# Inclusion of closed submanifolds of a manifold

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.

• There is something wrong with your question: as it stands, $S_k$ is not a subset of $M$. – abx Aug 28 '19 at 4:51
• 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). – Ryan Budney Aug 28 '19 at 5:39
• Ryan Budney: Can you clarify what $M$ is in your example, and what's $n$? – Ville Salo Aug 28 '19 at 6:42
• $M$ 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. – Ryan Budney Aug 28 '19 at 7:08
• 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. – 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 $$M$$.