Let $A\subset\mathbb{R}^{3}$ be the unit cube $[0,1]^3$ without that part which lies in the ball of radius $1$ and center $(1,1,1)$. Is there a good description of the sets $A+A = \{\alpha_{1}+\alpha_{2}:\alpha_{k}\in A\}$ and $A+A+A$ (defined similarly). In $\mathbb{R}^{2}$ this problem is not so hard. First, let me draw $A$:
Now I can draw $A+A$ and $A+A+A$ just by geometric intuition:
We can describe both of these sets as a union of unit squares and some copies of $A$. I originally thought that in three dimensions that these sets would be similarly described: a union of unit cubes and some copies of $A$. I have been furiously trying to draw this and nothing looks right (because this description is almost surely wrong). I am thinking that maybe it can be described as a union of unit cubes, some of which have the unit ball removed (with center at some appropriate corner).
It would be extremely helpful to my life to have a good description of these sets, preferably one which extends to arbitrary dimension. Thank you for your time.
Best Answer
Throughout this answer, when I speak of a unit cube or a translation of $A$ being "at" some point $(x,y,z)$, I will mean that the bottom-lower-left corner (0,0,0) is translated to that point. The details of this answer are fairly involved at some points, and so I will skip some details when they amount to just checking many cases that are all qualitatively similar.
I find that $A+A$ consists of unit cubes at $(0,0,0), (1,0,0), (0,1,0), (0,0,1)$ and translations of $A$ at $(1,1,0), (1,0,1), (0,1,1)$. Similarly, I find that $[0,1]^3 + A$ consists of unit cubes at $(0,0,0), (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1)$ and a translation of $A$ at $(1,1,1)$. It will follow by induction that all higher sums of $A$, such as $A+A+A$, are also unions of unit cubes and translates of $A$.
Any point $(x,y,z)$ with at most one of $x,y,z$ in $(1,2]$, say $x\in (1,2]$ and $y,z \in [0,1]$, can be reached as a sum like $(1,y,0) + (x-1,0,z)$. This shows that $A+A$ contains the unit cubes as claimed. Similarly, for $[0,1]^3 + A$, as long as one of $x,y,z\in [0,2]$ is at most 1, say $z\leq 1$ and $x,y\in [1,2]$, we can write something like $(x,y,z) = (1,1,z) + (x-1,y-1,0) \in [0,1]^3 + A$. So the unit cubes of $[0,1]^3 + A$ are as claimed.
Now, to show that there are translates of $A$ at the locations claimed requires two things: Showing that the Minkowski sum contains at least $A$ at that point is easy, e.g. to show that $A+A$ contains a translate of $A$ at $(1,1,0)$ follows immediately from the fact that $(1,1,0)\in A$, hence $(1,1,0)+A \subset A+A$. However, to show that $A + A$ contains nothing more within the unit cube at $(1,1,0)$ is more difficult, and really is the only hard step in this entire proof. Here goes:
We want to show that $A+A$ intersected with the unit cube at $(1,1,0)$ does not contain points within distance 1 of the top-upper-right corner $(2,2,1)$. This means that $||(2,2,1) - x - y||\geq 1$ for any $x,y\in A$. Equivalently, $||\left( (1,1,1) - x \right) + \left( (1,1,1) - y\right) - (0,0,1)|| \geq 1$. Now by the definition of $A$, both $(1,1,1) - x$ and $(1,1,1) - y$ are vectors with all nonnegative components whose length is at least 1. The extremal case is when both of these vectors lie on the unit sphere. Let's suppose that this is the case, and call $u=(1,1,1)-x$ and $v=(1,1,1)-y$. Then the preceding inequality can be written $||u+v-e_3||\geq 1$. This is in turn equivalent to $||u - (e_3-v)||\geq 1$, which has the following nice geometric interpretation: $u$ lies on the positive octant of the unit sphere centered at the origin, and $e_3 - v$ lies on the negative octant of the unit sphere centered at $e_3$, so we are just asserting that the distance between the positive octant of the unit sphere centered at the origin and the negative octant of the unit sphere centered at $e_3$ is at least 1.
In order for two points on the two spheres to be minimizers of distance, either they must be boundary points or the tangent planes at these two points must be perpendicular to the line connecting them, and in particular the tangent planes of the two points must be parallel to each other. I'm not going to go through all the cases here, but basically the latter case of parallel tangent planes can't be a minimizer of distance, because points with equal tangent planes would need to be opposite points on the respective spheres, and because the two spheres are offset from each other, such points' tangent planes are not perpendicular to the line connecting them. Similarly, when you work through all the boundary points, you find that the only possible minimizers are the vertices of the sphere octants, and checking all 9 cases you can see that the minimum distance is 1, as desired.
This proves the claim that $A+A$ is a union of cubes and translates of $A$. A similar, easier argument shows that $[0,1]^3 + A$ has a translate of $A$ at $(1,1,1)$.
Now both $A+A$ and $[0,1]^3 + A$ consist of a union of cubes and translates of $A$, which are arranged on a lattice. Inductively you can see that $A+A+A$ can be decomposed into a union of translates of $[0,1]^3 + A$ and $A+A$, and thus will itself consist of a union of cubes and translates of $A$. To figure out where they all go is a combinatorial problem. Adding $A$ to a cube at $(x,y,z)$ will result in cubes at $(x+1,y,z), (x,y+1,z),\dots$ and a translate of $A$ at $(x+1,y+1,z+1)$, and adding $A$ to a translate of $A$ at $(x,y,z)$ will result in cubes at $(x,y,z), (x+1,y,z), (x,y+1,z), (x,y,z+1)$ and translates of $A$ at $(x+1,y+1,z), (x+1,y,z+1),(x,y+1,z+1)$. Working it all out, I find that $A+A+A$ will have translates of $A$ at $(2,2,0),(2,0,2),(0,2,2),(2,1,1),(1,2,1),(1,1,2)$ and cubes at $(0,0,0),(1,0,0),(0,1,0),(0,0,1),(1,1,0),(1,0,1),(0,1,1),(1,1,1)(2,0,0),(0,2,0),(0,0,2),(2,1,0),(2,0,1),(1,2,0),(0,2,1)$.
Analogous results should hold in all dimensions. In general, $A+A$ will consist of translates of $A$ at all points of the form $(1,1,\dots,1) - e_k$ for some $k$, and cubes at all points of the form $(1,1,\dots,1) - e_{k_1} - e_{k_2} - \dots$, i.e. points with at least two 0 coordinates and the rest 1. Then higher sums like $A+A+A$ will consist of translates of $A$ at points like $(2,2,\dots,2) - e_{k_1}-e_{k_2}$, of which there are $\binom{2+n-1}{2}$ (with $n$ being the dimension of the space), along with cubes at all points "below", i.e. subtracting more $e_k$'s.