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Shape operator and path connected surface (global theory of surface)
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3.1 Theorem (O'neill) If its shape operator is identically zero, them M is part of a plane in $R^3$
The first part of Proof.
By the definition of shape operator, $S = 0$ means that any unit normal vector field $E_3$ on $M$ is Euclidean parallel, and hence can be identified with a point of $R^3$.
...
Question.
I don't know why S = 0 implies a parallel normal vector field.
I have tried the following:
Let $M$ be a orientable path-connected surface.
Since M is orientable, let Z denote the derivatable unit-normal vector field defined in M.
Claim. If $dZ = 0 $ (shape operator is identically zero), Z is parallel normal vector field. (Z is Gaussian mapping)
Try.
Since $dZ = 0$, Normal curvature is zero.
On the other hand, there is coordinate patch $mathtt x : D rightarrow M$ for each point $pin M$.
Since $R^3$ is locally path-connected space, there is path- connected open set $D'$. Thus, coordiante path $mathtt x : D' rightarrow M$ is a well - defined coordinate function.
Z$circ mathtt x (u,v)$ = $(x, y , z)$.
Since $Z_u = Z_u = 0$, $x_u = x_v = 0$. $D'$ is domain, Because $D'$ is path- connected open set.
Since gradient $x$ = 0, $Z$ is identically constant on $mathtt x(D')$.
In short, Z is a constant function in the appropriate neighborhood of each point. I can not go on since then. Maybe I should use path- connecte surface M, but I don't like it. Please help.
general-topology geometry surfaces path-connected
asked Jul 25 at 5:03
LeeHanWoong
398
add a comment |Â
up vote
0
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3.1 Theorem (O'neill) If its shape operator is identically zero, them M is part of a plane in $R^3$
The first part of Proof.
By the definition of shape operator, $S = 0$ means that any unit normal vector field $E_3$ on $M$ is Euclidean parallel, and hence can be identified with a point of $R^3$.
...
Question.
I don't know why S = 0 implies a parallel normal vector field.
I have tried the following:
Let $M$ be a orientable path-connected surface.
Since M is orientable, let Z denote the derivatable unit-normal vector field defined in M.
Claim. If $dZ = 0 $ (shape operator is identically zero), Z is parallel normal vector field. (Z is Gaussian mapping)
Try.
Since $dZ = 0$, Normal curvature is zero.
On the other hand, there is coordinate patch $mathtt x : D rightarrow M$ for each point $pin M$.
Since $R^3$ is locally path-connected space, there is path- connected open set $D'$. Thus, coordiante path $mathtt x : D' rightarrow M$ is a well - defined coordinate function.
Z$circ mathtt x (u,v)$ = $(x, y , z)$.
Since $Z_u = Z_u = 0$, $x_u = x_v = 0$. $D'$ is domain, Because $D'$ is path- connected open set.
Since gradient $x$ = 0, $Z$ is identically constant on $mathtt x(D')$.
In short, Z is a constant function in the appropriate neighborhood of each point. I can not go on since then. Maybe I should use path- connecte surface M, but I don't like it. Please help.
general-topology geometry surfaces path-connected
asked Jul 25 at 5:03
LeeHanWoong
398
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
3.1 Theorem (O'neill) If its shape operator is identically zero, them M is part of a plane in $R^3$
The first part of Proof.
By the definition of shape operator, $S = 0$ means that any unit normal vector field $E_3$ on $M$ is Euclidean parallel, and hence can be identified with a point of $R^3$.
...
Question.
I don't know why S = 0 implies a parallel normal vector field.
I have tried the following:
Let $M$ be a orientable path-connected surface.
Since M is orientable, let Z denote the derivatable unit-normal vector field defined in M.
Claim. If $dZ = 0 $ (shape operator is identically zero), Z is parallel normal vector field. (Z is Gaussian mapping)
Try.
Since $dZ = 0$, Normal curvature is zero.
On the other hand, there is coordinate patch $mathtt x : D rightarrow M$ for each point $pin M$.
Since $R^3$ is locally path-connected space, there is path- connected open set $D'$. Thus, coordiante path $mathtt x : D' rightarrow M$ is a well - defined coordinate function.
Z$circ mathtt x (u,v)$ = $(x, y , z)$.
Since $Z_u = Z_u = 0$, $x_u = x_v = 0$. $D'$ is domain, Because $D'$ is path- connected open set.
Since gradient $x$ = 0, $Z$ is identically constant on $mathtt x(D')$.
In short, Z is a constant function in the appropriate neighborhood of each point. I can not go on since then. Maybe I should use path- connecte surface M, but I don't like it. Please help.
general-topology geometry surfaces path-connected
asked Jul 25 at 5:03
LeeHanWoong
398
3.1 Theorem (O'neill) If its shape operator is identically zero, them M is part of a plane in $R^3$
The first part of Proof.
By the definition of shape operator, $S = 0$ means that any unit normal vector field $E_3$ on $M$ is Euclidean parallel, and hence can be identified with a point of $R^3$.
...
Question.
I don't know why S = 0 implies a parallel normal vector field.
I have tried the following:
Let $M$ be a orientable path-connected surface.
Since M is orientable, let Z denote the derivatable unit-normal vector field defined in M.
Claim. If $dZ = 0 $ (shape operator is identically zero), Z is parallel normal vector field. (Z is Gaussian mapping)
Try.
Since $dZ = 0$, Normal curvature is zero.
On the other hand, there is coordinate patch $mathtt x : D rightarrow M$ for each point $pin M$.
Since $R^3$ is locally path-connected space, there is path- connected open set $D'$. Thus, coordiante path $mathtt x : D' rightarrow M$ is a well - defined coordinate function.
Z$circ mathtt x (u,v)$ = $(x, y , z)$.
Since $Z_u = Z_u = 0$, $x_u = x_v = 0$. $D'$ is domain, Because $D'$ is path- connected open set.
Since gradient $x$ = 0, $Z$ is identically constant on $mathtt x(D')$.
In short, Z is a constant function in the appropriate neighborhood of each point. I can not go on since then. Maybe I should use path- connecte surface M, but I don't like it. Please help.
general-topology geometry surfaces path-connected
asked Jul 25 at 5:03
LeeHanWoong
398
edited Jul 25 at 6:08
asked Jul 25 at 5:03
LeeHanWoong
398
asked Jul 25 at 5:03
LeeHanWoong
398
asked Jul 25 at 5:03
LeeHanWoong
398
398
add a comment |Â
add a comment |Â
1 Answer
1
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1
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If $Z$ - (the Gauss map assigning to each point $p$ on the surface the (outer) unit normal at that point) - is constant in a neighborhood of $p$, then for every point $q$ in a neighborhood of $p$ on the surface, the tangent space is one and the same - the one that is orthogonal to the constant value of $Z$. This means that in a neighborhood of $p$, all points of the surface are orthogonal to the constant value of $Z$, hence belong to a plane.
answered Jul 25 at 5:45
uniquesolution
7,595721
Is Z the same for different p and q? I know that surface is plane in a neighborhood in p. I wonder if it is flat about the whole(M).
– LeeHanWoong
Jul 25 at 6:17
The shape operator is by its nature a local creature. Everything is about a neighborhood of $p$. A surface can be flat at some parts, and non-flat at others. For example, the shape operator of a cylinder at a point on one of its flat tops is zero, but it is not zero at points along its "round" parts. All you can deduce from the data is that the surface is a plane in a neighborhood of the point. HOWEVER, if the shape operator is identically zero at EVERY point of the surface, then ALL the points of the surface belong to the same plane, right?
– uniquesolution
Jul 25 at 6:33
I think that Z is constant in the neighborhood of each point, but may be different from each other. Please solve my ignorance
– LeeHanWoong
Jul 25 at 6:43
What is the exact referent to "Theorem 3.1" from O'neill? I am looking at the book now (second revised edition) and there is no Theorem 3.1. Also, can you please give me an example of a differentiable surface whose shape operator is identically zero which is not a plane? In other words, an example where $Z$ has one constant value at the nbd of some point $p$, and another constant value at the nbd of some other point $q$.
– uniquesolution
Jul 25 at 6:46
It may be 6.3.1. I can not see it in pdf. Thank you for your reference. I intuitively think that it is flat. I want to show that Z equals the value in the nbd of other points.
– LeeHanWoong
Jul 25 at 6:57
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If $Z$ - (the Gauss map assigning to each point $p$ on the surface the (outer) unit normal at that point) - is constant in a neighborhood of $p$, then for every point $q$ in a neighborhood of $p$ on the surface, the tangent space is one and the same - the one that is orthogonal to the constant value of $Z$. This means that in a neighborhood of $p$, all points of the surface are orthogonal to the constant value of $Z$, hence belong to a plane.
answered Jul 25 at 5:45
uniquesolution
7,595721
Is Z the same for different p and q? I know that surface is plane in a neighborhood in p. I wonder if it is flat about the whole(M).
– LeeHanWoong
Jul 25 at 6:17
The shape operator is by its nature a local creature. Everything is about a neighborhood of $p$. A surface can be flat at some parts, and non-flat at others. For example, the shape operator of a cylinder at a point on one of its flat tops is zero, but it is not zero at points along its "round" parts. All you can deduce from the data is that the surface is a plane in a neighborhood of the point. HOWEVER, if the shape operator is identically zero at EVERY point of the surface, then ALL the points of the surface belong to the same plane, right?
– uniquesolution
Jul 25 at 6:33
I think that Z is constant in the neighborhood of each point, but may be different from each other. Please solve my ignorance
– LeeHanWoong
Jul 25 at 6:43
What is the exact referent to "Theorem 3.1" from O'neill? I am looking at the book now (second revised edition) and there is no Theorem 3.1. Also, can you please give me an example of a differentiable surface whose shape operator is identically zero which is not a plane? In other words, an example where $Z$ has one constant value at the nbd of some point $p$, and another constant value at the nbd of some other point $q$.
– uniquesolution
Jul 25 at 6:46
It may be 6.3.1. I can not see it in pdf. Thank you for your reference. I intuitively think that it is flat. I want to show that Z equals the value in the nbd of other points.
– LeeHanWoong
Jul 25 at 6:57
add a comment |Â
up vote
1
down vote
If $Z$ - (the Gauss map assigning to each point $p$ on the surface the (outer) unit normal at that point) - is constant in a neighborhood of $p$, then for every point $q$ in a neighborhood of $p$ on the surface, the tangent space is one and the same - the one that is orthogonal to the constant value of $Z$. This means that in a neighborhood of $p$, all points of the surface are orthogonal to the constant value of $Z$, hence belong to a plane.
answered Jul 25 at 5:45
uniquesolution
7,595721
Is Z the same for different p and q? I know that surface is plane in a neighborhood in p. I wonder if it is flat about the whole(M).
– LeeHanWoong
Jul 25 at 6:17
The shape operator is by its nature a local creature. Everything is about a neighborhood of $p$. A surface can be flat at some parts, and non-flat at others. For example, the shape operator of a cylinder at a point on one of its flat tops is zero, but it is not zero at points along its "round" parts. All you can deduce from the data is that the surface is a plane in a neighborhood of the point. HOWEVER, if the shape operator is identically zero at EVERY point of the surface, then ALL the points of the surface belong to the same plane, right?
– uniquesolution
Jul 25 at 6:33
I think that Z is constant in the neighborhood of each point, but may be different from each other. Please solve my ignorance
– LeeHanWoong
Jul 25 at 6:43
What is the exact referent to "Theorem 3.1" from O'neill? I am looking at the book now (second revised edition) and there is no Theorem 3.1. Also, can you please give me an example of a differentiable surface whose shape operator is identically zero which is not a plane? In other words, an example where $Z$ has one constant value at the nbd of some point $p$, and another constant value at the nbd of some other point $q$.
– uniquesolution
Jul 25 at 6:46
It may be 6.3.1. I can not see it in pdf. Thank you for your reference. I intuitively think that it is flat. I want to show that Z equals the value in the nbd of other points.
– LeeHanWoong
Jul 25 at 6:57
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If $Z$ - (the Gauss map assigning to each point $p$ on the surface the (outer) unit normal at that point) - is constant in a neighborhood of $p$, then for every point $q$ in a neighborhood of $p$ on the surface, the tangent space is one and the same - the one that is orthogonal to the constant value of $Z$. This means that in a neighborhood of $p$, all points of the surface are orthogonal to the constant value of $Z$, hence belong to a plane.
answered Jul 25 at 5:45
uniquesolution
7,595721
If $Z$ - (the Gauss map assigning to each point $p$ on the surface the (outer) unit normal at that point) - is constant in a neighborhood of $p$, then for every point $q$ in a neighborhood of $p$ on the surface, the tangent space is one and the same - the one that is orthogonal to the constant value of $Z$. This means that in a neighborhood of $p$, all points of the surface are orthogonal to the constant value of $Z$, hence belong to a plane.
answered Jul 25 at 5:45
uniquesolution
7,595721
answered Jul 25 at 5:45
uniquesolution
7,595721
answered Jul 25 at 5:45
uniquesolution
7,595721
answered Jul 25 at 5:45
uniquesolution
7,595721
7,595721
Is Z the same for different p and q? I know that surface is plane in a neighborhood in p. I wonder if it is flat about the whole(M).
– LeeHanWoong
Jul 25 at 6:17
The shape operator is by its nature a local creature. Everything is about a neighborhood of $p$. A surface can be flat at some parts, and non-flat at others. For example, the shape operator of a cylinder at a point on one of its flat tops is zero, but it is not zero at points along its "round" parts. All you can deduce from the data is that the surface is a plane in a neighborhood of the point. HOWEVER, if the shape operator is identically zero at EVERY point of the surface, then ALL the points of the surface belong to the same plane, right?
– uniquesolution
Jul 25 at 6:33
I think that Z is constant in the neighborhood of each point, but may be different from each other. Please solve my ignorance
– LeeHanWoong
Jul 25 at 6:43
What is the exact referent to "Theorem 3.1" from O'neill? I am looking at the book now (second revised edition) and there is no Theorem 3.1. Also, can you please give me an example of a differentiable surface whose shape operator is identically zero which is not a plane? In other words, an example where $Z$ has one constant value at the nbd of some point $p$, and another constant value at the nbd of some other point $q$.
– uniquesolution
Jul 25 at 6:46
It may be 6.3.1. I can not see it in pdf. Thank you for your reference. I intuitively think that it is flat. I want to show that Z equals the value in the nbd of other points.
– LeeHanWoong
Jul 25 at 6:57
add a comment |Â
Is Z the same for different p and q? I know that surface is plane in a neighborhood in p. I wonder if it is flat about the whole(M).
– LeeHanWoong
Jul 25 at 6:17
The shape operator is by its nature a local creature. Everything is about a neighborhood of $p$. A surface can be flat at some parts, and non-flat at others. For example, the shape operator of a cylinder at a point on one of its flat tops is zero, but it is not zero at points along its "round" parts. All you can deduce from the data is that the surface is a plane in a neighborhood of the point. HOWEVER, if the shape operator is identically zero at EVERY point of the surface, then ALL the points of the surface belong to the same plane, right?
– uniquesolution
Jul 25 at 6:33
I think that Z is constant in the neighborhood of each point, but may be different from each other. Please solve my ignorance
– LeeHanWoong
Jul 25 at 6:43
What is the exact referent to "Theorem 3.1" from O'neill? I am looking at the book now (second revised edition) and there is no Theorem 3.1. Also, can you please give me an example of a differentiable surface whose shape operator is identically zero which is not a plane? In other words, an example where $Z$ has one constant value at the nbd of some point $p$, and another constant value at the nbd of some other point $q$.
– uniquesolution
Jul 25 at 6:46
It may be 6.3.1. I can not see it in pdf. Thank you for your reference. I intuitively think that it is flat. I want to show that Z equals the value in the nbd of other points.
– LeeHanWoong
Jul 25 at 6:57
Is Z the same for different p and q? I know that surface is plane in a neighborhood in p. I wonder if it is flat about the whole(M).
– LeeHanWoong
Jul 25 at 6:17
Is Z the same for different p and q? I know that surface is plane in a neighborhood in p. I wonder if it is flat about the whole(M).
– LeeHanWoong
Jul 25 at 6:17
The shape operator is by its nature a local creature. Everything is about a neighborhood of $p$. A surface can be flat at some parts, and non-flat at others. For example, the shape operator of a cylinder at a point on one of its flat tops is zero, but it is not zero at points along its "round" parts. All you can deduce from the data is that the surface is a plane in a neighborhood of the point. HOWEVER, if the shape operator is identically zero at EVERY point of the surface, then ALL the points of the surface belong to the same plane, right?
– uniquesolution
Jul 25 at 6:33
The shape operator is by its nature a local creature. Everything is about a neighborhood of $p$. A surface can be flat at some parts, and non-flat at others. For example, the shape operator of a cylinder at a point on one of its flat tops is zero, but it is not zero at points along its "round" parts. All you can deduce from the data is that the surface is a plane in a neighborhood of the point. HOWEVER, if the shape operator is identically zero at EVERY point of the surface, then ALL the points of the surface belong to the same plane, right?
– uniquesolution
Jul 25 at 6:33
I think that Z is constant in the neighborhood of each point, but may be different from each other. Please solve my ignorance
– LeeHanWoong
Jul 25 at 6:43
I think that Z is constant in the neighborhood of each point, but may be different from each other. Please solve my ignorance
– LeeHanWoong
Jul 25 at 6:43
What is the exact referent to "Theorem 3.1" from O'neill? I am looking at the book now (second revised edition) and there is no Theorem 3.1. Also, can you please give me an example of a differentiable surface whose shape operator is identically zero which is not a plane? In other words, an example where $Z$ has one constant value at the nbd of some point $p$, and another constant value at the nbd of some other point $q$.
– uniquesolution
Jul 25 at 6:46
What is the exact referent to "Theorem 3.1" from O'neill? I am looking at the book now (second revised edition) and there is no Theorem 3.1. Also, can you please give me an example of a differentiable surface whose shape operator is identically zero which is not a plane? In other words, an example where $Z$ has one constant value at the nbd of some point $p$, and another constant value at the nbd of some other point $q$.
– uniquesolution
Jul 25 at 6:46
It may be 6.3.1. I can not see it in pdf. Thank you for your reference. I intuitively think that it is flat. I want to show that Z equals the value in the nbd of other points.
– LeeHanWoong
Jul 25 at 6:57
It may be 6.3.1. I can not see it in pdf. Thank you for your reference. I intuitively think that it is flat. I want to show that Z equals the value in the nbd of other points.
– LeeHanWoong
Jul 25 at 6:57
add a comment |Â
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