How to write the Thom spectrum representing cobordism as an $\Omega$-spectrum?

Question

It is often said [e.g. Atiyah, "Bordism and Cobordism" (1961)] that the Thom spectrum $MSO(i)$ represents oriented cobordism, in the following sense: \begin{eqnarray} MSO^n(X,Y) &:=& \lim_{i \rightarrow \infty} \langle \Sigma^{i-n}(X/Y), MSO(i) \rangle\\ &=& \lim_{i \rightarrow \infty} \langle X/Y, \Omega^{i-n} MSO(i) \rangle\\ &=& \langle X/Y, \Omega^{i-n} MSO(i) \rangle, ~~\text{large}~i, \end{eqnarray} for finite CW pairs $(X,Y)$. where $\Sigma$ is the reduced suspension, $\Omega$ is the usual loop space functor, and $\langle-, -\rangle$ is the homotopy classes of pointed maps. The direct limit was taken with respect to the maps \begin{equation} \langle \Sigma^{i-n}(X/Y), MSO(i) \rangle \rightarrow \langle \Sigma^{i+1-n}(X/Y), \Sigma MSO(i) \rangle \xrightarrow{f_{i*}} \langle \Sigma^{i+1-n}(X/Y), MSO(i+1) \rangle. ~ (1) \end{equation} where $f_{i}:\Sigma MSO(i) \rightarrow MSO(i+1)$ is the natural map mentioned in Atiyah.

By the Brown representability theorem, one should be able to represent oriented cobordism in the usual sense that \begin{equation} MSO^n(X,Y) \stackrel{?}{\cong} \langle X/Y, K_n \rangle ~~~ (2) \end{equation} for some $\Omega$-spectrum $\{K_n\}$. So this is something like moving the direct limit inside $\langle -,- \rangle$.

My question is: If $K_n$ exists, then what is it? Or is it because the Brown representability theorem hypothesized a generalized cohomology theory on all CW pairs, that there isn't an $\Omega$-spectrum $\{K_n\}$ representing oriented cobordism, which is defined only for finite CW pairs?

I was able to show that (1) is actually the same, via adjunction, as the maps \begin{equation} \langle X/Y, \Omega^{i-n} MSO(i) \rangle \rightarrow \langle X/Y, \Omega^{i+1-n}\Sigma MSO(i) \rangle \rightarrow \langle X/Y, \Omega^{i+1-n} MSO(i+1) \rangle ~~~ (3) \end{equation} induced by \begin{equation} MSO(i) \xrightarrow{\eta_{MSO(i)}} \Omega \Sigma MSO(i) \xrightarrow{\Omega(f_i)} \Omega MSO(i+1), ~~~(4) \end{equation} where $\eta_Y:Y \rightarrow \Omega \Sigma Y$ is the unit of the adjunction $\Sigma \dashv \Omega$. Can we go from here to contruct $K_n$ out of $MSO(i)$?

Sorry for this potentially elementary question.

Answer 1

A concrete model for Ω^∞ applied to Thom spectra (which is all what we need because Thom spectra are connective) was given by Quinn in his thesis.

Very roughly, Ω^∞MG is a simplicial set whose n-simplices are n-dimensional G-manifolds with corners, with corners of codimension i assigned to simplicial faces of Δ^n of codimension i.

For precise definitions and statements see Quinn's paper “Assembly maps in bordism-type theories”, especially Sections 3 and 6.

Laures and McClure explain how to promote these simplicial sets to strictly commutative symmetric ring spectra. See their papers “Multiplicative properties of Quinn spectra” and “Commutativity properties of Quinn spectra”.