On the Spectral Gap and the Diameter of Cayley Graphs

Abstract

We obtain a new bound connecting the first nontrivial eigenvalue of the Laplace operator on a graph and the diameter of the graph. This bound is effective for graphs with small diameter as well as for graphs with the number of maximal paths comparable to the expected value.

Appendix

Here we collect further natural properties of Bohr sets and related notions which have well-known abelian analogs. We do this for the convenience of the reader interested in this particular form of Bohr sets; most of these results are more or less contained in [2, 14, 15].

In Section 5 we have used the connection of Bohr sets with the set of unitary representations $\rho$ such that $\| \widehat{A}{} (\rho) \| \ge (1-\varepsilon) |A|$ for a given set $A\subseteq {\mathbf G}$. Thus it is natural to give a more general

Definition 25.

Let $A\subseteq {\mathbf G}$ be a set and $\varepsilon \in [0,1]$ a real number. The spectrum $\mathrm{Spec} _\varepsilon(A)$ of $A$ is the set of unitary representations

$$\mathrm{Spec} _\varepsilon(A) = \bigl\{\rho \colon\, \,\| \widehat{A}{} (\rho)\| \ge \varepsilon |A|\bigr\}.$$

Using the arguments of the proof of Lemma 17, we obtain a non-abelian analog of the well-known result of Yudin [20].

Proposition 26.

Let $A\subseteq {\mathbf G}$ be a set and $\varepsilon_1, \varepsilon_2 \in [0,1]$ real numbers. Then

$$\mathrm{Spec} _{1-\varepsilon_1}(A) \cdot \mathrm{Spec} _{1-\varepsilon_2}(A) \subseteq \mathrm{Spec} _{1-\varepsilon_1 - \varepsilon_2}(A).$$

Proof.

As follows from the arguments of the proof of Lemma 17 (see estimate (5.2)), a unitary representation $\rho$ belongs to $\mathrm{Spec} _{1-\varepsilon}(A)$ if and only if

$$\Biggl\| \sum_{g\in {\mathbf G} } (A * A^{-1})(g) (I - \rho(g)) \Biggr\| \le (2 \varepsilon - \varepsilon^2) |A|^2 = \bigl(1-(1-\varepsilon)^2\bigr)|A|^2.$$

However,

$$ I-\rho_1(g) \rho_2(g) = (I-\rho_1(g)) \rho_2(g) + I - \rho_2(g)$$

(A.1)

and hence, by the triangle inequality for the operator norm, we get

$$\Biggl\|\sum_{g\in {\mathbf G} } (A * A^{-1})(g) (I - \rho_1(g) \rho_2(g))\Biggr\| \le \bigl(2 \varepsilon_1 - \varepsilon^2_1 + 2 \varepsilon_2 - \varepsilon^2_2\bigr) |A|^2 = \bigl(1-(1- \varepsilon_1-\varepsilon_2)^2\bigr) |A|^2,$$

as required. $\quad\Box$

Our next result shows that $\operatorname{Bohr} (\rho,\delta)$ has small product set and hence it suffices to check the condition of smallness of the quantity $\sigma^{(d)}_P (B) \le 1-\alpha$ in Theorem 18 only for sets with small product.

Proposition 27.

Let $\delta \in [0,2/5]$ be a real number and $\rho$ a unitary representation. Then

$$\bigl| \operatorname{Bohr} (\rho,\delta) \operatorname{Bohr} (\rho,\delta)\bigr| \le 2^{21 d_\rho^2/2} \mathopen| \operatorname{Bohr} (\rho,\delta)|$$

and there are sets $X,Y \subseteq {\mathbf G} ,$ $|X|,|Y|< 2^{25 d_\rho^2},$ such that

$$\operatorname{Bohr} (\rho,\delta) \subseteq \operatorname{Bohr} \biggl( \rho,\frac\delta 2 \biggr) X \qquad\textit{and}\qquad \operatorname{Bohr} (\rho,\delta) \subseteq Y \operatorname{Bohr} \biggl( \rho,\frac\delta 2 \biggr).$$

Proof.

Let $k=d_\rho$. In view of (5.1) it is enough to compare $\mathopen| \operatorname{Bohr} (\rho,\delta)|$ and $\mathopen| \operatorname{Bohr} (\rho,2\delta)|$. Further, one can check that $2(1-\cos \theta) \le \theta^2$ and $2(1-\cos \theta) \ge \theta^2/2$ for $|\theta|\le \sqrt{6}$. Put

$$\eta := \eta (\delta) = \frac{1}{2\pi} \arccos \biggl( 1 - \frac{\delta^2}{2} \biggr).$$

We have $\delta/(2\pi)\le \eta(\delta) \le \delta/\pi$. Let $U(\delta)$ be the set of unitary matrices $U$ such that $\|U - I\| \le \delta$. In [14, Lemma 17.4] it was proved that the Haar measure $\mu$ of $U(\delta)$ equals

$$ \begin{aligned} \, \mu(U(\delta)) &= \frac{1}{k!} \intop_{-\eta}^{\eta} \dots \intop_{-\eta}^{\eta} \prod_{1\le n<m \le k} \bigl|e^{2\pi i \theta_n} - e^{2\pi i \theta_m}\bigr|^2 \,d\theta_1 \dots d\theta_k \\[3pt] &= \frac{1}{k!} \intop_{-\eta}^{\eta} \dots \intop_{-\eta}^{\eta} \prod_{1\le n<m \le k} 2\bigl(1-\cos(2\pi(\theta_n-\theta_m))\bigr) \,d\theta_1 \dots d\theta_k. \end{aligned}$$

(A.2)

Put

$$F(k) = \frac{(2\pi)^{k(k-1)}}{k!} \intop_{-1}^1 \dots \intop_{-1}^1 \prod_{1\le n<m \le k} (\theta_n-\theta_m)^2 \,d\theta_1 \dots d\theta_k.$$

From (A.2) and our bounds for $2(1-\cos \theta)$, it follows that

$$2^{-k(k-1)/2}\, \eta^{k^2} F(k) \le \mu(U(\delta)) \le \eta^{k^2} F(k),$$

because $4\pi \eta(1) =2\pi/3 \le \sqrt{6}$. Using the assumption $\delta \le 2/5$ and the previous formula, we obtain

$$ \frac{\mu(U(5\delta/2))}{\mu(U(\delta/2))} \le 2^{k(k-1)/2} \,\frac{\eta(5\delta/2)^{k^2}}{\eta(\delta/2)^{k^2}} \le 2^{k(k-1)/2} \cdot 2^{10k^2} < 2^{21k^2/2}.$$

(A.3)

Now let $V = \rho ( {\mathbf G} )$. It is easy to see that $\mathopen| \operatorname{Bohr} (\rho,\delta)| \ge |V\cap U(\delta/2) u|$ for any unitary matrix $u$, because $U(\delta/2) u (U(\delta/2) u)^{-1} \subseteq U(\delta)$. Further, integrating with respect to the Haar measure, we get in view of (A.3)

$$\begin{aligned} \, \mathopen| \operatorname{Bohr} (\rho,2\delta)| &= \biggl(\mu\biggl(U\biggl( \frac\delta 2 \biggr)\!\biggr)\!\biggr)^{\!-1} \intop \biggl|\rho( \operatorname{Bohr} (\rho,2\delta)) \cap U\biggl( \frac\delta 2 \biggr) u\biggr|\,du \\[4pt] &\le \mathopen| \operatorname{Bohr} (\rho,\delta)| \frac{\mu(U(5\delta/2))}{\mu(U(\delta/2))} < 2^{21k^2/2} \kern1pt \mathopen| \operatorname{Bohr} (\rho,\delta)|. \end{aligned}$$

Now by the Ruzsa covering lemma (see, e.g., [19]) one finds $X$ (and similarly $Y$) such that

$$\operatorname{Bohr} (\rho,\delta) \subseteq \operatorname{Bohr} \biggl( \rho,\frac\delta 4 \biggr) \operatorname{Bohr} ^{-1}\biggl( \rho,\frac\delta 4 \biggr) \kern1pt X \subseteq \operatorname{Bohr} \biggl( \rho,\frac\delta 2 \biggr) \kern1pt X,$$

where, as above,

$$|X| \le \frac{\mathopen| \operatorname{Bohr} (\rho,5\delta/4)|}{\mathopen| \operatorname{Bohr} (\rho,\delta/4)|} \le \frac{\mu( \operatorname{Bohr} (\rho,11\delta/8))}{\mu( \operatorname{Bohr} (\rho,\delta/8))} \le 2^{k(k-1)/2}\, \frac{\eta(11\delta/8)^{k^2}}{\eta(\delta/8)^{k^2}} < 2^{25 k^2}.$$

This completes the proof. $\quad\Box$

Having a lower bound for the size of one-dimensional Bohr sets (see [14, Lemma 17.3] or the proposition above), one can obtain a lower bound for the size of Bohr sets with an arbitrary $\Gamma$.

Proposition 28.

Let $\operatorname{Bohr} (\rho_j,\delta_j),$ $j=1,\dots,k,$ be Bohr sets such that $\delta_1 \le \delta_2 \le \dots \le \delta_k$. Then

$$\bigl| \operatorname{Bohr} (\{\rho_1,\dots,\rho_k\}, \delta_k)\bigr| \ge | {\mathbf G} |^{-1} \prod_{j=1}^k \biggl| \operatorname{Bohr} \biggl(\rho_j,\frac{\delta_j}2 \biggr)\biggr|.$$

Proof.

Let $B = \operatorname{Bohr} (\{\rho_1,\dots,\rho_k\}, \delta_k)$ and $B_j = \operatorname{Bohr} (\rho_j, \delta_j/2)$, $j=1,2,\dots,k$. Clearly, for any $j$ one has $B_j B^{-1}_j \subseteq B$. Hence

$$ \begin{aligned} \, \sigma:= {}&\mathopen{}\sum_{x\in {\mathbf G} } (B_1 * B^{-1}_1)(x) \dots (B_k * B^{-1}_k)(x) = \sum_{x\in B} (B_1 * B^{-1}_1)(x) \dots (B_k * B^{-1}_k)(x) \\[4pt] \le{}& |B|\cdot |B_1|\cdot \dots\cdot |B_k|. \end{aligned}$$

(A.4)

On the other hand, in view of formulas (2.4) and (2.5), we get

$$ \sigma = \frac{1}{| {\mathbf G} |} \sum_{\rho} d_\rho \bigl< \widehat{B}{} _1 (\rho) \widehat{B}^*_1 (\rho) \dots \widehat{B}{} _{k-1} (\rho) \widehat{B}{} ^*_{k-1} (\rho), \widehat{B}{} _k (\rho) \widehat{B}{} ^*_k (\rho) \bigr> \ge \frac{|B_1|^2 \dots |B_k|^2}{| {\mathbf G} |},$$

(A.5)

because the operators $\widehat{B}{} _1 (\rho) \widehat{B}{} ^*_1 (\rho)$ are Hermitian and positive semidefinite. Comparing (A.4) and (A.5), we obtain the result. $\quad\Box$

A Bohr set $\operatorname{Bohr} (\rho,\delta)$ is said to be regular if

$$\bigl| \mathopen| \operatorname{Bohr} (\rho,(1+\kappa)\delta)| - \mathopen| \operatorname{Bohr} (\rho,\delta)| \bigr| \le 100 \kern1pt d^2_\rho |\kappa| \cdot \mathopen| \operatorname{Bohr} (\rho,\delta)|$$

whenever $|\kappa| \le 1/(100 d^2_\rho)$. Even in the abelian case it is easy to see that not every Bohr set is regular (see, e.g., [19, Sect. 4.4]). Nevertheless, it was shown in [2] that for ${\mathbf G} = {\mathbb Z}/N{\mathbb Z}$ one can find a regular Bohr set by slightly decreasing the parameter $\delta$. We show the same for general groups, repeating the arguments from [19, Lemma 4.25] (see also [14, Lemma 9.3]).

Proposition 29.

Let $\delta \in [0,1/2]$ be a real number and $\rho$ a unitary representation. Then there is a $\delta_1 \in [\delta,2\delta]$ such that $\operatorname{Bohr} (\rho,\delta_1)$ is regular.

Proof.

Consider the nondecreasing function $f \colon\, [0,1] \to {\mathbb R}$ defined as

$$f(a) := d^{-2}_\rho \log \mu( \operatorname{Bohr} (\rho,2^a \delta)).$$

By the first part of Proposition 27, we have $f(1)- f(0) \le \log (21/2)$. Clearly, if we could find $a\in [0.1, 0.9]$ such that $|f(a) - f(a')| \le 25|a-a'|$ for all $|a'-a| \le 0.1$, then the set $\operatorname{Bohr} (\rho,2^a \delta)$ would be regular. If there is no such $a$, then for every $a$ from this interval there is an interval $I_a$, $a\in I_a$, $|I_a| \le 0.1$, with $\intop_{I_a}df> 25|I_a|$. Obviously, these intervals cover $[0.1, 0.9]$, and by the Vitali covering lemma (see, e.g., [19, Sect. 4.4]) one can find a finite subcollection of disjoint intervals of total measure at least $0.8/5$, say. Then

$$\log \frac{21}2 \ge \intop_0^1\! df \ge 25 \cdot \frac{0.8}{5} = 4,$$

which is a contradiction. $\quad\Box$

Finally, let us say something nontrivial about the spectrum of regular Bohr sets.

Proposition 30.

Let $B= \operatorname{Bohr} (\rho,\delta)$ be a regular Bohr set and $B' = \operatorname{Bohr} (\rho,\delta'),$ where $\delta' \le \kappa \delta/(100 d_\rho^2)$ and $\kappa \in (0,1)$ is a real number. Then

$$\mathrm{Spec} _{\varepsilon} (B) \subseteq \mathrm{Spec} _{1-{2\kappa}/{\varepsilon}} (B').$$

Proof.

Let $\pi \in \mathrm{Spec} _{\varepsilon} (B)$. Let also $B^\pm = \operatorname{Bohr} (\rho,\delta\pm\delta')$. We have

$$ \begin{aligned} \, \varepsilon |B| \le \| \widehat{B}{} (\pi) \| &\le |B'|^{-1} \| \widehat{B}{} (\pi)\|\cdot\| \widehat{B}{} ' (\pi)\| + \Biggl\| \sum_x \bigl(B(x) - |B'|^{-1} (B*B')(x)\bigr) \pi(x) \Biggr\| \\[4pt] &\le |B'|^{-1} \| \widehat{B}{} (\pi)\|\cdot\| \widehat{B}{} ' (\pi)\| + \sum_x \bigl|B(x) - |B'|^{-1} (B*B')(x)\bigr|. \end{aligned}$$

(A.6)

It is easy to see that the summation in (A.6) is taken over $B^+\setminus B^-$. By the regularity of $B$ one can estimate this sum as $2\kappa |B|$. Hence

$$\| \widehat{B}{} ' (\pi)\| \ge |B'| \bigl(1-2\kappa |B|\cdot\| \widehat{B}{} (\pi)\|^{-1}\bigr) \ge |B'| (1-2\kappa \varepsilon^{-1})$$

or, in other words, $\pi \in \mathrm{Spec} _{1-2\kappa\varepsilon^{-1}} (B')$. This completes the proof. $\quad\Box$