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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.

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    $\begingroup$ You can't sum in a general manifold. $\endgroup$ – positrón0802 Dec 14 '18 at 16:28
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    $\begingroup$ What do you mean by $n-m$ for $n,m\in M$? $\endgroup$ – positrón0802 Dec 14 '18 at 16:29
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    $\begingroup$ 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? $\endgroup$ – Matt Samuel Dec 14 '18 at 16:56
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    $\begingroup$ But that just doesn't work. Where the charts overlap, the two different definitions of the subtraction might not agree. $\endgroup$ – Matt Samuel Dec 14 '18 at 17:12
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    $\begingroup$ 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 $p \neq q$. $\endgroup$ – mollyerin Dec 15 '18 at 1:11
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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.

Smoothness of Vector Sums of Plane Convex Sets

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).

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