High-girth near-Ramanujan graphs with localized eigenvectors

TL;DR

This paper constructs infinite families of high-girth, near-Ramanujan regular graphs with localized eigenvectors, revealing a discrete analogue of quantum scarring phenomena and advancing understanding of graph eigenvector localization.

Contribution

It introduces a method to build high-girth, near-Ramanujan graphs with localized eigenvectors, extending prior work and connecting to quantum ergodicity phenomena.

Findings

Existence of infinite high-girth, near-Ramanujan graphs with localized eigenvectors

Eigenvectors can be fully localized on small vertex sets

Girth scales logarithmically with graph size

Abstract

We show that for every prime $d$ and $\alpha\in (0,1/6)$, there is an infinite sequence of $(d+1)$-regular graphs $G=(V,E)$ with girth at least $2\alpha \log_{d}(|V|)(1-o_d(1))$, second adjacency matrix eigenvalue bounded by $(3/\sqrt{2})\sqrt{d}$, and many eigenvectors fully localized on small sets of size $O(|V|^\alpha)$. This strengthens the results of Ganguly-Srivastava, who constructed high girth (but not expanding) graphs with similar properties, and may be viewed as a discrete analogue of the "scarring" phenomenon observed in the study of quantum ergodicity on manifolds. Key ingredients in the proof are a technique of Kahale for bounding the growth rate of eigenfunctions of graphs, discovered in the context of vertex expansion and a method of Erd\H{o}s and Sachs for constructing high girth regular graphs.

Equations

$$|S|\ge {\Omega }_d({\epsilon }^2{d}^{2^{-7}{\epsilon }^2\mathrm{girth}(G)}),$$

$$|S|\ge \frac{\epsilon d^{\epsilon \mathrm{girth}(G)/4}}{2d^2},$$

$$\frac{1}{m}|\{v\in V_m:N_m(v,R_m)\text{ contains a cycle}\}|\le {\alpha }_m,$$

$$|{\lambda }_i(A_m)|<(d+1)(1-\beta ),$$

$$\frac{1}{m}\sum _{i\le m}|⟨{\psi }_i^{(m)},a_m{\psi }_i^{(m)}⟩{|}^2\lesssim {\beta }^{-2}\min \{R_m,\log (1/{\alpha }_m){\}}^{-1}\Vert a_m{\Vert }_2^2+{\alpha }_m^{1/2}\Vert a_m{\Vert }_{\infty }^2⟶0,$$

$$|⟨{\psi }_{i_m}^{(m)},a_m{\psi }_{i_m}^{(m)}⟩|⟶0.$$

$$\underset{S\subset V,|S|=\gamma |V|}{\max }\frac{N(S)}{|S|},$$

$$\lfloor 2{\log }_{2d-1}(n-1)\rfloor +2\text{leavevmode}(\text{leavevmode}>2{\log }_{2d-1}n).$$

$$m(r,s)=2\left(\genfrac{}{}{0ex}{}{s-1}{r-1}\right)(d-1{)}^r{d}^{s-r}$$

$$m(s)=\sum _{r=1}^{s}2\left(\genfrac{}{}{0ex}{}{s-1}{r-1}\right)(d-1{)}^r{d}^{s-r}=2(d-1)\sum _{j=0}^{s-1}(\genfrac{}{}{0ex}{}{s-1}{j})(d-1{)}^j{d}^{s-1-j}=2(d-1)(2d-1{)}^{s-1}.$$

$$1+\sum _{s=1}^{k}2(d-1)(2d-1{)}^{s-1}=1+2(d-1)\frac{(2d-1{)}^k-1}{2d-2}=(2d-1{)}^k.$$

$$\frac{2}{3}{\log }_{d}m\ge 4r.$$

$$2{\log }_{2d-1}((d+1)d^{r-1})\text{leavevmode}(\ge 2\alpha {\log }_{2d-1}(m)-O(1)).$$

$$\Omega ({\sin }^2\theta )=\frac{\Vert v_{S_i}{\Vert }_2^2+\Vert v_{S_{i+1}}{\Vert }_2^2}{\Vert v(r){\Vert }_2^2}=O(1/{\sin }^2\theta ).$$

$$A_G\nu =\lambda \nu .$$

$$(\frac{3d-1}{\sqrt{d(2d-1)}}+\epsilon )\sqrt{d}.$$

$$|f^T{A}_{T}f|\le 2\sqrt{d}\sum _{w\in W}f(w{)}^2+\sqrt{d}\sum _{v\in L}f(v{)}^2.$$

$$|f^T{A}_{T}f|$$

$$=\sum _{u\to v}2t|f(u)|\frac{|f(v)|}{t}\text{ for any choice of t>0}$$

$$\le \sum _{u\to v}\left(t^2{f}^2(u)+\frac{1}{t^2}{f}^2(v)\right)$$

$$=\frac{d+1}{t^2}{f}^2(r)+t^2\sum _{u\in L}{f}^2(u)+\sum _{u\in W-\{r\}}{f}^2(u)\left(t^2+\frac{d}{t^2}\right)$$

$$\le 2\sqrt{d}\sum _{u\in W}{f}^2(u)+\sqrt{d}\sum _{u\in L}{f}^2(u)\text{ by choosing t2=d}$$

$$\frac{{\sum }_{v\in X_h}g(v{)}^2}{{\sum }_{v\in X_h}s(v{)}^2}\ge \frac{{\sum }_{v\in X_{h-1}}g(v{)}^2}{{\sum }_{v\in X_{h-1}}s(v{)}^2}.$$

$$|g^T{A}_{H}g|\le 2\sqrt{d}\sum _{v\in V}{g}^2(v)+\frac{(d+1)2n}{m}.$$

$$|g^T{A}_{T_2}g|\le 2\sqrt{d}\sum _{v\in V_2}{g}^2(v)+\sqrt{d}\sum _{v\in L_1}{g}^2(v).$$

$$|g^T{A}_{T_3}g|\le 2\sqrt{d}\sum _{v\in V_3}{g}^2(v)+\sqrt{d}\sum _{v\in L_2}{g}^2(v).$$

$$|g^T{A}_{M}g|=|\sum _{i=1}^{n}2g(u_i)g(v_i)|\le \sum _{v\in L_1\cup L_2}{g}^2(v).$$

$$|\mu |=|g^{T}Ag|$$

$$=2\sqrt{d}+(\sqrt{d}+1)\sum _{u\in L_1\cup L_2}{g}^2(u)+\frac{(d+1)2n}{m}.$$

$$x_{i+1}=\min \{b+\epsilon -\frac{1}{x_i},c\}.$$

$$As(v)\le |\mu |s(v).$$

$$\frac{c}{\sqrt{d}}(d+1)\le |\mu |$$

$$|\mu |>b\sqrt{d}=(c+\frac{1}{c})\sqrt{d}\ge \frac{c}{\sqrt{d}}(d+1).$$

Full text

††This work was partially completed while SG and NS were at the Simons Institute for the Theory of Computing, as part of the “Geometry of Polynomials” program.

1 Introduction

We study the relationship between geometric properties of finite regular graphs, such as girth and expansion, and localization properties of their Laplacian / adjacency matrix eigenvectors. This line of work was initiated by Brooks and Lindenstrauss, who proved that the eigenvectors of high girth graphs cannot be too localized in the following sense (in fact, they studied graphs with few short cycles, but we will state the restriction of their results for high girth graphs for simplicity).

Theorem 1.1** ([BL13]).**

Suppose $G=(V,E)$ is a $(d+1)-$regular graph with adjacency matrix $A$. Then for any normalized $\ell_{2}$ eigenvector $v\in\mathbb{R}^{V}$ of $A$ and $S\subset V$ with $\|v_{S}\|_{2}^{2}:=\sum_{x\in S}v^{2}_{x}\geq\varepsilon,$

[TABLE]

where $\mathrm{girth}(G)$ denotes the length of the shortest cycle in $G$.

The recent work [GS18] improved (1) to

[TABLE]

under the same assumptions on $G$. Moreover, given any $\varepsilon\in(0,1),$ they proved that for infinitely many $m\in\mathbb{N}$ there is a $(d+1)-$regular graph $G_{m}$ with $m$ vertices, $\mathrm{girth}(G_{m})\geq(1/8)\log_{d}(m)$, and a localized eigenvector satisfying $\|v_{S}\|_{2}^{2}=\varepsilon$ and $|S|\leq O(d^{4\varepsilon\mathrm{girth}(G_{m})})$. This shows that (2) is sharp up to constants and a factor of $\varepsilon d^{-2}$ in the regime where the girth is logarithmic in the number of vertices.

In this work, we construct examples which improve on the above in three ways: (1) the graphs we construct are expanders with near-optimal spectral gap, (2) we improve the bounds on $\mathrm{girth}(G_{m})$ as well as the localization size $|S|$ by constant factors, (3) our constructions are explicit whereas [GS18] used the probabilistic method to show existence non-constructively.

Theorem 1.2**.**

For every $d=p+1$, $p$ prime, and parameter $\alpha\in(0,1/6)$ there are infinitely many integers $m$ such that there exists a $(d+1)-$regular graph $G_{m}=(V_{m},E_{m})$ on $m$ vertices with the following properties,

$|\lambda_{i}(A_{m})|\leq(3/\sqrt{2})\sqrt{d}$* for all nontrivial eigenvalues $i\neq 1$ of the adjacency matrices $A_{m}$.* 2. 2.

$\mathrm{girth}(G_{m})\geq 2\alpha\log_{2d-1}(m)-O(1)=2\alpha\log_{d}(m)\cdot(1-O(\log^{-1}(d))).$** 3. 3.

There is a set $S_{m}\subset V_{m}$ of size $O(m^{\alpha})$ such that $A_{m}$ has at least $\ell_{m}:=\lfloor\alpha\log_{d}(m)\rfloor$ eigenvalues $\lambda\in(-2\sqrt{d},2\sqrt{d})$ with corresponding eigenvectors $v:A_{m}v=\lambda v$ supported entirely on $S_{m}$.

Moreover, the set of eigenvalues $\lambda$ realized by the localized eigenvectors of $A_{m},$ over all such $m$ is dense in the interval $(-2\sqrt{d},2\sqrt{d})$.

Note that the number $3/\sqrt{2}\cdot\sqrt{d}\approx 2.121\sqrt{d}$ above is quite close to the best possible bound of $2\sqrt{d}$ for an infinite sequence of regular graphs [Nil91].

Remark 1.1**.**

(Partial Localization on Smaller Sets). In fact the proof of Theorem 1.2 produces eigenvectors $v,$ with the additional property that for $\varepsilon\in(0,1)$, there exists a subset $S$ of vertices with $|S|=O(m^{\Theta(\varepsilon)\alpha}),$ and $\|v_{S}\|_{2}^{2}\geq\Theta(\varepsilon)$.

Finally, we show how to modify our construction to produce many localized eigenvectors corresponding to eigenvalues with very high multiplicity.

Theorem 1.3** (Many Localized Eigenvectors).**

Theorem 1.2 is true with the last property replaced by: there are $\ell_{m}:=\alpha\log_{d}(m)$ eigenvalues $\lambda_{1},\ldots,\lambda_{\ell_{m}}$, each of multiplicity at least $\Omega(m^{1-4\alpha})$, such that each eigenspace has a basis of orthogonal eigenvectors supported on sets of size $O(m^{\alpha})$.

1.1 Implications for Quantum Ergodicity on Graphs

The additional property of expansion in our examples is relevant to the study of quantum ergodicity on graphs. Anantharaman and Le Masson proved that if a sequence of graphs has few short cycles and a spectral gap, then the eigenvectors must be equidistributed on average in a sense stronger than Theorem 1.1.

Theorem 1.4**.**

[ALM15, BLML15]** Suppose $G_{m}=(V_{m},E_{m})$ is a sequence of $(d+1)$-regular graphs on $m$ vertices with adjacency matrices $A_{m}$ satisfying:

1. (BST)

The sequence of graphs converges to a tree in the sense of Benjamini-Schramm, i.e., there exist $R_{m}\to\infty$ and $\alpha_{m}\to 0$, such that

[TABLE]

where $N_{m}(v,R)$ is the set of vertices at distance at most $R$ from $v,$ in $G_{m}$. Note that this condition is implied by $\mathrm{girth}(G_{m})\rightarrow\infty$. 2. (EXP)

There is a constant $\beta>0$ such that

[TABLE]

for all nontrivial eigenvalues $i\neq 1$.

Then for any sequence of test functions $a_{m}:V_{m}\rightarrow\mathbb{R}$ with $\sum_{x\in V_{m}}a_{m}(x)=0,\|a_{m}\|_{\infty}\leq 1$:

[TABLE]

where $\psi_{1}^{(m)},\ldots\psi_{m}^{(m)}$ is any eigenbasis of $A_{m}$.

The above may be viewed as a discrete analogue of the quantum ergodicity theorem of Shnirelman, Zelditch, and Colin de Verdiére [Shn74, DV85, Z*+*87], which states that if the geodesic flow on a compact manifold is ergodic, then it must have a dense subsequence of Laplacian eigenfunctions whose mass distribution converges weakly to the volume measure as the energy goes to infinity. In Theorem 1.4, the manifold has been replaced by a sequence of graphs, the condition of ergodic geodesic flow has been replaced by BST and EXP, and the notion of weak convergence involves a sequence of test functions on the graphs rather than a single test function on the manifold.

An even stronger notion of delocalization for the Laplacian on a manifold is Quantum Unique Ergodicity (QUE) (see e.g. [Sar12] for a detailed discussion), where instead of a dense subsequence of eigenfunctions, one requires that every subsequence of eigenfunctions becomes equidistributed. It is not completely clear what the correct analogous notion should be for a sequence of finite graphs. There are various proposals; one definition which appears in Anantaraman’s ICM survey [Ana18] and in [LMS*+*17, Question 1.3] in the context of sequences of manifolds is: for every sequence of test functions $a_{m}$ as in Theorem 1.4, and every sequence of eigenfunctions $\psi_{i_{m}}^{(m)}$, one has

[TABLE]

Since the graphs constructed in Theorem 1.2 satisfy BST and EXP, the theorem shows that these properties cannot imply unique ergodicity in the above sense: take the $\psi_{i_{m}}^{(m)}$ to be the localized eigenvectors of $G_{m}$, and let $a_{m}$ be the indicator functions of the sets $S_{m}$ on which they are localized, translated by a constant to have mean zero. It is then immediate that $\langle\psi_{i_{m}}^{(m)},a_{m}\psi_{i_{m}}^{(m)}\rangle=1-o(1)$ for the entire sequence.

The presence of localized eigenvectors is sometimes referred to as “scarring” (see e.g. [Ana18, HH10]), which may be partial or complete depending on whether a large fraction or all of the mass is localized on a small set. Theorem 1.2 and Remark 1.1 may be interpreted as saying that scarring can occur even under strong expansion and girth assumptions.

Remark 1.2** (QE over intervals).**

The works [ALM15, BLML15] also study a more refined version of quantum ergodicity on graphs, where the average (3) is taken over a spectral window $I\subset(-2\sqrt{d},2\sqrt{d})$ rather than the entire spectrum. These results hold on intervals $I$ of width roughly $1/\log(m)$, and it would be interesting to see whether our examples can prove a lower bound on the length of the smallest window that is possible. While Theorem 1.3 does produce many localized eigenvectors in a very small window (due to high multiplicity), the problem of controlling the other eigenvectors well enough to say that the average in a small window is not equidistributed is not pursued here and remains open.

1.2 Techniques and Vertex Expansion

The starting point of the proofs of Theorems 1.2 and 1.3 is a construction in the proof of [GS18, Theorem 1.6] which has the following ingredients:

(Pairing trees [GS18, Lemma 3.4]): A pair of trees is glued by randomly identifying their leaves, ensuring that the final graph has high girth. 2. 2.

(Degree-Correction [GS18, Lemma 3.5]): The above gluing yields an irregular graph where the identified leaf vertices have degree two. Each such vertex is identified with a particular vertex of degree $d-1$ in a degree-correcting gadget whose remaining vertices have degree $d+1$ thereby yielding a $d+1$ regular graph.

We modify this proof in two ways. First, we replace the random pairing in step (1) by a more efficient, simpler, and deterministic method. Second, in order to obtain the additional property of expansion, we replace the degree-correcting gadget in step (2) by a high girth Ramanujan graph [LPS88, Mur03]. To analyze the spectrum of the resulting graph, we must argue that its largest nontrivial eigenvector cannot have too much mass on the interface between the trees and the Ramanujan graph — once this is established, it is easy to analyze the contributions from the two pieces separately. We do this by employing a lemma of N. Kahale, which supplies a way to control the mass of eigenvectors on certain highly symmetric sets (such as our interface) by exhibiting certain appropriate super-harmonic test functions, and by a careful construction of such a function.

Kahale’s lemma originally appeared in the influential paper [Kah92] which showed that a $(d+1)$-regular graph $G=(V,E)$ with all nontrivial eigenvalues bounded by $2\sqrt{d}+o_{n}(1)$ must have linear expansion at least $(d+1)/2-o_{n}(1)$, where linear expansion is defined as:

[TABLE]

for a small constant $\gamma>0$ (in fact, he showed a more general inequality relating the parameters). As we discuss in Remark 4.1, this implies that our examples cannot have $|\lambda_{i}|\leq 2\sqrt{d}+o_{n}(1)$ since our gluing procedure produces a set with significantly smaller linear vertex expansion than $(d+1)/2$.

Note that it is possible to prove Theorem 1.2 with a weaker bound of $3\sqrt{d}$ without using Kahale’s lemma; however, since the bound we attain is quite close to optimal and we have not seen this technique appear in the quantum ergodicity literature, we believe it is valuable to present it.

2 Pairing trees

Our goal is to construct high girth almost-Ramanujan expanders with one or many localized eigenvectors. The starting point of the construction is the following lemma, improving the one from [GS18] and simplifying its proof. We refer to a finite tree in which all vertices except the leaves have degree $(d+1)$ and every leaf is at distance $D$ from the root as a d-ary tree of depth $D$.

Lemma 2.1** (Pairing of Trees).**

Suppose $T_{1}$ and $T_{2}$ are two $d-$ary trees of depth $D$, each with $n=(d+1)d^{D-1}$ leaf vertices $V_{1}$ and $V_{2}$. Then there is a bijection $\pi:V_{1}\rightarrow V_{2}$ such that the graph obtained from the vertex disjoint union of $T_{1}$ and $T_{2}$ by identifying $v$ and $\pi(v)$ for all $v$ has girth at least

[TABLE]

Proof.

We apply a variant of the method of Erdős and Sachs [ES63], (see also [ABGR18] for a similar argument). Let $g$ be the maximum possible girth of a graph obtained as above, and let $\pi:V_{1}\rightarrow V_{2}$ be a bijection for which the girth is $g$ and the number of cycles of length exactly $g$ is minimum. Note that $g$ is even, as the graph is bipartite. Let $G$ be the graph with the identified leaves obtained by $\pi$, and let $L$ denote the set of all $n$ vertices of degree $2$ in it, that is, all the identified leaves. Obviously every cycle of $G$ must contain vertices of $L$. Let $x\in L$ be a vertex contained in a shortest cycle $C$ of $G$.

Claim: For every $k\geq 0$ the number of vertices $y\in L$ of distance at most $2k$ from $x$ is at most $(2d-1)^{k}$.

Proof of claim:  Any shortest path of length precisely $2s\leq 2k$ between $x$ and another vertex $y\in L$ is a concatenation of some number $r$ of paths $P_{1},P_{2},\ldots,P_{r}$, where $P_{i}$ is a path from $x_{i-1}$ to $x_{i}$ with $x_{0},x_{1},\ldots,x_{r}\in L$, $x_{0}=x$, $x_{r}=y$, and either all even paths $P_{i}$ are in $T_{1}$ and all odd ones are in $T_{2}$ or vice versa. Let $2k_{i}$ be the length of $P_{i}$, then $\sum_{i=1}^{r}k_{i}=s\leq k$. Let $m(r,s)$ denote the number of paths as above with these values of $r$ and $s$. We next show that for all $1\leq r\leq s$

[TABLE]

The factor $2$ is for deciding if the first path $P_{1}$ is in $T_{1}$ or in $T_{2}$. The factor ${{s-1}\choose{r-1}}$ is the number of ways to choose the subset of $(r-1)$ elements $\{k_{1},k_{1}+k_{2},\ldots k_{1}+k_{2}+\cdots+k_{r-1}\}$ in the set $\{1,2,\ldots,s-1\}$. (This already determines $k_{r}=s-(k_{1}+k_{2}+\cdots+k_{r-1})$.) Once these choices are fixed, there is only one way for the edges numbers $1,2,\ldots,k_{i}$ of each path $P_{i}$, given the previous paths, as these edges go up the tree. There are $d-1$ possibilities for the edge number $k_{i+1}$ of this path, and there are $d^{k_{i}-1}$ choices for the remaining edges of $P_{i}$. The product of all these terms gives the expression in (5) for $m(r,s)$. For each fixed $s$, summing over all $1\leq r\leq s$ we conclude that the number $m(s)$ of paths of length exactly $2s$ starting in $x$ is

[TABLE]

Adding the trivial path of length [math] from $x$ to itself and summing over all $s$ from $1$ to $k$ we conclude that the total number of paths as above of length at most $2k$ starting at $x$ is

[TABLE]

The number above provides an upper bound for the number of vertices $y\in L$ that lie within distance $2k$ of $x$ (which may be smaller as several paths may lead to the same vertex). This completes the proof of the claim.

Returning to the proof of the lemma, define $k=\lfloor\log_{2d-1}(n-1)\rfloor$. Then $(2d-1)^{k}<n$ and hence there is a vertex $y\in L$ whose distance from $x$ in $G$ is larger than $2k$ (and hence at least $2k+2$). Let $u$ be the unique parent of $x$ in $T_{1}$ and let $u^{\prime}$ be the unique parent of $x$ in $T_{2}$. Similarly, let $v$ be the unique parent of $y$ in $T_{1}$ and let $v^{\prime}$ be the unique parent of $y$ in $T_{2}$. Change the bijection $\pi$ to the bijection $\pi^{\prime}$ obtained by swapping the images of $x$ and $y$ to get a graph $G^{\prime}$ obtained from $G$ by removing the edges $xu$ and $yv$ and by adding the edges $xv$ and $yu$. This swapping removes the shortest cycle $C$ of length $g$ in $G$ that contains $x$, and is not contained in $G^{\prime}$. Every new cycle contained in $G^{\prime}$ and not in $G$ must include at least one of the new edges $xv$, $yu$. If it contains exactly one of them, say $xv$, then it must also contain a path in $G$ from $x$ to $v$. The length of such a path is at least $2k+1$ (as the distance in $G$ from $x$ to $y$ is at least $2k+2$) showing that in this case the length of the new cycle is at least $2k+2$. If it contains both new edges $xv$ and $yu$ it must also contain either a path in $G$ from $x$ to $y$ (of length at least $2k+2)$) or a path in $G$ from $x$ to $u$ (of length at least $g-1$) and a path in $G$ from $y$ to $v$ (of length at least $g-1$). Therefore, in this case the length of the new cycle is either at least $2k+2+2=2k+4$ (in fact larger) or at least $2g-2+2=2g>g$. It follows that the only possibility to obtain a new cycle of length at most $g$ is if $g\geq 2k+2$. If the girth $g$ is smaller than $G^{\prime}$ has girth at least $g$ and the number of its cycles of length $g$ is smaller than that number in $G$, contradicting the choice of $G$. This shows that the girth $g$ satisfies $g\geq 2k+2=2\lfloor\log_{2d-1}(n-1)\rfloor+2$, completing the proof of the lemma. ∎

Remark 2.1**.**

For large fixed $d,$ the above graph has girth close to $2\log_{d}N$, where $N$ is the number of its vertices. This is strictly larger than the highest known girth of an $N$ vertex $(d+1)$-regular graph, which is roughly $\frac{4}{3}\log_{d}N$ (for some values of $d$). However, many of the vertices of our graph here have degree $2$, and suppressing them will not give graphs of girth larger than $(1+o(1))\log_{d}N$.

3 The construction

Let $d$ a prime and $\alpha\in(0,1/6)$ be given. Let $H=(V,E)$ be a $(d+1)$-regular non-bipartite Ramanujan graph with $m$ vertices and girth larger than $2/3\log_{d}(m)$; by [LPS88] such graphs exist for infinitely many $m$. Set $r=\lfloor\alpha\log_{d}(m)\rfloor$ and note that

[TABLE]

Fix a vertex $u$ of $H$. The induced subgraph on all vertices of distance at most $r$ from $u$ is a tree $T_{1}$ rooted at $u$. Let $n$ be the number of its leaves, let the set of leaves be $L_{1}=\{u_{1},..,u_{n}\}$ and let $V_{1}$ denote the set of all non-leaves of $T_{1}$. Take a matching from the set $L_{1}$ to the set of vertices $L_{2}=\{v_{1},..,v_{n}\}$, all at distance exactly $r+1$ from $u$, and remove the matching $u_{i}v_{i}$. Note that all $u_{i}$ are far from each other in the graph $H-V_{1}$ since the girth is significantly larger than $2r$. Similarly, all vertices $v_{i}$ are far from each other in $H-V_{1}$ for the same reason. Now take another $d$-tree $T_{2}$ isomorphic to $T_{1}$ on new vertices, and let $u^{\prime}$ denote its root. Identify the leaves of $T_{1}$ with these of $T_{2}$ using Lemma 2.1. Let $V_{2}$ denote the set of all non-leaves of $T_{2}$. As the vertices $u_{i}$ are far from each other in $H-V_{1}$ the girth stays as large as guaranteed by the lemma. Finally add a third tree $T_{3}$ with the same parameters on new vertices, rooted at $v^{\prime}$, identify its leaves with the vertices $v_{i}$ and let $V_{3}$ denote the set of its non-leaves. Call the resulting graph $G$.

The next sequence of lemmas will be needed to show that $G$ satisfies the claims of Theorem 1.2.

Lemma 3.1**.**

The girth of $G$ is at least

[TABLE]

Proof.

Follows from the above definition, Lemma 2.1 and that $\frac{2}{3}\log_{d}m>4r$. ∎

We now discuss the eigenvectors of $G$. We begin by recording some facts about eigenvalues and eigenvectors of rooted $d-$ary trees which also appear in [GS18] and will be critical to our construction. Recall that the eigenvalues of a $d-$ary tree are contained in the interval $(-2\sqrt{d},2\sqrt{d})$ [HLW06, Section 5]. For our purposes we will only consider eigenvalues corresponding to eigenvectors which are radial, which means that they assign the same value to vertices in a given level. We will refer to such eigenvalues as radial eigenvalues.

Lemma 3.2**.**

(Radial Eigenvalues)[GS18, Lemma 3.1] For any positive integer $D\geq 2,$ $A_{D}$ the adjacency matrix of $T_{D},$ a $d-$ary tree of depth $D,$ has exactly $D+1$ radial eigenvalues counting multiplicities.

Lemma 3.3**.**

(Eigenvalues of $d-$ary Trees)[GS18, Lemma 3.2] The set of all radial eigenvalues of any infinite sequence of distinct finite $d-$ary trees is dense in the interval $(-2\sqrt{d},2\sqrt{d})$.

Lemma 3.4** (Eigenvectors of $d-$ary Trees).**

[GS18, Lemma 3.3]** Assume $d\geq 2$ and let $T$ be a $d-$ary tree of depth $D$ with root $r$. Let $S_{0}=\{r\},S_{1},\ldots,S_{D}\subset T$ be the vertices at levels $0,1,\ldots,D$ of the tree and let $v$ be a radial eigenvector of its adjacency matrix with eigenvalue $\lambda=2\sqrt{d}\cos\theta\in(-2\sqrt{d},2\sqrt{d})$. Then every pair of adjacent levels has approximately the same total $\ell_{2}^{2}$ mass as the root:

[TABLE]

Given the above we have the following lemma about how radial tree eigenvectors can be used to construct eigenvectors of $G.$

Lemma 3.5**.**

For any radial eigenvalue $\lambda$ of the adjacency matrix of a tree of depth $r-1$, there exists an eigenvector $\nu$ supported on $V_{1}\cup V_{2}$ such that

[TABLE]

Proof.

For completeness we include the arguments that essentially appear in [GS18, Proof of Theorem 1.6]. Consider any such eigenvalue $\lambda$ and its corresponding radial eigenvector $f$. Now construct the function $\nu$ that equals $f$ on the top $r-1$ levels of $T_{1}$, i.e., $V_{1}$ and correspondingly $-f$ on $V_{2}$, and is zero elsewhere. We claim that $\nu$ is an eigenvector of $G$ with eigenvalue $\lambda$. To see this, note that the eigenvector equation is trivially satisfied on $V_{1}$ and $V_{2}$ because all new neighbors of those vertices are assigned a value of [math] in $\nu$. The remaining vertices where the eigenvector equation needs to be checked are the ones obtained by gluing $L_{1}$ to the leaves of $T_{2}.$ Now every such vertex $v,$ satisfies $\nu(v)=0$ and there exists two neighbors of $v$, say $u\in V_{1}$ and $w\in V_{2}$ with $\nu(u)=-\nu(w)$ and furthermore $\nu$ is [math] on every remaining neighbor, clearly implying the eigenvector equation at $v.$ ∎

We next show that $G$ is nearly Ramanujan.

4 The spectrum of $G$

Proposition 4.1**.**

For every fixed $\varepsilon>0$, if $m$ is sufficiently large then the absolute value of every nontrivial eigenvalue of $G$ is at most

[TABLE]

Remark 4.1**.**

For every fixed $d,$ the number $\frac{3d-1}{\sqrt{d(2d-1)}}$ is smaller than $\frac{3}{\sqrt{2}}=2.12132..$ For $d=2$ (cubic graphs) the number is $5/\sqrt{6}=2.04124..$. Therefore, the graph $G$ is close to being Ramanujan.

However, for all $r>1$ it is not quite Ramanujan. Indeed, it contains a set of vertices $Y$, namely the set of all vertices in levels $r-1,r-3,r-5,...$ of the two trees $T_{1},T_{2}$, that expands by a factor of less than $(d+1)/2$. By Theorem 4.1 in [Kah95] such a graph must contain a nontrivial eigenvalue of absolute value bigger than $2\sqrt{d}+\delta(d,r)$ for some positive $\delta(d,r)$.

To prove Proposition 4.1, we need the following simple lemma about the spectrum of finite $d$-ary trees.

Lemma 4.1**.**

Let $T$ be a finite $d$-ary tree, that is, a tree with a root $r$ of degree $d+1$ in which every non-leaf has $d$ children. Let $A_{T}$ denote the adjacency matrix of $T$, let $W$ be the set of its non-leaves and let $L$ be the set of its leaves. Then for any vector $f$ supported on $W\cup L$,

[TABLE]

Proof.

Orient all edges towards the root $r$. Then

[TABLE]

∎

We also need the following lemma of Kahale about growth rate of eigenfunctions.

Lemma 4.2**.**

*([Kah95, Lemma 5.1])

Consider a graph on a vertex set $U$, and let $A$ denote its adjacency matrix. Let $X$ be a set of vertices. Let $h$ be a positive integer and let $s$ be a function on $U$. Let $X_{i}$ be the set of all vertices at distance $i$ from $X$, and assume that the following conditions hold.*

For $h-1\leq i,j\leq h$ all vertices in $X_{i}$ have the same number of neighbors in $X_{j}$. 2. 2.

The function $s$ is constant on $X_{h-1}$ and on $X_{h}$. 3. 3.

The function $s$ is positive and $As(v)\leq|\mu|s(v)$ for every $v$ of distance at most $h-1$ from $X$, where $\mu$ is a nonzero real.

Then for any function $g$ on $U$ which satisfies $|Ag(u)|=|\mu||g(u)|$ for all vertices $u$ of distance at most $h-1$ from $X$, we have:

[TABLE]

Equipped with the above lemma we now proceed to proving Proposition 4.1.

Proof of Proposition 4.1.

The adjacency matrix $A_{G}$ of $G$ is the sum $A_{G}=A_{H}-A_{M}+A_{T_{2}}+A_{T_{3}}$, where $A_{H}$ is the adjacency matrix of $H$, $A_{M}$ is the adjacency matrix of the matching $u_{i}v_{i}$ and $A_{T_{2}},A_{T_{3}}$ are the adjacency matrices of the trees $T_{2}$ and $T_{3}$, respectively. It is convenient to view all the graphs $H,T_{2},T_{3},M$ as graphs on the set of all vertices of $G$, which is $U=V\cup V_{2}\cup V_{3}$. Thus all the matrices above have rows and columns indexed by the set $U$, and each of the corresponding graphs has many isolated vertices.

Put $b=\frac{3d-1}{\sqrt{d(2d-1)}}$. We have to show that every nontrivial eigenvalue $\mu$ of $G$ has absolute value at most $(b+\varepsilon)\sqrt{d}$ for any $\varepsilon>0$ provided $m$ is sufficiently large. Let $g:U\rightarrow R$ be an eigenvector of $\mu$ satisfying $\sum_{v\in U}g(u)^{2}=1$. As $g$ is orthogonal to the top eigenvector, $\sum_{v\in U}g(v)=0$. The total number of vertices in $V_{2}\cup V_{3}$ is smaller than $2n$, and therefore, by Cauchy-Schwartz, $|\sum_{v\in V_{2}\cup V_{3}}g(v)|\leq\sqrt{2n}$. It thus follows that $|\sum_{v\in V}g(v)|\leq\sqrt{2n}.$ Considering the projection of the restriction of $g$ to $V$ on the all ones vector and its complement we conclude that

[TABLE]

Recall that $L_{1}$ is the set of leaves of $T_{2}$ (and $T_{1}$). By Lemma 4.1

[TABLE]

Similarly

[TABLE]

The contribution of the omitted matching can be bounded as follows

[TABLE]

Combining (7), (8), (9), (10) we conclude that

[TABLE]

In order to complete the proof, it thus suffices to show that if $|\mu|\geq(b+\varepsilon)\sqrt{d}$, then, as $m$ tends to infinity, the sum $\sum_{u\in L_{1}\cup L_{2}}g^{2}(u)$ tends to zero. This is done using Lemma 4.2, as described next.

We first bound the sum $\sum_{u\in L_{2}}g^{2}(u)$. This is simple and works even if we only assume that $|\mu|\geq 2\sqrt{d}$. Indeed, starting with $X=\{v^{\prime}\}$, let $X_{i}$ be the set of vertices of distance $i$ from $X$. Define $s(v)=d^{-i/2}$ for all $v\in X_{i}$ for $0\leq i\leq r+t$, where $r+t$ is the largest integer smaller than half the girth of $H$. It is easy to check that the conditions of Lemma 4.2 hold. Thus by its conclusion the sum $\sum_{v\in X_{i}}g^{2}(v)$ is nondecreasing in $i$ for all $i\geq r$. Since this sum for $i=r$ is exactly $\sum_{v\in L_{2}}g^{2}(v)$ and $t$ tends to infinity with $m$, and since the sum over all $r\leq i\leq r+t$ is at most $1$, it follows that the sum for $i=r$ is negligible. 2. 2.

Bounding the sum $\sum_{u\in L_{1}}g^{2}(u)$ is harder. Here we use the assumption that $|\mu|\geq(b+\varepsilon)\sqrt{d}$ where $b=\frac{3d-1}{\sqrt{d(2d-1)}}$. Define $X=X_{0}=\{u,u^{\prime}\}$ and let $X_{i}$ denote the set of all vertices of $G$ of distance exactly $i$ from $X$. Put $c=\sqrt{(2d-1)/d}$ and note that $c+1/c=b$. Define a sequence of reals $s_{0},s_{1},s_{2},\ldots$ as follows. For $0\leq i\leq r$, $s_{i}=c^{i}d^{-i/2}$. For $i=r+1$, $s_{r+1}=c^{r-1}d^{-(r+1)/2}$ and for all $i\geq 1$, $s_{r+1+i}=\alpha_{i}d^{-(r+1+i)/2}$ where the numbers $\alpha_{i}$ are defined by setting $\alpha_{0}=c^{r-1}$ and $\alpha_{i}/\alpha_{i-1}=x_{i}$ for $i\geq 1$ with $x_{1}=1/c=\sqrt{d/(2d-1)}$ and for $i\geq 1$,

[TABLE]

Using the sequence $s_{i}$ define a function $s$ on the vertices in the union $\cup_{i\leq r+t}X_{i}$, where $r+t$ is smaller than half the girth of $H$, by putting $s(v)=s_{i}$ for all $v\in X_{i}$. We proceed to show that for every vertex $v\in\cup_{i<r+t}X_{i}$,

[TABLE]

For $v\in X=X_{0}$ this is equivalent to

[TABLE]

which is certainly true as

[TABLE]

For $v\in X_{i}$ with $1\leq i\leq r-1$ the required inequality is

[TABLE]

which follows from the fact that $1/c+c=b\leq b+\varepsilon$. For $v\in X_{r}$ the inequality is

[TABLE]

For this it suffices to check that

[TABLE]

which holds as the left hand side is equal to $b$. For $v\in X_{r+1}$ it suffices to check that

[TABLE]

which holds as $x_{2}\leq b+\varepsilon-1/x_{1}$ by its definition. Finally, for $v\in X_{r+1+i},i\geq 1$ the required inequality is equivalent to $\sqrt{d}\alpha_{i-1}+\sqrt{d}\alpha_{i+1}\leq|\mu|\alpha_{i},$ that is , $\sqrt{d}\frac{1}{x_{i}}+\sqrt{d}x_{i+1}\leq(b+\varepsilon)\sqrt{d}$ which again holds by the definition of $x_{i+1}$. This completes the proof of (12). Next we observe that $x_{1}=1/c>1/\sqrt{2}$, that $x_{i+1}\geq x_{i}$ for all $i$, and that if $x_{i}<c-\varepsilon$ then $x_{i+1}\geq x_{i}+\varepsilon$. Indeed $x_{1}=\sqrt{d/(2d-1)}>1/\sqrt{2}$, and $x_{i+1}\leq c$ for all $i$ by the definition of $x_{i+1}$. The function $g(x)=b-1/x$ is increasing and concave in the interval $[1/c,c]$ and as $g(x)=x$ at the endpoints of the interval, $g(x)\geq x$ for all $x$ in the interval implying that $x_{i+1}=\min\{g(x_{i})+\varepsilon,c\}\geq x_{i}$ for all $x_{i}\in[1/c,c]$ and that $x_{i+1}\geq x_{i}+\varepsilon$ if $x_{i}\leq c-\varepsilon$.

By the above discussion it follows that $x_{i}$ is at least $c>\sqrt{3/2}$ for all $i\geq 1/\varepsilon$ implying that the sequence $|X_{j}|s_{j}^{2}$ is increasing exponentially for $j>r+1/\varepsilon$. Between levels $r$ and $r+1/\varepsilon$ the terms of this sequence decrease by a factor larger than $1/c$ in each step, and it therefore follows that the term number $r$ of this sequence is negligible compared to any term number $r+\omega(1)$. This together with Lemma 4.2 completes the proof.

∎

Using the above ingredients we now verify all the claims in Theorem 1.2.

Proof of Theorem 1.2.

By the construction with described in Section 3, there exists a graph $G=(V,E)$ with the following properties (with $r=\lfloor\alpha\log_{d}(m)\rfloor$):

$|V|=M=m+2n$ where $n=\frac{d^{r+1}+d^{r}-2}{d-1}=d^{r}\left(1+O(\frac{1}{d})\right).$ 2. 2.

By Lemma 3.1, and the choice of $r,$ we have

[TABLE]

Thus the second claim in the theorem is verified by observing that $M=m+O(m^{\alpha}).$ 3. 3.

By Proposition 4.1,

[TABLE]

for all nontrivial eigenvalues $i\neq 1$ of the adjacency matrix $A_{M}$. 4. 4.

The third claim follows by Lemmas 3.2, 3.3, and 3.5, and noticing that $|V_{1}\cup V_{2}|=O(d^{r})=O(m^{\alpha}).$

The final claim in the statement of the theorem is an easy consequence of Lemma 3.3 which states that the set of finite $d-$ary tree eigenvalues is a dense subset of $(-2\sqrt{d},2\sqrt{d}).$ Furthermore, Remark 1.1 follows by choosing $S$ to be the top $\lfloor\varepsilon r\rfloor$ levels of $T_{1}.$ ∎

We conclude by explaining how to modify the construction to produce many localized eigenvectors.

Proof of Theorem 1.3.

The proof follows from the observation that in the construction described in Section 3, one can glue several trees to $H$ ‘far away’ from each other to maintain high girth and every other property mentioned in Theorem 1.2. More precisely, in the construction in Section 3, instead of considering a tree $T_{1}$ rooted at $u,$ consider trees $T^{(1)}_{1},T^{(2)}_{1},\ldots,T^{(k)}_{1}$ rooted at $u_{1},u_{2},\ldots,u_{k}$ and repeat the construction $k$ times with the corresponding trees

[TABLE]

rooted at $\{u^{\prime}_{1},u^{\prime}_{2},\ldots u^{\prime}_{k}\}$ and $\{v_{1},v_{2},\ldots,v_{k}\}$ respectively. The same arguments as before imply that the girth condition in Theorem 1.2 is satisfied as long as the graph distance in $H$ between any $u_{i}$ and $u_{j}$ is at least $4r.$ Lemma 4.3 below shows that one can take $k$ to be as large as $m^{1-4\alpha}$ where $r=\lfloor\alpha\log_{d}(m)\rfloor.$ Finally the proof of Proposition 4.1 in this case follows in exactly the same way after defining the set $X$ to be $\{v_{1},v_{2},\ldots,v_{k}\}$ and $\{u_{1},u^{\prime}_{1},u_{2},u^{\prime}_{2},\ldots,u_{k},u^{\prime}_{k}\}$ in 1. and 2. respectively instead of $\{v\}$ and $\{u,u^{\prime}\}.$ ∎

Lemma 4.3**.**

Given any $d+1$-regular graph $G=(V,E)$ of size $m,$ and any $k,$ there are at least $\frac{m(d-1)}{(d+1)d^{k}}$ vertices in $V,$ all of whose mutual distances are at least $k.$

Proof.

Consider a maximal set $S$ of such vertices. Simple considerations imply that every other vertex in $V$ must be at distance at most $k$ from the set $S.$ Now the total number of such vertices is at most $|S|{(d+1)d^{k}/(d-1)}$ which finishes the proof. ∎