Let $M$ be a smooth manifold. Can we say that $M - M = { n - m : n,m in M}$ is also a smooth manifold?
$begingroup$
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.
differential-geometry manifolds differential-topology smooth-manifolds
asked Dec 14 '18 at 16:25
Error 404Error 404
3,88321336
$endgroup$
|
show 17 more comments
$begingroup$
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.
differential-geometry manifolds differential-topology smooth-manifolds
asked Dec 14 '18 at 16:25
Error 404Error 404
3,88321336
$endgroup$
2
$begingroup$
You can't sum in a general manifold.
$endgroup$
– positrón0802
Dec 14 '18 at 16:28
2
$begingroup$
What do you mean by $n-m$ for $n,min M$?
$endgroup$
– positrón0802
Dec 14 '18 at 16:29
1
$begingroup$
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
3
$begingroup$
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
2
$begingroup$
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
|
show 17 more comments
$begingroup$
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.
differential-geometry manifolds differential-topology smooth-manifolds
asked Dec 14 '18 at 16:25
Error 404Error 404
3,88321336
$endgroup$
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.
differential-geometry manifolds differential-topology smooth-manifolds
differential-geometry manifolds differential-topology smooth-manifolds
asked Dec 14 '18 at 16:25
Error 404Error 404
3,88321336
asked Dec 14 '18 at 16:25
Error 404Error 404
3,88321336
edited Dec 14 '18 at 17:48
Error 404
asked Dec 14 '18 at 16:25
Error 404Error 404
3,88321336
asked Dec 14 '18 at 16:25
Error 404Error 404
3,88321336
asked Dec 14 '18 at 16:25
Error 404Error 404
3,88321336
3,88321336
2
$begingroup$
You can't sum in a general manifold.
$endgroup$
– positrón0802
Dec 14 '18 at 16:28
2
$begingroup$
What do you mean by $n-m$ for $n,min M$?
$endgroup$
– positrón0802
Dec 14 '18 at 16:29
1
$begingroup$
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
3
$begingroup$
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
2
$begingroup$
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
|
show 17 more comments
2
$begingroup$
You can't sum in a general manifold.
$endgroup$
– positrón0802
Dec 14 '18 at 16:28
2
$begingroup$
What do you mean by $n-m$ for $n,min M$?
$endgroup$
– positrón0802
Dec 14 '18 at 16:29
1
$begingroup$
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
3
$begingroup$
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
2
$begingroup$
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
2
2
$begingroup$
You can't sum in a general manifold.
$endgroup$
– positrón0802
Dec 14 '18 at 16:28
$begingroup$
You can't sum in a general manifold.
$endgroup$
– positrón0802
Dec 14 '18 at 16:28
2
2
$begingroup$
What do you mean by $n-m$ for $n,min M$?
$endgroup$
– positrón0802
Dec 14 '18 at 16:29
$begingroup$
What do you mean by $n-m$ for $n,min M$?
$endgroup$
– positrón0802
Dec 14 '18 at 16:29
1
1
$begingroup$
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
$begingroup$
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
3
3
$begingroup$
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
$begingroup$
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
2
2
$begingroup$
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
$begingroup$
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
|
show 17 more comments
1 Answer
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$begingroup$
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).
answered Dec 14 '18 at 18:00
BigbearZzzBigbearZzz
8,88821652
$endgroup$
add a comment |
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$begingroup$
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).
answered Dec 14 '18 at 18:00
BigbearZzzBigbearZzz
8,88821652
$endgroup$
add a comment |
$begingroup$
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).
answered Dec 14 '18 at 18:00
BigbearZzzBigbearZzz
8,88821652
$endgroup$
add a comment |
$begingroup$
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).
answered Dec 14 '18 at 18:00
BigbearZzzBigbearZzz
8,88821652
$endgroup$
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).
answered Dec 14 '18 at 18:00
BigbearZzzBigbearZzz
8,88821652
answered Dec 14 '18 at 18:00
BigbearZzzBigbearZzz
8,88821652
answered Dec 14 '18 at 18:00
BigbearZzzBigbearZzz
8,88821652
answered Dec 14 '18 at 18:00
BigbearZzzBigbearZzz
8,88821652
8,88821652
add a comment |
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2
$begingroup$
You can't sum in a general manifold.
$endgroup$
– positrón0802
Dec 14 '18 at 16:28
2
$begingroup$
What do you mean by $n-m$ for $n,min M$?
$endgroup$
– positrón0802
Dec 14 '18 at 16:29
1
$begingroup$
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
3
$begingroup$
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
2
$begingroup$
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