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.

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    $\begingroup$ There is something wrong with your question: as it stands, $S_k$ is not a subset of $M$. $\endgroup$ – abx Aug 28 '19 at 4:51
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    $\begingroup$ 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 '19 at 5:39
  • $\begingroup$ Ryan Budney: Can you clarify what $M$ is in your example, and what's $n$? $\endgroup$ – Ville Salo Aug 28 '19 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 '19 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 '19 at 7:14

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.


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