0
$\begingroup$

Consider a smooth compact manifold $M$ of dimension $n$, with or without boundary. Choose a submanifold $N$ of $M$ of dimension $k$, where $1 \leq k \leq n - 1$, such that $N$ is either without boundary, or $\partial N \subset \partial M$. My question is, are there well-known necessary/sufficient conditions which dictate whether $N$ can be extended to a $k + 1$ dimensional submanifold $N'$ such that $N \subset N'$, and $N'$ is either without boundary, or $\partial N' \subset \partial M$?

Note: Heavily edited after Ryan Budney's comment (the initial question had incorrect notation). Budney's comment also shows that an extension is not always possible. But I would like to understand some conditions which guarantee or disallow the existence of such extensions.

$\endgroup$
  • 2
    $\begingroup$ There is something wrong with your question: as it stands, $S_k$ is not a subset of $M$. $\endgroup$ – abx Aug 28 at 4:51
  • 2
    $\begingroup$ I suspect the question is intended to be: if $X$ is a compact submanifold of $M$ having dimension $k$ then can you find a submanifold $Y$ of $M$ having dimension $k+1$ with $X \subset Y$? i.e. can you extend your submanifolds, one dimension at a time. Hopefully the question-asker can clarify. If my interpretation is correct, the answer would be no. For example, take the $0$-section in $TS^2$, the tangent bundle to the $2$-sphere (or its associated disc bundle). $\endgroup$ – Ryan Budney Aug 28 at 5:39
  • $\begingroup$ Ryan Budney: Can you clarify what $M$ is in your example, and what's $n$? $\endgroup$ – Ville Salo Aug 28 at 6:42
  • $\begingroup$ $M$ is the unit disc bundle of $S^2$, so $n=4$. I'm suggesting try $k=2$, taking the $0$-section of this disc bundle over $S^2$ as the submanifold $X$. There can't be a $Y$ as the unit tangent bundle does not have any $1$-dimensional sub-bundles. $\endgroup$ – Ryan Budney Aug 28 at 7:08
  • $\begingroup$ Ok, somehow I thought you meant $0$-section as a one-point $0$-dimensional submanifold of the space of sections, which did not make any sense. Thanks for clarifying. $\endgroup$ – Ville Salo Aug 28 at 7:14
1
$\begingroup$

The first obstruction is that the normal bundle to $N$ in $M$ must have a $1$-dimensional subbundle. That is the obstruction I used in my example involving $TS^2$ in the above comment.

If the normal bundle has a $1$-dimensional sub-bundle, then you are close to done. For example, say the normal bundle to $N$ in $M$ has a trivial $2$-dimensional sub-bundle, then you could embed the "double" of the total space of the $1$-dimensional sub-bundle in $M$. i.e. you would have $N$ as the fibre of an embedded $S^1 \times N$ in $M$.

In general I suppose you could interpret this as a type of cobordism problem, so you could in principle find cohomological obstructions. Off the top of my head I don't see any non-trivial ones, but I'll give it a little thought.

$\endgroup$

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.