Mixing
Show if the set ${x,y,zinmathbb{R}^3:vert xvert +vert y vert=2}$ is a smooth surface
$begingroup$
My assumption is that this won't be a smooth surface because of the absolute values. Because if I look at the function $f(x,y)=vert xvert +vert yvert-2$ it has "edges". But not sure how to actually show something is not a surface.
differential-geometry
asked Feb 3 at 2:51
AColoredReptileAColoredReptile
401210
$endgroup$
|
show 2 more comments
$begingroup$
My assumption is that this won't be a smooth surface because of the absolute values. Because if I look at the function $f(x,y)=vert xvert +vert yvert-2$ it has "edges". But not sure how to actually show something is not a surface.
differential-geometry
asked Feb 3 at 2:51
AColoredReptileAColoredReptile
401210
$endgroup$
$begingroup$
What’s your definition of surface?
$endgroup$
– Seewoo Lee
Feb 3 at 3:31
$begingroup$
A subset $S$ of $mathbb{R}^3$ is a surface if for ever point $pin S$ there is an open set $U$ in $mathbb{R}^3$ and an open set $W$ in $mathbb{R}^3$ containing p such that $Scap W$ is homeomorphic to $U$.
$endgroup$
– AColoredReptile
Feb 3 at 3:40
$begingroup$
$Uinmathbb{R}^2$ sorry.
$endgroup$
– AColoredReptile
Feb 3 at 3:46
$begingroup$
In fact, it doesn’t matter whether it has ‘edge’ or ‘vertex’. For example, you can find homeomorphi between open disk in $mathbb{R}^2$ and small part of that rectangular cylinder near edge.
$endgroup$
– Seewoo Lee
Feb 3 at 5:35
$begingroup$
So it is a surface?
$endgroup$
– AColoredReptile
Feb 3 at 5:50
|
show 2 more comments
$begingroup$
My assumption is that this won't be a smooth surface because of the absolute values. Because if I look at the function $f(x,y)=vert xvert +vert yvert-2$ it has "edges". But not sure how to actually show something is not a surface.
differential-geometry
asked Feb 3 at 2:51
AColoredReptileAColoredReptile
401210
$endgroup$
My assumption is that this won't be a smooth surface because of the absolute values. Because if I look at the function $f(x,y)=vert xvert +vert yvert-2$ it has "edges". But not sure how to actually show something is not a surface.
differential-geometry
differential-geometry
asked Feb 3 at 2:51
AColoredReptileAColoredReptile
401210
asked Feb 3 at 2:51
AColoredReptileAColoredReptile
401210
edited Feb 3 at 6:53
AColoredReptile
asked Feb 3 at 2:51
AColoredReptileAColoredReptile
401210
asked Feb 3 at 2:51
AColoredReptileAColoredReptile
401210
asked Feb 3 at 2:51
AColoredReptileAColoredReptile
401210
401210
$begingroup$
What’s your definition of surface?
$endgroup$
– Seewoo Lee
Feb 3 at 3:31
$begingroup$
A subset $S$ of $mathbb{R}^3$ is a surface if for ever point $pin S$ there is an open set $U$ in $mathbb{R}^3$ and an open set $W$ in $mathbb{R}^3$ containing p such that $Scap W$ is homeomorphic to $U$.
$endgroup$
– AColoredReptile
Feb 3 at 3:40
$begingroup$
$Uinmathbb{R}^2$ sorry.
$endgroup$
– AColoredReptile
Feb 3 at 3:46
$begingroup$
In fact, it doesn’t matter whether it has ‘edge’ or ‘vertex’. For example, you can find homeomorphi between open disk in $mathbb{R}^2$ and small part of that rectangular cylinder near edge.
$endgroup$
– Seewoo Lee
Feb 3 at 5:35
$begingroup$
So it is a surface?
$endgroup$
– AColoredReptile
Feb 3 at 5:50
|
show 2 more comments
$begingroup$
What’s your definition of surface?
$endgroup$
– Seewoo Lee
Feb 3 at 3:31
$begingroup$
A subset $S$ of $mathbb{R}^3$ is a surface if for ever point $pin S$ there is an open set $U$ in $mathbb{R}^3$ and an open set $W$ in $mathbb{R}^3$ containing p such that $Scap W$ is homeomorphic to $U$.
$endgroup$
– AColoredReptile
Feb 3 at 3:40
$begingroup$
$Uinmathbb{R}^2$ sorry.
$endgroup$
– AColoredReptile
Feb 3 at 3:46
$begingroup$
In fact, it doesn’t matter whether it has ‘edge’ or ‘vertex’. For example, you can find homeomorphi between open disk in $mathbb{R}^2$ and small part of that rectangular cylinder near edge.
$endgroup$
– Seewoo Lee
Feb 3 at 5:35
$begingroup$
So it is a surface?
$endgroup$
– AColoredReptile
Feb 3 at 5:50
$begingroup$
What’s your definition of surface?
$endgroup$
– Seewoo Lee
Feb 3 at 3:31
$begingroup$
What’s your definition of surface?
$endgroup$
– Seewoo Lee
Feb 3 at 3:31
$begingroup$
A subset $S$ of $mathbb{R}^3$ is a surface if for ever point $pin S$ there is an open set $U$ in $mathbb{R}^3$ and an open set $W$ in $mathbb{R}^3$ containing p such that $Scap W$ is homeomorphic to $U$.
$endgroup$
– AColoredReptile
Feb 3 at 3:40
$begingroup$
A subset $S$ of $mathbb{R}^3$ is a surface if for ever point $pin S$ there is an open set $U$ in $mathbb{R}^3$ and an open set $W$ in $mathbb{R}^3$ containing p such that $Scap W$ is homeomorphic to $U$.
$endgroup$
– AColoredReptile
Feb 3 at 3:40
$begingroup$
$Uinmathbb{R}^2$ sorry.
$endgroup$
– AColoredReptile
Feb 3 at 3:46
$begingroup$
$Uinmathbb{R}^2$ sorry.
$endgroup$
– AColoredReptile
Feb 3 at 3:46
$begingroup$
In fact, it doesn’t matter whether it has ‘edge’ or ‘vertex’. For example, you can find homeomorphi between open disk in $mathbb{R}^2$ and small part of that rectangular cylinder near edge.
$endgroup$
– Seewoo Lee
Feb 3 at 5:35
$begingroup$
In fact, it doesn’t matter whether it has ‘edge’ or ‘vertex’. For example, you can find homeomorphi between open disk in $mathbb{R}^2$ and small part of that rectangular cylinder near edge.
$endgroup$
– Seewoo Lee
Feb 3 at 5:35
$begingroup$
So it is a surface?
$endgroup$
– AColoredReptile
Feb 3 at 5:50
$begingroup$
So it is a surface?
$endgroup$
– AColoredReptile
Feb 3 at 5:50
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
We only need to check at a point on edge. Let $P = (2, 0, 0) in S$. Consider the open neighborhood
$$
W = Scap {(x, y, z),:, x>0, y^{2} + z^{2}<1}
$$
then we can check that this is homeomorphic to the open unit disk $D$ by $phi:Wto D$ defined as
$$
phi(x, y, z) = (y, z).
$$
To show that it's smooth, you may think that the surface is not smooth since it is so edgy. However, that's not true. For example, you may observe that the above map $phi$ is just projection to a plane. Now consider two small open neighborhood of a given point of $S$ on an edge, and let $phi_{1}, phi_{2}$ be appropriate projections to a plane, which gives a chart. Now you can see that the transition map is an identity map on their intersection, which is definitely smooth!
We can start with easier example. Let $C = {(x, y),:, y = |x|}$ be a graph of $f(x) = |x|$, which seems to be a non-smooth curve (and it is, in some sense).
However, we can give this curve a smooth structure by the following way - just project it to the $x$-axis. It is each to check that $C$ is homeomorphic to $mathbb{R}$ by the projection, and we can just transport smooth structure on the real line ($x$-axis) to $C$.
answered Feb 3 at 6:49
Seewoo LeeSeewoo Lee
7,2642930
$endgroup$
$begingroup$
Sorry I forgot to put smooth surface in the original question.
$endgroup$
– AColoredReptile
Feb 3 at 6:54
$begingroup$
@AColoredReptile That's okay since actually, that surface is also smooth! I edited.
$endgroup$
– Seewoo Lee
Feb 3 at 7:28
add a comment |
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1 Answer
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$begingroup$
We only need to check at a point on edge. Let $P = (2, 0, 0) in S$. Consider the open neighborhood
$$
W = Scap {(x, y, z),:, x>0, y^{2} + z^{2}<1}
$$
then we can check that this is homeomorphic to the open unit disk $D$ by $phi:Wto D$ defined as
$$
phi(x, y, z) = (y, z).
$$
To show that it's smooth, you may think that the surface is not smooth since it is so edgy. However, that's not true. For example, you may observe that the above map $phi$ is just projection to a plane. Now consider two small open neighborhood of a given point of $S$ on an edge, and let $phi_{1}, phi_{2}$ be appropriate projections to a plane, which gives a chart. Now you can see that the transition map is an identity map on their intersection, which is definitely smooth!
We can start with easier example. Let $C = {(x, y),:, y = |x|}$ be a graph of $f(x) = |x|$, which seems to be a non-smooth curve (and it is, in some sense).
However, we can give this curve a smooth structure by the following way - just project it to the $x$-axis. It is each to check that $C$ is homeomorphic to $mathbb{R}$ by the projection, and we can just transport smooth structure on the real line ($x$-axis) to $C$.
answered Feb 3 at 6:49
Seewoo LeeSeewoo Lee
7,2642930
$endgroup$
$begingroup$
Sorry I forgot to put smooth surface in the original question.
$endgroup$
– AColoredReptile
Feb 3 at 6:54
$begingroup$
@AColoredReptile That's okay since actually, that surface is also smooth! I edited.
$endgroup$
– Seewoo Lee
Feb 3 at 7:28
add a comment |
$begingroup$
We only need to check at a point on edge. Let $P = (2, 0, 0) in S$. Consider the open neighborhood
$$
W = Scap {(x, y, z),:, x>0, y^{2} + z^{2}<1}
$$
then we can check that this is homeomorphic to the open unit disk $D$ by $phi:Wto D$ defined as
$$
phi(x, y, z) = (y, z).
$$
To show that it's smooth, you may think that the surface is not smooth since it is so edgy. However, that's not true. For example, you may observe that the above map $phi$ is just projection to a plane. Now consider two small open neighborhood of a given point of $S$ on an edge, and let $phi_{1}, phi_{2}$ be appropriate projections to a plane, which gives a chart. Now you can see that the transition map is an identity map on their intersection, which is definitely smooth!
We can start with easier example. Let $C = {(x, y),:, y = |x|}$ be a graph of $f(x) = |x|$, which seems to be a non-smooth curve (and it is, in some sense).
However, we can give this curve a smooth structure by the following way - just project it to the $x$-axis. It is each to check that $C$ is homeomorphic to $mathbb{R}$ by the projection, and we can just transport smooth structure on the real line ($x$-axis) to $C$.
answered Feb 3 at 6:49
Seewoo LeeSeewoo Lee
7,2642930
$endgroup$
$begingroup$
Sorry I forgot to put smooth surface in the original question.
$endgroup$
– AColoredReptile
Feb 3 at 6:54
$begingroup$
@AColoredReptile That's okay since actually, that surface is also smooth! I edited.
$endgroup$
– Seewoo Lee
Feb 3 at 7:28
add a comment |
$begingroup$
We only need to check at a point on edge. Let $P = (2, 0, 0) in S$. Consider the open neighborhood
$$
W = Scap {(x, y, z),:, x>0, y^{2} + z^{2}<1}
$$
then we can check that this is homeomorphic to the open unit disk $D$ by $phi:Wto D$ defined as
$$
phi(x, y, z) = (y, z).
$$
To show that it's smooth, you may think that the surface is not smooth since it is so edgy. However, that's not true. For example, you may observe that the above map $phi$ is just projection to a plane. Now consider two small open neighborhood of a given point of $S$ on an edge, and let $phi_{1}, phi_{2}$ be appropriate projections to a plane, which gives a chart. Now you can see that the transition map is an identity map on their intersection, which is definitely smooth!
We can start with easier example. Let $C = {(x, y),:, y = |x|}$ be a graph of $f(x) = |x|$, which seems to be a non-smooth curve (and it is, in some sense).
However, we can give this curve a smooth structure by the following way - just project it to the $x$-axis. It is each to check that $C$ is homeomorphic to $mathbb{R}$ by the projection, and we can just transport smooth structure on the real line ($x$-axis) to $C$.
answered Feb 3 at 6:49
Seewoo LeeSeewoo Lee
7,2642930
$endgroup$
We only need to check at a point on edge. Let $P = (2, 0, 0) in S$. Consider the open neighborhood
$$
W = Scap {(x, y, z),:, x>0, y^{2} + z^{2}<1}
$$
then we can check that this is homeomorphic to the open unit disk $D$ by $phi:Wto D$ defined as
$$
phi(x, y, z) = (y, z).
$$
To show that it's smooth, you may think that the surface is not smooth since it is so edgy. However, that's not true. For example, you may observe that the above map $phi$ is just projection to a plane. Now consider two small open neighborhood of a given point of $S$ on an edge, and let $phi_{1}, phi_{2}$ be appropriate projections to a plane, which gives a chart. Now you can see that the transition map is an identity map on their intersection, which is definitely smooth!
We can start with easier example. Let $C = {(x, y),:, y = |x|}$ be a graph of $f(x) = |x|$, which seems to be a non-smooth curve (and it is, in some sense).
However, we can give this curve a smooth structure by the following way - just project it to the $x$-axis. It is each to check that $C$ is homeomorphic to $mathbb{R}$ by the projection, and we can just transport smooth structure on the real line ($x$-axis) to $C$.
answered Feb 3 at 6:49
Seewoo LeeSeewoo Lee
7,2642930
edited Feb 3 at 7:28
answered Feb 3 at 6:49
Seewoo LeeSeewoo Lee
7,2642930
answered Feb 3 at 6:49
Seewoo LeeSeewoo Lee
7,2642930
answered Feb 3 at 6:49
Seewoo LeeSeewoo Lee
7,2642930
7,2642930
$begingroup$
Sorry I forgot to put smooth surface in the original question.
$endgroup$
– AColoredReptile
Feb 3 at 6:54
$begingroup$
@AColoredReptile That's okay since actually, that surface is also smooth! I edited.
$endgroup$
– Seewoo Lee
Feb 3 at 7:28
add a comment |
$begingroup$
Sorry I forgot to put smooth surface in the original question.
$endgroup$
– AColoredReptile
Feb 3 at 6:54
$begingroup$
@AColoredReptile That's okay since actually, that surface is also smooth! I edited.
$endgroup$
– Seewoo Lee
Feb 3 at 7:28
$begingroup$
Sorry I forgot to put smooth surface in the original question.
$endgroup$
– AColoredReptile
Feb 3 at 6:54
$begingroup$
Sorry I forgot to put smooth surface in the original question.
$endgroup$
– AColoredReptile
Feb 3 at 6:54
$begingroup$
@AColoredReptile That's okay since actually, that surface is also smooth! I edited.
$endgroup$
– Seewoo Lee
Feb 3 at 7:28
$begingroup$
@AColoredReptile That's okay since actually, that surface is also smooth! I edited.
$endgroup$
– Seewoo Lee
Feb 3 at 7:28
add a comment |
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$begingroup$
What’s your definition of surface?
$endgroup$
– Seewoo Lee
Feb 3 at 3:31
$begingroup$
A subset $S$ of $mathbb{R}^3$ is a surface if for ever point $pin S$ there is an open set $U$ in $mathbb{R}^3$ and an open set $W$ in $mathbb{R}^3$ containing p such that $Scap W$ is homeomorphic to $U$.
$endgroup$
– AColoredReptile
Feb 3 at 3:40
$begingroup$
$Uinmathbb{R}^2$ sorry.
$endgroup$
– AColoredReptile
Feb 3 at 3:46
$begingroup$
In fact, it doesn’t matter whether it has ‘edge’ or ‘vertex’. For example, you can find homeomorphi between open disk in $mathbb{R}^2$ and small part of that rectangular cylinder near edge.
$endgroup$
– Seewoo Lee
Feb 3 at 5:35
$begingroup$
So it is a surface?
$endgroup$
– AColoredReptile
Feb 3 at 5:50