Mixing
![How do you know what part of the semicircle it is when $argdfrac{z-2}{z+2} = dfracpi2$ [closed]](https://lh3.googleusercontent.com/-fAwP2lED17E/AAAAAAAAAAI/AAAAAAAAAAA/AGDgw-hLZjtCxpq2bWrtdUKAXoN2hJoAWA/s72-c-mo/photo.jpg?sz=32)
Torus, manifolds
$begingroup$
I have some trouble with the following questions:
$mathbb{R}^3$ has standard coördinates $(x, y, z)$. Regard in the plane $x=0$ the circle with centre $(x,y,z) = (0,0,b)$ and radius $a$, $0<a<b$. The area that arise when you turn the circle around the y-axis is called T.
1A. Give the equation of T and prove that it's a manifold of dimension 2.
I thought the following:
$$T= int_{C} pi (f(y))^2 dy $$ where C is the circle described above and $f(y)= sqrt(a-y^2+2zb-b^2)$ But now I don't know how to continue, cause I don't really have any boundaries.
B. Regard now $mathbb{S}^1 subseteq mathbb{R}^2$. Write for the standard 2-Torus $mathbb{T}^2= mathbb{S}^1 times mathbb{S}^1$, then $mathbb{T}^2 subseteq mathbb{R}^4$ is a two dimensional manifold. Prove that $mathbb{T}^2$ and $T$ are diffeomorph.
manifolds
asked Feb 17 '14 at 18:35
LeslieLeslie
315
$endgroup$
add a comment |
$begingroup$
I have some trouble with the following questions:
$mathbb{R}^3$ has standard coördinates $(x, y, z)$. Regard in the plane $x=0$ the circle with centre $(x,y,z) = (0,0,b)$ and radius $a$, $0<a<b$. The area that arise when you turn the circle around the y-axis is called T.
1A. Give the equation of T and prove that it's a manifold of dimension 2.
I thought the following:
$$T= int_{C} pi (f(y))^2 dy $$ where C is the circle described above and $f(y)= sqrt(a-y^2+2zb-b^2)$ But now I don't know how to continue, cause I don't really have any boundaries.
B. Regard now $mathbb{S}^1 subseteq mathbb{R}^2$. Write for the standard 2-Torus $mathbb{T}^2= mathbb{S}^1 times mathbb{S}^1$, then $mathbb{T}^2 subseteq mathbb{R}^4$ is a two dimensional manifold. Prove that $mathbb{T}^2$ and $T$ are diffeomorph.
manifolds
asked Feb 17 '14 at 18:35
LeslieLeslie
315
$endgroup$
$begingroup$
$T$ is a subset of $mathbb{R}^3$, why would it be equal to an integral?
$endgroup$
– Najib Idrissi
Aug 2 '18 at 12:51
add a comment |
$begingroup$
I have some trouble with the following questions:
$mathbb{R}^3$ has standard coördinates $(x, y, z)$. Regard in the plane $x=0$ the circle with centre $(x,y,z) = (0,0,b)$ and radius $a$, $0<a<b$. The area that arise when you turn the circle around the y-axis is called T.
1A. Give the equation of T and prove that it's a manifold of dimension 2.
I thought the following:
$$T= int_{C} pi (f(y))^2 dy $$ where C is the circle described above and $f(y)= sqrt(a-y^2+2zb-b^2)$ But now I don't know how to continue, cause I don't really have any boundaries.
B. Regard now $mathbb{S}^1 subseteq mathbb{R}^2$. Write for the standard 2-Torus $mathbb{T}^2= mathbb{S}^1 times mathbb{S}^1$, then $mathbb{T}^2 subseteq mathbb{R}^4$ is a two dimensional manifold. Prove that $mathbb{T}^2$ and $T$ are diffeomorph.
manifolds
asked Feb 17 '14 at 18:35
LeslieLeslie
315
$endgroup$
I have some trouble with the following questions:
$mathbb{R}^3$ has standard coördinates $(x, y, z)$. Regard in the plane $x=0$ the circle with centre $(x,y,z) = (0,0,b)$ and radius $a$, $0<a<b$. The area that arise when you turn the circle around the y-axis is called T.
1A. Give the equation of T and prove that it's a manifold of dimension 2.
I thought the following:
$$T= int_{C} pi (f(y))^2 dy $$ where C is the circle described above and $f(y)= sqrt(a-y^2+2zb-b^2)$ But now I don't know how to continue, cause I don't really have any boundaries.
B. Regard now $mathbb{S}^1 subseteq mathbb{R}^2$. Write for the standard 2-Torus $mathbb{T}^2= mathbb{S}^1 times mathbb{S}^1$, then $mathbb{T}^2 subseteq mathbb{R}^4$ is a two dimensional manifold. Prove that $mathbb{T}^2$ and $T$ are diffeomorph.
manifolds
manifolds
asked Feb 17 '14 at 18:35
LeslieLeslie
315
asked Feb 17 '14 at 18:35
LeslieLeslie
315
asked Feb 17 '14 at 18:35
LeslieLeslie
315
asked Feb 17 '14 at 18:35
LeslieLeslie
315
asked Feb 17 '14 at 18:35
LeslieLeslie
315
315
$begingroup$
$T$ is a subset of $mathbb{R}^3$, why would it be equal to an integral?
$endgroup$
– Najib Idrissi
Aug 2 '18 at 12:51
add a comment |
$begingroup$
$T$ is a subset of $mathbb{R}^3$, why would it be equal to an integral?
$endgroup$
– Najib Idrissi
Aug 2 '18 at 12:51
$begingroup$
$T$ is a subset of $mathbb{R}^3$, why would it be equal to an integral?
$endgroup$
– Najib Idrissi
Aug 2 '18 at 12:51
$begingroup$
$T$ is a subset of $mathbb{R}^3$, why would it be equal to an integral?
$endgroup$
– Najib Idrissi
Aug 2 '18 at 12:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
a)
Try $T=lbrace (sqrt{y^2+z^2} - b)^2 + z^2 = a, (x,y,z) in mathbb{R}^3 rbrace$, i think it gives you the required parametrization of your torus, embedded in the 3 dimensional euclidian space.
It's a 2 manifold because it's the zeros of the submersion
$f:mathbb{R}^3 to mathbb{R}, (x,y,z) mapsto (sqrt{y^2+z^2} - b)^2 + z^2 - a$
(you can check it easily).
b) For showing it's diffeomorphic to the product of two circles, you need to give first a two dimensional parametrization of T:
$phi : mathbb{R}^2 to mathbb{R}^3, (theta, psi) mapsto left(
matrix
{
cos 2pi theta&0&-sin 2pi theta \
0&1&0 \
sin 2pi theta&0&cos 2pi theta
}
right)left(
matrix
{
0 \
sqrt{a} cos 2pipsi \
b+sqrt{a} sin 2pi psi
}
right) $
Note that the vectr of the right defines just your circle C, when the matrix gives you the rotation around the y axis.
Then you have to show that
_this map is smooth
_it's $mathbb{Z}^2$ periodic, so induces a map $(mathbb{R/Z})^2 to mathbb{R}^3$
_this new map is the diffeomorphism you need!!
answered Feb 17 '14 at 20:34
LéoLéo
78937
$endgroup$
$begingroup$
Thank you for your reply! I still don't get both questions but I will first concentrate on A now. How did you get that formula for T? I know that a formula for a Torus is: $$ (c - sqrt(x^2 - y^2))^2 + z^2 = a^2$$ where $c$ is the the radius from the center of the hole to the center of the torus tube and $a$ is the radius of the tube. So in this case: $a=a$ and $c=b$, but you do not have $a^2$ on the right hand side. Also, do you think I have to derive the formula for a Torus in this question, or is something I should have known?
$endgroup$
– Leslie
Feb 18 '14 at 9:40
$begingroup$
And for B how can I see that te first matrix gives rotation around the y-axis?
$endgroup$
– Leslie
Feb 18 '14 at 15:02
$begingroup$
By the way in the torus I meant $x^2 + y^2$. Further I want to know why I have to prove that it's periodic? Because for a diffeomorphism it must be smooth, a bijection and it's inverse must be differentiable as wel
$endgroup$
– Leslie
Feb 18 '14 at 21:07
$begingroup$
Sorry, it's an $a^2$, as well... So it's exactly what you write, right? For understand this formula, i think you have to draw it!
$endgroup$
– Léo
Feb 19 '14 at 15:51
$begingroup$
First, when $z=+-a$, you find your circle (the meridian), and when $z$ moves, you have to make the other circle (the longitude).
$endgroup$
– Léo
Feb 19 '14 at 15:53
|
show 7 more comments
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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active
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votes
$begingroup$
a)
Try $T=lbrace (sqrt{y^2+z^2} - b)^2 + z^2 = a, (x,y,z) in mathbb{R}^3 rbrace$, i think it gives you the required parametrization of your torus, embedded in the 3 dimensional euclidian space.
It's a 2 manifold because it's the zeros of the submersion
$f:mathbb{R}^3 to mathbb{R}, (x,y,z) mapsto (sqrt{y^2+z^2} - b)^2 + z^2 - a$
(you can check it easily).
b) For showing it's diffeomorphic to the product of two circles, you need to give first a two dimensional parametrization of T:
$phi : mathbb{R}^2 to mathbb{R}^3, (theta, psi) mapsto left(
matrix
{
cos 2pi theta&0&-sin 2pi theta \
0&1&0 \
sin 2pi theta&0&cos 2pi theta
}
right)left(
matrix
{
0 \
sqrt{a} cos 2pipsi \
b+sqrt{a} sin 2pi psi
}
right) $
Note that the vectr of the right defines just your circle C, when the matrix gives you the rotation around the y axis.
Then you have to show that
_this map is smooth
_it's $mathbb{Z}^2$ periodic, so induces a map $(mathbb{R/Z})^2 to mathbb{R}^3$
_this new map is the diffeomorphism you need!!
answered Feb 17 '14 at 20:34
LéoLéo
78937
$endgroup$
$begingroup$
Thank you for your reply! I still don't get both questions but I will first concentrate on A now. How did you get that formula for T? I know that a formula for a Torus is: $$ (c - sqrt(x^2 - y^2))^2 + z^2 = a^2$$ where $c$ is the the radius from the center of the hole to the center of the torus tube and $a$ is the radius of the tube. So in this case: $a=a$ and $c=b$, but you do not have $a^2$ on the right hand side. Also, do you think I have to derive the formula for a Torus in this question, or is something I should have known?
$endgroup$
– Leslie
Feb 18 '14 at 9:40
$begingroup$
And for B how can I see that te first matrix gives rotation around the y-axis?
$endgroup$
– Leslie
Feb 18 '14 at 15:02
$begingroup$
By the way in the torus I meant $x^2 + y^2$. Further I want to know why I have to prove that it's periodic? Because for a diffeomorphism it must be smooth, a bijection and it's inverse must be differentiable as wel
$endgroup$
– Leslie
Feb 18 '14 at 21:07
$begingroup$
Sorry, it's an $a^2$, as well... So it's exactly what you write, right? For understand this formula, i think you have to draw it!
$endgroup$
– Léo
Feb 19 '14 at 15:51
$begingroup$
First, when $z=+-a$, you find your circle (the meridian), and when $z$ moves, you have to make the other circle (the longitude).
$endgroup$
– Léo
Feb 19 '14 at 15:53
|
show 7 more comments
$begingroup$
a)
Try $T=lbrace (sqrt{y^2+z^2} - b)^2 + z^2 = a, (x,y,z) in mathbb{R}^3 rbrace$, i think it gives you the required parametrization of your torus, embedded in the 3 dimensional euclidian space.
It's a 2 manifold because it's the zeros of the submersion
$f:mathbb{R}^3 to mathbb{R}, (x,y,z) mapsto (sqrt{y^2+z^2} - b)^2 + z^2 - a$
(you can check it easily).
b) For showing it's diffeomorphic to the product of two circles, you need to give first a two dimensional parametrization of T:
$phi : mathbb{R}^2 to mathbb{R}^3, (theta, psi) mapsto left(
matrix
{
cos 2pi theta&0&-sin 2pi theta \
0&1&0 \
sin 2pi theta&0&cos 2pi theta
}
right)left(
matrix
{
0 \
sqrt{a} cos 2pipsi \
b+sqrt{a} sin 2pi psi
}
right) $
Note that the vectr of the right defines just your circle C, when the matrix gives you the rotation around the y axis.
Then you have to show that
_this map is smooth
_it's $mathbb{Z}^2$ periodic, so induces a map $(mathbb{R/Z})^2 to mathbb{R}^3$
_this new map is the diffeomorphism you need!!
answered Feb 17 '14 at 20:34
LéoLéo
78937
$endgroup$
$begingroup$
Thank you for your reply! I still don't get both questions but I will first concentrate on A now. How did you get that formula for T? I know that a formula for a Torus is: $$ (c - sqrt(x^2 - y^2))^2 + z^2 = a^2$$ where $c$ is the the radius from the center of the hole to the center of the torus tube and $a$ is the radius of the tube. So in this case: $a=a$ and $c=b$, but you do not have $a^2$ on the right hand side. Also, do you think I have to derive the formula for a Torus in this question, or is something I should have known?
$endgroup$
– Leslie
Feb 18 '14 at 9:40
$begingroup$
And for B how can I see that te first matrix gives rotation around the y-axis?
$endgroup$
– Leslie
Feb 18 '14 at 15:02
$begingroup$
By the way in the torus I meant $x^2 + y^2$. Further I want to know why I have to prove that it's periodic? Because for a diffeomorphism it must be smooth, a bijection and it's inverse must be differentiable as wel
$endgroup$
– Leslie
Feb 18 '14 at 21:07
$begingroup$
Sorry, it's an $a^2$, as well... So it's exactly what you write, right? For understand this formula, i think you have to draw it!
$endgroup$
– Léo
Feb 19 '14 at 15:51
$begingroup$
First, when $z=+-a$, you find your circle (the meridian), and when $z$ moves, you have to make the other circle (the longitude).
$endgroup$
– Léo
Feb 19 '14 at 15:53
|
show 7 more comments
$begingroup$
a)
Try $T=lbrace (sqrt{y^2+z^2} - b)^2 + z^2 = a, (x,y,z) in mathbb{R}^3 rbrace$, i think it gives you the required parametrization of your torus, embedded in the 3 dimensional euclidian space.
It's a 2 manifold because it's the zeros of the submersion
$f:mathbb{R}^3 to mathbb{R}, (x,y,z) mapsto (sqrt{y^2+z^2} - b)^2 + z^2 - a$
(you can check it easily).
b) For showing it's diffeomorphic to the product of two circles, you need to give first a two dimensional parametrization of T:
$phi : mathbb{R}^2 to mathbb{R}^3, (theta, psi) mapsto left(
matrix
{
cos 2pi theta&0&-sin 2pi theta \
0&1&0 \
sin 2pi theta&0&cos 2pi theta
}
right)left(
matrix
{
0 \
sqrt{a} cos 2pipsi \
b+sqrt{a} sin 2pi psi
}
right) $
Note that the vectr of the right defines just your circle C, when the matrix gives you the rotation around the y axis.
Then you have to show that
_this map is smooth
_it's $mathbb{Z}^2$ periodic, so induces a map $(mathbb{R/Z})^2 to mathbb{R}^3$
_this new map is the diffeomorphism you need!!
answered Feb 17 '14 at 20:34
LéoLéo
78937
$endgroup$
a)
Try $T=lbrace (sqrt{y^2+z^2} - b)^2 + z^2 = a, (x,y,z) in mathbb{R}^3 rbrace$, i think it gives you the required parametrization of your torus, embedded in the 3 dimensional euclidian space.
It's a 2 manifold because it's the zeros of the submersion
$f:mathbb{R}^3 to mathbb{R}, (x,y,z) mapsto (sqrt{y^2+z^2} - b)^2 + z^2 - a$
(you can check it easily).
b) For showing it's diffeomorphic to the product of two circles, you need to give first a two dimensional parametrization of T:
$phi : mathbb{R}^2 to mathbb{R}^3, (theta, psi) mapsto left(
matrix
{
cos 2pi theta&0&-sin 2pi theta \
0&1&0 \
sin 2pi theta&0&cos 2pi theta
}
right)left(
matrix
{
0 \
sqrt{a} cos 2pipsi \
b+sqrt{a} sin 2pi psi
}
right) $
Note that the vectr of the right defines just your circle C, when the matrix gives you the rotation around the y axis.
Then you have to show that
_this map is smooth
_it's $mathbb{Z}^2$ periodic, so induces a map $(mathbb{R/Z})^2 to mathbb{R}^3$
_this new map is the diffeomorphism you need!!
answered Feb 17 '14 at 20:34
LéoLéo
78937
answered Feb 17 '14 at 20:34
LéoLéo
78937
answered Feb 17 '14 at 20:34
LéoLéo
78937
answered Feb 17 '14 at 20:34
LéoLéo
78937
78937
$begingroup$
Thank you for your reply! I still don't get both questions but I will first concentrate on A now. How did you get that formula for T? I know that a formula for a Torus is: $$ (c - sqrt(x^2 - y^2))^2 + z^2 = a^2$$ where $c$ is the the radius from the center of the hole to the center of the torus tube and $a$ is the radius of the tube. So in this case: $a=a$ and $c=b$, but you do not have $a^2$ on the right hand side. Also, do you think I have to derive the formula for a Torus in this question, or is something I should have known?
$endgroup$
– Leslie
Feb 18 '14 at 9:40
$begingroup$
And for B how can I see that te first matrix gives rotation around the y-axis?
$endgroup$
– Leslie
Feb 18 '14 at 15:02
$begingroup$
By the way in the torus I meant $x^2 + y^2$. Further I want to know why I have to prove that it's periodic? Because for a diffeomorphism it must be smooth, a bijection and it's inverse must be differentiable as wel
$endgroup$
– Leslie
Feb 18 '14 at 21:07
$begingroup$
Sorry, it's an $a^2$, as well... So it's exactly what you write, right? For understand this formula, i think you have to draw it!
$endgroup$
– Léo
Feb 19 '14 at 15:51
$begingroup$
First, when $z=+-a$, you find your circle (the meridian), and when $z$ moves, you have to make the other circle (the longitude).
$endgroup$
– Léo
Feb 19 '14 at 15:53
|
show 7 more comments
$begingroup$
Thank you for your reply! I still don't get both questions but I will first concentrate on A now. How did you get that formula for T? I know that a formula for a Torus is: $$ (c - sqrt(x^2 - y^2))^2 + z^2 = a^2$$ where $c$ is the the radius from the center of the hole to the center of the torus tube and $a$ is the radius of the tube. So in this case: $a=a$ and $c=b$, but you do not have $a^2$ on the right hand side. Also, do you think I have to derive the formula for a Torus in this question, or is something I should have known?
$endgroup$
– Leslie
Feb 18 '14 at 9:40
$begingroup$
And for B how can I see that te first matrix gives rotation around the y-axis?
$endgroup$
– Leslie
Feb 18 '14 at 15:02
$begingroup$
By the way in the torus I meant $x^2 + y^2$. Further I want to know why I have to prove that it's periodic? Because for a diffeomorphism it must be smooth, a bijection and it's inverse must be differentiable as wel
$endgroup$
– Leslie
Feb 18 '14 at 21:07
$begingroup$
Sorry, it's an $a^2$, as well... So it's exactly what you write, right? For understand this formula, i think you have to draw it!
$endgroup$
– Léo
Feb 19 '14 at 15:51
$begingroup$
First, when $z=+-a$, you find your circle (the meridian), and when $z$ moves, you have to make the other circle (the longitude).
$endgroup$
– Léo
Feb 19 '14 at 15:53
$begingroup$
Thank you for your reply! I still don't get both questions but I will first concentrate on A now. How did you get that formula for T? I know that a formula for a Torus is: $$ (c - sqrt(x^2 - y^2))^2 + z^2 = a^2$$ where $c$ is the the radius from the center of the hole to the center of the torus tube and $a$ is the radius of the tube. So in this case: $a=a$ and $c=b$, but you do not have $a^2$ on the right hand side. Also, do you think I have to derive the formula for a Torus in this question, or is something I should have known?
$endgroup$
– Leslie
Feb 18 '14 at 9:40
$begingroup$
Thank you for your reply! I still don't get both questions but I will first concentrate on A now. How did you get that formula for T? I know that a formula for a Torus is: $$ (c - sqrt(x^2 - y^2))^2 + z^2 = a^2$$ where $c$ is the the radius from the center of the hole to the center of the torus tube and $a$ is the radius of the tube. So in this case: $a=a$ and $c=b$, but you do not have $a^2$ on the right hand side. Also, do you think I have to derive the formula for a Torus in this question, or is something I should have known?
$endgroup$
– Leslie
Feb 18 '14 at 9:40
$begingroup$
And for B how can I see that te first matrix gives rotation around the y-axis?
$endgroup$
– Leslie
Feb 18 '14 at 15:02
$begingroup$
And for B how can I see that te first matrix gives rotation around the y-axis?
$endgroup$
– Leslie
Feb 18 '14 at 15:02
$begingroup$
By the way in the torus I meant $x^2 + y^2$. Further I want to know why I have to prove that it's periodic? Because for a diffeomorphism it must be smooth, a bijection and it's inverse must be differentiable as wel
$endgroup$
– Leslie
Feb 18 '14 at 21:07
$begingroup$
By the way in the torus I meant $x^2 + y^2$. Further I want to know why I have to prove that it's periodic? Because for a diffeomorphism it must be smooth, a bijection and it's inverse must be differentiable as wel
$endgroup$
– Leslie
Feb 18 '14 at 21:07
$begingroup$
Sorry, it's an $a^2$, as well... So it's exactly what you write, right? For understand this formula, i think you have to draw it!
$endgroup$
– Léo
Feb 19 '14 at 15:51
$begingroup$
Sorry, it's an $a^2$, as well... So it's exactly what you write, right? For understand this formula, i think you have to draw it!
$endgroup$
– Léo
Feb 19 '14 at 15:51
$begingroup$
First, when $z=+-a$, you find your circle (the meridian), and when $z$ moves, you have to make the other circle (the longitude).
$endgroup$
– Léo
Feb 19 '14 at 15:53
$begingroup$
First, when $z=+-a$, you find your circle (the meridian), and when $z$ moves, you have to make the other circle (the longitude).
$endgroup$
– Léo
Feb 19 '14 at 15:53
|
show 7 more comments
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$begingroup$
$T$ is a subset of $mathbb{R}^3$, why would it be equal to an integral?
$endgroup$
– Najib Idrissi
Aug 2 '18 at 12:51