Spectra on finite Schreier graphs

Since the Schreier graphsΓnare finite, let us start by computing the matrix of∆n. The set of vertices ofΓ_nisXⁿ, let us order them lexicographically. The matrix of∆_nis a matrix of sizedⁿ×dⁿ. We will usually write it as ad×dblock matrix where each block is a matrix of sizedⁿ⁻¹×dⁿ⁻¹. A block denoted by a scalar is the corresponding multiple of the identity matrixI_dⁿ⁻¹.

Proof. In order to write the adjacency matrix ofΓ_n, let us first write the adjacency matrices associated with each of the generators inSwe consider. For a generators∈S, we denote its associated adjacency matrix forΓ_nbys_n, and ifu, v∈Xⁿ, its coefficient(u, v)is1if s(u) =vand0otherwise.

Recall thatais the generator ofA, and it permutes the subtrees of the first level cyclically.

This means that, for everyi∈Xandv∈Xⁿ⁻¹,a(iv) = (i+ 1)v. We can therefore write the adjacency matrix of this generator as

a₀ = 1, a_n=

Now, for anyb∈Bandk≥0, if we write the matrix

Recall that these matrices have sizedⁿ×dⁿ, so every block is itself a matrix of size dⁿ⁻¹×dⁿ⁻¹, and scalars denote the corresponding multiple of the identity matrixI_dⁿ⁻¹.

Now notice that, for everyn≥0, The former is clear, and, for the latter, we have, for everyk≥0,

X

The sum in the first block does not depend on k, since ω_k is an epimorphism and all elements ofAhave exactlyd^m−1preimages inB. Hence we can inductively conclude that

P

If we now try to find the characteristic polynomial of∆_n, we will not find any explicit relation with that of∆n−1. Instead, we consider the matrix

Qn(p, q) :=Bn+pAn−q.

These additional parameters will allow us to find a relation between the determinant of Q_n(p, q)and Qn−1(p⁰, q⁰), for some differentp⁰ andq⁰. According to Lemma 4.1.1, by settingp = 1, q = 0, we recover the matrix of∆n, so more specifically we want to find sp(∆_n) ={q| |Q_n(1, q)|= 0}.

As mentioned above, the strategy consists of two steps. First, we will provide the relation between the determinants ofQn(p, q)andQn−1(p⁰, q⁰)(Proposition 4.1.4). Second, we will solve this recurrence to find a factorization of the determinant ofQ_n(p, q)(Proposition 4.1.6).

Our computations involve matrices of the formrAn+s, withr, s∈R, so let us start with the following result, which will be useful later on.

Lemma 4.1.2. Letr, s, r⁰, s⁰∈R. Then, diagonal andd−2ones for the rest, which shows the claim.

For(2), we have For(3), we can verify, using(1), that

(rAn+s) (rAn−(d−2)r−s) =r²A²_n−(d−2)r²An−(d−2)rs−s²=

=r²((d−2)An+d−1)−(d−2)r²An−(d−2)rs−s²= (r−s)(s+ (d−1)r).

Claim(4)can be checked directly, again using(1).

Proposition 4.1.3. Forn= 0andn= 1, we have

Proposition 4.1.4. Forn≥2, we have

|Q_n(p, q)|= (αβ^d2−3d+1γ^d−1)^dn−2

Qn−1(p⁰, q⁰) , withαandβas in Proposition 4.1.3,

p⁰ := d^m−1β

Proof. We start computing the determinant of|Q_n(p, q)|directly, performing elementary transformations of rows and columns in determinants.

|Q_n(p, q)|=

=

We continue the computation of the determinant by taking the first Schur complement.

Namely, whenever a matrixP is invertible, we have equality

Let us now compute these two determinants. First, using Lemma 4.1.2 with r = d^m−1(α−p)ands=γ+d^m−1(α−p), we obtain

|C_n|=h

(s−r)^d−1(s+ (d−1)r)idⁿ⁻¹

= (αβγ^d−1)^dn−1, as well as

C_n⁻¹= −1 αβγ

d^m−1(α−p)(An−(d−1))−γ . Similarly, again by Lemma 4.1.2 but now with

r= (d−2)d^m−1, s= (d−1) β−(d−2)d^m−1 , r⁰ =d^m−1(α−p), s⁰ =−(γ+ (d−1)d^m−1(α−p)), we find

D_nC_n⁻¹ = −1 αγ

d^m−1βA_n−(d−1)δ . Indeed,

(d−2)rr⁰+rs⁰+r⁰s=

= (d−2)rr⁰−r(γ+ (d−1)r⁰) +r⁰(d−1)(β−r) =

=−drr⁰−rγ+ (d−1)βr⁰ =

=d^m−1[(d−1)β(α−p)−(d−2)(γ+d^m(α−p))] =

=d^m−1β[α−(d−1)p] =d^m−1β², and

(d−1)rr⁰+ss⁰ =

= (d−1)rr⁰−(d−1)(β−r)(γ+ (d−1)r⁰) =

= (d−1)(drr⁰+rγ−β(γ+ (d−1)r⁰)) =

= (d−1)(r(γ+dr⁰)−β(γ+ (d−1)r⁰)) =

= (d−1)(rαβ−β(γ+ (d−1)r⁰)) =

=−(d−1)β(γ+ (d−1)d^m−1(α−p)−(d−2)d^m−1α) =

=−(d−1)β(γ+d^m−1(α+ (d−1)p)) =

=−(d−1)β(γ+d^m−1β) =−(d−1)βδ.

Therefore,

Bn−1−q−p²Dn−1C_n−1⁻¹ = Bn−1−q+ p²

αγ

d^m−1βAn−1−(d−1)δ

=

Bn−1+d^m−1β

αγ p²An−1−

q+(d−1)δ αγ p²

=

=Bn−1+p⁰An−1−q⁰ =

=Qn−1(p⁰, q⁰).

Finally, we conclude the computation of the determinant ofQn(p, q):

|Q_n(p, q)|= (β^d−3)^dn−1|C_n−1|

Bn−1−q−p²Dn−1C_n−1⁻¹ =

= (β^d−3)^dn−1(αβγ^d−1)^dn−2

Qn−1(p⁰, q⁰) =

= (αβ^d2−3d+1γ^d−1)^dn−2

Qn−1(p⁰, q⁰) .

This concludes the first part of the strategy, finding a recurrence relation between the de-terminants ofQn(p, q)andQn−1(p⁰, q⁰). For the next part, we need to unfold this recurrence relation to get a factorization of|Q_n(p, q)|. Proposition 4.1.3 provides it forn= 0,1. Let us inductively compute it forn≥2.

Proposition 4.1.5. Forn= 2, we have

|Q₂(p, q)|= (α+p)β^(d−2)d+1H₀^d−1, where

Hx :=Hx(p, q) =q²−((d−2)p+d^m−2)q−((d−1)p²+ (d^m−1x+d−2)p+d^m−1).

Proof. Letα⁰ :=α(p⁰, q⁰)andβ⁰ :=β(p⁰, q⁰)following the definition in Proposition 4.1.3.

Then, by that Proposition and Proposition 4.1.4,

|Q₂(p, q)|=αβ^d2−3d+1γ^d−1

Q₁(p⁰, q⁰)

=αβ^d2−3d+1γ^d−1(α⁰+p⁰)β^0d−1. We can verify the following relations

α⁰+p⁰= β

α(α+p), β⁰ = β γH₀. Therefore,

|Q₂(p, q)|=αβ^d2−3d+1γ^d−1β

α(α+p) β

γH₀ d−1

=

= (α+p)β^(d−2)d+1H₀^d−1.

The motivation for the definition of the polynomials H_x from Proposition 4.1.5 will become apparent in Proposition 4.1.6. They form a family of polynomials inpandqindexed by the pointx∈R. For different values ofx ∈R, the equationH_x = 0defines different hyperbolas inpandq.

Proposition 4.1.6. For anyn≥2, we have the factorization

|Q_n(p, q)|= (α+p)β^{(d−2)dn−1+1}

withαandβas in Proposition 4.1.3,H_xas in Proposition 4.1.5 andF being the map F(x) =x²−d(d−1).

Proof. The case n = 2is shown in Proposition 4.1.5. We use again the recurrence in Proposition 4.1.4 to show the result forn≥3inductively. LetH_x⁰ :=H_x(p⁰, q⁰). We can

The relation between the determinants ofQ_n(p, q) and Qn−1(p⁰, q⁰) is given by the substitutionp7→p⁰,q 7→q⁰. ForQ2, one of the factors of the determinant is the polynomial we calledH₀. To compute the determinant ofQ₃, we have to developH₀⁰. It is in this analysis that the polynomialsH_xand the mapF arise. They are the link betweenH_x⁰ and Hy that allows us to unfold the recurrence.

From the factorization in Proposition 4.1.6 we can extractsp(∆_n), as we mentioned above, by settingp= 1. Recall that|S|=d^m+d−2.

Theorem 4.1.7. LetG_ωbe a spinal group withd≥2,m≥1andω∈Ω_d,m, and let∆_nbe the adjacency operator for the spinal generating setSon the Schreier graphΓ_n, forn≥0.

We have

sp(∆₀) ={|S|}, sp(∆1) ={|S|,|S| −d}, and, forn≥2,

sp(∆_n) ={|S|,|S| −d} ∪ψ⁻¹

n−2

[

k=0

F^−k(0)

! ,

whereF(x) =x²−d(d−1)andψ(x) = _dm−1¹ (x²−(|S| −2)x−(|S|+d−2)).

Proof. We already established thatsp(∆n) ={q| |Q_n(1, q)|= 0}. By Propositions 4.1.3 and 4.1.6, the determinant vanishes in the following cases:

• α+ 1 = 0. Then|S| ∈sp(∆n)with multiplicity1, for everyn≥0.

• β = 0. Thend^m−2 =|S| −d∈sp(∆_n)with multiplicity(d−2)dⁿ⁻¹+ 1, ifn≥1.

• Hx = 0, for somex∈F^−k(0)with0≤k≤n−2. This implies that

|S| −2

2 ±

s

|S| −2 2

2

+|S|+d−2 +d^m−1x∈sp(∆_n)

each with multiplicity(d−2)d^n−k−2+1. These two eigenvalues are the two preimages ofxby the mapψdefined above.

Remark 4.1.8. Note that, for spinal groups withm = 1, we have the equalityψ(x) = F(x−(d−2)). In that case, we can rewrite

sp(∆_n) ={|S|} ∪

n−1

[

k=0

G^−k(d−2), ∀n≥1, withG(x) =ψ(x) +d−2 =F(x−(d−2)) +d−2.

−2 2

−2 2

0

F⁻¹(0)

−2 2

−2 2

F⁻¹(0)

F⁻²(0)

Figure 4.1: Construction of the preimages of0byF ford= 2. The set of all preimages is dense in the interval[−2,2].

−3 3

−3 3

0

F⁻¹(0) −3−3 3

3

F⁻¹(0)

F⁻²(0)

Figure 4.2: Construction of the preimages of0byF ford= 3. The set of all preimages accumulates on a Cantor set.

Remark 4.1.9. The mapψis symmetric about its minimal point ^|S|−2₂ , and satisfies

ψ⁻¹(d) ={|S|,−2}, ψ⁻¹(−d) =







|S| −2

2 ±

s

|S| −2 2

2

+ 2(d−2)





 ,

ψ⁻¹(−d(d−1)) ={|S| −d, d−2}.

The preimages of 0 by the mapF are contained in[−d, d], as shown in Figures 4.1 and 4.2 ford= 2,3, and they accumulate on its Julia set, which is the entire interval[−2,2]

ifd= 2or a Cantor set ifd >2.

Dans le document Structural and Spectral Properties of Schreier Graphs of Spinal Groups (Page 71-81)