Brouwer's conjecture holds asymptotically almost surely

TL;DR

This paper proves that Brouwer's conjecture is almost surely true for large random graphs and extends this result to weighted graphs with negative weights, showing it holds with high probability as the number of vertices grows.

Contribution

It demonstrates that Brouwer's conjecture holds asymptotically almost surely for large random and weighted graphs, including those with negative weights.

Findings

Brouwer's conjecture holds with probability tending to one as vertices increase.

The result extends to weighted graphs with negative weights.

Most graphs with large vertices satisfy Brouwer's conjecture.

Abstract

We show that for a sequence of random graphs Brouwer's conjecture holds true with probability tending to one as the number of vertices tends to infinity. Surprisingly, it was found that a similar statement holds true for weighted graphs with possible negative weights as well. For graphs with a fixed number of vertices, the result implies that there are constants $C>0$ and $n_{0}$ such that if $n\geq n_{0}$ then among all $2^{{n \choose 2}}$ graphs with $n$ vertices, at least $\left(1-\exp\left(-Cn\right)\right)2^{{n \choose 2}}$ graphs satisfy Brouwer's conjecture.

Equations

$$\sum _{i=1}^k{\lambda }_i\le \left|E\right|+\left(\genfrac{}{}{0ex}{}{k+1}{2}\right),$$

$$\sum _{i=1}^k{\lambda }_i\le e(G)+\left(\genfrac{}{}{0ex}{}{k+1}{2}\right),$$

$$L_{n}=\left[\begin{array}[]{cccc}\sum_{j\neq 1}\xi_{1j}^{(n)}&-\xi_{12}^{(n)}&\cdots&-\xi_{1n}^{(n)}\\ -\xi_{21}^{(n)}&\sum_{j\neq 2}\xi_{2j}^{(n)}&&-\xi_{2n}^{(n)}\\ \vdots&&\ddots&\vdots\\ -\xi_{n1}^{(n)}&\cdots&&\sum_{j\neq n}\xi_{nj}^{(n)}\end{array}\right],$$

$$\underset{i,j,n}{\sup }\mathbb{E}\left[{\left|\left({\xi }_{ij}^{(n)}-{\mu }_n\right)/{\sigma }_n\right|}^p\right]<\infty$$

$$\mathbb{P}\left[\sum _{i=1}^k{\lambda }_i\le e(G_n)+\left(\genfrac{}{}{0ex}{}{k+1}{2}\right)\right]\ge 1-\exp \left(-C\text{μn}n\right),$$

$$\mathbb{P}\left[\sum _{i=1}^k{\lambda }_i\le e(G_n)+\left(\genfrac{}{}{0ex}{}{k+1}{2}\right)\right]$$

$$k(1+ϵ)\text{μnn}<\text{μn}\left(1-\delta \right)\left(\genfrac{}{}{0ex}{}{n}{2}\right)+\left(\genfrac{}{}{0ex}{}{k+1}{2}\right)$$

$$\Delta =n^2{\mu }_n(\delta -1+{\mu }_n(ϵ+1{)}^2)-n{\mu }_n(\delta +ϵ)+1/4.$$

$$\Delta \le n^2{\mu }_n(\delta -1+(1-\gamma )(ϵ+1{)}^2)-n{\mu }_n(\delta +ϵ)+1/4.$$

$$\Delta$$

$$=n^2{\mu }_n(\delta -{\gamma }^2)+1/4$$

$$=-n^2{\mu }_n{\gamma }^2/2+1/4.$$

$$\mathbb{P}\left[\sum _{i=1}^k{\lambda }_i\le e(G_n)+\left(\genfrac{}{}{0ex}{}{k+1}{2}\right)\right]$$

$$\ge \mathbb{P}\left[\sum _{i=1}^k{\lambda }_i\le k\left(1+ϵ\right){\mu }_{n}n\right]+\mathbb{P}\left[{\mu }_n(1-\delta )\left(\genfrac{}{}{0ex}{}{n}{2}\right)\le e(G_n)\right]-1.$$

$$\mathbb{P}\left[\underset{n\to \infty }{\lim }\frac{{\lambda }_{\text{max}}\left(L_n\right)}{n{\mu }_n}=1\right]=1.$$

$$\mathbb{P}\left[\sum _{i=1}^k{\lambda }_i\le k\left(1+ϵ\right){\mu }_{n}n\right]=1.$$

$$\mathbb{P}\left[\sum _{i=1}^k{\lambda }_i\le e(G_n)+\left(\genfrac{}{}{0ex}{}{k+1}{2}\right)\right]\ge \mathbb{P}\left[{\mu }_n(1-\delta )\left(\genfrac{}{}{0ex}{}{n}{2}\right)\le e(G_n)\right].$$

$$\mathbb{P}\left[e(G_n)\le (1-\delta )\text{μn}\left(\genfrac{}{}{0ex}{}{n}{2}\right)\right]\le \exp -\frac{{\delta }^2{\left(\text{μn}\left(\genfrac{}{}{0ex}{}{n}{2}\right)\right)}^2}{\left(\genfrac{}{}{0ex}{}{n}{2}\right)C^2}=\exp -\frac{{\delta }^2\text{μn2}\left(\genfrac{}{}{0ex}{}{n}{2}\right)}{C^2}$$

$$\mathbb{P}\left[(1-\delta )\text{μn}\left(\genfrac{}{}{0ex}{}{n}{2}\right)\le e(G_n)\right]\ge 1-\exp -\frac{{\delta }^2\text{μn2}\left(\genfrac{}{}{0ex}{}{n}{2}\right)}{C^2}.$$

$$\mathbb{P}\left[\sum _{i=1}^k{\lambda }_i\le e(G_n)+\left(\genfrac{}{}{0ex}{}{k+1}{2}\right)\right]\ge 1-\exp \left(-C_0\text{μn}n\right).$$

$$k(2+ϵ)\text{σn}\sqrt{n\log n}<\text{μn}\left(1-\delta \right)\left(\genfrac{}{}{0ex}{}{n}{2}\right)+\left(\genfrac{}{}{0ex}{}{k+1}{2}\right)$$

$$\text{μn}\left(1-\delta \right)\left(\genfrac{}{}{0ex}{}{n}{2}\right)+\left(\genfrac{}{}{0ex}{}{k+1}{2}\right)-k(2+ϵ)\text{σn}\sqrt{n\log n}>\text{Cμn}\left(1-\delta \right)n^2+\frac{k^2}{2}-k(2+ϵ)\text{σn}\sqrt{n\log n}.$$

$$\Delta$$

$${\sigma }_n^2\frac{\log n}{{\mu }_{n}n}<\frac{2C\left(1-\delta \right)}{{\left(2+ϵ\right)}^2},$$

$$\mathbb{P}\left[\sum _{i=1}^k{\lambda }_i\le e(G_n)+\left(\genfrac{}{}{0ex}{}{k+1}{2}\right)\right]$$

$$\ge \mathbb{P}\left[\sum _{i=1}^k{\lambda }_i\le k(2+ϵ)\text{σn}\sqrt{n\log n}\right]+\mathbb{P}\left[{\mu }_n(1-\delta )\left(\genfrac{}{}{0ex}{}{n}{2}\right)\le e(G_n)\right]-1.$$

$$\mathbb{P}\left[\underset{n\to \infty }{\mathrm{limsup}}\frac{{\lambda }_{\text{max}}\left(L_n\right)}{{\sigma }_n\sqrt{n\log }n}=2\right]=1.$$

$$\mathbb{P}\left[{\lambda }_{\text{max}}\left(L_n\right)\le \left(2+ϵ\right){\sigma }_n\sqrt{n\log }n\right]=1.$$

$$\mathbb{P}\left[\sum _{i=1}^k{\lambda }_i\le k\left(2+ϵ\right){\sigma }_n\sqrt{n\log }n\right]=1.$$

$$\mathbb{P}\left[\sum _{i=1}^k{\lambda }_i\le e(G_n)+\left(\genfrac{}{}{0ex}{}{k+1}{2}\right)\right]\ge \mathbb{P}\left[{\mu }_n(1-\delta )\left(\genfrac{}{}{0ex}{}{n}{2}\right)\le e(G_n)\right].$$

Full text

Rocha was supported by the Czech Science Foundation, grant number GJ16-07822Y

1. Introduction

Brouwer’s conjecture [1] states that any graph $G=\left(E,V\right)$ with Laplacian matrix $L$ and eigenvalues $\lambda_{1}\geq\ldots\geq\lambda_{n}$ satisfies

[TABLE]

for $k=1,\ldots,n$. There have been many partial progresses on this conjecture using particular methods from matrix theory. This conjecture seems to be a difficult problem and to this date it remains open. In this paper we present an approach to this problem using methods from random matrices and random graphs. Here we address the following question: for how many graphs inequality (1.1) holds? In contrast to what previous investigations have been focused so far, instead of approaching the problem for graphs enjoying a prescribed structure, we show that a great proportion of graphs satisfy the conjecture.

In this paper it is shown that Brouwer’s conjecture holds asymptotically almost surely for random graphs under general conditions to be specified later. That is to say that graphs that potentially do not satisfy the inequality in Brouwer’s conjecture are rare, in a precise measure theoretic sense. That suggests a change of focus in the research that has been done on this problem so far. Instead of searching for new families of graphs for which the conjecture holds, one should attempt to understand these rare cases for which the conjecture potentially do not hold.

It comes as a surprise that this approach reveals that a similar statement holds true for weighted graphs, and even with possible negative weights. To state the result precisely some notation is needed. First, we denote by $w_{uv}$ the weight of an edge $uv\in E$ and set $w_{uv}=0$ in case $uv\notin E$. Then, we define $e(G)=\sum_{uv\in E}w_{uv}$. The Laplacian matrix of a weighted graph have the number $-w_{uv}$ in the off-diagonal entry $uv$ and $\sum_{j\neq u}w_{uj}$ in the diagonal entry $uu$. As usual, an unweighted graph can be seen as a $\{0,1\}$ weighted graph with its standard Laplacian matrix. Clearly, in this case $\left|E\right|=e(G)$. We show that for a weighted graph $G$ and for $k=1,\ldots,n$ we have

[TABLE]

asymptotically almost surely.

A random weighted graph $G_{n}$ is a graph with $n$ vertices and edge weights given by a random variable $\xi_{ij}^{(n)}$ for each $ij\in[n]$. Throughout this paper a sequence of random weighted graphs is under the general condition described as follow:

(Condition $1$) Let $\left\{G_{1},G_{2},G_{3},\ldots\right\}$ be a sequence of random weighted graph with $n$ vertices and Laplacian matrix given by

[TABLE]

where for $i<j$ we have that $\xi_{ij}^{(n)}$ are bounded random variables on the same probability space and independent for each $n$ (not necessarily identically distributed) with $\xi_{ij}^{(n)}=\xi_{ji}^{(n)}$, $\mathbb{E}\left[\xi_{ij}^{(n)}\right]=\mu_{n}$, $\text{Var}\left[\xi_{ij}^{(n)}\right]=\sigma_{n}^{2}$ , and

[TABLE]

for some $p>6$.

We have the setup to state the main result of this paper which implies Brouwer’s conjecture asymptotically almost surely.

Theorem 1**.**

Assume $\left\{G_{1},G_{2},G_{3},\ldots\right\}$ are independent random graphs as in Condition $1$. If $\mu_{n}\in(0,1-\gamma]$ for some $\gamma\in(0,1)$ and $\frac{\mu_{n}}{\sigma_{n}}\left(\frac{n}{\log n}\right)^{1/2}\rightarrow\infty$ as $n\rightarrow\infty$, then for $n$ large enough, we have that

[TABLE]

for some constant $C>0$ that depends only on $\gamma$.

It is interesting to understand what this result is saying for the set of graphs with fixed number of vertices. For such graphs this result is implying a strong statement that quantifies the number of graphs satisfying Brouwer’s conjecture. In fact, the most simple instance of Theorem 1 is the most meaningful for the conjecture. To see that consider the Erdős–Rényi random graph with probability $1/2$ and distribution $\mathbb{G}(n,1/2)$, i.e., a graph drawn from this distribution has $n$ vertices where each pair of vertices has an edge with probability $1/2$ independently at random. It is a basic fact that any graph with $n$ vertices is equally likely in the distribution $\mathbb{G}(n,1/2)$. Therefore, Theorem 1 implies that for $n\geq n_{0}$ among all $2^{{n\choose 2}}$ graphs with $n$ vertices that exist, at least $\left(1-\exp-Cn\right)2^{{n\choose 2}}$ graphs satisfy Brouwer’s conjecture.

The hypothesis $\frac{\mu_{n}}{\sigma_{n}}\left(\frac{n}{\log n}\right)^{1/2}\rightarrow\infty$ in Theorem 1 ensures that $\mu_{n}/\sigma_{n}$ is not approaching zero too fast. That is necessary because of the concentration of the largest eigenvalue around $\mu_{n}n$ and because there is a threshold phenomena happening here. When $\mu_{n}/\sigma_{n}$ goes to zero fast, the largest eigenvalue concentrates around $\sigma_{n}\sqrt{n\log}n$. In that case we still can apply the same method, but the difference in concentration requires a different analysis. For this reason we provide a separate theorem for this range, where $\mu_{n}/\sigma_{n}$ goes to zero in such a way that $\frac{\mu_{n}}{\sigma_{n}^{2}}\left(\frac{n}{\log n}\right)\rightarrow\infty$.

Theorem 2**.**

Assume $\left\{G_{1},G_{2},G_{3},\ldots\right\}$ are independent random graphs as in Condition $1$. If $\frac{\mu_{n}}{\sigma_{n}}\left(\frac{n}{\log n}\right)^{1/2}\rightarrow 0$ and $\sigma_{n}^{2}\frac{\log n}{\mu_{n}n}\rightarrow 0$ as $n\rightarrow\infty$, then for $n$ large enough, we have that

[TABLE]

for some constant $C>0$.

We get into the detailed proof in the next section, where we first present the method from a general perspective. Our intention is to give an insight on how such ideas can be used to provide bounds for the partial sum of eigenvalues of random matrices from different ensembles.

2. Idea and proofs

The main idea is straightforward and it consists in finding functions $f$ and $g$ depending on $n$ satisfying:

• •

There exists $n_{0}$ such that $f(n)\leq g(n)$ for $n\geq n_{0}$

• •

$\mathbb{P}\left[\sum_{i=1}^{k}\lambda_{i}\leq f\right]\rightarrow 1$ as $n\rightarrow\infty$

• •

$\mathbb{P}\left[g\leq e(G_{n})+{k+1\choose 2}\right]\rightarrow 1$ as $n\rightarrow\infty$

Once we figure out what $f$ and $g$ should be, by Bonferroni’s inequality we clearly have

[TABLE]

And that finishes the proof. It is also clear that the technical challenge here is to find such functions. To do that, we need some information about the spectrum as $n$ increases. For instance, if the limiting spectral distribution is known, then we can obtain a candidate for $f$, which will be an approximation of the limit. For $g$ we can use some Chernoff-type bound for the random variable that amounts the total edge weight.

We remark that there is nothing particular in this idea about the Laplacian matrix. Apart from the fact that Brouwer’s conjecture claims what the correct bound should be, this method can be applied to any matrix ensemble for which we know the limiting spectral distribution.

Next, we proceed with the proof of the main results. Our analysis relies on the following technical lemma which fully describes the aforementioned relevant functions $f$ and $g$.

Lemma 3**.**

If $\mu_{n}\in(0,1-\gamma]$ for some $\gamma\in(0,1)$, then there are $\epsilon>0,$ \text{\delta>0}, and $n_{0}$ such that for all $n\geq n_{0}$ we have

[TABLE]

for all $k\in\mathbb{R}$.

Proof.

First, we define a polynomial in $k$ by f(k):=\text{\mu_{n}}\left(1-\delta\right){n\choose 2}+{k+1\choose 2}-k(1+\epsilon)\text{\mu_{n}n}. That has discriminant

[TABLE]

It suffices to find \epsilon,\text{\delta}, and $n_{0}$ such that the discriminant is negative for all $n\geq n_{0}$. To this end, we use the upper bound on $\mu_{n}$ to obtain

[TABLE]

Now fix $\delta=\gamma^{2}/2$ and $\epsilon>0$ such that $(\epsilon+1)^{2}=1+\gamma$. That allow us to bound the discriminant by

[TABLE]

Clearly, for $n$ large enough the last expression is dominated by the term $-n^{2}\mu_{n}\gamma^{2}/2$ and therefore there exists a $n_{0}$ such that for all $n\geq n_{0}$ we have that $f(k)>0$ for all $k$. That proves the lemma. ∎

To estimate the largest eigenvalue $\lambda_{\text{max}}\left(L_{n}\right)$ we use Corollary 1.1 (b1) and (b2) from [2].

Lemma 4**.**

Assume $\left\{G_{1},G_{2},G_{3},\ldots\right\}$ are independent random graphs as in Condition $1$.

1. (1)

If  \lim_{n\rightarrow\infty}\text{\frac{\mu_{n}}{\sigma_{n}}\left(\frac{n}{\log n}\right)^{1/2}=\infty} and $\mu_{n}>0$, then $\mathbb{P}\left[\lim_{n\rightarrow\infty}\frac{\lambda_{\text{max}}\left(L_{n}\right)}{n\mu_{n}}=1\right]=1.$ 2. (2)

If  \lim_{n\rightarrow\infty}\text{\frac{\mu_{n}}{\sigma_{n}}\left(\frac{n}{\log n}\right)^{1/2}=0}, then $\mathbb{P}\left[\limsup_{n\rightarrow\infty}\frac{\lambda_{\text{max}}\left(L_{n}\right)}{\sigma_{n}\sqrt{n\log}n}=2\right]=1.$

Now, we are ready to proceed with the main proof which was roughly sketched in the beginning of this section.

Proof of Theorem 1.

From now on, we fix \epsilon,\text{\delta}, and $n_{0}$ given by Lemma 3 to obtain that for all $n\geq n_{0}$ we have

[TABLE]

By part (1) from Lemma 4

[TABLE]

Thus, for $\epsilon$ given by Lemma 3 we can find $m_{0}$ such that $n\geq m_{0}$ implies $\mathbb{P}\left[\lambda_{\text{max}}\left(L_{n}\right)\leq\left(1+\epsilon\right)\mu_{n}n\right]=1.$ That implies

[TABLE]

Thus, inequality (2.3) together with (2.2) gives us that for all $n\geq\min\left\{m_{0},n_{0}\right\}$ we have

[TABLE]

To bound this probability, we use that the expected number of edges in $G_{n}$ is $\mu_{n}{n\choose 2}$. We remark that $\left|\xi_{ij}^{(n)}\right|<C$ for all $\xi_{ij}^{(n)}$. Thus, Hoeffding´s inequality implies that

[TABLE]

for all $\delta\in(0,1)$. Equivalently, we have

[TABLE]

Finally, this last inequality and inequality (2.4) ensures that for all $n\geq\min\left\{m_{0},n_{0}\right\}$ we have

[TABLE]

for some constant $C_{0}>0$. That finishes the proof. ∎

Now, the proof of Theorem 2 is basically the same, only the details in the analysis change. For that we need a different version of Lemma 3 given bellow.

Lemma 5**.**

If $\sigma_{n}^{2}\frac{\log n}{\mu_{n}n}\rightarrow 0$ as $n\rightarrow\infty$, then there are $\epsilon>0,$ \text{\delta>0}, and $n_{0}$ such that for all $n\geq n_{0}$ we have

[TABLE]

for all $k$.

Proof.

First, there is a constant $C>0$ such that for $n$ large, we have

[TABLE]

We proceed to show that there are $\epsilon>0$ and $\delta>0$, where \text{C\mu_{n}}\left(1-\delta\right)n^{2}+\frac{k^{2}}{2}-k(2+\epsilon)\text{\sigma_{n}}\sqrt{n\log n}>0 for $n$ large enough and that finishes the proof.

To show it we define a polynomial in $k$ by f(k):=\frac{k^{2}}{2}-k(2+\epsilon)\text{\sigma_{n}}\sqrt{n\log n}+\text{C\mu_{n}}\left(1-\delta\right)n^{2}. This polynomial has discriminant

[TABLE]

It suffices to find \epsilon,\text{\delta}, and $n_{0}$ such that the discriminant is negative for all $n\geq n_{0}$. In fact, it is enough to choose any $\epsilon>0$ and $\delta\in(0,1)$. Notice that $\Delta<0$ is equivalent to

[TABLE]

which is true for $n$ large enough, as required. ∎

Proof of Theorem 2.

First, we fix \epsilon,\text{\delta}, and $n_{0}$ given by Lemma 5 to obtain that for all $n\geq n_{0}$ we have

[TABLE]

Part (2) from Lemma 4 provides us with

[TABLE]

Again, for $\epsilon$ given by Lemma 5 we can find $m_{0}$ such that $n\geq m_{0}$ implies

[TABLE]

That gives us

[TABLE]

Thus, inequality (2.6) together with (2.7) gives us that for all $n\geq\min\left\{m_{0},n_{0}\right\}$ we have

[TABLE]

The rest of the proof is the same as in Theorem 1 and that finishes the proof. ∎