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Changing topics, but still obsessed by Fourier analysis, let us discuss now, following , some sharp results in the circulant graph case. Let us start with:

Definition

Associated to any circulant graph [math]X[/math] having [math]N[/math] vertices are:

• The set [math]S\subset\mathbb Z_N[/math] given by [math]i-_Xj\iff j-i\in S[/math].
• The group [math]E\subset\mathbb Z_N^*[/math] consisting of elements [math]g[/math] such that [math]gS=S[/math].
• The number [math]k=|E|[/math], called type of [math]X[/math].

The interest in the type [math]k[/math] is that this is the good parameter measuring the complexity of the spectral theory of [math]X[/math], as we will soon see. To start with, here are a few basic examples, and properties of the type, with [math]\varphi[/math] being the Euler function:

(1) The type can be [math]2,4,6,8,\ldots[/math] This is because [math]\{\pm 1\}\subset E[/math].

(2) [math]C_N[/math] is of type [math]2[/math]. Indeed, we have [math]S=\{\pm 1\}[/math], [math]E=\{\pm 1\}[/math].

(3) [math]K_N[/math] is of type [math]\varphi(N)[/math]. Indeed, here [math]S=\emptyset[/math], [math]E=\mathbb Z_N^*[/math].

Let us first discuss the spectral theory of the circulant graphs. In what follows [math]X[/math] will be a circulant graph having [math]p[/math] vertices, with [math]p[/math] prime. We denote by [math]\xi[/math] the column vector [math](1,w,w^2,\ldots ,w^{p-1})[/math], where [math]w=e^{2\pi i/p}[/math]. With this convention, we have:

Theorem

The eigenspaces of [math]d[/math] are given by [math]V_0={\mathbb C}1[/math] and

[[math]] V_x=\bigoplus_{a\in E}{\mathbb C}\,\xi^{xa} [[/math]]

with [math]x\in \mathbb Z_p^*[/math]. Moreover, we have [math]V_x=V_y[/math] if and only if [math]xE=yE[/math].

Show Proof

Since [math]d[/math] is circulant, we have [math]d(\xi^x)=f(x)\xi^x[/math], with [math]f:{\mathbb Z}_p\to{\mathbb C}[/math] being:

[[math]] f(x)=\sum_{t\in S}w^{xt} [[/math]]

Let [math]K={\mathbb Q}(w)[/math] and let [math]H[/math] be the Galois group of the Galois extension [math]\mathbb Q \subset K[/math]. We have then a well-known group isomorphism, as follows:

[[math]] \mathbb Z_p^*\simeq H\quad,\quad x\to s_x=[w\to w^x] [[/math]]

Also, we know from a standard theorem of Dedekind that the family [math]\{s_x\mid x\in{\mathbb Z}_p^*\}[/math] is free in [math]{\rm End}_{\mathbb Q}(K)[/math]. Now for [math]x,y\in \mathbb Z_p^*[/math] consider the following operator:

[[math]] L = \sum_{t \in S} s_{xt} - \sum_{t \in S} s_{yt} \in End_{\mathbb Q}(K) [[/math]]

We have [math]L({w}) = f(x)-f(y)[/math], and since [math]L[/math] commutes with the action of [math]H[/math], we have:

[[math]] L=0 \iff L({w}) =0 \iff f(x)=f(y) [[/math]]

By linear independence of the family [math]\{s_x\mid x\in \mathbb Z_p^*\}[/math] we get:

[[math]] f(x) = f(y) \iff xS=yS \iff xE=yE [[/math]]

It follows that [math]d[/math] has precisely [math]1+(p-1)/k[/math] distinct eigenvalues, the corresponding eigenspaces being those in the statement.

■

Consider now a commutative ring [math](R,+,\cdot)[/math]. We denote by [math]R^*[/math] the group of invertibles, and we assume [math]2\in R^*[/math]. A subgroup [math]G\subset R^*[/math] is called even if [math]-1\in G[/math]. We have:

Definition

An even subgroup [math]G\subset R^*[/math] is called [math]2[/math]-maximal if, inside [math]G[/math]:

[[math]] a-b=2(c-d)\implies a=\pm b [[/math]]

We call [math]a=b,c=d[/math] trivial solutions, and [math]a=-b=c-d[/math] hexagonal solutions.

To be more precise, in what regards our terminology, consider the group [math]G\subset{\mathbb C}[/math] formed by [math]k[/math]-th roots of unity, with [math]k[/math] even. An equation of the form [math]a-b=2(c-d)[/math] with [math]a,b,c,d\in G[/math] says that the diagonals [math]a-b[/math] and [math]c-d[/math] must be parallel, and that the first one is twice as much as the second one. But this can happen only when [math]a,c,d,b[/math] are consecutive vertices of a regular hexagon, and here we have [math]a+b=0[/math].

The relation with our quantum symmetry considerations will come from:

Proposition

Assume that [math]R[/math] has the property [math]3\neq 0[/math], and consider a [math]2[/math]-maximal subgroup [math]G\subset R^*[/math]. Then, the following happen:

• [math]2,3\not\in G[/math].
• [math]a+b=2c[/math] with [math]a,b,c\in G[/math] implies [math]a=b=c[/math].
• [math]a+2b=3c[/math] with [math]a,b,c\in G[/math] implies [math]a=b=c[/math].

Show Proof

All these assertions are elementary, as follows:

(1) This follows from the following formulae, which cannot hold in [math]G[/math]:

[[math]] 4-2=2(2-1)\quad,\quad 3-(-1)=2(3-1) [[/math]]

Indeed, the first one would imply [math]4=\pm 2[/math], and the second one would imply [math]3=\pm 1[/math]. But from [math]2\in R^*[/math] and [math]3\neq 0[/math] we get [math]2,4,6\neq 0[/math], contradiction.

(2) We have [math]a-b=2(c-b)[/math]. For a trivial solution we have [math]a=b=c[/math], and for a hexagonal solution we have [math]a+b=0[/math], hence [math]c=0[/math], hence [math]0\in{G}[/math], contradiction.

(3) We have [math]a-c=2(c-b)[/math]. For a trivial solution we have [math]a=b=c[/math], and for a hexagonal solution we have [math]a+c=0[/math], hence [math]b=-2a[/math], hence [math]2\in{G}[/math], contradiction.

■

As a first result now, coming from this study, we have:

Theorem

A circulant graph [math]X[/math], on [math]p\geq 5[/math] prime vertices, such that

[[math]] E\subset {\mathbb Z}_p [[/math]]

is [math]2[/math]-maximal, has no quantum symmetry, [math]G^+(X)=G(X)[/math].

Show Proof

This comes from the above results, via a long algebraic study, as follows:

(1) We use Proposition 15.24, which ensures that [math]V_1,V_2,V_3[/math] are eigenspaces of [math]d[/math]. By [math]2[/math]-maximality of [math]E[/math], these eigenspaces are different. From the eigenspace preservation in Theorem 15.22 we get formulae of the following type, with [math]r_a,r_a',r_a''\in{\mathcal A}[/math]:

[[math]] \alpha (\xi)=\sum_{a\in E}\xi^a\otimes r_a\quad,\quad \alpha (\xi^2)=\sum_{a\in E}\xi^{2a}\otimes r_a'\quad,\quad \alpha (\xi^3)=\sum_{a\in E}\xi^{3a}\otimes r_a'' [[/math]]

(2) We take the square of the first relation, we compare with the formula of [math]\alpha(\xi^2)[/math], and we use [math]2[/math]-maximality. We obtain in this way the following formula:

[[math]] \alpha(\xi^2)=\sum_{c\in E}\xi^{2c}\otimes r_c^2 [[/math]]

(3) We multiply now this relation by the formula of [math]\alpha(\xi)[/math], we compare with the formula of [math]\alpha(\xi^3)[/math], and we use [math]2[/math]-maximality. We obtain the following formula:

[[math]] \alpha(\xi^3)=\sum_{b\in E}\xi^{3b}\otimes r_b^3 [[/math]]

(4) As a conclusion, the three formulae in the beginning are in fact as follows:

[[math]] \alpha (\xi)=\sum_{a\in E}\xi^a\otimes r_a\quad,\quad \alpha(\xi^2)=\sum_{a\in E}\xi^{2a}\otimes r_a^2\quad,\quad \alpha(\xi^3)=\sum_{a\in E}\xi^{3a}\otimes r_a^3 [[/math]]

(5) Our claim now is that for [math]a\neq b[/math], we have the following “key formula”:

[[math]] r_ar_b^3=0 [[/math]]

Indeed, in order to prove this claim, consider the following equality:

[[math]] \left(\sum_{a\in E}\xi^a\otimes r_a\right) \left(\sum_{b\in E}\xi^{2b}\otimes r_b^2\right)=\sum_{c\in E}\xi^{3c}\otimes r_c^3 [[/math]]

By eliminating all [math]a=b[/math] terms, which produce the sum on the right, we get:

[[math]] \sum\left\{r_ar_b^2\Big| a,b\in E,\,a\neq b,\,a+2b=x\right\}=0 [[/math]]

(6) We fix now elements [math]a,b\in E[/math] satisfying [math]a\neq b[/math]. We know from [math]2[/math]-maximality that the equation [math]a+2b=a'+2b'[/math] with [math]a',b'\in E[/math] has at most one non-trivial solution, namely the hexagonal one, given by [math]a'=-a[/math] and [math]b'=a+b[/math]. Now with [math]x=a+2b[/math], we get that the above equality is in fact one of the following two equalities:

[[math]] r_ar_b^2=0\quad,\quad r_ar_b^2+r_{-a}r_{a+b}^2=0 [[/math]]

(7) In the first case, we are done. In the second case, we know that [math]a_1=b[/math] and [math]b_1=a+b[/math] are distinct elements of [math]E[/math]. So, consider the following equation, over [math]E[/math]:

[[math]] a_1+2b_1=a_1'+2b_1' [[/math]]

The hexagonal solution of this equation, given by [math]a_1'=-a_1[/math] and [math]b_1'=a_1+b_1[/math], cannot appear, because [math]b_1'=a_1+b_1[/math] can be written as [math]b_1'=a+2b[/math], and by [math]2[/math]-maximality we get [math]b_1'=-a=b[/math], which contradicts [math]a+b\in E[/math]. Thus the equation [math]a_1+2b_1=a_1'+2b_1'[/math] with [math]a_1',b_1'\in E[/math] has only trivial solutions, and with [math]x=a_1+2b_1[/math] in the above, we get:

[[math]] r_{a_1}r_{b_1}^2=0 [[/math]]

Now remember that this follows by identifying coefficients in [math]\alpha(\xi)\alpha(\xi^2)=\alpha(\xi^3)[/math]. The same method applies to the formula [math]\alpha(\xi^2)\alpha(\xi)=\alpha(\xi^3)[/math], and we get as well:

[[math]] r_{b_1}^2r_{a_1}=0 [[/math]]

We have now all ingredients for finishing the proof of the key formula, as follows:

[[math]] r_ar_b^3 =r_ar_b^2r_b =-r_{-a}r_{a+b}^2r_b =-r_{-a}r_{b_1}^2r_{a_1} =0 [[/math]]

(8) We come back now to the following formula, proved for [math]s=1,2,3[/math]:

[[math]] \alpha(\xi^s)=\sum_{a\in E}\xi^{sa}\otimes r_a^s [[/math]]

By using the key formula, we get by recurrence on [math]s[/math] that this holds in general.

(9) In particular with [math]s=p-1[/math] we get the following formula:

[[math]] \alpha(\xi^{-1})=\sum_{a\in E}\xi^{-a}\otimes r_a^{p-1} [[/math]]

On the other hand, from [math]\xi^*=\xi^{-1}[/math] we get the following formula:

[[math]] \alpha(\xi^{-1})=\sum_{a\in E}\xi^{-a}\otimes r_a^* [[/math]]

But this gives [math]r_a^* = r_a^{p-1}[/math] for any [math]a[/math]. Now by using the key formula we get:

[[math]] (r_ar_b)(r_ar_b)^* =r_ar_br_b^*r_a^* =r_ar_b^pr_a^* =(r_ar_b^3)(r_b^{p-3}r_a^*) =0 [[/math]]

(10) But this gives [math]r_ar_b=0[/math]. Thus, we have the following equalities:

[[math]] r_ar_b=r_br_a=0 [[/math]]

On the other hand, [math]{\mathcal A}[/math] is generated by coefficients of [math]\alpha[/math], which are in turn powers of elements [math]r_a[/math]. It follows that [math]{\mathcal A}[/math] is commutative, and we are done.

■

Still following , we can now formulate a main result, as follows:

Theorem

A type [math]k[/math] circulant graph having [math]p \gt \gt k[/math] vertices, with [math]p[/math] prime, has no quantum symmetry.

Show Proof

This follows from Theorem 15.25 and some arithmetics, as follows:

(1) Let [math]k[/math] be an even number, and consider the group of [math]k[/math]-th roots of unity [math]G=\{1,w,\ldots,w^{k-1}\}[/math], where [math]w=e^{2\pi i/k}[/math]. By standard arithmetics, [math]G[/math] is [math]2[/math]-maximal in [math]{\mathbb C}[/math].

(2) As a continuation of this, again by some standard arithmetics, for [math]p \gt 6^{\varphi(k)}[/math], with [math]\varphi[/math] being the Euler function, any subgroup [math]E\subset{\mathbb Z}_p^*[/math] of order [math]k[/math] is [math]2[/math]-maximal.

(3) But this proves our result. Indeed, by using (2), we can apply Theorem 15.25 provided that we have [math]p \gt 6^{{\varphi(k)}}[/math], and our graph has no quantum symmetry, as desired.

■

We should mention that the above result, from , is quite old. The challenge is to go beyond this, with results for the graphs having an abelian group action, [math]A\curvearrowright X[/math].

General references

Banica, Teo (2024). "Graphs and their symmetries". arXiv:2406.03664 [math.CO].