How is the differential in complex cobordism defined? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T10:32:47Z http://mathoverflow.net/feeds/question/67169 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67169/how-is-the-differential-in-complex-cobordism-defined How is the differential in complex cobordism defined? David White 2011-06-07T18:08:06Z 2011-06-08T00:16:55Z <p>This is my first MO question...hopefully it's not a bad one...</p> <p>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 $MU$.</p> <p>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 <em>Complex Cobordism and Stable Homotopy</em> (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:</p> <p>(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.</p> <p>(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 $MU^n(X)$)</p> <p>(3) Mention Thom's theorem that $\Omega_* \cong \pi_*(MU)$, and proceed from there.</p> <p>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?</p> <p>$^!$: 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.</p> http://mathoverflow.net/questions/67169/how-is-the-differential-in-complex-cobordism-defined/67173#67173 Answer by Fernando Muro for How is the differential in complex cobordism defined? Fernando Muro 2011-06-07T18:19:43Z 2011-06-07T18:19:43Z <p>In (2) you define the differential $d$ as $d=0$. The group $C_n(X)$ is exactly $MU_n(X)$.</p> http://mathoverflow.net/questions/67169/how-is-the-differential-in-complex-cobordism-defined/67183#67183 Answer by Neil Strickland for How is the differential in complex cobordism defined? Neil Strickland 2011-06-07T21:20:48Z 2011-06-07T21:35:38Z <p>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 <code>$\partial[M,f]=[\partial M,f|_{\partial M}]$</code>, 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 <a href="http://neil-strickland.staff.shef.ac.uk/talks/durham.pdf" rel="nofollow">http://neil-strickland.staff.shef.ac.uk/talks/durham.pdf</a></p>