Commit 160be9f7 authored by Jannes Bantje's avatar Jannes Bantje

add important example

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\section{Introduction -- Julian} % (fold)
\section{Introduction: Genera and Multiplicative Sequences -- Julian} % (fold)
The story start by looking at bordism invariants.
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\item and higher versions thereof.\todo{specify}
We start by recalling some facts from bordism theory starting with an example of a bordism invariant:
Let $M$ be a closed, $4n$-dimensional oriented smooth manifold.
The cup product gives rise to a symmetric bilinear form on $H^{2n}(M;\mathbb{Q})$.
......@@ -38,14 +40,12 @@ In studying such invariant, we will
is defined as the number of positive Eigenvalues minus the number of negative ones of this symmetric bilinear form.
As the signature is invariant under oriented cobordism, we get a ring homomorphism $\operatorname{sign} \colon \Omega_*^{\SO} \to \mathbb{Z}$.
This bordism invariant has the nice feature, that if $M$ is closed, spin and $\dim M \equiv 4 \mod 8$ then
\operatorname{sign}(M) \equiv 0 \mod 16
One of the main players of bordism theory are the \Index{Thom spectra}, which arise as the sequence of Thom spaces of the universal bundles over $\B G(n)$.
It is denoted by $MG$ and we have
......@@ -53,7 +53,7 @@ It is denoted by $MG$ and we have
\todo{a reference sounds reasonable}
The oriented bordism ring is given by
\Omega^{\SO}_* \otimes \mathbb{Q} = \mathbb{Q} \benbraceX[\big]{y_{4k} \given k \in \mathbb{N}}
......@@ -61,7 +61,45 @@ It is denoted by $MG$ and we have
where $y_{4k}= \benbrace*{\mathbb{C}\mathbb{P}^{2k}} \in \Omega^{\SO}_{4k}$.
Furthermore two manifolds represent the same class precisely if all their Prontryagin numbers and Stiefel-Whitney numbers coincide.\todo{reference}
Furthermore two manifolds represent the same class precisely if all their Prontryagin numbers and Stiefel-Whitney numbers coincide.\todo{reference? -- Julian just mentioned Pontryagin numbers?}
Now we can define, what is meant by \enquote{genus} in this context:
Let $R$ be an integral domain over $\mathbb{Q}$.
Then a \Index{genus} is a unital ring homomorphism
\varphi \colon \Omega^{\SO}_* \otimes \mathbb{Q} \To{} R
Since we know $\Omega^{\SO}_*$ by \cref{thm:oriented-bordism_ring}, we study these genera by less abstract, more computational means, as we are going to explain now:
The general idea is to consider Chern resp. Pontryagin classes formally as elementary symmetric functions, \cite[Sec.~1.4]{hirzebruch_modularforms}.
Let $E = E_1 \oplus \ldots \oplus E_n$ be a direct sum of complex line bundles and set $x_i = c_1(E_i) \in H^2(X;\mathbb{Z})$.
By \cref{thm:axioms_chern} we have
c(E) = (1 + x_1) \cdots (1+ x_n) = 1 + c_1 + \ldots + c_r
where the Chern class $c_r= \sigma(x_1, \ldots ,x_n)$ is the $r$-th \Index{elementary symmetric function} in the $x_i$.
By assuming the existence of Hermitian structures on the $E_1$, we can reduce their structure groups from $\mathbb{C}^\times$ to $T^1=S^1\cong\Unitary(1)$ and therefore the one of $E$ to
T^n = \set*{A \in \Unitary(n) \given A = \operatorname{diag} \enbrace*{e^{2 \pi i \varphi_1},\ldots , e^{2 \pi i \varphi_n}}, \varphi_i \in \mathbb{R}}
which is a maximal torus of the connected Lie group $\Unitary(n)$.
All such maximal tori can be conjugated into each other and such a conjugation by, say, $g \in \Unitary(n)$ permutes the $S^1$ factors of $T^n$ but still yields an isomorphism of the $\Unitary(n)$-bundle (but not necessarily of the $T^n$-bundle).
Since an explicit expression of genus has to be invariant under such $\Unitary(n)$-isomorphisms, it must remain fixed under permutations of the $x_i$.
And therefore it has to be a polynomial in the Chern classes (which we identified as elementary symmetric polynomials) by the fundamental theorem on symmetric polynomials!
A similar story can be told for the Pontryagin classes, see \cite[8]{hirzebruch_modularforms}.
There is a one-to-one correspondence of genera $\varphi$ and multiplicative sequences $Q$.
% section introduction_by_julian (end)
\section{Modular Forms -- Jens} % (fold)
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