Commit 615204cf authored by Jannes Bantje's avatar Jannes Bantje

complete talk summar

parent 40b31a08
Pipeline #60074 passed with stages
in 1 minute and 55 seconds
......@@ -616,8 +616,65 @@ Integrating over the fibre (if $X$ is oriented) can rid us of the cotangent bund
\[
\indeX D = (-1)^{\frac{n(n+1)}{2}} \enbrace*{\pi_! \ch(\sigma(D)) \cdot \Ahat(X)^2}[X]
\]
These formulas become even more intriguing in the special case, that the structure group of $X^{2n}$ is a connected subgroup of $\SO(2n)$, as the operator does not show up on the right hand side anymore!
\newpage
\todo[inline]{\textcite[Thm.~III.13.13]{lawson_spin} is the closest thing to the Atiyah--Singer formula in \cite{hirzebruch_modularforms} I found so far.}
These formulas become even more intriguing in the special case, that the structure group of $X^{2n}$ is a connected subgroup $G \subset \SO(2n)$ and the bundles $E$ and $F$ are associated to unitary representations $\rho_E$ and $\rho_F$.
Here the operator $D \colon \Gamma(E) \to \Gamma(F)$ (or its symbol) does not show up on the right hand side anymore!
This can be found in \cite[Thm.~13.13]{lawson_spin}.
A version of this variant for elliptic complexes is, what \textcite[Sec.~5.2]{hirzebruch_modularforms} state as \enquote{the} Atiyah--Singer index theorem:
\begin{theorem}
For a closed, oriented smooth manifold of dimension $2n$ and an elliptic complex $(D_i)$ associated to the tangent bundle over $X$ we have
\[
\indeX((D_i)) = (-1)^n \enbrace*{\enbrace*{\frac{1}{e(\Tang^* X)} \sum_{i=0}^m (-1)^i \cdot \ch(E_i)} \td(\Tang X \otimes \mathbb{C})}[X]
\]
\end{theorem}
\subsection*{What is \enquote{the} index theorem good for?}
The most basic observation is, that the left hand side is an integer by design and therefore the right hand side is as well -- which is not at all obvious without the context of the index theorem.
This alone is not that helpful, since the right hand side lacks any (geometric) meaning so far.
Therefore applications often arise by considering concrete differential operators where either of the sides of the formula carries a meaning.
One of the most famous applications is the following:
\begin{example}
Let $X$ be spin of dimension $4n$.
Then the Atiyah--Singer operator $\Dirac^+$ (the plus sign refers to a certain grading) satisfies the following index formula
\[
\indeX(\Dirac^+) = \Ahat(X)
\]
This implies, that the \Ahat-genus for $4n$-dimensional spin manifolds is integral.
See \cite[Thm.~13.10]{lawson_spin} for the computations necessary for this.
This establishes the \enquote{second stage} on the way to the Witten genus mentioned in \cref{sec:introduction_julian}.
\end{example}
A much more basic example is:
\begin{example}
Consider the de Rham complex $E_i = \Lambda^i \enbrace*{\Tang^*X \otimes \mathbb{C}}$ with differential operators given by the exterior derivative $d$.
Following the definitions and de Rham's theorem then easily yield
\[
\indeX (d) = e(X)
\]
Pluggin everything into the index formula yields the same result.
\end{example}
Form both of these examples we see a theme emerging: Many characteristic numbers can be realised as the index of an differential operator!
\begin{example}{\cite[Sec.~5.5]{hirzebruch_modularforms}}
Let $X$ be closed, oriented, $4k$-dimensional and equipped with a Riemannian metric on the tangent bundle.
Hodge theory tells us, that we can write the signature as
\[
\operatorname{sign} X = \dim H^{2k}_+ - \dim H^{2k}_-
\]
with respect to a decomposition of $H^{2k}$ using the intersection form.
This difference of dimensions seems familiar \ldots and in fact equals the index of $d + d^*$.
Applying the index theorem and some further computations then yield
\[
\operatorname{sign} X = \indeX(d+ d^*) = \enbrace*{\prod_{j=1}^{2k} \frac{x_j}{\tanh(x_j)}} [X]
\]
But this is exactly the formula for the $L$-genus and Hirzebruch's signature theorem follows from Atiyah--Singer!
\end{example}
% section index_theory (end)
% chapter elliptic_genera_phd_seminar (end)
\ No newline at end of file
......@@ -189,6 +189,7 @@
\settoheight{\Fredsize}{$\Fred$}
\DeclareMathOperator{\indeX}{ind}
\DeclareMathOperator{\IndeX}{Ind}
\DeclareMathOperator{\topind}{top-ind}
\newcommand{\kkprod}{\otimes}
\newcommand{\FredRed}{\raisebox{0pt}[\Fredsize][\depth]{$\widetilde{\raisebox{0pt}[5.8pt][\depth]{$\Fred$}}$}}
......@@ -272,6 +273,8 @@
\newcommand{\thoM}{M\mkern-3mu}
\DeclareMathOperator{\td}{td}
\DeclareMathOperator{\ch}{ch}
\DeclareMathOperator{\Thom}{Th}
\newcommand{\cpt}{\mathrm{cpt}}
\newcommand{\tmf}{\ensuremath{\mathrm{tmf}}}
\DeclareMathOperator{\Map}{Map}
\newcommand{\sa}{\mathrm{sa}}
......@@ -328,6 +331,11 @@
\newcommand{\gradedMR}{\raisebox{0pt}[\Msize][\depth]{$\widehat{\raisebox{0pt}[5.8pt][\depth]{\ensuremath{\MR}}}$}}
% ======================================================================================
%-- Dirac operators and so on
% ======================================================================================
\newcommand{\Dirac}{\slashed{D}}
% ======================================================================================
%-- algebra stuff
% ======================================================================================
......
......@@ -65,6 +65,7 @@
\usepackage{xfrac}
\usepackage{mathdots} % Verbesserung von Punkten wie zB \ldots
\usepackage{centernot}
\usepackage{slashed}
\usepackage{stackrel}
\usepackage{nicematrix}
\DeclareSymbolFont{bbold}{U}{bbold}{m}{n}
......
Markdown is supported
0%
or
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment