# Difference between bordism and cobordism

I have looked around for hours and although I have seen many definitions of bordism and cobordism (for some authors these two coincide and for some other not (without mentioning explicitly what's the difference)), I wasn't able to understand how they defined and what's the difference (if exists).

I know one definition, of iso. classes of $n$-manifolds over a manifold, say $X$ modulo co(bordism), but I have seen this definition to mention both bordism and cobordism too. So I don't know to what really corresponds.

Moreover, when looking of cobordism (co)-homology theory thinks are getting more complicated. For instance, I am aware of the Thom's theorem which associates the cobordism ring with $\pi_{*}(MU)$, therefore I understand the terminology. However, the complex cobordism theory many authors say that is represented by the Thom complex, but the latter is not $\Omega$-spetrum, thus it cannot represent a cohomology theory (corresponds to such a theory but not represents). What struggles me is the usage of terminology, and how the authors use them interchangeably. Could you please explain me what's the difference, and how the Thom spectrum represents the coborism theory?

P.S.1 By Brown's representability theorem, even though the Thom spectrum is not representing as I said, but constructs a cohomology theory, we know that this theory is represented by some $\Omega$-spectrum. So an additional (natural) question is, what's this representing $\Omega$-spectrum that represents the cobordism theory?

P.S.2 I have read many other related question MSE but no answer replies really to this question. So hope to finally someone write out an answer to sort out th above!

Thank you!

• Short answer for the first part: Bordism and cobordism are generally synonymous (the "co-" prefix for the latter refers to sharing a boundary (French 'bord'), not a contravariant version of another theory), but I've also seen them refer to the unoriented and complex theories, respectively. The difference between those two is that the latter imposes an additional structure on the normal bundle. (You can also talk about oriented cobordism, in which the cobordant manifold also has to have the right orientation. Then the resulting ring is not just 2-torsion.) – anomaly Aug 22 '18 at 14:58
• I'm confused by your question. $MU$ is a spectrum and thus represents a cohomology theory. You can form its fibrant replacement if you want with the usual telescope construction. – JHF Aug 22 '18 at 15:00
• @anomaly thank you for sharing that. So the "co" is not referring to some dual notion of bordism, hence two manifolds are cobordant, when they are bordant + the normal bundle structure, right? – user430191 Aug 22 '18 at 15:10
• That's right, though it's also common to just use the two terms interchangeably. – anomaly Aug 22 '18 at 15:11
• I think there is some confusion regarding what a morphism of spectra is. In some basic models of spectra, $[\Sigma^\infty X, MU]$ could be defined as the colimit of $[\Sigma^k X, \Sigma^k MU_n]$ as $k \to \infty$. Note that as soon as $k \geq 2$, the domain $\Sigma^k X$ is a double suspension and thus an abelian cogroup. Hence the set of homotopy classes of maps acquires the structure of an abelian group. In fact, $[E, F]$ is always an abelian group for any spectra $E$ and $F$. – JHF Aug 22 '18 at 16:22