cobordism spectrum

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cobordism spectrum

Form the set \(\Omega_n^O\) of cobordism classes of \(n{\hbox{-}}\)manifolds.

The \(O\) denotes that fact that any bundle over a manifold \(M\) admits a Riemannian metric, so embedding \(M\) into \({\mathbf{R}}^\infty\) yields a Reduction of structure group.md) to $O(N to \(O(N)\) for some \(N\).

Define addition as disjoint union and multiplication by the Cartesian product to form a graded ring \(\Omega^O\), and it turns out that there is a spectrum \({\operatorname{MO}}\) such that \begin{align*} \Omega^O_n \cong_{{\mathsf{Grp}}} \pi_n MO \end{align*}

Idea: any manifold is determined by its embedding into \({\mathbf{R}}^\infty\), take the normal bundle \(\nu\), form the Thom space \(M^\nu\) by collapsing the complement of the normal bundle. This yields a map \(S^{?} \to M^\nu\). Now use the fact that \(\nu\) is an \(N{\hbox{-}}\)dimensional bundle and is classified by a map \(M \to BO(N)\), the classifying space for \(O(N){\hbox{-}}\)bundles with universal bundle \(\gamma_N\),. We take Thom spaces (?) to get a map \(M^\nu \to BO(N)^{\gamma_N}\), then take \begin{align*} \lim_{N\to \infty} [S^{N+n}, BO(N)^{\gamma_N}]_{\text{unstable?}} \end{align*} This is independent of the embedding and only depends on the cobordism class of \(M\), so we define \(MO(n) \coloneqq BO(N)^{\gamma_N}\).

Equivariant structure

See orthogonal spectra. attachments/Pasted%20image%2020220510185755.png attachments/Pasted%20image%2020220510185834.png attachments/Pasted%20image%2020220510185854.png

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