# Let $M$ be a smooth manifold. Can we say that $M - M = \{ n - m : n,m \in M\}$ is also a smooth manifold?

I was thinking about this for a while. The definition I use for the smooth manifold is the same as per Wikipedia. Let $$\{(U_k,\phi_k)\}$$ is a smooth atlas of $$M$$. Then the natural atlas which was coming in my mind for $$M-M$$ was $$\{(U_i - U_j, \phi_i - \phi_j)\}$$ where $$(\phi_i - \phi_j)(u)=\phi_i(u) - \phi_j(u) \; \forall u \; \in U_i - U_j$$. Is my approach correct?

By $$M - M$$ I just mean formal difference of two sets where an element $$x$$ of $$M-M$$ can be written as $$n-m$$ for some $$n,m \in M$$. Note that "difference" in $$M-M$$ has no meaning but I am seeking a suitable atlas for this set so that when I am in some $$\Bbb R^n$$, I will do subtraction as per the addition is done in the group $$\Bbb R^n$$.

As a beginner in learning the subject, I am not confident in writing down the details. Thank you.

EDIT: After a recent comment by @MikeMiller, I realized that I was actually working in $$M \times M$$. So I thought to change my definition of $$M - M$$. I see now $$M - M$$ as a set of equivalence classes where the equivalence relation is such that any to pairs $$(m,n)$$ and $$(p,q)$$ (or $$m-n$$ and $$p-q$$) are equivalent if we have a suitable atlas for $$M-M$$ such that in local coordinates, $$m-n=p-q \in \Bbb R^n$$. The problem is I want to know whether such an atlas exists.

• You can't sum in a general manifold. – positrón0802 Dec 14 '18 at 16:28
• What do you mean by $n-m$ for $n,m\in M$? – positrón0802 Dec 14 '18 at 16:29
• I don't see how your atlas would work exactly. An atlas is not only bijections of the subsets with $\mathbb{R}^n$, but also transition maps to piece them together. How would you define those? – Matt Samuel Dec 14 '18 at 16:56
• But that just doesn't work. Where the charts overlap, the two different definitions of the subtraction might not agree. – Matt Samuel Dec 14 '18 at 17:12
• If $M$ is path connected and the atlas on $M$ is maximal, it seems like it should be possible to put any two points $p, q$ of $M$ in the same chart (by "gluing" charts along a path), and then, via scaling & rotating, there should exist a chart with $\varphi(p) = 0$ and $\varphi(q) = 1$ -- i.e. it seems to me that your "equivalence class" definition of $M - M$ has only the "same" and "different" classes -- the classes represented by $(p, p)$ and $(p, q)$ for any $p \neq q$. – mollyerin Dec 15 '18 at 1:11

This is not an answer to your question but I feel it's too long to be included in a comment. I just want to mention that there's an interesting result regarding the Minkowski sum of $$2$$ convex sets with smooth (of class $$C^\infty$$) boundary.
The main result is that the sum of $$2$$ convex sets with $$C^\infty$$ need not have $$C^\infty$$ boundary. In fact we only get the smoothness of the boundary up to class $$C^{20/3}$$. This might suggest that the answer to your question could be negative (thought I am not sure since your question right now is not well-defined).