Geometric interpretation of torsion homology classes 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 $M$?
 A: 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


*

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

*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.
