6
$\begingroup$

I asked the following question on StackExchange but received no response:

Let $M$ be a smooth $n$-dimensional compact manifold with boundary. Let $U$ be an open neighborhood of $\partial M$. Assume that we have a fixed immersion $\bar f : U \to \mathbb{R}^{2n-1}$. Can we always find an immersion $f : M \to \mathbb{R}^{2n-1}$ such that $f |_{\partial M} = \bar f |_{\partial M}$?

I believe this should work in $\mathbb{R}^{2n}$. Since $\bar f$ is an immersion, $d\bar f$ is a bundle monomorphism $TU \to U \times \mathbb{R}^{2n}$ covering $\bar f$. We can think of this as a section of the associated bundle over $U$ whose fibers are the Stiefel manifold $V_{n}(\mathbb{R}^{2n})$. Since $\pi_{n-1}\left(V_{n}(\mathbb{R}^{2n})\right) = 0$, we can extend the section from $\partial M$ to $M$. Now we can apply Smale-Hirsch to get the desired immersion $f : M \to \mathbb{R}^{2n}$.

Does the statement hold in $\mathbb{R}^{2n-1}$?

EDIT: To incorporate Oscar's comment, let's assume that we can perturb $\bar f$, if necessary. We can also assume that $U$ is connected.

$\endgroup$
  • $\begingroup$ It doesn't hold when $n=1$, and I doubt this is a special case. $\endgroup$ – Oscar Randal-Williams Dec 5 '16 at 14:56
  • $\begingroup$ @OscarRandal-Williams Yes, fair enough; I've edited. Not sure if it will make a difference.. $\endgroup$ – Michael Harrison Dec 5 '16 at 15:17
3
$\begingroup$

No, you can't generally do this even with the added assumptions. The bundle $\tau = TS^{n-1} \to S^{n-1}$ is non-trivial for $n-1 \neq 1,3,7$, but it is stable after (one) trivialisation. Hence $\tau \oplus TD^n\vert_{S^{n-1}}$ is a trivial bundle over $S^{n-1}$.

By Smale--Hirsch this means there is an immersion $i: S^{n-1} \times [0,\epsilon] \hookrightarrow \mathbb{R}^{2n-1}$ whose ($(n-1)$-dimensional) normal bundle is isomorphic to (the pullback of) $\tau$. If $i$ extended to an immersion of $D^n$ then the normal bundle would extend to $D^n$ and hence be trivial, but it is not.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.