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

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 $$\langle \Sigma^{i-n}(X/Y), MSO(i) \rangle \rightarrow \langle \Sigma^{i+1-n}(X/Y), \Sigma MSO(i) \rangle \xrightarrow{f_{i*}} \langle \Sigma^{i+1-n}(X/Y), MSO(i+1) \rangle. ~ (1)$$ 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 $$MSO^n(X,Y) \stackrel{?}{\cong} \langle X/Y, K_n \rangle ~~~ (2)$$ 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 $$\langle X/Y, \Omega^{i-n} MSO(i) \rangle \rightarrow \langle X/Y, \Omega^{i+1-n}\Sigma MSO(i) \rangle \rightarrow \langle X/Y, \Omega^{i+1-n} MSO(i+1) \rangle ~~~ (3)$$ induced by $$MSO(i) \xrightarrow{\eta_{MSO(i)}} \Omega \Sigma MSO(i) \xrightarrow{\Omega(f_i)} \Omega MSO(i+1), ~~~(4)$$ 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.

• You pretty much wrote the answer. $K_n = \mathrm{colim}\, \Omega^{i-n}MSO(i)$. If $X/Y$ is a finite CW-space, then it is a compact object in the category of spaces, and its stabilization $\Sigma^{\infty}(X/Y)$ is a compact object in the category of spectra. $D$ is comact if $Hom(D,\cdot)$ preserves filtered colimits, in particular colimits of towers. If $X/Y$ is infinite then the formula you wrote doesn't define a cohomology theory and the only correct way to define it is via the representing object given above. – Anton Fetisov Apr 8 '15 at 23:00
• @Anton Fetisov Thanks a lot! I'm yet to understand your comment, but I figured we could give a pedestrian proof as follows, which is probably a translation of yours. Observe that the image of $f:X/Y\rightarrow \lim_{i\rightarrow \infty}\Omega^{i-n}MSO(i)$ is compact (recall $X$, $Y$ are finite CW complexes), and that the $\Omega^{i-n}MSO(i)$'s are Hausdorff. – user46652 Apr 26 '15 at 5:06
• (cont'd) From the fact that a compact subspace of the colimit of a sequence of Hausdorff spaces is contained in one of the spaces, the surjectivity of the obvious map $$\lim_{i\rightarrow \infty}\langle X/Y, \Omega^{i-n} MSO(i) \rangle \rightarrow \langle X/Y, \lim_{i\rightarrow \infty} \Omega^{i-n} MSO(i) \rangle$$ follows. The same argument but for a homotopy $f_t:X/Y\rightarrow \lim_{i\rightarrow \infty}\Omega^{i-n}MSO(i)$ shows the injectivity. – user46652 Apr 26 '15 at 5:07