in these papers “symmetrictrees” for a reason we will explain later.We will give here another proof of Oscar Rojo’s theorems, by looking at theeigenfunctions of Laplacian, and using the result of Bıyıkoglu’s study on sign
Email addresses: [email protected] (Hao Chen), [email protected] (J¨urgenJost)
Preprint submitted to Elsevier October 22, 2018 raphs [4]. By our approach, we can see that because of the symmetry prop-erties of the trees in question, the eigenfunctions of Laplacian have a stratifiedstructure, which will be very useful for the proof.The main goal of this paper is to prove the theorem 1. The case studied in[3] will be briefly discussed at the end to show that our approach can be easilyextended.
2. Preparation
Before we start, we shall agree on the notations and the terminology, which,for our convenience, are not the same as in Oscar Rojo’s papers. We shall alsostudy the symmetry of the trees in question to justify the name we gave tothem.In a rooted tree, every edge links a parent to one of its child. The root isthe only vertex without parent, and the leaves are the vertices without child.We denote C v the set of children of vertex v , and label the children by integers1 , . . . , | C v | .The level of a vertex v is defined by its distance to the root, denoted by l v .Therefore, unlike in Oscar Rojo’s papers, for a k -level tree, root will be at level0, and leaves at level k −
1. If there exists a path ( v , ..., v n ) such that v = p , v n = q and v i is the child of v i − for all 1 ≤ i ≤ n , we call p ancestor of q and q descendant of p . A vertex v and all its descendants form the maximal subtreerooted at v , denoted by T v .There is one and only one path from the root to any non-root vertex. Avertex can thus be uniquely identified by listing in order the labels of non-rootvertices on this path. For example, a vertex with identity { , , , } means thatit’s the 1 st child of the 4 th child of the 2 nd child of the 3 rd child of the root. Wecall this list of label “identity” of v and denote I v . The root’s identity is definedas an empty list {} . The length of I v equals its level l v . We denote I iv the i th number in I v .If a vertex v have n children, we denote S v the symmetric group S n actingon C v by relabeling the children of v . The action of S v on the vertex set V T (orby abuse of language, on the tree T ) is defined by its action on C v . That is, if avertex u with label i is a child of v , it will be relabeled by σ ∈ S v to σ ( i ). Labelsof other vertices are not changed. In the words of identity, σ ∈ S v changes I l v u to σ ( I l v u ) if u ∈ T v , and does not touch other numbers in I u or identities of othervertices.The trees studied in the papers of Oscar Rojo are such that vertices at thesame level l have the same number of children c ( l ), i.e. are of the same degree d ( l ) (see figure 1). Let the total number of vertices at level l be n ( l ), we have n (0) = 1 and c ( l ) = d ( l ) − n ( l ) = n ( l − c ( l −
1) = Q l − i =0 c ( i ) for l > S v for all vertices v . Here “invariant” implies that the vertex set V T is invariant. Since vertices are identified by their identities, the invariancemeans that for any vertex, its identity can be found in the vertex set of the treeafter the action of S v . In this sense, the action of S v on a symmetric tree isactually a permutation of the subtrees rooted at the children of v . See figure 2for a tree not invariant under the action of S v .Let F be the set of real-valued functions on the set of vertices V T (or byabuse of language, on the tree T ). If T is symmetric, because of the invariance,2 Pajek
Figure 1: A typical symmetric tree. All the non-root vertices are labeled. we can define the action of S v on F by ( σ ◦ f )( T ) = f ( σ ◦ T ), where f ∈ F and σ ∈ S v .
3. Sign Graph of Trees
Our proof benifits from the studies on sign graphs. The sign graph (strongdiscrete nodal domain) is a discrete version of Courant’s nodal domain. Consider G = ( V, E ) and a real-valued function f on V . A positive (resp. negative) signgraph is a maximal, connected subgraph of G with vertex set V ′ , such that f | V ′ > f | V ′ < M is called generalized Laplacian matrix of graph G = ( V, E ) if M has nonpos-itive off-diagonal entries, and M ( u, v ) < E connecting u and v . Obviously, the Laplacian L studied in Oscar Rojo’s papers L ( u, v ) = deg u if u = v − u, v ) ∈ E Dirichlet Laplacian L Ω on a vertex set Ω, is an operator defined on F Ω ,the set of real-valued functions on Ω. It is defined by L Ω f = ( L ˜ f ) | Ω (3)where ˜ f ∈ F vanishes on V − Ω and equals to f on Ω. It can be regarded asa Laplacian defined on a subgraph with limit condition, and has many proper-ties similar to those of Laplacian. A Dirichlet Laplacian is also a generalizedLaplacian.[5, 6, 7] have proved the following discrete analogues of Courant’s NodalDomain Theorem for generalized Laplacian:3 }{1} {2}{1,2} {} {1}{2}{2,2} Pajek
Figure 2: An example showing a non-symmetric tree which is not invariant under the actionof S root , the identity { , } disappears and { , } appears after relabeling the children of root. Theorem 2.
Let G be a connected graph and A be generalized Laplacian of G ,let the eigenvalues of A be non-decreasingly ordered, and λ k be a eigenvalue ofmultiplicity r , i.e. λ ≤ · · · ≤ λ k − < λ k = · · · = λ k + r − < λ k + r ≤ · · · ≤ λ n (4) Then a λ k -eigenvalue has at most k + r − sign graphs. And, [4] has studied the nodal domain theory of trees. Instead of inequality,we have kind of equality on trees. But we have to study two cases
Theorem 3 (T¨urker Biyikoglu) . Let T be a tree, let A be a generalized Laplacianof T . If f is a λ k -eigenfunction without a vanishing coordinate (vertex where f = 0 ), then λ k is simple and f has exactly k sign graphs. Theorem 4 (T¨urker Biyikoglu) . Let T be a tree, let A be a generalized Laplacianof T . Let λ be an eigenvalue of A all of whose eigenfunctions have vanishingcoordinates. Then eigenfunctions of λ have at least one common vanishing coordinates. Let Z be the set of all common vanishing points, G − Z is then a forest withcomponent T , . . . , T m , Let A , . . . , A m be restriction of A to T , . . . , T m ,then λ is a simple eigenvalue of A , . . . , A m , and A i has a λ -eigenfunctionwithout vanishing coordinates, for i = 1 , . . . , m . Let k , . . . , k m be the positions of λ in the spectra of A , . . . , A m in nondecreasing order. Then the number of sign graphs of an eigenfunctionof λ is at most k + . . . + k m , and there exists an λ -eigenfunction with k + . . . + k m sign graphs. In this theorem, if A is the Laplacian, A i in the second point are in fact theDirichlet Laplacians on T i . 4 . Proof We first prove the following theorem.
Theorem 5.
Consider a symmetric tree T rooted at r . If f is a λ -eigenfunctionof L such that f ( r ) = 0 , then we can find a λ -eigenfunction ˜ f such that l u = l v ⇒ ˜ f ( u ) = ˜ f ( v ) , i.e. ˜ f takes a same value on vertices at the same level. Wecall ˜ f the stratified eigenfunction (see figure 3). Figure 3: An example of stratified structure on the whole tree. The size of vertex indicatesthe absolute value of ˜ f , the color of vertex indicates the sign, gray for negative, black forpositive, white for 0. Proof.
We first consider the case where f has no vanishing coordinate, and provethat f is itself a stratified eigenfunction. By the theorem 3 we know that λ issimple.Assume that l u = l v and f ( u ) = f ( v ). There is a unique path connecting u and v . Choose on this path a vertex p with the lowest level (closest to the root),it is the first common ancestor of u and v , so I l p +1 u and I l p +1 v are the labels ofthe last different ancestors of u and v . Take σ = τ ( I l p +1 u , I l p +1 v ) ∈ S p where τ ( i, j ) is the transposition of i and j , σ ◦ f is then another λ -eigenfunction bysymmetry, but σ ◦ f = f since f ( u ) = f ( v ), which violates the simplicity of λ .So f ( u ) = f ( v ) and f itself is a the stratified eigenfunction. This is also truefor Dirichlet Laplacian.We study then the case with vanishing points. By theorem 4, we know thatall λ -eigenfunctions have at least one common vanishing coordinate. Let v bea common vanishing coordinate, a vertex u of the same level must also be one.If it is not the case, a same argument as above will lead to a contradiction thatthere is another λ -eigenfunction who does not vanish on v . We can now use theterm “vanishing level”. There can be no two consecutive vanishing levels, if ithappens, all the lower levels must vanish, so is the root, which is assumed to benon zero.The common vanishing coordinates divide the tree into components withoutvanishing points. By theorem 4, λ is a simple eigenvalue of Dirichlet Laplacian5n each component. Components between two vanishing levels are identical, sothey have a same eigenfunction (up to a factor) for Dirichelt Laplacian.We can construct ˜ f as following: Choose a vanishing vertex v on whosechildren f takes different values. Symmetrize the children of v by taking f ′ = 1 | C v | ! X σ ∈ S v σ ◦ f (5)If f ′ is a stratified eigenfunction, ˜ f = f ′ and the construction is finished. Ifnot, there must be another vertex whose children are not symmetrized, then let f = f ′ and repeat the procedure.This procedure will end in a finite number of steps because the symmetriza-tion can not be undone and the tree is finite. When all the vertices have theirchildren symmetrized, the result must be stratified, because λ is simple as aneigenvalue of Dirichlet Laplacian on each non-vanishing component.This theorem is also true for Dirichlet Laplacian on a rooted subtree (seefigure 4). Together with the following theorem, we can find all the eigenfunctionsof Laplacian. Figure 4: An example of stratified structure on subtrees.
Theorem 6.
Consider a symmetric tree T with root r . Let v be a non-rootvertex. If λ is an eigenvalue of L T v whose eigenfunctions do not vanish at v ,then λ is also an eigenvalue of L , whose multiplicity is at least n ( l v ) − n ( l v − .Proof. Let u be a non-root vertex of a symmetric tree T and p its parent. Wefirst notice that, if λ is an eigenvalue of Dirichlet Laplacian L| T u , λ is also aneigenvalue of Laplacian on L| T p . Because for a λ -eigenfunction f of DirichletLaplacian L| T u , let i be the label of u , ∀ j = i , the action of τ ( i, j ) ∈ S p on f will help constructing another λ -eigenfunction f ′ as, f ′ = ( f − τ ( i, j ) ◦ f if f ( u ) = 0 τ ( i, j ) ◦ f if f ( u ) = 0 (6)6e see that the multiplicity of λ as an eigenvalue of L| T p is | C p | − f ( u ) = 0, because f ′ constructed by differenct j are independent. If f ( u ) = 0, f itself is also an eigenfunction of L| T p so the multiplicity is | C p | .If f ( v ) = 0, let p be the parent of v . We conclude by a recursive argument,that the multiplicity of λ as an eigenvalue of L = L| T r is( | C p |− l p − Y i =0 c ( i ) = l p Y i =0 c ( i ) − l p − Y i =0 c ( i ) = n ( l p +1) − n ( l p ) = n ( l v ) − n ( l v −
1) (7)It is obvious that a stratified eigenfunction of (Dirichlet) Laplacian can notvanish on the root of (sub)tree, otherwise it will vanish everywhere. So for avertex v of a k -level symmetric tree, we can find k − l v stratified independenteigenfunctions of L| T v (because T v have k − l v levels), and none of them vanishesat v . The eigenfunctions who vanishes at v are independent to these stratifiedeigenfunctions, because of the “-” in the f ′ given in equation 6.Let’s count how many independent eigenfunctions have been found. For avertex v , we find k − l v stratified eigenfunction of L| T v . For each of them, thereare n ( l v ) − n ( l v −
1) independent eigenfunctions of L with the same eigenvalue,who are in fact stratified on maximal subtrees rooted on vertices at level l v . Asa special case, the root have k stratified eigenfunctions. The total number ofindependent eigenfunctions is then (by summation by parts) k − X i =1 ( k − i )[ n ( i ) − n ( i − k = − ( k − n (0)+ k X i =1 n ( i )+ k = k X i =0 n ( i ) = | V T | (8)Which means that all the eigenfunctions have been found.Now we can prove Oscar Rojo’s theorem 1 Proof of Theorem 1.
We search for stratified eigenfunctions of the (Dirichlet)Laplacians on symmetric (sub)trees. In fact, knowing that the eigenfunctionsare stratified, we can regard them as a function of level, and write the Laplacianequation for stratified eigenfunctions as L f ( l ) = d ( l ) f ( l ) − f ( l − − c ( l ) f ( l +1) = λf . For a subtree rooted at level l , λ is an eigenvalue of matrix S = d ( l ) − c ( l ) 0 · · · − d ( l + 1) . . . ...0 . . . . . . . . . 0... . . . d ( k − − c ( k − · · · − d ( k − (9)Let R be a ( k − l ) × ( k − l ) diagonal matrix whose i th element is ( − i p n ( i − l ) , ≤ i ≤ k − l , we have T = RSR − = d ( l ) p c ( l ) 0 · · · p c ( l ) d ( l + 1) . . . ...0 . . . . . . . . . 0... . . . d ( k − p c ( k − · · · p c ( k − d ( k − (10)7hich is similar to S . Note that d ( k −
1) = 1, the matrix given by OscarRojo is recognized (in a different notation) when l = 0, as well as its principalsubmatrices when l >
5. Combinition of two symmetric trees
The tree T studied in [3] are obtained by identifying the roots of two sym-metric trees T and T . We can still use the terminology “level” by allowingnegative levels, and use the sign of level to distinguish the two symmetric sub-trees. So a vertex has always a bigger absolute level than its parent. The treeis still invariant under S v , except for the root, where we must distinguish thechildren at +1 level and at − S + r and S − r . The stratified structure is still valid.Let’s count again how many eigenfunctions can be found with the help ofstratified structure. An eigenfunction f of L| T who vanishes at the root is alsoan eigenfunction of T . From the previous analysis, we can find | V T | − k sucheigenfunctions, where k is the number of levels in T . Same for T . Then wesearch for the stratified eigenfunctions of L , there are k + k −
1. We havefound in all | V T | + | V T | − | V T | independent eigenfunctions. That is all theeigenfunctions.So by the same argument as before, with the knowledge of stratified struc-ture, we can write the Laplacian equation for stratified eigenfunctions, and theresult of Oscar Rojo is immediate. Acknowledgment
This work was done at Max-Planck-Institut f¨ur Mathematik in den Naturwis-senschaften as an internship at the end of my third year in Ecole Polytechnique.I’d like to thank Frank Bauer for the very helpful discussions and suggestions.The figures in this paper are generated by Pajek, a program for analysing graphs.