Il problema è che l’ho scritto, ma non sembra funzionare. Se non carico l’ultimo capitolo va tutto a meraviglia, ma con l’ultimo capitolo ho questo problema.
Scusate, pensavo di aver allegato i files, ma evidentemente non l’ho fatto.
Questo è il file principale:
`\documentclass[a4paper, 11pt, headinclude, footinclude, titlepage, drafting]{scrbook} %manca il frontespizio, da fare!
\input{PackagesThesis.tex}
\input{SettingsThesis.tex}
\begin{document}
\frontmatter
\dominitoc
\tableofcontents
\mainmatter
\part{Prerequisites}
% \input{Chapters/1_Homology_Cohomology.tex}
% \input{Chapters/2_Vector_Bundles.tex}
% \input{Chapters/3_CharClasses.tex}
\part{The Fedosov Index Formula}
\input{Chapters/IndexManifold.tex}
\backmatter
% \appendix
%
\input{Others/BiblioBibTeX.tex}
\end{document}`
e questo è il file BiblioBibTeX:
`\cleardoublepage
\addcontentsline{toc}{chapter}{\tocEntry{\bibname}} % \tocEntry serve per far comparire bibliography nello stesso stile degli altri capitoli (nell'indice)
\nocite{*}
%\printbibliography % with biblatex
\bibliography{Others/BiblioDB}`
I files per caricare i pacchetti e le impostazioni sono i seguenti:
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\usepackage{enumerate}
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\usepackage{xypdf} % per migliorare i diagrammi di xypic in output
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\usepackage{mparhack} % finezze tipografiche
\usepackage{fixltx2e} % finezze tipografiche
\usepackage{relsize} % finezze tipografiche
%\usepackage{graphicx} % per le figure
\usepackage{emptypage} % per avere le pagine bianche senza numeri di pagina né testatine
%\usepackage[babel]{csquotes} % consigliato con biblatex
%\usepackage[style=numeric-comp,hyperref]{biblatex} % per la bibliografia (cfr classicthesis)
%\usepackage{makeidx} % per l'indice analitico
\usepackage[eulerchapternumbers, parts, pdfspacing]{classicthesis}
\usepackage{caption} % altrimenti c'è un errore in arsclassica
\usepackage[tight, english]{minitoc} % per i mini-indici nei capitoli
\usepackage{arsclassica}
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% Altro
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\usepackage{eucal} % per il mathcal + figo
\usepackage{mathrsfs} % per il math script
%usepackage{xspace} % per il comando \xspace
\usepackage{xcolor} % per i colori
%\usepackage{lettrine} % per le iniziali di capitolo grandi
%\usepackage[colorlinks=true, breaklinks=true]{hyperref} % per i links (colorlinks: colorati)
%\usepackage[toc]{glossaries} % per l'indice dei simboli
%\usepackage[lowtilde]{url} % per indirizzi internet, già caricato da hyperref in classicthesis`
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%\makeglossaries`
In ultimo vi scrivo il codice sorgente per l’ultimo capitolo, anche se dubito che sia utile…
`\chapter{Index of an elliptic system on a compact manifold}
\minitoc
\mtcskip
\thispagestyle{empty}
% ————————————————
\section{Setting and Background}
% ————————————————
We assume throughout this chapter that every cited object is smooth, except when differently specified.
We consider a compact manifold $M$ and a vector bundle over it, identified with a family $P \in \mathrm{M}^{n \times n}\big(\smooth{M}\big)$ of projectors; the description is through local
frames and coframes as in Section \vref{framecoframe}. The vector bundle is endowed here with its \textcolor{red}{standard Hermitian metric}%
\marginpar{Define Hermitian metric on vector bundles in the appropriate Chapter}
and we shall deal with orthogonal projection operators. In this case, in fact, i.e. when the matrices $P(x)$ are Hermitian, for a given frame $e_\alpha(x)$ defined in a neighbourhood
$U_\alpha$, the dual coframe is uniquely determined by the formula
\[
\eta_\alpha(x) = \big(e_\alpha(x)^\dag e_\alpha(x)\big)^{-1}e_\alpha(x)^\dag,
\]
where the symbol $\dag$ denotes Hermitian conjugation. Moreover, this choice will not affect the generality of the subject, by virtue of the following lemma. We omit henceforth the dependence
on the point $x$ (with the understanding that each object is actually a matrix) and also on the neighbourhood $U_\alpha$.
\begin{lemma}
Any projector $P$ is homotopic to an orthogonal projector $P'$.
\end{lemma}
\begin{proof}
Given a frame $e$, let $P = e \eta$; we construct an orthogonal projector $P'$ by putting
\[
P' \= e(e^\dag e)^{-1}e^\dag.
\]
We have then that $PP' = P'$ and $P'P = P$:
\begin{gather*}
PP' = e\eta e(e^\dag e)^{-1}e^\dag = e(e^\dag e)^{-1}e^\dag = P', \\
P'P = e(e^\dag e)^{-1}e^\dag e \eta = e \eta = P.
\end{gather*}
Thus the convex combination $P(t) \= tP' – (1 – t)P$ is a projector for all $t \in [0,1]$, and this is the required (straight-line) homotopy between $P$ and $P'$.
\end{proof}
When taking into consideration the curvature matrix associated with a connection on the vector bundle $P$, we shall denote the multiplication of two matrices whose elements are differential
forms on $M$ just as the usual matrix multiplication, i.e. without an interposed sign. By the way, if $A \= (a_i^j)$ is a matrix of $p$-forms and $B \= (b_i^j)$ is a matrix of $q$-forms, the generic
element of their product is given by $c_i^j \= a_i^k \wedge b_k^j$. We shall also denote by $\d A$ the matrix with entries $\d a_i^j$.
The following identities follow directly from the properties of exterior derivative, trace and Hermitian conjugation:
\begin{gather}
\d(AB) = \d A\, B + (-1)^p A\,\d B, \\
\tr(AB) = (-1)^{pq}\tr(BA), \label{tr}\\
(AB)^\dag = (-1)^{pq}(BA)^\dag.
\end{gather}
The Levi-Civita connection on $T^*M$ is obtained%
\marginpar{Show this? On $TM$ or $T^*M$? See Park, where to find the correlation algebraic–differential?}
by embedding the manifold $M$ in $\C^N$ for some appropriate natural number $N$ by Whitney theorem, then applying the ordinary
directional derivative (which is a connection in $\C^N$) and finally projecting the result onto the tangent bundle, seen as a subbundle of $\Theta^N(M)$.
\begin{defn}\label{CovDiff}
A \dfn{vector form}%
\footnote{We call a \dfn{vector form} a (column) vector of differential forms.}
$\vartheta$ is said to be \dfn{invariant under $P$} if
\[
P\vartheta = \vartheta.
\]
Similarly, a \dfn{matrix form}%
\footnote{We call a \dfn{matrix form} a (square) matrix of differential forms.}
$A$ is said to be \dfn{invariant under $P$} if
\[
PA = AP = A.
\]
\end{defn}
With the notation adopted thus far, we give the following definition.
\begin{defn}
The \dfn{covariant differential} of a vector form $\vartheta$ invariant under $P$ is %
\marginpar{Show this? Compare algebraic definition (Park, p.~188) with geometric definition (Ch. 2)}
\[
\nabla \vartheta \= P\, \d\vartheta,
\]
and that of a matrix form $A$ invariant under $P$
\[
\nabla A \= P\, \d A\, P.
\]
\end{defn}
The curvature of this Levi-Civita connection is then given by
\begin{align*}
\nabla^2 \vartheta & = P\, \d (P\, \d\vartheta) = P\, \d P\, \d(P\vartheta) \\
& = P\, \d P\, \d P\, \vartheta + P\, \d P\, P \d\vartheta.
\end{align*}
In order to simplify this expression, we make use of the following lemma.
\begin{lemma}\label{PdPP}
We have the equality
\begin{equation} \label{PdPP:eq}
P\, \d P\, P = 0.
\end{equation}
Furthermore, the matrices $P$ and $\d P\, \d P$ commute.
\end{lemma}
\begin{proof}
By differentiating the idempotence relation $P^2 = P$, we obtain:
\[
\d P\, P + P\, \d P = \d P
\]
and multiplying by $P$ on the left (the same result is anyhow achieved by multiplication on the right) we get
\[
P\, \d P\, P + P^2\, \d P = P\, \d P,
\]
which simplifies to \eqref{PdPP}. The commutation property follows immediately from differentiation of \eqref{PdPP}:
\[
\d P\, \d P\, P – P\, \d P\, \d P = 0. \qedhere
\]
\end{proof}
The curvature now reads
\[
\nabla^2 \vartheta = P\, \d P\, \d P\, \vartheta \eq \Omega\vartheta,
\]
where $\Omega$ denotes the curvature matrix of $\nabla$. From the previous lemma we deduce that any integer power $\Omega^m$ is invariant under $P$: indeed, because of the
commutativity of $P$ and $\d P\, \d P$, $\Omega$ can be represented in the forms
\[
\Omega \= P\, \d P\, \d P = P^2 \d P\, \d P = P\, \d P\, \d P\, P = \d P\, \d P\, P,
\]
whence $P\Omega = \Omega P = \Omega$.
\begin{lemma}
The covariant differential of $\Omega^m$ vanishes for all $m \in \N$.
\end{lemma}
\begin{proof} \marginpar{See Park}
We have, for $m = 1$:
\[
\d\Omega = \d(P\, \d P\, \d P\, P) = \d P\, \Omega + \Omega\, \d P,
\]
whence, by Lemma \ref{PdPP} and Definition \ref{CovDiff},
\[
\nabla\Omega = P\, \d P\, \Omega P + P \Omega\, \d P\, P = P\, \d P\, P \Omega + \Omega P\, \d P\, P = 0.
\]
Note that it is sufficient to show that $\nabla\Omega = 0$ (the second Bianchi identity \veqref{Bianchi}), because of the invariance of $\Omega$ under $P$:
\begin{align*}
\nabla\Omega^2 & \= P\, \d\Omega^2\, P \\
& = P(\d\Omega\, \Omega + \Omega\, \d\Omega)P \\
& = P(\d P\, \Omega + \Omega\, \d P)\Omega P + P\Omega(\d P\, \Omega + \Omega\, \d P)P \\
& = P\, \d P\, \Omega^2 P + P\Omega\, \d P\, \Omega P + P\Omega\, \d P\, \Omega P + P \Omega^2 \, \d P\, P \\
& = P\, \d P\, P \Omega^2 + 2 \Omega P\, \d P\, P \Omega + \Omega^2 P\, \d P\, P \\
& = 0
\end{align*}
and by induction one achieves the result.
\end{proof}
\begin{lemma}\label{lemma:dnabla}
Let $A$ be a matrix form invariant under $P$. Then
\[
\d (\tr A) = \tr (\nabla A).
\]
\end{lemma}
\begin{proof} \marginpar{See Park}
By virtue of Lemma \ref{PdPP} again and property \veqref{tr} we have
\begin{align*}
\d (\tr A) & = \d[\tr(PAP)] = \tr[\d(PAP)] \\
& = \tr[\d(PA)P + PA\, \d P] \\
& = \tr(\d P\, AP + P\, \d A\, P + PA\, \d P) \\
& = \tr(\nabla A) + \tr(P\, \d P\, A + A\, \d P\, P) \\
& = \tr(\nabla A) + \tr(P\, \d P\, PA + AP\, \d P\, P) \\
& = \tr(\nabla A). \qedhere
\end{align*}
\end{proof}
We now introduce the $2k$-forms%
% \footnote{The reason for which we divide $\Omega$ by the factor $2\pi i$ is to have integer coefficients in the cohomology classes.}
\[
\varphi_k \= \tr \left( – \frac{\Omega}{2\pi i} \right)^k.
\]
A direct application of the previous two lemmas shows that each $\varphi_k$ is closed, and therefore they define some cohomology classes on $M$. The following result shows that these
last ones do not depend on the vector bundle $P$, but only on its homotopy class.
\begin{lemma}
If $P_1$ and $P_2$ are equivalent vector bundles, then the corresponding forms $\varphi_k^{(1)}$ and $\varphi_k^{(2)}$ are cohomologous.
\end{lemma}
\begin{proof}
Let $P \= P(x, t)$ be a homotopy between $P_1$ and $P_2$. If $\Omega \= \Omega(t)$ denotes the curvature of $P(x, t)$ then we have, by the Fundamental Theorem of Calculus:%
\footnote{Actually, $\Omega$ depends also on $x$, but we already pointed out that $\nabla\Omega^k = 0$.}
\[
\tr\Omega_2^k – \tr\Omega_1^k = \int_0^1 \frac{d}{dt} \tr\Omega^k(t)\, dt.
\]
The aim is to show that the integral on the right-hand side is zero, or, more precisely, that the integrand is an exact matrix form. We have
\[
\tr \frac{d}{dt} \Omega^k(t) = k \tr(\Omega^{k – 1}\dot{\Omega}),
\]
where the dot denotes the differentiation with respect to the parameter $t$. Besides,
\begin{align*}
\dot{\Omega} & = \dot{P}\, \d P\, \d P\,P + P(\dot{\d P}\, \d P + \d P\,\dot{\d P})P + P\, \d P\, \d P\,\dot{P} \\
& = \dot{P}\Omega + P\, \d(\dot{P}\, \d P – \d P\,\dot{P})\,P + \Omega\dot{P}.
\end{align*}
Substituting in the above equation, the first and last term vanish, because $P\dot{P}P = 0$ (by Lemma \ref{PdPP}, with differentiation with respect to $t$ in place of exterior differentiation),
yielding
\[
\tr \frac{d}{dt} \Omega^k(t) = k \tr\{\Omega^{k – 1}P\, \d(\dot{P}\, \d P – \d P\,\dot{P})\,P\}.
\]
Now observe that $P$ commutes with $\dot{P}\, \d P$ and with $\d P\, \dot{P}$: by differentiating $P\, \d P\,P = 0$ with respect to $t$ we obtain%
\marginpar{I can't show this commutativity! By the way, what is the purpose of showing this?}
\[
\dot{P}\, \d P\, P + P\, \dot{\d P}\, P + P\, \d P\, \dot{P} = 0
\]
and applying exterior differentiation to $P\dot{P}P = 0$ gives
\[
\d P\, \dot{P}P + P\, \d\dot{P}\, P + P\dot{P}\, \d P = 0.
\]
Subtracting the two equations and rearranging terms hands: \marginpar{This shows only that $P$ commutes with the difference $\dot{P}\, \d P – \d P\,\dot{P}$}
\[
P(\dot{P}\, \d P – \d P\, \dot{P}) = (\dot{P}\, \d P\, – \d P\, \dot{P})P,
\]
whence it follows that \marginpar{Why should $\dot{P}\, \d P – \d P\, \dot{P} = P(\dot{P}\, \d P – \d P\,\dot{P})P$?}
\[
P\, \d(\dot{P}\, \d P – \d P\,\dot{P})\,P = P\, \d[P(\dot{P}\, \d P – \d P\,\dot{P})P]\,P = \nabla[P(\dot{P}\, \d P – \d P\,\dot{P})P].
\]
Therefore, recalling that $\nabla\Omega = 0$, that $\Omega$ is invariant under $P$ and using Lemma \ref{lemma:dnabla}:
\begin{align*}
\frac{d}{dt}\tr\Omega^k & = k \tr \{\nabla[P(\dot{P}\, \d P – \d P\,\dot{P})P]\Omega^{k – 1}\} \\
& = k \tr \{\nabla[P(\dot{P}\, \d P – \d P\,\dot{P})P\Omega^{k – 1}]\} \\
& = k\, \d\{ \tr [P(\dot{P}\, \d P – \d P\,\dot{P})\Omega^{k – 1}]\},
\end{align*}
which shows the exactness of the integrand.
\end{proof}
From this lemma it follows that the forms $\varphi_k$ define actually \emph{real} cohomology classes on $M$, because in each class of equivalent bundles there exists an orthogonal projector,
which makes the matrix $\Omega$ Hermitian.
The matrix forms $\varphi_k$ can also be expressed via the local curvature matrices $\Omega_j$, each of which is defined in a neighbourhood $U_j$ of a finite open cover $\{U_j\}$ (existing
because of compactness) of $M$. Let $e_j$ and $\eta_j$ be a frame and a coframe in $U_j$, so that $P_j = e_j\eta_j$; then
\begin{align*}
\Omega & \= P\, \d P\, \d P = e_j\eta_j\, \d(e_j\eta_j)\, \d(e_j\eta_j) \\
& = e_j\eta_j(\d e_j\, \eta_j + e_j\, \d\eta_j)(\d e_j\, \eta_j + e_j\, \d\eta_j) \\
& = e_j\eta_j(\d e_j\, \eta_j\, \d e_j\, \eta_j + \d e_j\, \eta_j e_j\, \d\eta_j + e_j\, \d\eta_j\, \d e_j\, \eta_j + e_j\, \d\eta_j\, e_j\, \d\eta_j) \\
& = e_j[(\eta_j\, \d e_j)^2\eta_j + \eta_j\, \d e_j\, \d\eta_j + \d\eta_j\, \d e_j\, \eta_j + \d\eta_j\, e_j\, \d\eta_j] \\
& = e_j[(\eta_j\, \d e_j)^2 + \d\eta_j\, \d e_j]\eta_j + e_j[\d(\eta_j e_j)\d\eta_j] \\
& = e_j[(\eta_j\, \d e_j)^2 + \d\eta_j\, \d e_j]\eta_j \\
& \eq e_j\Omega_j\eta_j,
\end{align*}
where
\[
\Omega_j \= (\eta_j\, \d e_j)^2 + \d\eta_j\, \d e_j = (\eta_j\, \d e_j)^2 + \d(\eta_j\, \d e_j)
\]
is precisely the local curvature matrix in the given frame and coframe in $U_j$. Using property \eqref{tr} of the trace \vpageref{tr}, we obtain the following local expression for the forms
$\varphi_k$:
\begin{equation}\label{phik:local}
\varphi_k = \tr\left( – \frac{\Omega_j}{2\pi i} \right)^k = \left( – \frac{1}{2\pi i} \right)^k \tr[(\eta_j\, \d e_j)^2 + \d(\eta_j\, \d e_j)]^k.
\end{equation}
% —————————————————
\section{The Formula for the Index}
% —————————————————
Let $M$ be an $m$-dimensional Riemannian manifold, let $T^*M$ be its cotangent bundle and denote by $S^*M$ the sphere bundle in $T^*M$. A point in the cotangent bundle can thus
be described by a pair $(s, t)$, where $s \in S^*M$ and $t$ is a non-negative real number. In other words, we split the description of a (co)vector in each fiber into its direction and length. All
points of the kind $(s, 0)$ form the zero section of $T^*M$; for each of these we observe that $s$ is not uniquely defined. It is also convenient to consider the so-called \dfn{infinite section}, that
is the set of all “infinitely distant points'' $(s, +\infty)$, which belong to the Alexandrov compactification \marginpar{Talk about Alexandrov compactification in the appendix?} $\overline{T^*M}$
of the cotangent bundle.
Consider a cover of $\overline{T^*M}$ made up of two open sets only, $U_0$ and $U_\infty$; the former including all “finitely distant points'' (i.e. $U_0 \= T^*M$) and the latter consisting of
all points except for those forming the zero section (i.e. $U_\infty \= \overline{T^*M} \setminus \{(s, t) \in T^*M : t = 0\}$).
We now construct a vector bundle of rank $n \geq m$ over $\overline{T^*M}$, by means of frames and coframes on these two neighbourhoods, in such a way that the transition function on
$S^*M$ coincides with a given non-degenerate matrix-valued function $\tau \colon S^*M \to \GL_n(\C)$.
Let $f_1$ and $f_2$ be smooth real functions of the non-negative real variable $t$, such that
\begin{itemize}
\item $f_1^2 + f_2^2 = 1$;
\item $f_1(t) \= 0$ for all $t \in \big(0, \frac{1}{2}\big)$;
\item $f_1(t) \= 1$ for all $t > 1$
\end{itemize}
(we do not specify their values elsewhere). Define then on $U_0$ the matrices $e_0(s, t) \in \mathrm{M}^{2n \times n}(\C)$ and $\eta_0(s, t) \in \mathrm{M}^{n \times 2n}(\C)$ by setting
\[
e_0(s, t) \= \begin{pmatrix}
f_2(t)\, \Id_{\C^n} \\
f_1(t) \tau(s)
\end{pmatrix}, \qquad
\eta_0(s, t) \= \Big( f_2(t)\, \Id_{\C^n}, f_1(t) \big(\tau(s)\big)^{-1} \Big).
\]
Similarly, on $U_\infty$ we put
\[
e_\infty(s, t) \= \begin{pmatrix}
f_2(t) \big(\tau(s)\big)^{-1} \\
f_1(t)\, \Id_{\C^n}
\end{pmatrix}, \qquad
\eta_\infty(s, t) \= \big( f_2(t) \tau(s), f_1(t)\, \Id_{\C^n} \big).
\]
It is immediate to check that $\eta_0 e_0 = \eta_\infty e_\infty = \Id_{\C^n}$. In the intersection $U_0 \cap U_\infty$, which corresponds to the positive real semi-axis $0 < t < +\infty$, we have the
relations
\[
e_0(s, t) = e_\infty(s, t) \tau(s), \quad \eta_0(s, t) = \big( \tau(s) \big)^{-1} \eta_\infty(s, t),
\]
which make $e_0$, $\eta_0$, $e_\infty$ and $\eta_\infty$ frames and coframes of a vector bundle over $U_0$ and $U_\infty$ respectively, with transition function $\tau$ (see \veqref{transf}).
The family $P$ of projectors associated with the vector bundle has therefore the form
\begin{align}\label{vbundle:constrP}
P(s, t) & \= e_0(s, t)\eta_0(s, t) = e_\infty(s, t)\eta_\infty(s, t) \\
& = \begin{pmatrix}
f_2^2(t)\, \Id_{\C^n} & f_1(t) f_2(t) \big(\tau(s)\big)^{-1} \\
f_1(t) f_2(t) \tau(s) & f_1^2(t)\, \Id_{\C^n}
\end{pmatrix}.
\end{align}
In order to find the expression for the forms $\varphi_k$, we use formula \veqref{phik:local}. We then compute, removing the dependence on $s$ and $t$ but still bearing it in mind,
\[
\d e_0 = \begin{pmatrix}
\d f_2\, \Id_{\C^n} \\
\d f_1\, \tau + f_1\, \d\tau
\end{pmatrix},
\]
whence
\[
\eta_0\, \d e_0 = f_2\, \d f_2\, \Id_{\C^n} + f_1\, \d f_1\, \Id_{\C^n} + f_1^2\, \tau^{-1} \d\tau = f_1^2\, \tau^{-1} \d\tau,
\]
because $f_1\, \d f_1 + f_2\, \d f_2 = 0$, as obtained by differentiating the first property of the functions $f_1$ and $f_2$. Therefore,
\begin{align*}
(\eta_0\, \d e_0)^2 + \d(\eta_0\, \d e_0) & = (f_1^2\, \tau^{-1} \d\tau)^2 + \d(f_1^2\, \tau^{-1} \d\tau) \\
& = f_1^4(\tau^{-1} \d\tau)^2 + \d f_1^2\, \tau^{-1} \d\tau + f_1^2\, \d\tau^{-1} \d\tau \\
& = f_1^4(\tau^{-1} \d\tau)^2 + \d f_1^2\, \tau^{-1} \d\tau - f_1^2 (\tau^{-1}\d\tau)^2 \\
& = f_1^2(f_1^2 - 1) (\tau^{-1}\d\tau)^2 + \d f_1^2\, \tau^{-1} \d\tau \\
& = - f_1^2f_2^2 (\tau^{-1}\d\tau)^2 + \d f_1^2\, \tau^{-1} \d\tau,
\end{align*}
where, in the third equality, we used the fact that $\d\tau^{-1} = - \tau^{-1}\d\tau\, \tau^{-1}$, as is readily seen by differentiating the identity $\tau\tau^{-1} = \Id_{\C^n}$.
Thus, observing that all powers of $\d f_1^2$ higher than $1$ are annihilated and using the binomial formula, we have
\begin{align*}
\varphi_k & = \left( -\frac{1}{2\pi i} \right)^k \tr[ - f_1^2 f_2^2 (\tau^{-1}\d\tau)^2 + \d f_1^2\, \tau^{-1} \d\tau]^k \\
& = \left( \frac{1}{2\pi i} \right)^k [(f_1f_2)^{2k} \tr(\tau^{-1}\d\tau)^{2k} - k (f_1f_2)^{2k - 2} \d f_1^2 \tr(\tau^{-1} \d\tau)^{2k - 1}].
\end{align*}
This expression can be further simplified if we note that, by associativity of matrix product and property \veqref{tr},
\begin{align*}
\tr(\tau^{-1}\d\tau)^{2k} & = \tr[(\tau^{-1}\d\tau)^{2k - 1} (\tau^{-1}\d\tau)] \\
& = \tr[(\tau^{-1}\d\tau)(\tau^{-1}\d\tau)^{2k - 1}] \\
& = - \tr[(\tau^{-1}\d\tau)^{2k - 1} (\tau^{-1}\d\tau)],
\end{align*}
which in its turn implies that this term must vanish. Moreover, the form $\tr(\tau^{-1} \d\tau)^{2k - 1}$ is closed, because
\[
\d[\tr(\tau^{-1} \d\tau)^{2k - 1}] = \tr[\d(\tau^{-1} \d\tau)^{2k - 1}] = - \tr(\tau^{-1} \d\tau)^{2k} = 0,
\]
the second equality easily following by induction and the expression for $\d\tau^{-1}$ exploited before.
With all this information, the forms $\varphi_k$ can now be written as
\[
\varphi_k = - \frac{k}{(2\pi i)^k}\, \d \left(\int_0^t {\big(f_1(s)f_2(s)\big)^{2k - 2}\, \d f_1^2(s)} \cdot \tr(\tau^{-1} \d\tau)^{2k - 1} \right).
\]
For $t > 1$ we have
\[
\int_0^t {\big(f_1(s)f_2(s)\big)^{2k – 2}\, \d f_1^2(s)} = \int_0^1 {[\xi(1 – \xi)]^{k – 1}\, \d \xi} = B(k, k) = \frac{[(k – 1)!]^2}{(2k – 1)!},
\]
where $B$ is the beta function. Then, setting
\[
h_k(t) \= \frac{(2k – 1)!}{[(k – 1)!]^2}\int_0^t {\big(f_1(s)f_2(s)\big)^{2k – 2}\, \d f_1^2(s)},
\]
the forms $\varphi_k$ now read
\[
\varphi_k = – \frac{1}{(2\pi i)^k}\frac{k!(k – 1)!}{(2k – 1)!}\, \d[h_k\tr(\tau^{-1} \d\tau)^{2k – 1}]
\]
and are compactly supported in $(0,1]$, because $h_k(t) = 1$ for all $t > 1$.
\begin{rmk}
We could replace \marginpar{Show this?} $h_k$ by a smooth function $h$, which vanishes in a neighbourhood of the origin and is equal to $1$ for $t > 1$, without affecting the
cohomology class of $\varphi_k$.
\end{rmk}
We then introduce the closed forms on $S^*M$
\[
\psi_k \= \frac{1}{(2\pi i)^k} \frac{(k – 1)!}{(2k – 1)!} \tr(\tau^{-1}\d\tau)^{2k – 1},
\]
so that the Chern character of the vector bundle $P$ has the expression
\[
\ch P = n – \d\! \left( h(t) \sum_{k = 1}^m \psi_k \right).
\]
\begin{rmk}
Since the family $P$ of projectors is constant in a neighbourhood of the zero and infinite sections of $\overline{T^*M}$, we may assume that $P$ is defined on the suspension
\marginpar{Talk about suspension in the appendix? Collapsing construction?} $S(S^*M)$ of $S^*M$. We nowhere used the fact that $S^*M$ is a sphere bundle and therefore everything
we said keeps being true if $S^*M$ is replaced by any (compact?) manifold and $\overline{T^*M}$ by the suspension $S(S^*M)$.
In particular, a given map $\theta \colon \S^{2m – 1} \to \GL_{2m – 1}(\C)$ \marginpar{Is $\GL_{2m – 1}(\C)$ correct?} defines a vector bundle $P$ (as in \veqref{vbundle:constrP}) over the
sphere $\S^{2m}$, which is indeed the suspension of $\S^{2m – 1}$. From the Bott periodicity theorem \marginpar{State this theorem? Where? K-Theory or appendix?} it follows that the
only stable homotopy invariant of $\theta$ is the value of $\ch P$ on $\S^{2m}$. Integrating $\d(h\psi_m)$ over $\S^{2m}$ and applying the Stokes theorem we find that in this case
\[
\ch P = \pm \frac{1}{(2\pi i)^m} \frac{(m – 1)!}{(2m – 1)!} \int_{\S^{2m – 1}} \tr(\theta^{-1}\d\theta)^{2m – 1}.
\]
\end{rmk}
We now consider an elliptic differential operator $L \colon \smoothim{M}{\C^n} \to \smoothim{M}{\C^n}$. Its symbol $\sigma$, considered on $S^*M$, is a non-degenerate matrix-valued
function and therefore it defines a vector bundle of rank $n$ represented through a family $P$ of projectors on $T^*M$, which is constant outside some compact subset of $T^*M$.
The differential form $\ch \sigma$, which has compact support on $T^*M$, is defined as
\[
\ch\sigma \= \ch P – n,
\]
where $n$ is the rank of $P$, as before.
The Atiyah-Singer theorem \marginpar{Write about it? Here, before, in the appendix, as an aside in another chapter,\dots?} gives now an expression for the index of $L$, relating it to the value
of the $2m$-dimensional component of the product $(-1)^m\ch\sigma \cdot \mathscr{T}(M)$ on the fundamental cycle \marginpar{What is the fundamental cycle?} on $T^*M$:
\[
\ind L = (-1)^m \int_{T^*M} \big(\ch\sigma \cdot \mathscr{T}(M) \big)_{2m}.
\]
The integrand is recovered through the expression of $\ch P$ and the Todd class \textcolor{red}{(insert reference to the Todd class here)}:
\[
\big(\ch\sigma \cdot \mathscr{T}(M) \big)_{2m} = – \d\left\{ h(t) \left( \psi_m + \sum_{k = 1}^{\lfloor m/4 \rfloor} \psi_{m – 2k} \mathscr{T}_k \right) \right\},
\]
We observe that this form vanishes for $t > 1$; therefore, we can integrate it over the bundle $B^*M$ of unit balls of $T^*M$, i.e. over the set of all $t \leq 1$, and this yields
\[
\ind L = (-1)^{m + 1} \int_{B^*M} \d\left\{ h(t) \left( \psi_m + \sum_{k = 1}^{\lfloor m/4 \rfloor} \psi_{m – 2k} \mathscr{T}_k \right) \right\}.
\]
Applying the Stokes theorem we finally obtain
\[
\ind L = (-1)^{m + 1} \int_{S^*M} \left( \psi_m + \sum_{k = 1}^{\lfloor m/4 \rfloor} \psi_{m – 2k} \mathscr{T}_k \right).
\]
In the case where all the $\mathscr{T}_k$ vanish \marginpar{Insert conditions for the Todd class to vanish, explain intuitively what it means.} we have the simpler form
\[
\ind L = \frac{(-1)^{m + 1}}{(2\pi i)^m} \frac{(m – 1)!}{(2m – 1)!} \int_{S^*M} \tr(\sigma^{-1}\d\sigma)^{2m – 1}.
\]`