Renormalisation of Pair Correlation Measures for Primitive Inflation Rules and Absence of Absolutely Continuous Diffraction

Abstract

The pair correlations of primitive inflation rules are analysed via their exact renormalisation relations. We introduce the inflation displacement algebra that is generated by the Fourier matrix of the inflation and deduce various consequences of its structure. Moreover, we derive a sufficient criterion for the absence of absolutely continuous diffraction components, as well as a necessary criterion for its presence. This is achieved via estimates for the Lyapunov exponents of the Fourier matrix cocycle of the inflation rule. We also discuss some consequences for the spectral measures of such systems. While we develop the theory first for the classic setting in one dimension, we also present its extension to primitive inflation rules in higher dimensions with finitely many prototiles up to translations.

Appendix: The (Skew) Kronecker Product Algebra

The structure of the correlation measures relies on some properties of the Kronecker product matrices

$$\begin{aligned} \varvec{A} (k) \, {:}{=}\, B(k) \otimes \overline{B(k)} , \end{aligned}$$

(36)

defined for $k\in \mathbb {R}$. Obviously, one has $\overline{\varvec{A} (k)} = \varvec{A} (-k)$ and $\det (\varvec{A} (k)) = |\det (B(k))|^{2 n_{a}}$.

In view of the structure of Eq. (36), let us now consider the $\mathbb {R}$-algebra $\varvec{\mathcal {A}}$ that is generated by the matrix family $\{ \varvec{A} (k) \mid k\in \mathbb {R}\}$. Due to the Kronecker product structure, $\varvec{\mathcal {A}}$ fails to be irreducible, no matter what the structure of the IDA $\mathcal {B}$ is. Let us explore this in some more detail. Let $V=\mathbb {C}^{n_{a}}$ and consider $W{:}{=}V \otimes ^{}_{\mathbb {C}} V$, the (complex) tensor product, which is a vector space over $\mathbb {C}$ of dimension $n_{a}^{2}$, but also one over $\mathbb {R}$, then of dimension $2n_{a}^{2}$. In the latter setting, consider the involution $C \! : \, W \!\longrightarrow W$ defined by

$$\begin{aligned} x \otimes y \; \longmapsto \; C (x \otimes y) \, {:}{=}\, \overline{y \otimes x} \, = \, {\bar{y}} \otimes {\bar{x}} \end{aligned}$$

together with its unique extension to an $\mathbb {R}$-linear mapping on W. Observe that there is no $\mathbb {C}$-linear extension, because $C \bigl (a (x \otimes y)\bigr ) = {\bar{a}}\; C (x\otimes y)$ for $a \in \mathbb {C}$. With this definition of C, one finds for an arbitrary $k\in \mathbb {R}$ that

$$\begin{aligned} \begin{aligned} \varvec{A} (k)\, C(x\otimes y) \,&= \, \bigl (B(k) \otimes \,\overline{\! B(k)\! }\, \bigr ) ( {\bar{y}} \otimes {\bar{x}} ) \, = \, \bigl (B(k){\bar{y}}\bigr ) \otimes \bigl (\, \overline{\! B(k)x } \, \bigr ) \\&= \, C \bigl ( \bigl (B(k) \otimes \, \overline{\! B(k)\! }\, \bigr ) (x \otimes y ) \bigr ) \, = \, C \bigl ( \varvec{A} (k) \, ( x \otimes y ) \bigr ), \end{aligned} \end{aligned}$$

so C commutes with the linear map defined by $\varvec{A} (k)$, for any $k\in \mathbb {R}$. The $\mathbb {R}$-linear mapping C has eigenvalues $\pm 1$ and is diagonalisable, as follows from the unique splitting of an arbitrary $w\in W$ as $w = \frac{1}{2} \bigl ( w + C(w)\bigr ) + \frac{1}{2} \bigl ( w - C(w)\bigr )$. So, our vector space splits as $W \! = W_{\! +} \oplus W_{\! -}$ into real vector spaces that are eigenspaces of C. Their dimensions are

$$\begin{aligned} \dim ^{}_{\mathbb {R}} (W_{\! +}) \, = \, \dim ^{}_{\mathbb {R}} (W_{\! -}) \, = \, n_{a}^{2} \end{aligned}$$

since $W_{\! -} = \mathrm {i}W_{\! +}$ with $W_{\! +} \cap W_{\! -} = \{ 0 \}$. It is thus clear that $W_{\! +}$ and $W_{\! -}$ are invariant (real) subspaces for the $\mathbb {R}$-algebra $\varvec{\mathcal {A}}$.

Observe next that we have

$$\begin{aligned} \varvec{A} (k) \, = \, B(k) \otimes \overline{B (k)} \; = \! \sum _{x,y \in S_{T}}\! \mathrm {e}^{2 \pi \mathrm {i}k (x-y)} \, D_{x} \otimes D_{y} \; = \sum _{z\in \triangle _{T}} \mathrm {e}^{2 \pi \mathrm {i}k z}\, F_{z} \end{aligned}$$

where $\triangle _{T} {:}{=}S_{T} - S_{T}$ is the Minkowski difference, with $-\triangle _{T}=\triangle _{T}$, and

$$\begin{aligned} F_{z} \; = \sum _{\begin{array}{c} x,y \in S_{T} \\ x-y = z \end{array}} D_{x} \otimes D_{y} . \end{aligned}$$

(37)

In analogy to before, $F_{z} = F_{z'}$ is possible for $z\ne z'$. For instance, if $z=x-y$ with $x\ne y$ and $D_{x} = D_{y}$, one can get $F_{z} = F_{-z}$ if there is no other way to write z as a difference of two numbers in $S_{T}$.

For $a\in \mathbb {C}$, one easily checks that $C \circ (a F_{z}) = {\bar{a}} C \circ F_{z} = {\bar{a}} F_{-z} \circ C$, which implies $[C, \varvec{A}(k)]=0$ for all $k\in \mathbb {R}$, in line with our previous derivation. It is immediate that $\varvec{\mathcal {A}}$ is contained in the $\mathbb {R}$-algebra $\varvec{\mathcal {A}}_{F}$ that is generated by the matrices $\{ F_{z}+F_{-z} \mid 0 \leqslant z\in \triangle _{T}\}$ together with $\{ \mathrm {i}( F_{z} - F_{-z}) \mid 0 \leqslant z\in \triangle _{T}\}$, and an argument similar to the one previously used for $\mathcal {B}$ shows that $\varvec{\mathcal {A}}_{F} \subseteq \varvec{\mathcal {A}}$, hence $\varvec{\mathcal {A}} = \varvec{\mathcal {A}}_{F}$. Since $\dim ^{}_{\mathbb {C}} (\mathcal {B}) \leqslant n_{a}^{2}$, and since we generate the real algebra only after taking the Kronecker product, one has

$$\begin{aligned} \dim ^{}_{\mathbb {R}} (\varvec{\mathcal {A}}) \, \leqslant \, n_{a}^{4} , \end{aligned}$$

which is also clear from $\dim ^{}_{\mathbb {R}} (W_{\! +}) = n_{a}^{2}$. Moreover, one has the following result.

Lemma 5.15

Let $\varrho$ be a primitive inflation rule on an alphabet with $n_{a}$ letters, and assume that the IDA $\mathcal {B}$ of $\varrho$ is irreducible over $\mathbb {C}$. Then, the induced $\mathbb {R}$-algebra $\varvec{\mathcal {A}}$ is isomorphic with ${{\,\mathrm{Mat}\,}}(n_{a}^2, \mathbb {R})$, and its action on the subspace $W_{\! +}$ is irreducible as well, this time over $\mathbb {R}$.

Proof

Here, $\mathcal {B}$ irreducible over $\mathbb {C}$ means $\mathcal {B}= {{\,\mathrm{Mat}\,}}(n_{a}, \mathbb {C})$. With $\varGamma {:}{=}\mathcal {B}\otimes ^{}_{\mathbb {C}} \mathcal {B}$, where $\otimes ^{}_{\mathbb {C}}$ denotes the tensor product over $\mathbb {C}$, one has $\varGamma \simeq {{\,\mathrm{Mat}\,}}(n_{a}^{2}, \mathbb {C})$ by standard arguments. Clearly, $\varGamma$ is a $\mathbb {C}$-algebra of dimension $n_{a}^{4}$, but also an $\mathbb {R}$-algebra, then of dimension $2 n_{a}^{4}$. Now, using the Kronecker product as representation of the tensor product, the mapping defined by $M\otimes N \mapsto \overline{N}\otimes \overline{M}$ has a unique extension to an automorphism $\sigma$ of $\varGamma$ as an $\mathbb {R}$-algebra.

Our $\mathbb {R}$-algebra $\varvec{\mathcal {A}}$ consists of all fixed points of $\sigma$, so $\varvec{\mathcal {A}} = \{ Q \in \varGamma \mid \sigma (Q) = Q \}$. Employing the elementary matrices $E_{ij}$ from ${{\,\mathrm{Mat}\,}}(n_{a}, \mathbb {R})$ with $E_{ij,k\ell } {:}{=}E_{ik} \otimes E_{j \ell }$, we can give a basis of $\varvec{\mathcal {A}}$, seen as a vector space over $\mathbb {R}$, by

$$\begin{aligned} \big \{ \tfrac{1}{2} (E_{ij,k\ell } + E_{k\ell ,ij}) \mid (i,j) \leqslant (k,\ell ) \big \} \cup \big \{ \tfrac{\mathrm {i}}{2} (E_{ij,k\ell } - E_{k\ell ,ij}) \mid (i,j) < (k,\ell ) \big \} , \end{aligned}$$

where lexicographic ordering is used for the double indices. Note that the cardinalities are $\frac{1}{2} n_{a}^{2} (n_{a}^{2} + 1)$ and $\frac{1}{2} n_{a}^{2} (n_{a}^{2} - 1)$, which add up to $\dim ^{}_{\mathbb {R}} (\varvec{\mathcal {A}}) = n_{a}^{4}$.

Next, observe that we can get $E_{ij,k\ell }$ and $E_{k\ell ,ij}$ by a simple (complex) linear combination of $\tfrac{1}{2} (E_{ij,k\ell } + E_{k\ell ,ij})$ and $\tfrac{\mathrm {i}}{2} (E_{ij,k\ell } - E_{k\ell ,ij})$, and vise versa. Put together, this defines a (complex) inner automorphism of $\varGamma$. Observing that ${{\,\mathrm{Mat}\,}}(n_{a}^{2}, \mathbb {R}) = \{ Q \in \varGamma \mid {\overline{Q}} = Q \}$, this construction can now be used to show that $\varvec{\mathcal {A}} \simeq {{\,\mathrm{Mat}\,}}(n_{a}^{2}, \mathbb {R})$, which is a central simple algebra. Since $\varvec{\mathcal {A}} W_{\! +} \subseteq W_{\! +}$ and $\dim ^{}_{\mathbb {R}} (W_{\! +}) = n_{a}^{2}$, the claimed irreducibility over the reals follows. $\quad \square$

Note that all $F_{z}$ are non-negative, integer matrices. They clearly satisfy the relation $\sum _{z\in \triangle _{T}} F_{z} = \varvec{A} (0) = M_{\varrho } \otimes M_{\varrho }$. Moreover, under some mild conditions, the spectral radius of $F^{}_0$ is $\lambda$, while the other matrices $F_z$ have smaller spectral radius.

Let us come back to Eq. (36), which implies

$$\begin{aligned} \Vert \varvec{A} (k) \Vert ^{}_{\mathrm {F}} \, = \, \Vert B (k) \Vert ^{2}_{\mathrm {F}} . \end{aligned}$$

If we consider the matrix cocycle defined by $\varvec{A}^{(n)} (k) = B^{(n)} (k) \otimes \overline{B^{(n)} (k)}$, it is immediate that the maximal Lyapunov exponents, for all $k\in \mathbb {R}$, are related by

$$\begin{aligned} \chi ^{\varvec{A}} (k) \, = \, 2 \chi ^{B} (k) , \end{aligned}$$

(38)

which also holds for the higher-dimensional case with $k\in \mathbb {R}^d$. Clearly, one can now reformulate Theorems 3.28 and 5.7 in terms of $\chi ^{\varvec{A}}$. In particular, one has the following reformulation of Theorem 3.34 and Corollary 3.35 and their higher-dimensional analogues.

Corollary 5.16

Let $\varrho$ be a primitive inflation rule in $\mathbb {R}^d$ with finitely many translational prototiles and expansive map Q. Let $B^{(n)} (.)$ be its Fourier matrix cocycle, with $\det (B(k))\ne 0$ for at least one $k\in \mathbb {R}^d$, and $\varvec{A}^{(n)} (.) = B^{(n)} (.) \otimes \overline{B^{(n)} (.)}$ the corresponding Kronecker product cocycle. If the diffraction measure of the hull defined by $\varrho$ contains a non-trivial absolutely continuous component, one has $\chi ^{\varvec{A}} (k) = \log |\det (Q) |$ for a subset of $\mathbb {R}^d$ of positive measure, which has full measure when $\chi ^{\varvec{A}} (k)$ is constant for a.e. $k\in \mathbb {R}^d$. $\quad \square$