Embedded (framed) cobordisms - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T04:49:13Z http://mathoverflow.net/feeds/question/79176 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79176/embedded-framed-cobordisms Embedded (framed) cobordisms algori 2011-10-26T17:52:14Z 2011-10-28T11:22:30Z <p>[The title initially was "Actions of gauge groups on framed cobordisms. This has been changed.]</p> <p>This question is a follow-up to my answer to <a href="http://mathoverflow.net/questions/78209/when-is-a-submanifold-of-mathbf-rn-given-by-global-equations" rel="nofollow">http://mathoverflow.net/questions/78209/when-is-a-submanifold-of-mathbf-rn-given-by-global-equations</a></p> <p>Suppose $M$ is a smooth compact $d$-dimensional submanifold of $\mathbb{R}^n$ given as the transversal zero locus of $k=n-d$ functions. The normal bundle of $M$ in $\mathbb{R}^n$ is framed and moreover, it turns out that $M$ is <strike>framed cobordant to 0</strike> bounds a submanifold of $\mathbb{R}^n$, unless $d=0$. (At first I thought this was a consequence of Sard's lemma; now I think this is not quite so obvious but true nonetheless.)</p> <p>Q1. I would like to ask: is there a submanifold $M$ of $\mathbb{R}^n$ with trivial normal bundle such that no framing of this bundle makes $M$ framed cobordant to 0? <strike>A positive answer to this would mean that the answer to the above-mentioned question is negative.</strike> [This question still stands, but I don't think it is directly related to the above mentioned question in the other thread.]</p> <p>Q2. Is there a manifold $M\subset\mathbb{R}^n$ with trivial normal bundle such that no $N$ with boundary $M$ can be embedded in $\mathbb{R}^n$? Presumably this is more difficult than Q1. [But if the answer is positive, this would mean that there are submanifolds of $\mathbb{R}^n$ with trivial normal bundles that can't be given by global equations.]</p> <p>In general, if $M$ is embeddable in $\mathbb{R}^n$, there seems to be no reason any of the manifolds bounded by $M$ should be. However I do not know of any obstructions or counter-examples.</p> <p>Q3. What if we drop the condition that the normal bundle is trivial in Q2 and replace it with the weaker condition that $M$ is cobordant to 0, i.e., that all Stiefel-Whitney numbers of $M$ are 0? </p> http://mathoverflow.net/questions/79176/embedded-framed-cobordisms/79184#79184 Answer by AndrĂ© Henriques for Embedded (framed) cobordisms AndrĂ© Henriques 2011-10-26T19:19:54Z 2011-10-28T11:22:30Z <p>The Lie group $SU(2)\cong S^3$, with its left-invariant framing represents a generator of the cobordinsm group of stably-framed 3-manifolds: $$[(S^3,\text{left-invariant framing})]=\bar 1\in \mathbb Z/24\cong\Omega^\text{fr}_3.$$</p> <p>If you change the framing on $S^3$, you can reach anything in the set $$\{\bar 1,\bar 3, \bar 5, \bar 7,\ldots,\bar {23}\}\subset\mathbb Z/24,$$ but not the other elements. <hr> Oops!<br> My answer deals with tangential framings, whereas the question was about normal framings.<br> I guess I'll leave it here as it might be of independent interest...</p>