Am i correct in the way I am deciding whether or not subsets of $Bbb R^3$ are surfaces or not?
$begingroup$
I just want to make sure that there isn't any gaps in my reasoning ( or flat out mistakes!) before I try to learn anymore about classifying subsets of $Bbb R^3$ as surfaces, so to that end .....
Consider the subsets of $Bbb R^3$
i) $S^2={(x,y,z)in Bbb R ^3 | x^2+y^2+z^2=1}$
ii) $A={(x,y,z)in Bbb R^3|x^2+z^2=y^2}$
In deciding whether or not these are surfaces I know we can use a corollary of the implicit function from calculus.
Theorem :suppose $f:Bbb R^3 rightarrow Bbb R$ is smooth and c is a regular value of $f$, with $f^{-1}(c)neq(0,0,0)$, then $f^{-1}(c)$ is a surface in $Bbb R^3$.
Here is my attempt at using the corollary:
.
.
i)$S^2={(x,y,z) in Bbb R^3 |x^2+y^2+z^2=1}$
let $f(x,y,z)=x^2+y^2+z^2$, clearly f is smooth and maps $Bbb R^3 rightarrow Bbb R$.
$Df=(partial f/partial x, partial f/partial y,partial f/partial z)=(2x,2y,2z)$
so every point except the origin is a regular point.
$Rightarrow f(0,0,0)=0$ is the only critical value of f and so $f^{-1}(1)=x^2+y^2+z^2$
describes a surface as 1 is a regular value , and so $S^2$ is a surface.
.
.
ii)let $f(x)=x^2+z^2-y^2$
again this is smooth with the proper mapping needed.
$Df=(2x,2z,-2y)$
so again the only critical point is the origin
$Rightarrow f(0,0,0)=0$ is the only critical value
so $f^{-1}(0)=x^2+z^2-y^2$ does not describe a surface so A is not a surface.
differential-geometry surfaces
asked Dec 6 '18 at 16:13
can'tcauchycan'tcauchy
999417
$endgroup$
|
show 1 more comment
$begingroup$
I just want to make sure that there isn't any gaps in my reasoning ( or flat out mistakes!) before I try to learn anymore about classifying subsets of $Bbb R^3$ as surfaces, so to that end .....
Consider the subsets of $Bbb R^3$
i) $S^2={(x,y,z)in Bbb R ^3 | x^2+y^2+z^2=1}$
ii) $A={(x,y,z)in Bbb R^3|x^2+z^2=y^2}$
In deciding whether or not these are surfaces I know we can use a corollary of the implicit function from calculus.
Theorem :suppose $f:Bbb R^3 rightarrow Bbb R$ is smooth and c is a regular value of $f$, with $f^{-1}(c)neq(0,0,0)$, then $f^{-1}(c)$ is a surface in $Bbb R^3$.
Here is my attempt at using the corollary:
.
.
i)$S^2={(x,y,z) in Bbb R^3 |x^2+y^2+z^2=1}$
let $f(x,y,z)=x^2+y^2+z^2$, clearly f is smooth and maps $Bbb R^3 rightarrow Bbb R$.
$Df=(partial f/partial x, partial f/partial y,partial f/partial z)=(2x,2y,2z)$
so every point except the origin is a regular point.
$Rightarrow f(0,0,0)=0$ is the only critical value of f and so $f^{-1}(1)=x^2+y^2+z^2$
describes a surface as 1 is a regular value , and so $S^2$ is a surface.
.
.
ii)let $f(x)=x^2+z^2-y^2$
again this is smooth with the proper mapping needed.
$Df=(2x,2z,-2y)$
so again the only critical point is the origin
$Rightarrow f(0,0,0)=0$ is the only critical value
so $f^{-1}(0)=x^2+z^2-y^2$ does not describe a surface so A is not a surface.
differential-geometry surfaces
asked Dec 6 '18 at 16:13
can'tcauchycan'tcauchy
999417
$endgroup$
5
$begingroup$
The implicit function theorem is not an if and only if theorem. So your proof of i) is correct because it uses the logical direction that is stated in the implicit function theorem, but your proof of ii) is invalid because it uses the other logical direction which is not stated.
$endgroup$
– Lee Mosher
Dec 6 '18 at 16:34
4
$begingroup$
For example, if we take $f(x,y,z)=z^2$ then $f(x,y,z)=0$ does define a surface, namely the $x$,$y$ plane given by the equation $z=0$, and this is so even though $0$ is a critical value of $f$ and every point $(x,y,0)$ on the surface is a critical point.
$endgroup$
– Lee Mosher
Dec 6 '18 at 16:39
$begingroup$
See, among other answers, this important characterization of smooth surfaces or submanifolds as specific graphs.
$endgroup$
– Ted Shifrin
Dec 6 '18 at 23:30
$begingroup$
@LeeMosher for number ii then would the following argument work : if we take the coordinate patch to be the function $bar x (x,z)rightarrow (x,y,z)$ but then note that $y=-sqrt{x^2+z^2}$ meaning that the partial derivatives of y are $tfrac{-z}{sqrt{x^2+z^2}}$ and $tfrac{-x}{sqrt{x^2+z^2}}$ which are not differentiable at (0,0) and so we can't have a surface if the domain of the co-ordinate patch contains this point ? I feel like my argument may be a little shaky but is that the right idea to move forward with ?
$endgroup$
– can'tcauchy
Dec 9 '18 at 20:05
2
$begingroup$
That's a profitable idea, I think, but the trouble is that you do not have the freedom to work with just a single convenient coordinate chart. If you wish to do a proof by examination of coordinate charts then you have to prove that no possible coordinate chart could work.
$endgroup$
– Lee Mosher
Dec 9 '18 at 21:20
|
show 1 more comment
$begingroup$
I just want to make sure that there isn't any gaps in my reasoning ( or flat out mistakes!) before I try to learn anymore about classifying subsets of $Bbb R^3$ as surfaces, so to that end .....
Consider the subsets of $Bbb R^3$
i) $S^2={(x,y,z)in Bbb R ^3 | x^2+y^2+z^2=1}$
ii) $A={(x,y,z)in Bbb R^3|x^2+z^2=y^2}$
In deciding whether or not these are surfaces I know we can use a corollary of the implicit function from calculus.
Theorem :suppose $f:Bbb R^3 rightarrow Bbb R$ is smooth and c is a regular value of $f$, with $f^{-1}(c)neq(0,0,0)$, then $f^{-1}(c)$ is a surface in $Bbb R^3$.
Here is my attempt at using the corollary:
.
.
i)$S^2={(x,y,z) in Bbb R^3 |x^2+y^2+z^2=1}$
let $f(x,y,z)=x^2+y^2+z^2$, clearly f is smooth and maps $Bbb R^3 rightarrow Bbb R$.
$Df=(partial f/partial x, partial f/partial y,partial f/partial z)=(2x,2y,2z)$
so every point except the origin is a regular point.
$Rightarrow f(0,0,0)=0$ is the only critical value of f and so $f^{-1}(1)=x^2+y^2+z^2$
describes a surface as 1 is a regular value , and so $S^2$ is a surface.
.
.
ii)let $f(x)=x^2+z^2-y^2$
again this is smooth with the proper mapping needed.
$Df=(2x,2z,-2y)$
so again the only critical point is the origin
$Rightarrow f(0,0,0)=0$ is the only critical value
so $f^{-1}(0)=x^2+z^2-y^2$ does not describe a surface so A is not a surface.
differential-geometry surfaces
asked Dec 6 '18 at 16:13
can'tcauchycan'tcauchy
999417
$endgroup$
I just want to make sure that there isn't any gaps in my reasoning ( or flat out mistakes!) before I try to learn anymore about classifying subsets of $Bbb R^3$ as surfaces, so to that end .....
Consider the subsets of $Bbb R^3$
i) $S^2={(x,y,z)in Bbb R ^3 | x^2+y^2+z^2=1}$
ii) $A={(x,y,z)in Bbb R^3|x^2+z^2=y^2}$
In deciding whether or not these are surfaces I know we can use a corollary of the implicit function from calculus.
Theorem :suppose $f:Bbb R^3 rightarrow Bbb R$ is smooth and c is a regular value of $f$, with $f^{-1}(c)neq(0,0,0)$, then $f^{-1}(c)$ is a surface in $Bbb R^3$.
Here is my attempt at using the corollary:
.
.
i)$S^2={(x,y,z) in Bbb R^3 |x^2+y^2+z^2=1}$
let $f(x,y,z)=x^2+y^2+z^2$, clearly f is smooth and maps $Bbb R^3 rightarrow Bbb R$.
$Df=(partial f/partial x, partial f/partial y,partial f/partial z)=(2x,2y,2z)$
so every point except the origin is a regular point.
$Rightarrow f(0,0,0)=0$ is the only critical value of f and so $f^{-1}(1)=x^2+y^2+z^2$
describes a surface as 1 is a regular value , and so $S^2$ is a surface.
.
.
ii)let $f(x)=x^2+z^2-y^2$
again this is smooth with the proper mapping needed.
$Df=(2x,2z,-2y)$
so again the only critical point is the origin
$Rightarrow f(0,0,0)=0$ is the only critical value
so $f^{-1}(0)=x^2+z^2-y^2$ does not describe a surface so A is not a surface.
differential-geometry surfaces
differential-geometry surfaces
asked Dec 6 '18 at 16:13
can'tcauchycan'tcauchy
999417
asked Dec 6 '18 at 16:13
can'tcauchycan'tcauchy
999417
asked Dec 6 '18 at 16:13
can'tcauchycan'tcauchy
999417
asked Dec 6 '18 at 16:13
can'tcauchycan'tcauchy
999417
asked Dec 6 '18 at 16:13
can'tcauchycan'tcauchy
999417
999417
5
$begingroup$
The implicit function theorem is not an if and only if theorem. So your proof of i) is correct because it uses the logical direction that is stated in the implicit function theorem, but your proof of ii) is invalid because it uses the other logical direction which is not stated.
$endgroup$
– Lee Mosher
Dec 6 '18 at 16:34
4
$begingroup$
For example, if we take $f(x,y,z)=z^2$ then $f(x,y,z)=0$ does define a surface, namely the $x$,$y$ plane given by the equation $z=0$, and this is so even though $0$ is a critical value of $f$ and every point $(x,y,0)$ on the surface is a critical point.
$endgroup$
– Lee Mosher
Dec 6 '18 at 16:39
$begingroup$
See, among other answers, this important characterization of smooth surfaces or submanifolds as specific graphs.
$endgroup$
– Ted Shifrin
Dec 6 '18 at 23:30
$begingroup$
@LeeMosher for number ii then would the following argument work : if we take the coordinate patch to be the function $bar x (x,z)rightarrow (x,y,z)$ but then note that $y=-sqrt{x^2+z^2}$ meaning that the partial derivatives of y are $tfrac{-z}{sqrt{x^2+z^2}}$ and $tfrac{-x}{sqrt{x^2+z^2}}$ which are not differentiable at (0,0) and so we can't have a surface if the domain of the co-ordinate patch contains this point ? I feel like my argument may be a little shaky but is that the right idea to move forward with ?
$endgroup$
– can'tcauchy
Dec 9 '18 at 20:05
2
$begingroup$
That's a profitable idea, I think, but the trouble is that you do not have the freedom to work with just a single convenient coordinate chart. If you wish to do a proof by examination of coordinate charts then you have to prove that no possible coordinate chart could work.
$endgroup$
– Lee Mosher
Dec 9 '18 at 21:20
|
show 1 more comment
5
$begingroup$
The implicit function theorem is not an if and only if theorem. So your proof of i) is correct because it uses the logical direction that is stated in the implicit function theorem, but your proof of ii) is invalid because it uses the other logical direction which is not stated.
$endgroup$
– Lee Mosher
Dec 6 '18 at 16:34
4
$begingroup$
For example, if we take $f(x,y,z)=z^2$ then $f(x,y,z)=0$ does define a surface, namely the $x$,$y$ plane given by the equation $z=0$, and this is so even though $0$ is a critical value of $f$ and every point $(x,y,0)$ on the surface is a critical point.
$endgroup$
– Lee Mosher
Dec 6 '18 at 16:39
$begingroup$
See, among other answers, this important characterization of smooth surfaces or submanifolds as specific graphs.
$endgroup$
– Ted Shifrin
Dec 6 '18 at 23:30
$begingroup$
@LeeMosher for number ii then would the following argument work : if we take the coordinate patch to be the function $bar x (x,z)rightarrow (x,y,z)$ but then note that $y=-sqrt{x^2+z^2}$ meaning that the partial derivatives of y are $tfrac{-z}{sqrt{x^2+z^2}}$ and $tfrac{-x}{sqrt{x^2+z^2}}$ which are not differentiable at (0,0) and so we can't have a surface if the domain of the co-ordinate patch contains this point ? I feel like my argument may be a little shaky but is that the right idea to move forward with ?
$endgroup$
– can'tcauchy
Dec 9 '18 at 20:05
2
$begingroup$
That's a profitable idea, I think, but the trouble is that you do not have the freedom to work with just a single convenient coordinate chart. If you wish to do a proof by examination of coordinate charts then you have to prove that no possible coordinate chart could work.
$endgroup$
– Lee Mosher
Dec 9 '18 at 21:20
5
5
$begingroup$
The implicit function theorem is not an if and only if theorem. So your proof of i) is correct because it uses the logical direction that is stated in the implicit function theorem, but your proof of ii) is invalid because it uses the other logical direction which is not stated.
$endgroup$
– Lee Mosher
Dec 6 '18 at 16:34
$begingroup$
The implicit function theorem is not an if and only if theorem. So your proof of i) is correct because it uses the logical direction that is stated in the implicit function theorem, but your proof of ii) is invalid because it uses the other logical direction which is not stated.
$endgroup$
– Lee Mosher
Dec 6 '18 at 16:34
4
4
$begingroup$
For example, if we take $f(x,y,z)=z^2$ then $f(x,y,z)=0$ does define a surface, namely the $x$,$y$ plane given by the equation $z=0$, and this is so even though $0$ is a critical value of $f$ and every point $(x,y,0)$ on the surface is a critical point.
$endgroup$
– Lee Mosher
Dec 6 '18 at 16:39
$begingroup$
For example, if we take $f(x,y,z)=z^2$ then $f(x,y,z)=0$ does define a surface, namely the $x$,$y$ plane given by the equation $z=0$, and this is so even though $0$ is a critical value of $f$ and every point $(x,y,0)$ on the surface is a critical point.
$endgroup$
– Lee Mosher
Dec 6 '18 at 16:39
$begingroup$
See, among other answers, this important characterization of smooth surfaces or submanifolds as specific graphs.
$endgroup$
– Ted Shifrin
Dec 6 '18 at 23:30
$begingroup$
See, among other answers, this important characterization of smooth surfaces or submanifolds as specific graphs.
$endgroup$
– Ted Shifrin
Dec 6 '18 at 23:30
$begingroup$
@LeeMosher for number ii then would the following argument work : if we take the coordinate patch to be the function $bar x (x,z)rightarrow (x,y,z)$ but then note that $y=-sqrt{x^2+z^2}$ meaning that the partial derivatives of y are $tfrac{-z}{sqrt{x^2+z^2}}$ and $tfrac{-x}{sqrt{x^2+z^2}}$ which are not differentiable at (0,0) and so we can't have a surface if the domain of the co-ordinate patch contains this point ? I feel like my argument may be a little shaky but is that the right idea to move forward with ?
$endgroup$
– can'tcauchy
Dec 9 '18 at 20:05
$begingroup$
@LeeMosher for number ii then would the following argument work : if we take the coordinate patch to be the function $bar x (x,z)rightarrow (x,y,z)$ but then note that $y=-sqrt{x^2+z^2}$ meaning that the partial derivatives of y are $tfrac{-z}{sqrt{x^2+z^2}}$ and $tfrac{-x}{sqrt{x^2+z^2}}$ which are not differentiable at (0,0) and so we can't have a surface if the domain of the co-ordinate patch contains this point ? I feel like my argument may be a little shaky but is that the right idea to move forward with ?
$endgroup$
– can'tcauchy
Dec 9 '18 at 20:05
2
2
$begingroup$
That's a profitable idea, I think, but the trouble is that you do not have the freedom to work with just a single convenient coordinate chart. If you wish to do a proof by examination of coordinate charts then you have to prove that no possible coordinate chart could work.
$endgroup$
– Lee Mosher
Dec 9 '18 at 21:20
$begingroup$
That's a profitable idea, I think, but the trouble is that you do not have the freedom to work with just a single convenient coordinate chart. If you wish to do a proof by examination of coordinate charts then you have to prove that no possible coordinate chart could work.
$endgroup$
– Lee Mosher
Dec 9 '18 at 21:20
|
show 1 more comment
1 Answer
1
active
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$begingroup$
An intuitive argument as to why $S^{2}$ is a surface can be made as such: it can be covered with 6 co-ordinate patches (up,down,right,left,front & back), and thus each point on it lies in the image of one of the 6 coordinate patches.
As for the second part, your main error is using the implicit function theorem as an If and only if statement.
answered Dec 9 '18 at 21:13
notabobnotabob
294
$endgroup$
$begingroup$
I would like to add that it was User Lee Mosher who had initially stated the second part of my answer, so credit to him.
$endgroup$
– notabob
Dec 9 '18 at 21:23
add a comment |
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$begingroup$
An intuitive argument as to why $S^{2}$ is a surface can be made as such: it can be covered with 6 co-ordinate patches (up,down,right,left,front & back), and thus each point on it lies in the image of one of the 6 coordinate patches.
As for the second part, your main error is using the implicit function theorem as an If and only if statement.
answered Dec 9 '18 at 21:13
notabobnotabob
294
$endgroup$
$begingroup$
I would like to add that it was User Lee Mosher who had initially stated the second part of my answer, so credit to him.
$endgroup$
– notabob
Dec 9 '18 at 21:23
add a comment |
$begingroup$
An intuitive argument as to why $S^{2}$ is a surface can be made as such: it can be covered with 6 co-ordinate patches (up,down,right,left,front & back), and thus each point on it lies in the image of one of the 6 coordinate patches.
As for the second part, your main error is using the implicit function theorem as an If and only if statement.
answered Dec 9 '18 at 21:13
notabobnotabob
294
$endgroup$
$begingroup$
I would like to add that it was User Lee Mosher who had initially stated the second part of my answer, so credit to him.
$endgroup$
– notabob
Dec 9 '18 at 21:23
add a comment |
$begingroup$
An intuitive argument as to why $S^{2}$ is a surface can be made as such: it can be covered with 6 co-ordinate patches (up,down,right,left,front & back), and thus each point on it lies in the image of one of the 6 coordinate patches.
As for the second part, your main error is using the implicit function theorem as an If and only if statement.
answered Dec 9 '18 at 21:13
notabobnotabob
294
$endgroup$
An intuitive argument as to why $S^{2}$ is a surface can be made as such: it can be covered with 6 co-ordinate patches (up,down,right,left,front & back), and thus each point on it lies in the image of one of the 6 coordinate patches.
As for the second part, your main error is using the implicit function theorem as an If and only if statement.
answered Dec 9 '18 at 21:13
notabobnotabob
294
answered Dec 9 '18 at 21:13
notabobnotabob
294
answered Dec 9 '18 at 21:13
notabobnotabob
294
answered Dec 9 '18 at 21:13
notabobnotabob
294
294
$begingroup$
I would like to add that it was User Lee Mosher who had initially stated the second part of my answer, so credit to him.
$endgroup$
– notabob
Dec 9 '18 at 21:23
add a comment |
$begingroup$
I would like to add that it was User Lee Mosher who had initially stated the second part of my answer, so credit to him.
$endgroup$
– notabob
Dec 9 '18 at 21:23
$begingroup$
I would like to add that it was User Lee Mosher who had initially stated the second part of my answer, so credit to him.
$endgroup$
– notabob
Dec 9 '18 at 21:23
$begingroup$
I would like to add that it was User Lee Mosher who had initially stated the second part of my answer, so credit to him.
$endgroup$
– notabob
Dec 9 '18 at 21:23
add a comment |
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5
$begingroup$
The implicit function theorem is not an if and only if theorem. So your proof of i) is correct because it uses the logical direction that is stated in the implicit function theorem, but your proof of ii) is invalid because it uses the other logical direction which is not stated.
$endgroup$
– Lee Mosher
Dec 6 '18 at 16:34
4
$begingroup$
For example, if we take $f(x,y,z)=z^2$ then $f(x,y,z)=0$ does define a surface, namely the $x$,$y$ plane given by the equation $z=0$, and this is so even though $0$ is a critical value of $f$ and every point $(x,y,0)$ on the surface is a critical point.
$endgroup$
– Lee Mosher
Dec 6 '18 at 16:39
$begingroup$
See, among other answers, this important characterization of smooth surfaces or submanifolds as specific graphs.
$endgroup$
– Ted Shifrin
Dec 6 '18 at 23:30
$begingroup$
@LeeMosher for number ii then would the following argument work : if we take the coordinate patch to be the function $bar x (x,z)rightarrow (x,y,z)$ but then note that $y=-sqrt{x^2+z^2}$ meaning that the partial derivatives of y are $tfrac{-z}{sqrt{x^2+z^2}}$ and $tfrac{-x}{sqrt{x^2+z^2}}$ which are not differentiable at (0,0) and so we can't have a surface if the domain of the co-ordinate patch contains this point ? I feel like my argument may be a little shaky but is that the right idea to move forward with ?
$endgroup$
– can'tcauchy
Dec 9 '18 at 20:05
2
$begingroup$
That's a profitable idea, I think, but the trouble is that you do not have the freedom to work with just a single convenient coordinate chart. If you wish to do a proof by examination of coordinate charts then you have to prove that no possible coordinate chart could work.
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– Lee Mosher
Dec 9 '18 at 21:20