Dtyjlui

Suppose I have a homology class $x in H_1(M)$ which is torsion of order $k$ say. Suppose furthermore that $M$ has Dimension big enough, such that every element of $H_1$ and $H_2$ can be relalized as submanifold of $M$.

Now what is the geometric picture of this torsion class? If $kx=0$, then are there $k$ submanifolds $N_i$ of $M$ each which represents $x$ and such that there is a $2$-dimensional submanifold $Sigma$ such that its boundary is equal to $bigcup_i N_i$ ?

Does this also mean that $N_1,...,N_k$ are bordant to a submanifold which is the boundary of a proper embedded 2-disk of $M$?

I haven't thought enough to give a full answer, but hope this partial answer may be helpful.

Let me restrict to the case that $mathrm{dim}(M)=3$ and $M$ is connected. (The condition on the connectedness of $M$ is not a real restriction, as you can do it componentwise. For the dimension part, I don't have a full understanding, but I believe that the argument here may be extended without much effort to give an affirmative answer as in dimension 3.)

Then the problem is asking about

1. the representability of the class $kx$ as an embedding of $k$ disjoint union of circles,


2. the existence of an embedded surface that bounds the link from above.

First, it is well-known (if it is not obvious...) that you can represent $x$ by an embedded circle (ie, knot), say $K$. As the knot $K$ admits a tubular neighborhood, it is a trivial matter to obtain an embedded representative of $kx$ by shifting copies of the knot $K$ slightly inside the tubular neighborhood. This embedded representative(a link in $M$), say $L$, will be the $k$ submanifolds that you desire.

Now, you can also find $k-1$ bands connecting the components of the link $L$ which is pairwise disjoint, only intersecting the link $L$ on the boundary, and the union of $L$ and the bands are connected. (There are many methods to achieve this; my favorite is choosing $k-1$ paths connecting the components and taking a band neighborhood of the paths.) Taking a band sum of $L$ along these bands, you can obtain a knot $K'$ which still represents $kx$. As $kx$ is a trivial class, you can find a Seifert surface of $K'$; ie, there is a connected, orientable, embedded surface, say $S$, whose boundary $partial S'$ is $K'$. (This is standard in knot theory; a standard proof uses Morse theory.)

Now we are almost there; On the boundary $partial S'=K'$, there are distinguished arcs which consist of the boundary of the band. Then you can find a collection of disjoint arcs in $S'$ connecting a distinguished arc to the distinguished arc on the other side of the band. Removing a band neighborhood of these arcs from $S'$, the resulting surface $S$ bounds $L$(or a link isotopic to $L$; but it is a trivial matter to choose a proper band neighborhood to bound exactly $L$, not up to isotopy).

And for the second question, you can make a puncture on the interior of the surface $S$ to get a bordism to a boundary of a disk, and actually we showed better; the bordism can be made embedded.