\section{(9 February 2009)}
\subsection*{Cut \( + \) Project Tilings}
Canonical cut + project tiling.
\[ 
\underset{= E^{\perp}}{\mathbb{R}^{n}}
\overset{\text{proj.}}{\longleftarrow} \underset{\supset
\mathbb{Z}^{N} \text{ lattice}}{\mathbb{R}^{N}}
\overset{\text{proj.}}{\longrightarrow} \mathbb{R}^{d} =: E^{\prime
\prime} \quad (d < N)
\]
\( \tilde{\mathcal{T}} \) is a \( \mathbb{Z}^{N} \)-periodic tiling 
of \( \mathbb{R}^{N} \) such that there are finitely many tiles up to 
translation, each of which is of the form \( \mathsf{k} = 
\underset{\text{poly'n in }E^{\perp}}{\mathsf{k}^{\perp}} \times 
\underset{\text{poly'n in }E^{\prime \prime}}{\mathsf{k}^{\prime \prime}} \).
(``\( \mathsf{k} \)'' for the German ``klotz'')
\begin{figure*}[ht]
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	\qbezier(1.41421, -2)(0, 0)(-2, 2.82843)
	\thinlines
	\put(10.4, 7.3){\( E^{\prime \prime} \)}
	\put(-3, 3){\( E^{\perp} \)}
\end{picture}
\end{center}
\begin{center}
    Cut and Project: \( N = 2 \,,\,\, d = 1 \)
\end{center}
\end{figure*}


\begin{figure*}[ht]
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% 	\multiput(-2, -2)(2, 0){7}{\circle*{0.2}}
% 	\multiput(-2, 0)(2, 0){7}{\circle*{0.2}}
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	\put(-3, 3){\( E^{\perp} \)}
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\begin{center}
    Cut and Project with \(  (\epsilon + E^{\prime \prime}) \cap \mathbb{Z}^{2} 
= \emptyset \)
\end{center}
\end{figure*}

Choose \( \epsilon \in \mathbb{R}^{N} \): look at intersection between 
\[ (\epsilon + E^{\prime \prime}) \cap \tilde{\mathcal{T}} \]


Assumption \( (\epsilon + E^{\prime \prime}) \cap \mathbb{Z}^{2} 
= \emptyset \)

if \( \epsilon + E^{\prime \prime} \) cuts through \( 
\mathbb{Z}^{2} \) then \( \epsilon + E^{\prime \prime} \) cuts 
through the boundary of two neighboring tiles; want to avoid this so 
that's why we put the \( \epsilon \) in.

Try to understand the inverse limit space.

\[ 
f \in C_{0\text{-}\mathcal{P}\text{-e}}(\overset{E^{\prime 
\prime}}{\mathbb{R}}) \text{ defines a function } \tilde{f} \text{ on 
a dense set of } \mathbb{R}^{2} / \mathbb{Z}^{2} \cong \mathbb{T}^{2}
\]

\[ 
C_{0\text{-}\mathcal{P}\text{-e}}(\mathbb{R}) \overset{i}{\subset} 
C(\underset{0\text{-cut}}{\mathbb{T}^{2}_{0\text{-c}}})
\]
\[ 
\underset{\cong \mathbb{T}^{2}_{0\text{-c}}}{\underbrace{
\mathrm{Spec} \left(C(\mathbb{T}^{2}_{0\text{-c}})\right)}}
\overset{i^{\ast}|_{\mathrm{Spec}}}{ \longrightarrow
}\mathrm{Spec}\left(C_{0 \text{-} \mathcal{P} \text{-e}} (\mathbb{R})
\right) \, , \quad x \in \mathbb{T}^{2}_{0\text{-c}} \mapsto \mathrm{ev}_{x}
\]
\[ 
i^{\ast}\mathrm{ev}_{x} = \mathrm{ev}_{x} \circ i
\]
\[ 
i^{\ast}\mathrm{ev}_{x} = i^{\ast}\mathrm{ev}_{y} \text{ if } x
\underset{\text{blue line} }{\sim} y \text{ or } x \underset{\text{green line}}{\sim} y
\]

Delone set of \( \mathcal{P} \) given by the vertices is
\[ 
(\epsilon + E^{\prime \prime}) \cap (\mathbb{Z}^{2} + 
\bigcup_{\mathsf{k}\text{ tile in } 
\tilde{\mathcal{T}}}\mathsf{k}^{\perp})
\]

Canonical choice: \( \pi^{\perp}(\text{unit cube}) \)
\[ 
\pi^{\perp} : \mathbb{R}^{N} \rightarrow E^{\perp} \text{projection 
along } E^{\prime \prime}
\]

The thing I think should be canonical

\emph{monical}: \( \mathsf{k}^{\perp} \) polyhedra s.th. \( \partial 
\mathsf{k}^{\perp} = \bigcup_{\text{faces}}f_{i} \)

Look at the stabilizer 
\[ 
\mathrm{Stab}_{\pi^{\perp}(\mathbb{Z}^{N})}(\text{affine space spanned 
by all }f_{i}>0)
\]
The rank of the stabilizer determines the complexity of the tiling.

Back to the construction of the algebra.
\[ 
\mathcal{A}_{\mathcal{P}} = \underset{(\alpha_{\xi}(f))(x) = f(x - \xi)}{C_{\mathcal{P}}(\mathbb{R}^{n}) 
\rtimes_{\alpha} \mathbb{R}^{n} } \cong \underset{\phi \text{ induced 
by } \omega \in \Omega_{\mathcal{P}} \mapsto \omega - x}{C(\Omega_{\mathcal{P}}) 
\rtimes_{\phi} \mathbb{R}^{n}}  
\]
Fix isomorphism \( C(\Omega_{\mathcal{P}}) 
\overset{\sigma}{\rightarrow} C_{\mathcal{P}}(\mathbb{R}^{n}) \), a 
\( \ast \)-isomorphism of C*-algebras: \( \sigma(\tilde{f})(x) := 
\tilde{f}(\mathcal{P} - x) \)

We can see why \( C_{\mathcal{P}}(\mathbb{R}^{n}) 
\rtimes_{\alpha} \mathbb{R}^{n} \) is a good algebra.

\begin{exa*}
    \[ 
    \mathcal{P} = \{x\} \,,\,\, C_{\{x\}}(\mathbb{R}^{n}) \cong 
    C_{0}(\mathbb{R}^{n})^{{+}^{\text{(adjoin } 1)}} = 
    C_{0}(\mathbb{R}^{n}) + \underset{\text{multiples of }1}{\mathbb{C}}
    \]
    \[ 
    \mathcal{A}_{\{x\}} = \underset{\text{quant'n of a classical 
    particle}}{C_{0}(\mathbb{R}^{n}) \rtimes_{\alpha} \mathbb{R}^{n}} + 
    \mathbb{C} \rtimes \mathbb{R}^{n}
    \]
    
    Recall.  We took the representation induced by \( \mathrm{ev}_{0} 
    : C_{\{x\}}(\mathbb{R}^{n}) \rightarrow \mathbb{C} \).  This gave 
    the Schr\"{o}dinger representation for the first part of \( 
    \mathcal{A}_{\{x\}} \).    
    \[ 
    \underset{\text{character 
    rep'n}}{C_{\mathcal{P}}(\mathbb{R}^{n})} \rtimes 
    \underset{\text{induced rep'n}}{\mathbb{R}^{n}}
    \]
    
    The representations on \( \mathcal{A}_{\mathcal{P}} \) we look at 
    are induced representations of the character irreducible 
    representations on \( C_{\mathcal{P}} \).
    
    So \( \forall \omega \in \Omega_{\mathcal{P}} = 
    \mathrm{Spec}\left(C_{\mathcal{P}}(\mathbb{R}^{n})\right) \) we 
    have an irreducible representation \( \mathrm{ev}_{\omega} \circ 
    \sigma^{-1} \) and therefore an irreducible representation \( 
    \pi_{\omega} \) of \( \mathcal{A}_{\mathcal{P}} \).
    
    Let \( F \in \mathcal{A}_{\mathcal{P}} \) of the form \( F : 
    \mathbb{R}^{n} \rightarrow C_{\mathcal{P}}(\mathbb{R}^{n}) \)
    \[ 
    \begin{split}
	    \tilde{F}(\xi) & = \sigma^{-1} \circ F(\xi)  \\
	    \tilde{F} & : \mathbb{R}^{n} \rightarrow 
	    C(\Omega_{\mathcal{P}}) \\
	    \pi_{\omega} & : \mathcal{A}_{\mathcal{P}} \rightarrow 
	    B(L^{2}(\mathbb{R}^{n}))  \\
	    (\pi_{\omega}(\tilde{F}) \psi)(x) & = \int \mathrm{d}y 
	    \underset{\text{integral kernel of 
	    }\pi_{\omega}(\tilde{F})}{\underbrace{\tilde{F}(x - y)(\omega + x)}} \psi(y)
	\end{split}
    \]
\end{exa*}

First important consequence of this:

Consider \( \omega = \mathcal{P} \) and 
\[ 
H = \underset{\text{in the Sobolev rep'n}}{\frac{\hat{p}^{2}}{2m} + V(\hat{q})} \,,\, V \in 
C_{\mathcal{P}}(\mathbb{R}^{n}, \mathbb{R})
\]

Then \( \forall g \in C_{0}(\mathbb{R}) \), 
\[ 
\underset{\text{spec calc}}{g(H)} \in 
\pi_{\mathcal{P}}(\mathcal{A}_{\mathcal{P}})
\]

Use Laplace transform.