The critical values of a map from a Torus to $mathbb{R}$
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Let $f(x,y,z)=z$ be the height function on a Torus in $mathbb{R}^3$.
Computing the representation of the differential of $f$ with respect to these coordinates, we get $f_{*,p}=left(begin{array}{c c c}0& 0& 1end{array}right)$, which has rank $1$ (and this rank is coordinate independent). Therefore, every point in the Torus should be a regular point. However, Loring Tu's book states that there are 4 critical points... Where did I go wrong?
Addendum for future reference:
In page 91, the proposition 8.11 is written with respect to the charts of the Torus. The Torus is of dimension 2, see wikipedia link, so our charts with domain the Torus are not $x,y,z$ but $phi(x,y,z)=(theta,varphi)$, such that $theta,varphi$ parametrize the Torus as in the wikipedia link. (We must switch the z with x in the wiki's parametrization to get the vertical Torus in Tu's book, i.e. $z=(R+rcos(theta))cos(varphi)$ ). We must think with respect to the parametrization to apply proposition 8.11.
differential-geometry
asked Dec 18 '18 at 12:43
An old man in the sea.An old man in the sea.
1,65211135
$endgroup$
add a comment |
$begingroup$
Let $f(x,y,z)=z$ be the height function on a Torus in $mathbb{R}^3$.
Computing the representation of the differential of $f$ with respect to these coordinates, we get $f_{*,p}=left(begin{array}{c c c}0& 0& 1end{array}right)$, which has rank $1$ (and this rank is coordinate independent). Therefore, every point in the Torus should be a regular point. However, Loring Tu's book states that there are 4 critical points... Where did I go wrong?
Addendum for future reference:
In page 91, the proposition 8.11 is written with respect to the charts of the Torus. The Torus is of dimension 2, see wikipedia link, so our charts with domain the Torus are not $x,y,z$ but $phi(x,y,z)=(theta,varphi)$, such that $theta,varphi$ parametrize the Torus as in the wikipedia link. (We must switch the z with x in the wiki's parametrization to get the vertical Torus in Tu's book, i.e. $z=(R+rcos(theta))cos(varphi)$ ). We must think with respect to the parametrization to apply proposition 8.11.
differential-geometry
asked Dec 18 '18 at 12:43
An old man in the sea.An old man in the sea.
1,65211135
$endgroup$
$begingroup$
Could you give me the page of the book?
$endgroup$
– positrón0802
Dec 18 '18 at 12:51
$begingroup$
@positrón0802 page 97
$endgroup$
– An old man in the sea.
Dec 18 '18 at 12:52
add a comment |
$begingroup$
Let $f(x,y,z)=z$ be the height function on a Torus in $mathbb{R}^3$.
Computing the representation of the differential of $f$ with respect to these coordinates, we get $f_{*,p}=left(begin{array}{c c c}0& 0& 1end{array}right)$, which has rank $1$ (and this rank is coordinate independent). Therefore, every point in the Torus should be a regular point. However, Loring Tu's book states that there are 4 critical points... Where did I go wrong?
Addendum for future reference:
In page 91, the proposition 8.11 is written with respect to the charts of the Torus. The Torus is of dimension 2, see wikipedia link, so our charts with domain the Torus are not $x,y,z$ but $phi(x,y,z)=(theta,varphi)$, such that $theta,varphi$ parametrize the Torus as in the wikipedia link. (We must switch the z with x in the wiki's parametrization to get the vertical Torus in Tu's book, i.e. $z=(R+rcos(theta))cos(varphi)$ ). We must think with respect to the parametrization to apply proposition 8.11.
differential-geometry
asked Dec 18 '18 at 12:43
An old man in the sea.An old man in the sea.
1,65211135
$endgroup$
Let $f(x,y,z)=z$ be the height function on a Torus in $mathbb{R}^3$.
Computing the representation of the differential of $f$ with respect to these coordinates, we get $f_{*,p}=left(begin{array}{c c c}0& 0& 1end{array}right)$, which has rank $1$ (and this rank is coordinate independent). Therefore, every point in the Torus should be a regular point. However, Loring Tu's book states that there are 4 critical points... Where did I go wrong?
Addendum for future reference:
In page 91, the proposition 8.11 is written with respect to the charts of the Torus. The Torus is of dimension 2, see wikipedia link, so our charts with domain the Torus are not $x,y,z$ but $phi(x,y,z)=(theta,varphi)$, such that $theta,varphi$ parametrize the Torus as in the wikipedia link. (We must switch the z with x in the wiki's parametrization to get the vertical Torus in Tu's book, i.e. $z=(R+rcos(theta))cos(varphi)$ ). We must think with respect to the parametrization to apply proposition 8.11.
differential-geometry
differential-geometry
asked Dec 18 '18 at 12:43
An old man in the sea.An old man in the sea.
1,65211135
asked Dec 18 '18 at 12:43
An old man in the sea.An old man in the sea.
1,65211135
edited Dec 20 '18 at 10:36
An old man in the sea.
asked Dec 18 '18 at 12:43
An old man in the sea.An old man in the sea.
1,65211135
asked Dec 18 '18 at 12:43
An old man in the sea.An old man in the sea.
1,65211135
asked Dec 18 '18 at 12:43
An old man in the sea.An old man in the sea.
1,65211135
1,65211135
$begingroup$
Could you give me the page of the book?
$endgroup$
– positrón0802
Dec 18 '18 at 12:51
$begingroup$
@positrón0802 page 97
$endgroup$
– An old man in the sea.
Dec 18 '18 at 12:52
add a comment |
$begingroup$
Could you give me the page of the book?
$endgroup$
– positrón0802
Dec 18 '18 at 12:51
$begingroup$
@positrón0802 page 97
$endgroup$
– An old man in the sea.
Dec 18 '18 at 12:52
$begingroup$
Could you give me the page of the book?
$endgroup$
– positrón0802
Dec 18 '18 at 12:51
$begingroup$
Could you give me the page of the book?
$endgroup$
– positrón0802
Dec 18 '18 at 12:51
$begingroup$
@positrón0802 page 97
$endgroup$
– An old man in the sea.
Dec 18 '18 at 12:52
$begingroup$
@positrón0802 page 97
$endgroup$
– An old man in the sea.
Dec 18 '18 at 12:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $mathbb{T}$ be the torus. Your matrix $begin{pmatrix}0&0&1end{pmatrix}$ is the differential of the map $pi:mathbb{R}^3to mathbb{R}$ given by $(x,y,z)mapsto z$ (which is linear hence equals its own differential). Your function $f:mathbb{T}to mathbb{R}$ is the composition of the inclusion $iota:mathbb{T}tomathbb{R}^3$ and $pi.$ Hence $df_p=picirc diota_p:T_pmathbb{T}to mathbb{R}.$ So $df_p$ fail to be surjective when $diota_p(T_pmathbb{R})$ is contained in the kernel of $pi,$ which is the span of $(1,0,0)$ and $(0,1,0),$ i.e. the $xy$ plane. So $p$ is critical when its tangent plane is horizontal.
answered Dec 18 '18 at 13:04
positrón0802positrón0802
4,513520
$endgroup$
$begingroup$
positrón, many thanks for your answer. To accept your answer, could you just tell me how one would obtain $diota_p$? I don't understand how one defines $iota$ on a Torus...
$endgroup$
– An old man in the sea.
Dec 19 '18 at 9:43
$begingroup$
I'm considering the torus $mathbb{T}$ as the "doughnut" surface in $mathbb{R}^3$ as drawn in page 97 in Tu's book, so $iota:mathbb{T}to mathbb{R}^3$ is just the inclusion and $diota_p:T_pmathbb{R}^3to mathbb{R}^3$ would be the inclusion of the tangent plane of the torus at $p$ into $mathbb{R}^3.$
$endgroup$
– positrón0802
Dec 19 '18 at 20:50
$begingroup$
Positrón, I still have some doubts. I'm sorry if they are somewhat 'dumb' on my part, but I'm self-studying this. My two questions are: 1) How is it different to consider just $mathbb{T}$ from $mathbb{T} subset mathbb{R}^3$? 2) I've tried drawing the tangent plane, and for the 2 points in the 'outer circle' I get the tangent plane is only in the xy plane. However, why is the tangent plane for the 2 points in the 'inner circle' also in the xy plane? I would really appreciate your help, Positrón.
$endgroup$
– An old man in the sea.
Dec 20 '18 at 8:47
1
$begingroup$
Don't worry at all. 1) $mathbb{T}$ here is defined as a subset of $mathbb{R}^3$ (a surface), sometimes when considering a submanifold of $mathbb{R}^n$ it is convenient to "forget" about the ambient space $mathbb{R}^n$ in which it lies, but not in this problem. 2) Take the draw and try to see how the tangent plane changes while one moves around the 'inner circle' from say 'left' to 'right', it starts being vertical and then the "slope" is being reduced to the point of being horizontal and then the "slope" increases again to the point of being vertical again in the 'right' side.
$endgroup$
– positrón0802
Dec 20 '18 at 13:27
add a comment |
$begingroup$
You have to look at the restriction of the differential at the tangent point of $xin T^2$, $T_xT^2$ is a 2-dimensional subspace which may be horizontal and for such a point $x$ $df_x$ is critical.
answered Dec 18 '18 at 12:54
Tsemo AristideTsemo Aristide
59.9k11446
$endgroup$
$begingroup$
Thanks for the answer, however I'm not getting it. Could you please elaborate a bit more. The tangent space at those points will be 2-dim. So you're saying that in every point of Torus, the basis of the tangent space is just ${frac{partial}{partial x}, frac{partial}{partial y}}$? Also, how do we reconcile that with the condition that all the partial derivatives must be equal to zero?
$endgroup$
– An old man in the sea.
Dec 18 '18 at 13:03
$begingroup$
yes, this is true, the best way to see this draw a picture or see a picture of a torus of lines you will see that, you can also compute the equation of the tangent space.
$endgroup$
– Tsemo Aristide
Dec 18 '18 at 13:05
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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votes
$begingroup$
Let $mathbb{T}$ be the torus. Your matrix $begin{pmatrix}0&0&1end{pmatrix}$ is the differential of the map $pi:mathbb{R}^3to mathbb{R}$ given by $(x,y,z)mapsto z$ (which is linear hence equals its own differential). Your function $f:mathbb{T}to mathbb{R}$ is the composition of the inclusion $iota:mathbb{T}tomathbb{R}^3$ and $pi.$ Hence $df_p=picirc diota_p:T_pmathbb{T}to mathbb{R}.$ So $df_p$ fail to be surjective when $diota_p(T_pmathbb{R})$ is contained in the kernel of $pi,$ which is the span of $(1,0,0)$ and $(0,1,0),$ i.e. the $xy$ plane. So $p$ is critical when its tangent plane is horizontal.
answered Dec 18 '18 at 13:04
positrón0802positrón0802
4,513520
$endgroup$
$begingroup$
positrón, many thanks for your answer. To accept your answer, could you just tell me how one would obtain $diota_p$? I don't understand how one defines $iota$ on a Torus...
$endgroup$
– An old man in the sea.
Dec 19 '18 at 9:43
$begingroup$
I'm considering the torus $mathbb{T}$ as the "doughnut" surface in $mathbb{R}^3$ as drawn in page 97 in Tu's book, so $iota:mathbb{T}to mathbb{R}^3$ is just the inclusion and $diota_p:T_pmathbb{R}^3to mathbb{R}^3$ would be the inclusion of the tangent plane of the torus at $p$ into $mathbb{R}^3.$
$endgroup$
– positrón0802
Dec 19 '18 at 20:50
$begingroup$
Positrón, I still have some doubts. I'm sorry if they are somewhat 'dumb' on my part, but I'm self-studying this. My two questions are: 1) How is it different to consider just $mathbb{T}$ from $mathbb{T} subset mathbb{R}^3$? 2) I've tried drawing the tangent plane, and for the 2 points in the 'outer circle' I get the tangent plane is only in the xy plane. However, why is the tangent plane for the 2 points in the 'inner circle' also in the xy plane? I would really appreciate your help, Positrón.
$endgroup$
– An old man in the sea.
Dec 20 '18 at 8:47
1
$begingroup$
Don't worry at all. 1) $mathbb{T}$ here is defined as a subset of $mathbb{R}^3$ (a surface), sometimes when considering a submanifold of $mathbb{R}^n$ it is convenient to "forget" about the ambient space $mathbb{R}^n$ in which it lies, but not in this problem. 2) Take the draw and try to see how the tangent plane changes while one moves around the 'inner circle' from say 'left' to 'right', it starts being vertical and then the "slope" is being reduced to the point of being horizontal and then the "slope" increases again to the point of being vertical again in the 'right' side.
$endgroup$
– positrón0802
Dec 20 '18 at 13:27
add a comment |
$begingroup$
Let $mathbb{T}$ be the torus. Your matrix $begin{pmatrix}0&0&1end{pmatrix}$ is the differential of the map $pi:mathbb{R}^3to mathbb{R}$ given by $(x,y,z)mapsto z$ (which is linear hence equals its own differential). Your function $f:mathbb{T}to mathbb{R}$ is the composition of the inclusion $iota:mathbb{T}tomathbb{R}^3$ and $pi.$ Hence $df_p=picirc diota_p:T_pmathbb{T}to mathbb{R}.$ So $df_p$ fail to be surjective when $diota_p(T_pmathbb{R})$ is contained in the kernel of $pi,$ which is the span of $(1,0,0)$ and $(0,1,0),$ i.e. the $xy$ plane. So $p$ is critical when its tangent plane is horizontal.
answered Dec 18 '18 at 13:04
positrón0802positrón0802
4,513520
$endgroup$
$begingroup$
positrón, many thanks for your answer. To accept your answer, could you just tell me how one would obtain $diota_p$? I don't understand how one defines $iota$ on a Torus...
$endgroup$
– An old man in the sea.
Dec 19 '18 at 9:43
$begingroup$
I'm considering the torus $mathbb{T}$ as the "doughnut" surface in $mathbb{R}^3$ as drawn in page 97 in Tu's book, so $iota:mathbb{T}to mathbb{R}^3$ is just the inclusion and $diota_p:T_pmathbb{R}^3to mathbb{R}^3$ would be the inclusion of the tangent plane of the torus at $p$ into $mathbb{R}^3.$
$endgroup$
– positrón0802
Dec 19 '18 at 20:50
$begingroup$
Positrón, I still have some doubts. I'm sorry if they are somewhat 'dumb' on my part, but I'm self-studying this. My two questions are: 1) How is it different to consider just $mathbb{T}$ from $mathbb{T} subset mathbb{R}^3$? 2) I've tried drawing the tangent plane, and for the 2 points in the 'outer circle' I get the tangent plane is only in the xy plane. However, why is the tangent plane for the 2 points in the 'inner circle' also in the xy plane? I would really appreciate your help, Positrón.
$endgroup$
– An old man in the sea.
Dec 20 '18 at 8:47
1
$begingroup$
Don't worry at all. 1) $mathbb{T}$ here is defined as a subset of $mathbb{R}^3$ (a surface), sometimes when considering a submanifold of $mathbb{R}^n$ it is convenient to "forget" about the ambient space $mathbb{R}^n$ in which it lies, but not in this problem. 2) Take the draw and try to see how the tangent plane changes while one moves around the 'inner circle' from say 'left' to 'right', it starts being vertical and then the "slope" is being reduced to the point of being horizontal and then the "slope" increases again to the point of being vertical again in the 'right' side.
$endgroup$
– positrón0802
Dec 20 '18 at 13:27
add a comment |
$begingroup$
Let $mathbb{T}$ be the torus. Your matrix $begin{pmatrix}0&0&1end{pmatrix}$ is the differential of the map $pi:mathbb{R}^3to mathbb{R}$ given by $(x,y,z)mapsto z$ (which is linear hence equals its own differential). Your function $f:mathbb{T}to mathbb{R}$ is the composition of the inclusion $iota:mathbb{T}tomathbb{R}^3$ and $pi.$ Hence $df_p=picirc diota_p:T_pmathbb{T}to mathbb{R}.$ So $df_p$ fail to be surjective when $diota_p(T_pmathbb{R})$ is contained in the kernel of $pi,$ which is the span of $(1,0,0)$ and $(0,1,0),$ i.e. the $xy$ plane. So $p$ is critical when its tangent plane is horizontal.
answered Dec 18 '18 at 13:04
positrón0802positrón0802
4,513520
$endgroup$
Let $mathbb{T}$ be the torus. Your matrix $begin{pmatrix}0&0&1end{pmatrix}$ is the differential of the map $pi:mathbb{R}^3to mathbb{R}$ given by $(x,y,z)mapsto z$ (which is linear hence equals its own differential). Your function $f:mathbb{T}to mathbb{R}$ is the composition of the inclusion $iota:mathbb{T}tomathbb{R}^3$ and $pi.$ Hence $df_p=picirc diota_p:T_pmathbb{T}to mathbb{R}.$ So $df_p$ fail to be surjective when $diota_p(T_pmathbb{R})$ is contained in the kernel of $pi,$ which is the span of $(1,0,0)$ and $(0,1,0),$ i.e. the $xy$ plane. So $p$ is critical when its tangent plane is horizontal.
answered Dec 18 '18 at 13:04
positrón0802positrón0802
4,513520
answered Dec 18 '18 at 13:04
positrón0802positrón0802
4,513520
answered Dec 18 '18 at 13:04
positrón0802positrón0802
4,513520
answered Dec 18 '18 at 13:04
positrón0802positrón0802
4,513520
4,513520
$begingroup$
positrón, many thanks for your answer. To accept your answer, could you just tell me how one would obtain $diota_p$? I don't understand how one defines $iota$ on a Torus...
$endgroup$
– An old man in the sea.
Dec 19 '18 at 9:43
$begingroup$
I'm considering the torus $mathbb{T}$ as the "doughnut" surface in $mathbb{R}^3$ as drawn in page 97 in Tu's book, so $iota:mathbb{T}to mathbb{R}^3$ is just the inclusion and $diota_p:T_pmathbb{R}^3to mathbb{R}^3$ would be the inclusion of the tangent plane of the torus at $p$ into $mathbb{R}^3.$
$endgroup$
– positrón0802
Dec 19 '18 at 20:50
$begingroup$
Positrón, I still have some doubts. I'm sorry if they are somewhat 'dumb' on my part, but I'm self-studying this. My two questions are: 1) How is it different to consider just $mathbb{T}$ from $mathbb{T} subset mathbb{R}^3$? 2) I've tried drawing the tangent plane, and for the 2 points in the 'outer circle' I get the tangent plane is only in the xy plane. However, why is the tangent plane for the 2 points in the 'inner circle' also in the xy plane? I would really appreciate your help, Positrón.
$endgroup$
– An old man in the sea.
Dec 20 '18 at 8:47
1
$begingroup$
Don't worry at all. 1) $mathbb{T}$ here is defined as a subset of $mathbb{R}^3$ (a surface), sometimes when considering a submanifold of $mathbb{R}^n$ it is convenient to "forget" about the ambient space $mathbb{R}^n$ in which it lies, but not in this problem. 2) Take the draw and try to see how the tangent plane changes while one moves around the 'inner circle' from say 'left' to 'right', it starts being vertical and then the "slope" is being reduced to the point of being horizontal and then the "slope" increases again to the point of being vertical again in the 'right' side.
$endgroup$
– positrón0802
Dec 20 '18 at 13:27
add a comment |
$begingroup$
positrón, many thanks for your answer. To accept your answer, could you just tell me how one would obtain $diota_p$? I don't understand how one defines $iota$ on a Torus...
$endgroup$
– An old man in the sea.
Dec 19 '18 at 9:43
$begingroup$
I'm considering the torus $mathbb{T}$ as the "doughnut" surface in $mathbb{R}^3$ as drawn in page 97 in Tu's book, so $iota:mathbb{T}to mathbb{R}^3$ is just the inclusion and $diota_p:T_pmathbb{R}^3to mathbb{R}^3$ would be the inclusion of the tangent plane of the torus at $p$ into $mathbb{R}^3.$
$endgroup$
– positrón0802
Dec 19 '18 at 20:50
$begingroup$
Positrón, I still have some doubts. I'm sorry if they are somewhat 'dumb' on my part, but I'm self-studying this. My two questions are: 1) How is it different to consider just $mathbb{T}$ from $mathbb{T} subset mathbb{R}^3$? 2) I've tried drawing the tangent plane, and for the 2 points in the 'outer circle' I get the tangent plane is only in the xy plane. However, why is the tangent plane for the 2 points in the 'inner circle' also in the xy plane? I would really appreciate your help, Positrón.
$endgroup$
– An old man in the sea.
Dec 20 '18 at 8:47
1
$begingroup$
Don't worry at all. 1) $mathbb{T}$ here is defined as a subset of $mathbb{R}^3$ (a surface), sometimes when considering a submanifold of $mathbb{R}^n$ it is convenient to "forget" about the ambient space $mathbb{R}^n$ in which it lies, but not in this problem. 2) Take the draw and try to see how the tangent plane changes while one moves around the 'inner circle' from say 'left' to 'right', it starts being vertical and then the "slope" is being reduced to the point of being horizontal and then the "slope" increases again to the point of being vertical again in the 'right' side.
$endgroup$
– positrón0802
Dec 20 '18 at 13:27
$begingroup$
positrón, many thanks for your answer. To accept your answer, could you just tell me how one would obtain $diota_p$? I don't understand how one defines $iota$ on a Torus...
$endgroup$
– An old man in the sea.
Dec 19 '18 at 9:43
$begingroup$
positrón, many thanks for your answer. To accept your answer, could you just tell me how one would obtain $diota_p$? I don't understand how one defines $iota$ on a Torus...
$endgroup$
– An old man in the sea.
Dec 19 '18 at 9:43
$begingroup$
I'm considering the torus $mathbb{T}$ as the "doughnut" surface in $mathbb{R}^3$ as drawn in page 97 in Tu's book, so $iota:mathbb{T}to mathbb{R}^3$ is just the inclusion and $diota_p:T_pmathbb{R}^3to mathbb{R}^3$ would be the inclusion of the tangent plane of the torus at $p$ into $mathbb{R}^3.$
$endgroup$
– positrón0802
Dec 19 '18 at 20:50
$begingroup$
I'm considering the torus $mathbb{T}$ as the "doughnut" surface in $mathbb{R}^3$ as drawn in page 97 in Tu's book, so $iota:mathbb{T}to mathbb{R}^3$ is just the inclusion and $diota_p:T_pmathbb{R}^3to mathbb{R}^3$ would be the inclusion of the tangent plane of the torus at $p$ into $mathbb{R}^3.$
$endgroup$
– positrón0802
Dec 19 '18 at 20:50
$begingroup$
Positrón, I still have some doubts. I'm sorry if they are somewhat 'dumb' on my part, but I'm self-studying this. My two questions are: 1) How is it different to consider just $mathbb{T}$ from $mathbb{T} subset mathbb{R}^3$? 2) I've tried drawing the tangent plane, and for the 2 points in the 'outer circle' I get the tangent plane is only in the xy plane. However, why is the tangent plane for the 2 points in the 'inner circle' also in the xy plane? I would really appreciate your help, Positrón.
$endgroup$
– An old man in the sea.
Dec 20 '18 at 8:47
$begingroup$
Positrón, I still have some doubts. I'm sorry if they are somewhat 'dumb' on my part, but I'm self-studying this. My two questions are: 1) How is it different to consider just $mathbb{T}$ from $mathbb{T} subset mathbb{R}^3$? 2) I've tried drawing the tangent plane, and for the 2 points in the 'outer circle' I get the tangent plane is only in the xy plane. However, why is the tangent plane for the 2 points in the 'inner circle' also in the xy plane? I would really appreciate your help, Positrón.
$endgroup$
– An old man in the sea.
Dec 20 '18 at 8:47
1
1
$begingroup$
Don't worry at all. 1) $mathbb{T}$ here is defined as a subset of $mathbb{R}^3$ (a surface), sometimes when considering a submanifold of $mathbb{R}^n$ it is convenient to "forget" about the ambient space $mathbb{R}^n$ in which it lies, but not in this problem. 2) Take the draw and try to see how the tangent plane changes while one moves around the 'inner circle' from say 'left' to 'right', it starts being vertical and then the "slope" is being reduced to the point of being horizontal and then the "slope" increases again to the point of being vertical again in the 'right' side.
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– positrón0802
Dec 20 '18 at 13:27
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Don't worry at all. 1) $mathbb{T}$ here is defined as a subset of $mathbb{R}^3$ (a surface), sometimes when considering a submanifold of $mathbb{R}^n$ it is convenient to "forget" about the ambient space $mathbb{R}^n$ in which it lies, but not in this problem. 2) Take the draw and try to see how the tangent plane changes while one moves around the 'inner circle' from say 'left' to 'right', it starts being vertical and then the "slope" is being reduced to the point of being horizontal and then the "slope" increases again to the point of being vertical again in the 'right' side.
$endgroup$
– positrón0802
Dec 20 '18 at 13:27
add a comment |
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You have to look at the restriction of the differential at the tangent point of $xin T^2$, $T_xT^2$ is a 2-dimensional subspace which may be horizontal and for such a point $x$ $df_x$ is critical.
answered Dec 18 '18 at 12:54
Tsemo AristideTsemo Aristide
59.9k11446
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Thanks for the answer, however I'm not getting it. Could you please elaborate a bit more. The tangent space at those points will be 2-dim. So you're saying that in every point of Torus, the basis of the tangent space is just ${frac{partial}{partial x}, frac{partial}{partial y}}$? Also, how do we reconcile that with the condition that all the partial derivatives must be equal to zero?
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– An old man in the sea.
Dec 18 '18 at 13:03
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yes, this is true, the best way to see this draw a picture or see a picture of a torus of lines you will see that, you can also compute the equation of the tangent space.
$endgroup$
– Tsemo Aristide
Dec 18 '18 at 13:05
add a comment |
$begingroup$
You have to look at the restriction of the differential at the tangent point of $xin T^2$, $T_xT^2$ is a 2-dimensional subspace which may be horizontal and for such a point $x$ $df_x$ is critical.
answered Dec 18 '18 at 12:54
Tsemo AristideTsemo Aristide
59.9k11446
$endgroup$
$begingroup$
Thanks for the answer, however I'm not getting it. Could you please elaborate a bit more. The tangent space at those points will be 2-dim. So you're saying that in every point of Torus, the basis of the tangent space is just ${frac{partial}{partial x}, frac{partial}{partial y}}$? Also, how do we reconcile that with the condition that all the partial derivatives must be equal to zero?
$endgroup$
– An old man in the sea.
Dec 18 '18 at 13:03
$begingroup$
yes, this is true, the best way to see this draw a picture or see a picture of a torus of lines you will see that, you can also compute the equation of the tangent space.
$endgroup$
– Tsemo Aristide
Dec 18 '18 at 13:05
add a comment |
$begingroup$
You have to look at the restriction of the differential at the tangent point of $xin T^2$, $T_xT^2$ is a 2-dimensional subspace which may be horizontal and for such a point $x$ $df_x$ is critical.
answered Dec 18 '18 at 12:54
Tsemo AristideTsemo Aristide
59.9k11446
$endgroup$
You have to look at the restriction of the differential at the tangent point of $xin T^2$, $T_xT^2$ is a 2-dimensional subspace which may be horizontal and for such a point $x$ $df_x$ is critical.
answered Dec 18 '18 at 12:54
Tsemo AristideTsemo Aristide
59.9k11446
answered Dec 18 '18 at 12:54
Tsemo AristideTsemo Aristide
59.9k11446
answered Dec 18 '18 at 12:54
Tsemo AristideTsemo Aristide
59.9k11446
answered Dec 18 '18 at 12:54
Tsemo AristideTsemo Aristide
59.9k11446
59.9k11446
$begingroup$
Thanks for the answer, however I'm not getting it. Could you please elaborate a bit more. The tangent space at those points will be 2-dim. So you're saying that in every point of Torus, the basis of the tangent space is just ${frac{partial}{partial x}, frac{partial}{partial y}}$? Also, how do we reconcile that with the condition that all the partial derivatives must be equal to zero?
$endgroup$
– An old man in the sea.
Dec 18 '18 at 13:03
$begingroup$
yes, this is true, the best way to see this draw a picture or see a picture of a torus of lines you will see that, you can also compute the equation of the tangent space.
$endgroup$
– Tsemo Aristide
Dec 18 '18 at 13:05
add a comment |
$begingroup$
Thanks for the answer, however I'm not getting it. Could you please elaborate a bit more. The tangent space at those points will be 2-dim. So you're saying that in every point of Torus, the basis of the tangent space is just ${frac{partial}{partial x}, frac{partial}{partial y}}$? Also, how do we reconcile that with the condition that all the partial derivatives must be equal to zero?
$endgroup$
– An old man in the sea.
Dec 18 '18 at 13:03
$begingroup$
yes, this is true, the best way to see this draw a picture or see a picture of a torus of lines you will see that, you can also compute the equation of the tangent space.
$endgroup$
– Tsemo Aristide
Dec 18 '18 at 13:05
$begingroup$
Thanks for the answer, however I'm not getting it. Could you please elaborate a bit more. The tangent space at those points will be 2-dim. So you're saying that in every point of Torus, the basis of the tangent space is just ${frac{partial}{partial x}, frac{partial}{partial y}}$? Also, how do we reconcile that with the condition that all the partial derivatives must be equal to zero?
$endgroup$
– An old man in the sea.
Dec 18 '18 at 13:03
$begingroup$
Thanks for the answer, however I'm not getting it. Could you please elaborate a bit more. The tangent space at those points will be 2-dim. So you're saying that in every point of Torus, the basis of the tangent space is just ${frac{partial}{partial x}, frac{partial}{partial y}}$? Also, how do we reconcile that with the condition that all the partial derivatives must be equal to zero?
$endgroup$
– An old man in the sea.
Dec 18 '18 at 13:03
$begingroup$
yes, this is true, the best way to see this draw a picture or see a picture of a torus of lines you will see that, you can also compute the equation of the tangent space.
$endgroup$
– Tsemo Aristide
Dec 18 '18 at 13:05
$begingroup$
yes, this is true, the best way to see this draw a picture or see a picture of a torus of lines you will see that, you can also compute the equation of the tangent space.
$endgroup$
– Tsemo Aristide
Dec 18 '18 at 13:05
add a comment |
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Could you give me the page of the book?
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– positrón0802
Dec 18 '18 at 12:51
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@positrón0802 page 97
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– An old man in the sea.
Dec 18 '18 at 12:52