\section{(1 May 2009)}
\subsection*{Another Gap Labelling\footnotemark}
\footnotetext{Thanks to Carl Olimb for the notes.}
Rotation Numbers (Moser, \( 1 \)-D).

Integrated Density of States (any dimension).

Now \( K_{n} \)-gap labelling (any dimension).

\( 1 \)-D differential operator \( H = -\partial^{2} + V \) (unit \( 
\hbar^{2} / 2 m  =  1 \)), where \( V \) is a differentiable function 
on \( \mathbb{R} \).

Consider \( E \in \mathbb{R} \),
\[ 
H \psi = E \psi \tag{\( \ast \)} \, , \, \psi : \mathbb{R} 
\rightarrow \mathbb{C} \quad (\psi \text{ non-zero})
\]
Always solvable for \( \psi \neq 0 \) but the solution depends on \( 
E \); this is where Moser started.
\begin{itemize}
    \item[\( 1) \)]  If \( E \) is an eigenvalue \( \exists \, \psi \in 
    L^{2}(\mathbb{R}) \) solving \( (\ast) \).

    \item[\( 2) \)]  If \( E \in \sigma(H) \) but not an eigenvalue:
    \begin{itemize}
        \item[]  Simple if \( \sigma(H) = 
	\sigma_{\underset{\text{cont}}{\text{abs}}}(H) \cup 
	\sigma_{\text{pure point}}(H) \); then \( \psi \) is bounded 
	\( \leadsto \) linear combinations of these to construct wave 
	packets.  This is difficult.
    
        \item[]  If \( E \in \sigma_{\text{sc}}(H) \) ``\( \psi \) is 
	critical''.  This was the devil until the 80's; then became 
	fashionable.
	
	Remark: If \( V = 0 \), then \( \sigma = \mathbb{R} = 
	\sigma_{\text{ac}} \) then use Fourier transforms to 
	construct wave packets.  If \( V \neq 0 \) then?
    \end{itemize}

    \item[\( 3) \)]  If \( E \notin \sigma(H) \) then \( \exists ! \) 
    (up to multiplicative constants) solutions \( \psi_{+} , \psi_{-} 
    \) which satisfy \( \lim_{x \rightarrow \pm \infty} \psi_{\pm}(x) 
    = 0 \) and \( x \rightarrow \mp \infty \) leads to exponential 
    increase of \( \psi_{\pm} \) (see Fig. \ref{fig:notspectral}). 
\begin{figure}[ht]
\setlength{\unitlength}{1em}
\begin{center}
\begin{picture}(20, 8)(-10, -4)
%     
%     \multiput(0, -4)(0, 1){9}{\multiput(-10, 0)(1, 0){21}{\circle*{0.2}}}
%         
\put(-10, 0){\vector(1, 0){20}}
\thicklines
\qbezier(-10, 1)(-8.75, 1)(-7.5, 0)
\qbezier(-7.5, 0)(-5.73223, -1.41421)(-5, -1.41421)
\qbezier(-5, -1.41421)(-3.75, -1.41421)(-2.5, 0)
\qbezier(-2.5, 0)(-0.73223, 2)(0, 2)
\qbezier(0, 2)(1.25, 2)(2.5, 0)
\qbezier(2.5, 0)(4.26777, -2.82843)(5, -2.82843)
\qbezier(5, -2.82843)(6.25, -2.82843)(7.5, 0)
\qbezier(7.5, 0)(9.26777, 4)(10, 4)
\thinlines
\put(2.5, 0){\circle*{0.3}}
\put(2.7, 0.4){\( \xi_{0} \)}
\end{picture}
\end{center}
    \caption{\( \psi_{-} \)}
    \label{fig:notspectral}
\end{figure}
This excludes any possible interpretation of these particles.  The 
only way is to restrict the system.
\end{itemize}
Rotation Numbers: \( E \notin \sigma(H) \); \( \alpha(E) = \) ``density of zeros'' of \( 
\psi_{-} \) = 2 density of rotation of \( \psi(x) + \mi \psi'(x) \).  
\( \alpha(E) \) is constant on gaps.

What is the density in this context?  We mean a sequence of intervals 
subdivided and take rotation numbers.  So may depend on this sequence.
\[ 
[ a_{n} , b_{n} ] \subset [ a_{n +1} , b_{n + 1} ] \subset \cdots 
\rightarrow \mathbb{R}
\]
The number of zeros in the interval is equal to the number of density 
states. Then
\[ 
\alpha(E) = \text{IDS}(E) = \langle \underset{\ni \tau}{0\text{-Hull}} \, | \, [ P(H) 
]_{0} \rangle \, , \, K_{0}\text{-label}.
\]
Now \( H_{\xi} = -\partial^{2} + V_{\xi} \), \( V_{\xi}(x) = V(x + 
\xi) \).  \( H_{\xi} \underset{^{\text{unitary}}}{\sim} H \), so \( 
\sigma(H_{\xi}) = \sigma(H) \).  If \( E^{\text{(fixed)}} \notin 
\sigma(H) \), \( H_{\xi} \psi_{\xi -} = E \psi_{\xi -} \) with \( 
\psi_{\xi -}(x) \overset{_{x \rightarrow -\infty}}{\longrightarrow 0} 
\).  Then \( \psi_{\xi -}(x) = \psi_{-}(x + \xi) \).  Consider Fig. 
\ref{fig:notspectral} again:
\begin{gather*}
    \xi_{0} \text{ is a zero of } \psi_{-} \text{}  \\
     \Leftrightarrow 0 \text{ is a zero of } 
\psi_{\xi_{0} -}  \\
    \Leftrightarrow E \text{ is an eigenvalue of } H_{\xi}\left| 
    _{\mathbb{R}^{-}} \right. \\
    \hspace{8em} \text{ with Dirichlet boundary 
    condition at } 0
\end{gather*}
Call \( E \) a \emph{Dirichlet value of} \( H_{\xi_{0}} \).

Notation:
\[ 
\hat{H}_{\xi_{0}} : = H_{\xi_{0}}\left| _{\mathbb{R^{-}}} \right. 
\text{ with Dirichlet boundary condition.}
\]

Write \( 
D_{\xi}(0) = D_{\xi} = \{ \text{Dirichlet values of } H_{\xi} \text{ 
in } \Delta \}
\), where \( \Delta \) is a gap in \( \sigma(H_{\xi}) \) \( = \sigma(H) \).

General remarks:
\begin{itemize}
    \item[i)] \( \sigma_{\text{ac}} ( \hat{H}_{\xi} ) \subset
    \sigma_{\text{ac}} ( H_{\xi} ) \).  So, \( 
    \sigma_{\text{ac}} ( \hat{H}_{\xi} ) \subset \sigma (H) \, \cup \,
    \bigcup_{\text{gaps } \Delta} D_{\xi}(\Delta)
    \)

    \item[ii)]  If \( \alpha(\Delta) \neq 0 \), then \( \abs { D_{\xi} 
    (\Delta) } \leq 1 \). ``No second eigenvalue in the same gap.''
\end{itemize}
Let \( \mu(\xi) = \) EV of \( \hat{H}_{\xi} \) in \( \Delta \).
\[ 
\{ \text{zeros of } \psi_{-} \} = \{ \xi \, | \, 
\abs{D_{\xi}(\Delta)} = 1 \} \tag{\( \ast\ast \)}
\]

Now vary \( \xi \,\); we get curves as in Fig. \ref{fig:mu}.
% 
\begin{figure}[ht]
\setlength{\unitlength}{1em}
\begin{center}
\begin{picture}(25, 14)(-15, -4)
%         
\put(-10, 0){\vector(1, 0){20}}
\thicklines
\qbezier(-10, 1)(-8.75, 1)(-7.5, 0)
\qbezier(-7.5, 0)(-5.73223, -1.41421)(-5, -1.41421)
\qbezier(-5, -1.41421)(-3.75, -1.41421)(-2.5, 0)
\qbezier(-2.5, 0)(-0.73223, 2)(0, 2)
\qbezier(0, 2)(1.25, 2)(2.5, 0)
\qbezier(2.5, 0)(4.26777, -2.82843)(5, -2.82843)
\qbezier(5, -2.82843)(6.25, -2.82843)(7.5, 0)
\qbezier(7.5, 0)(9.26777, 4)(10, 4)
\thinlines
\put(-7.5, 0){\circle*{0.3}}
\put(-2.5, 0){\circle*{0.3}}
\put(2.5, 0){\circle*{0.3}}
\put(2.7, 0.4){\( \xi_{0} \)}
\put(7.5, 0){\circle*{0.3}}
% 
\put(-10, 2){\vector(0, 1){8}}
\put(-10.2, 6){\line(1, 0){20.2}}
\put(-7.5, 6){\circle*{0.3}}
\put(-2.5, 6){\circle*{0.3}}
\put(2.5, 6){\circle*{0.3}}
\put(7.5, 6){\circle*{0.3}}
% 
\multiput(-10.1, 7)(0, 0.2){11}{\line(1, 0){0.2}}
\multiput(-10.1, 3)(0, 0.2){11}{\line(1, 0){0.2}}
\put(-10.5, 6){\vector(0, 1){1}}
\put(-10.5, 6){\vector(0, -1){1}}
\put(-15, 5.5){gap \( \Delta \leadsto \)}
\put(-13, 8){\( \sigma(H) \)}
\put(-13, 3.5){\( \sigma(H) \)}
\put(-9, 3.7){\( \mu(\xi) \)}
\put(-9.6, 5){\( \nwarrow \)}
% 
\qbezier(-8, 7)(-8, 6.5)(-7.5, 6)
\qbezier(-7.5, 6)(-7, 5.5)(-7, 5)
\qbezier(-3, 7)(-3, 6.5)(-2.5, 6)
\qbezier(-2.5, 6)(-2, 5.5)(-2, 5)
\qbezier(2, 7)(2, 6.5)(2.5, 6)
\qbezier(2.5, 6)(3, 5.5)(3, 5)
\qbezier(7, 7)(7, 6.5)(7.5, 6)
\qbezier(7.5, 6)(8, 5.5)(8, 5)
% 
\put(-4, 8.7){neg. slopes by Lemma \ref{negslopemu}}
\put(-2, 7.3){\( \swarrow \)}
\put(0.7, 7.3){\( \searrow \)}
\end{picture}
\end{center}
    \caption{\( \mu \)}
    \label{fig:mu}
\end{figure}

\begin{lemma}
    \( \mu'(\xi) < 0 \)
    \label{negslopemu}
\end{lemma}
\begin{proof}
    \( \mu(\xi) = \langle \hat{\psi}_{\xi} \, | 
    \underset{^{-\partial^{2} + V_{\xi}}}{\hat{H}_{\xi}} | 
    \, \hat{\psi}_{\xi} \rangle \)
    \begin{align*}
	\frac{\partial \mu ( \xi )}{\partial \xi} & = 
	\underset{\mu(\xi) ( \langle \hat{\psi}_{\xi} , 
	\hat{\psi}'_{\xi} \rangle + \langle \hat{\psi}'_{\xi} , 
	\hat{\psi}_{\xi} \rangle ) = \frac{\partial}{\partial \xi} 
	\abs{\hat{\psi}_{\xi}}^{2} = 0}{\underbrace{\langle
	\hat{\psi}'_{\xi} \, | \, \hat{H}_{\xi} \, | \,
	\hat{\psi}_{\xi} \rangle + \langle
	\hat{\psi}_{\xi} \, | \, \hat{H}_{\xi} \, | \,
	\hat{\psi}'_{\xi} \rangle}} + \underset{\uparrow}{\underbrace{\langle
	\hat{\psi}_{\xi} \, | \, V'_{\xi} \, | \,
	\hat{\psi}_{\xi} \rangle}} \\
         & ^{\hspace{10em}\text{expectation of mechanical force of this wave 
	 funtion}}  \\
         & = \int_{-\infty}^{0} \hat{\psi}_{\xi} V'_{\xi} 
	 \hat{\psi}_{\xi} \diff x  =  - \int_{-\infty}^{0}  \left( 
	 \hat{\psi}'_{\xi} V_{\xi} \hat{\psi}_{\xi} + 
	 \hat{\psi}_{\xi} + \hat{\psi}_{\xi} V_{\xi} 
	 \hat{\psi}'_{\xi} \right) \diff x  \\
         & = \int_{-\infty}^{0} \left(\hat{\psi}'_{\xi} \partial^{2} 
	 \hat{\psi}_{\xi} + \hat{\psi}_{\xi} \partial^{2} 
	 \hat{\psi}'_{\xi}\right) \diff x  \\
         & = -\int_{-\infty}^{0} \frac{\partial}{\partial x} ( 
	 \hat{\psi}'_{\xi} \hat{\psi}'_{\xi} ) \diff x = - 
	 \abs{\hat{\psi}'_{\xi}(0)}^{2} < 0
    \end{align*}
\end{proof}
 Two important results: \( - \abs{\hat{\psi}'_{\xi}(0)}^{2} < 0 \) 
 and the resultant force.
 
 In our Fig. \ref{fig:mu} the curves cannot cross since then we would 
 have a degenerate eigenvalue.  Now look at the ``line'' created by 
 \( \mu(\xi) \) on a circle.
 \[ 
 \tilde{\mu} : \mathbb{R} \rightarrow \{z \in \mathbb{C} \, | \, 
 \abs{z} = 1\} \, , \text{ where } \tilde{\mu}(\xi) = \exp\left( 2 \pi \mi 
 \, \frac{\mu(\xi) - E_{0}}{\abs{\Delta}} \right)  \, , \, E_{0} = \inf 
 \Delta
 \]
 where \( \mu(\xi) = E_{0} \) is no eigenvalue at \( \xi \).
 
 \underline{Final Result}.  Moser rotation number is the same as 
 another rotation number.
 \begin{align*}
     \alpha(E)  & = \text{ density of zeros of } \psi_{-}  \tag{\( \ast\ast \)}  \\
      & = \text{ density of intersections of } \mu \text{ with } E  \\
      & = \text{ density of } \tilde{\mu} = \beta(E) = \beta(\Delta)
 \end{align*}
\begin{align*}
    \beta ( \Delta ) &  = \lim_{[ a_{n} , b_{n} ] \nearrow \, \mathbb{R}} \, 
 \frac{\Delta}{\abs{b_{n} - a_{n}}} \frac{\Delta}{2 \pi \mi} 
 \int_{a_{n}}^{b_{n}} \mu^{\ast} ( \xi ) \mu ( \xi ) \diff \xi
  \\
     & = \lim_{[ a_{n} , b_{n} ] \nearrow \, \mathbb{R}} \, 
     \frac{\Delta}{\abs{\Delta}} \int_{a_{n}}^{b_{n}} \tr ( 
     -V'_{\xi}P_{\Delta} ( \hat{H}_{\xi} ) ) \diff \xi
\end{align*}
where \( P_{\Delta} (\hat{H}) \) is the spectral projection to states 
in \( \Delta \) (actually only one); so \( P_{\Delta} (\hat{H}) = 
\abs{\, \psi_{\xi} \rangle \langle \hat{\psi}_{\xi} \,} \).  This can 
be interpreted as a pairing of a \( 1 \)-trace with a \( K_{1} 
\)-class.
\begin{align*}
    U_{\xi} & = \mathrm{e}^{2 \pi \mi \frac{\hat{H}_{\xi} - 
E_{0}}{\abs{\Delta}}} P_{\Delta} ( \hat{H}_{\xi} ) + P_{\Delta} ( 
\hat{H}_{\xi} )^{\perp} \in \mathcal{K} ( L^{2} ( \mathbb{R} ) )^{+}  \\
    U_{\Delta} & : = ( \xi \mapsto U_{\xi} ) \in C_{0} ( \mathbb{R} , 
    \mathcal{K} ( L^{2} ( \mathbb{R} ) )^{+} ) = \s \mathbb{C} 
    \otimes \mathcal{K}^{+} = C ( \s^{1} ) \otimes \mathcal{K}^{+}
\end{align*}
so \( [ U_{\Delta} ] \in K_{1} ( C_{0} ( \mathbb{R} ) ) \).

Now \( 1 \)-trace = character of \( ( \negthickspace \negthickspace
\underset{^{\text{DeRahm forms}}}{\Omega ( \mathbb{R} )}
\negthickspace \negthickspace , \, d \, ,
\textstyle{\int}_{\mathbb{R}} ) \).  Then \( \langle \eta \, | \, [ 
U_{\Delta} ] \rangle = \beta ( \Delta ) \).

To finish, \( \alpha = \beta \) is an index theorem.  Why?
\begin{align*}
    [ U_{\Delta} ]_{1} & = \exp [ P_{\leq E} ( H ) ]_{0} \text{ and 
    additional work}  \\
     & \hspace{4em} \uparrow  \\
     & \hspace{2em} ^{\text{energy values } \leq E}
\end{align*}