Unuseful prequel
Let $M$ be a (compact, oriented, differentiable) manifold. Before knowing anything about homology theory a naif but clever mathematician may want to measure the holes in $M$ by the following procedure. One lets $B_k(M)$ be the abelian group generated by embedded submanifolds of $M$ of dimension $k$ with border. There is an obvious border operator $\delta \colon B_{k}(M) \to B_{k-1}(M)$ and trivially $\delta^2 = 0$ since the border of a manifold with border is a closed manifold. So one may define $H^K_{naif}(M)$ by taking the quotient of cycles modulo boundaries.
Ok, this does not really work, for at least two reasons. The first there is no way to make this functorial without considering more degenerate objects (singular simplices or currents being two possible choices). The second is that even if one adjusts the definition to get, for instance, the bordism groups of $M$, these will be highly nontrivial for differential geometric reasons, rather than for the topology of $M$ (for instance they will be nontrivial for a point).
Anyway, one finally chooses a working (co)homology theory, for instance singular, and then can use the fundamental class of submanifolds to link this theory to the naif one. Two questions naturally arise:
- Are fundamental classes of manifolds enough to generate the cohomology? This is nicely answered here.
- Are submanifolds with border enough to generate homology relations? This is the purpose of the present question.
The actual question
Let $M$ be a (compact, oriented, differentiable) manifold and let $N \subset M$ be a (closed) submanifold. The problem is to test whether $N$ is the boundary of submanifold with boundary of $M$. There are two obvious obstructions:
- $N$ should be bordant as an abstract variety
- The class $[N] \in H_{*}(M, \mathbb{Z})$ should be $0$.
Assume 1 and 2 hold. Are there any conditions on $M$, $N$ or the embedding $N \to M$ which guarantee that $N$ is the boundary of submanifold with boundary of $M$?
There are a few classical cases which come to mind:
- If $N$ and $M$ are spheres, 1 and 2 are vacuous and the thesis is true by the (generalized) Jordan curve theorem.
- If $N = S^1$ and $M = \mathbb{R}^3$, the thesis is still true by the existence of Seifert surfaces.
- If $\dim N = 1$ and $\dim M = 2$ the result seems true to me by the classification of surfaces as quotients of polygons and the usual Jordan curve theorem (but I did not check the details).
On the other hand I'm sure there are plenty of negative example even in $3$-dimensional topology. Is there anything nontrivial which can be said about this question, whether on the positive or negative side?
Best Answer
This was supposed to be a comment to Jeff's answer, but wouldn't fit.
If you can solve the bordism problem you get a smooth map $F: W \to M$, but $F$ is not homotopic to an immersion in general. The obstruction to being one is given by the Hirsch-Smale classification of immersions, which reduces it to a problem in homotopy theory: $F$ is homotopic to an immersion if and only if $F : W \to M$ admits an "injective formal differential" that restricts to the actual differential on $\partial W$.
If it is representable by an immersion, then it can be represented by a self-transverse immersion, and if $dim(M)-dim(W) > 2$ then I think a sort of Whitney trick can be used to remove double points.
Thus I think $N \subset M$ bounds a submanifold on $M$ if an only if it bounds an abstract manifold which can be immersed into $M$.