Indecomposable integral currents Let $\mathbf{I}_k(\mathbb{R}^n)$ denote the space of $k$-dimensional integral currents in $\mathbb{R}^n$ with finite mass. It is said that $T\in \mathbf{I}_k(\mathbb{R}^n)$ is indecomposable if there does not exist a nonzero $R\in \mathbf{I}_k(\mathbb{R}^n)$ such that $\mathbf{M}(R) + \mathbf{M}(T-R) = \mathbf{M}(T)$ and $\mathbf{M}(\partial R) + \mathbf{M}(\partial (T-R)) = \mathbf{M}(\partial T)$. It is known that indecomposable $1$-dimensional integral currents correspond to injective Lipschitz curves in $\mathbb{R}^n$. For $k\geq 2$, can you give me an example of an indecomposable integral current $T \in \mathbf{I}_k(\mathbb{R}^n)$ with $\partial T = 0$, compact support and which has not multiplicity $1$ almost-everywhere?
 A: I think the following might be an example, though it will require a bit of work if you want to make it more precise:
Take an immersion of a sphere, which is injective except for one cap at each pole, which instead overlap with the same orientation. You'll need four dimensions to avoid self-intersection, but otherwise such a construction should not be too hard to do. Obviously, the corresponding current $T$ has compact support and no boundary. I claim it also is indecomposable.
Assume this is not the case and there is a nontrivial $R$, which wlog. includes part of the overlap. Now since $R$ cannot have a boundary, we can progressively extend it: If it has part of the overlap, it needs to include all of it with multiplicity of at least 1, otherwise there would be boundary. By the same argument, it needs to include a piece of the sphere adjacent to one of the caps, which then implies it includes all of the injective part¹, which finally implies that it needs a second copy of the overlap and thus that $T=R$.
¹Ostensively, this is the difference to the one dimensional result. Here, choosing how to continue from the first copy of the overlap anywhere determines how we need to continue everywhere, as its boundary is connected. In one dimension this would no longer work. If you try the same with a circle, then you can just choose a continuation for one side of the overlap, follow that until you loop back to the other side and get a decomposition.
