This is my first MO question...hopefully it's not a bad one...

Background: As a stable homotopy theorist, I like to think of complex cobordism $MU$ as a ring spectrum. If I needed to get my hands dirty I could look at the representing spaces or go through the Thom construction of $MU$.

I would like to give a talk aimed at a more general mathematical audience discussing how formal group laws and Quillen's Theorem get involved with the story of complex cobordism. To do this I'm going to want to introduce complex cobordism in the more classical way, e.g. following Poincare's original definition or the comment in Ravenel's Complex Cobordism and Stable Homotopy (page 10) that $MU_{*}(X)$ can be defined completely analogously to $H_{*}(X)$. So here's what the start of the talk would look like:

(1) Define $\Omega_n = \mathcal{M}^n/\sim$ where $\mathcal{M}^n$ is a particular collection of $n$-dimensional manifolds$^!$ and $\sim$ is the cobordism relation.

(2) Define $C_n(X) = \langle \sigma:M^n\rightarrow X \;|\; M^n\in \mathcal{M}^n \rangle /\sim$ where $\sim$ is the bordism relation. Then there must be some differential $C_n(X)\rightarrow C_{n-1}(X)$ which gives rise to a homology theory $MU_n(X)$ (or dually to $MU^n(X)$)

(3) Mention Thom's theorem that $\Omega_* \cong \pi_*(MU)$, and proceed from there.

My question is, how is that differential in (2) defined? Is it easy to show that $d^2 = 0$? More generally, do people think this is a reasonable way to introduce $MU$ to an audience containing no stable homotopy theorists?

$^!$: By manifold here I mean even dimensional real manifold with a map $J:TM\rightarrow TM$ such that $p\circ J \simeq p$ and $J^2 = -1$. So $J$ gives an $\mathbb{R}$-linear action of $i$ on $T_xM$ for all $x$. This seems to capture a larger class of manifolds than just complex manifolds, and I can't see any way to get a larger class than this.

  • $\begingroup$ I believe "complex manifold" here means a manifold with a complex structure on the stable normal bundle. $\endgroup$ Jun 7, 2011 at 18:20
  • $\begingroup$ @aaron: I think that is equivalent. Also, you can just say that the map classifying the tangent bundle lifts over $BU \to BO(2n)$. $\endgroup$ Sep 12, 2011 at 3:40

2 Answers 2


For a general audience it is much better to treat $MO$ rather than $MU$, because the complex orientation creates many unpleasant subtleties. (See papers of Buchstaber and Ray for interesting examples where these subtleties make a concrete and computable difference.) Let us say that a geometric chain of dimension n in X is an equivalence class of pairs $(M,f)$, where M is a compact smooth manifold (possibly with boundary) and $f:M\to X$ is a continuous map. Here $(M,f)$ is equivalent to $(M',f')$ if there is a diffeomorphism $u:M\to M'$ with $f'u=f$. We write $GC_{\ast}(X)$ for the graded abelian monoid of geometric chains. This has a differential $\partial[M,f]=[\partial M,f|_{\partial M}]$, and the homology is $MO_{\ast}(X)$. (This needs a few remarks about homology of complexes of monoids, but there is nothing very subtle going on.) I have a general audience talk about $MO$ at http://neil-strickland.staff.shef.ac.uk/talks/durham.pdf

  • $\begingroup$ Thanks, I think that will clear things up a lot in my talk. Your images in that link are amazing! Do you mind if I ask how you made them? $\endgroup$ Jun 8, 2011 at 0:25
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    $\begingroup$ @David: The images were mostly produced using Maple. If you change .PDF to .mw in the above URL then you can download the Maple worksheet. $\endgroup$ Jun 8, 2011 at 7:00
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    $\begingroup$ @Gjergji: thanks for fixing the LaTeX. $\endgroup$ Jun 8, 2011 at 7:30
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    $\begingroup$ A silly joke re first sentence: another reason that it is much better to treat MO rather than MU is that this answer definitely fits here much better than on mathunderflow. $\endgroup$ Jul 5, 2016 at 11:56

In (2) you define the differential $d$ as $d=0$. The group $C_n(X)$ is exactly $MU_n(X)$.

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    $\begingroup$ Thanks. I guess it was a rather stupid question after all. I voted you up, but I decided to accept the other answer because it's going to help me considerably in my talk. $\endgroup$ Jun 8, 2011 at 0:27

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