## Embedded (framed) cobordisms

[The title initially was "Actions of gauge groups on framed cobordisms. This has been changed.]

This question is a follow-up to my answer to http://mathoverflow.net/questions/78209/when-is-a-submanifold-of-mathbf-rn-given-by-global-equations

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 framed cobordant to 0 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.)

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? A positive answer to this would mean that the answer to the above-mentioned question is negative. [This question still stands, but I don't think it is directly related to the above mentioned question in the other thread.]

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.]

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.

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?

-
Regarding your 2nd paragraph, consider the map $\mathbb R^4 \to \mathbb R^3$ which you obtain from $S^3 \to S^2$ by "coning off" to get a map $D^4 \to D^3$, then removing the boundary to give a map $\mathbb R^4 \to \mathbb R^3$. The pre-image of any point other than the origin is a Hopf fibre, but stabilized to be considered as a submanifold of $\mathbb R^4$. This represents $2$-torsion in the framed cobordism, since suspension $\mathbb Z \simeq \pi_3 S^2 \to \pi_4 S^3 \simeq \mathbb Z_2$ is reduction mod 2. – Ryan Budney Oct 26 2011 at 18:19
Ryan -- true, the "Hopf" framed circle represents 2-torsion, but after changing the framing we get 0 – algori Oct 26 2011 at 18:28
Perhaps you meant to include the change-of-framing statement in the 2nd paragraph ? The way I'm reading the statement, it appears false for the reasons given in my first comment. – Ryan Budney Oct 26 2011 at 20:09
Ryan -- I think you are right and the statement is false. Moreover, I'm not sure that even a change of framing would save it (nor that it won't). – algori Oct 26 2011 at 23:40

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: $$[(S^3,\text{left-invariant framing})]=\bar 1\in \mathbb Z/24\cong\Omega^\text{fr}_3.$$

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.

Oops!
My answer deals with tangential framings, whereas the question was about normal framings.
I guess I'll leave it here as it might be of independent interest...

-
 Andre -- thanks, that is interesting, but yes, this is not exactly what I was asking. – algori Oct 26 2011 at 23:47