Mixing
How to prove a map is a local diffeomorphism
Let $F:U subset R^2 rightarrow R^3$ by given by F(u,v) = $(usin{(alpha)}cos{(v)},usin{(alpha)}sin{(v)},ucos{(alpha}))$,
(u,v) $in U$ = {(u,v) $in R^2 space; u > 0$} with $alpha$ a constant.
Show F is a local diffeomorphism of U onto a cone C with the vertex at the origin and 2$alpha$ as the angle of the vertex.
- From Do Carmo, problem 4.2.1
I was thinking of using the Inverse function theorem. F is differentiable. We can compute the jacobian I suppose and show it's non-zero. I'm not sure how to construct this so it's a map onto a cone with the above properties.
Recall that the Inverse function theorem says if we have a function from $R^n rightarrow R^n$ that is continuously differentiable on some open set containing a point p, and that the determinate of Jf(a) $ne 0$ then there is some open set W containing f(a) such that $f:V rightarrow W$ has a continuous inverse $f^{-1}:W rightarrow V$ which is differentiable for all $y in W$.
Basically it says f is a diffeomorphism from V to W given it satisfies the above hypothesis.
Now above it wants us to prove it's a local diffeomorphism, not globally a diffeomorphism. That would just mean given any point p in the domain of F, there exists an open set U $subset$ dom(F) containing p such that $f:U rightarrow f(U)$ is a diffeomorphism.
differential-geometry
asked Apr 8 '15 at 9:54
Adam StaplesAdam Staples
526321
add a comment |
Let $F:U subset R^2 rightarrow R^3$ by given by F(u,v) = $(usin{(alpha)}cos{(v)},usin{(alpha)}sin{(v)},ucos{(alpha}))$,
(u,v) $in U$ = {(u,v) $in R^2 space; u > 0$} with $alpha$ a constant.
Show F is a local diffeomorphism of U onto a cone C with the vertex at the origin and 2$alpha$ as the angle of the vertex.
- From Do Carmo, problem 4.2.1
I was thinking of using the Inverse function theorem. F is differentiable. We can compute the jacobian I suppose and show it's non-zero. I'm not sure how to construct this so it's a map onto a cone with the above properties.
Recall that the Inverse function theorem says if we have a function from $R^n rightarrow R^n$ that is continuously differentiable on some open set containing a point p, and that the determinate of Jf(a) $ne 0$ then there is some open set W containing f(a) such that $f:V rightarrow W$ has a continuous inverse $f^{-1}:W rightarrow V$ which is differentiable for all $y in W$.
Basically it says f is a diffeomorphism from V to W given it satisfies the above hypothesis.
Now above it wants us to prove it's a local diffeomorphism, not globally a diffeomorphism. That would just mean given any point p in the domain of F, there exists an open set U $subset$ dom(F) containing p such that $f:U rightarrow f(U)$ is a diffeomorphism.
differential-geometry
asked Apr 8 '15 at 9:54
Adam StaplesAdam Staples
526321
Also I'm guessing F is not a local isometry since it does not include the corner point (0,0)?
– Adam Staples
Apr 8 '15 at 10:53
The corner point is not in the domain.
– Crostul
Apr 8 '15 at 11:06
yes, Do Carmo. I'll fix it now.
– Adam Staples
Apr 8 '15 at 11:41
add a comment |
Let $F:U subset R^2 rightarrow R^3$ by given by F(u,v) = $(usin{(alpha)}cos{(v)},usin{(alpha)}sin{(v)},ucos{(alpha}))$,
(u,v) $in U$ = {(u,v) $in R^2 space; u > 0$} with $alpha$ a constant.
Show F is a local diffeomorphism of U onto a cone C with the vertex at the origin and 2$alpha$ as the angle of the vertex.
- From Do Carmo, problem 4.2.1
I was thinking of using the Inverse function theorem. F is differentiable. We can compute the jacobian I suppose and show it's non-zero. I'm not sure how to construct this so it's a map onto a cone with the above properties.
Recall that the Inverse function theorem says if we have a function from $R^n rightarrow R^n$ that is continuously differentiable on some open set containing a point p, and that the determinate of Jf(a) $ne 0$ then there is some open set W containing f(a) such that $f:V rightarrow W$ has a continuous inverse $f^{-1}:W rightarrow V$ which is differentiable for all $y in W$.
Basically it says f is a diffeomorphism from V to W given it satisfies the above hypothesis.
Now above it wants us to prove it's a local diffeomorphism, not globally a diffeomorphism. That would just mean given any point p in the domain of F, there exists an open set U $subset$ dom(F) containing p such that $f:U rightarrow f(U)$ is a diffeomorphism.
differential-geometry
asked Apr 8 '15 at 9:54
Adam StaplesAdam Staples
526321
Let $F:U subset R^2 rightarrow R^3$ by given by F(u,v) = $(usin{(alpha)}cos{(v)},usin{(alpha)}sin{(v)},ucos{(alpha}))$,
(u,v) $in U$ = {(u,v) $in R^2 space; u > 0$} with $alpha$ a constant.
Show F is a local diffeomorphism of U onto a cone C with the vertex at the origin and 2$alpha$ as the angle of the vertex.
- From Do Carmo, problem 4.2.1
I was thinking of using the Inverse function theorem. F is differentiable. We can compute the jacobian I suppose and show it's non-zero. I'm not sure how to construct this so it's a map onto a cone with the above properties.
Recall that the Inverse function theorem says if we have a function from $R^n rightarrow R^n$ that is continuously differentiable on some open set containing a point p, and that the determinate of Jf(a) $ne 0$ then there is some open set W containing f(a) such that $f:V rightarrow W$ has a continuous inverse $f^{-1}:W rightarrow V$ which is differentiable for all $y in W$.
Basically it says f is a diffeomorphism from V to W given it satisfies the above hypothesis.
Now above it wants us to prove it's a local diffeomorphism, not globally a diffeomorphism. That would just mean given any point p in the domain of F, there exists an open set U $subset$ dom(F) containing p such that $f:U rightarrow f(U)$ is a diffeomorphism.
differential-geometry
differential-geometry
asked Apr 8 '15 at 9:54
Adam StaplesAdam Staples
526321
asked Apr 8 '15 at 9:54
Adam StaplesAdam Staples
526321
edited Apr 8 '15 at 11:41
Adam Staples
asked Apr 8 '15 at 9:54
Adam StaplesAdam Staples
526321
asked Apr 8 '15 at 9:54
Adam StaplesAdam Staples
526321
asked Apr 8 '15 at 9:54
Adam StaplesAdam Staples
526321
526321
Also I'm guessing F is not a local isometry since it does not include the corner point (0,0)?
– Adam Staples
Apr 8 '15 at 10:53
The corner point is not in the domain.
– Crostul
Apr 8 '15 at 11:06
yes, Do Carmo. I'll fix it now.
– Adam Staples
Apr 8 '15 at 11:41
add a comment |
Also I'm guessing F is not a local isometry since it does not include the corner point (0,0)?
– Adam Staples
Apr 8 '15 at 10:53
The corner point is not in the domain.
– Crostul
Apr 8 '15 at 11:06
yes, Do Carmo. I'll fix it now.
– Adam Staples
Apr 8 '15 at 11:41
Also I'm guessing F is not a local isometry since it does not include the corner point (0,0)?
– Adam Staples
Apr 8 '15 at 10:53
Also I'm guessing F is not a local isometry since it does not include the corner point (0,0)?
– Adam Staples
Apr 8 '15 at 10:53
The corner point is not in the domain.
– Crostul
Apr 8 '15 at 11:06
The corner point is not in the domain.
– Crostul
Apr 8 '15 at 11:06
yes, Do Carmo. I'll fix it now.
– Adam Staples
Apr 8 '15 at 11:41
yes, Do Carmo. I'll fix it now.
– Adam Staples
Apr 8 '15 at 11:41
add a comment |
2 Answers
2
active
oldest
votes
1) The Jacobian $Jac(F)(u,v)$ has rank $2$ at every point $(u,v)in U$, so that $F$ is an immersion, but certainly not an embedding: see point 3) below.
The image $F(U)$ is exactly the subset $zgt0, x^2+y^2=z^2tan ^2(alpha)$ of $mathbb R^3$.
2) This subset is indeed the intersection $Csubset mathbb R^3$ of the upper half-space $zgt 0$ with the cone of half-angle $alpha$ with vertex at the origin and axis along the $z$-axis.
That (blunted !) half-cone $C$ is a locally closed submanifold of $mathbb R^3$.
3) The coinduced morphism of differential manifolds $F_0:Uto F(U)=C$ is the universal covering map of $C$, and its restriction to any vertical line ${c}times mathbb Rsubset U ;(cgt 0)$ is the universal covering map of the horizontal circle $Ccap{z=ccos alpha} $ of $C$ .
An opinion
All in all, this exercise could have been written in a slightly more explicit way ...
answered Apr 8 '15 at 12:51
Georges ElencwajgGeorges Elencwajg
118k7180329
add a comment |
You have to do two things:
- Show that the image of $F$ is the required cone
- Show that $F$ is a local diffeomorphism
As for point 1, you can see that $F(u,v)=u (sin alpha cos v , sin alpha sin v, cos alpha)$, so it is clearly a cone (if you fix the variable $v$, and let $u in mathbb{R}$ you get a straight line through $0$, while if you fix the variable $u$ you get circles). To see that the angle of the vertex is $2 alpha$, you can use the fact that
$$F(u,v) cdot (0,0,1) = ||F(u,v)|| ||(0,0,1)|| cos( mbox{angle between $F(u,v)$ and $(0,0,1)$ } ) $$
where $cdot$ denotes the scalar product. LHS gives you $u cos alpha$, while RHS gives you $u cos( mbox{angle between $F(u,v)$ and $(0,0,1)$ } )$, so (supposing that $alpha in left( 0,pi right)$)
$$mbox{angle between $F(u,v)$ and $(0,0,1)$ } = alpha$$
As for point 2, compute the Jacobian
$$J(u,v) = left( begin{matrix}
sin alpha cos v & sin alpha sin v & cos alpha \
-u sin alpha sin v & usin alpha cos v & 0
end{matrix} right)
$$
Since $u >0$ (this hypothesis is important otherwise you would have the second row vanishing), you can see that this Jacobian has always rank $2$, because
$$
left| begin{matrix}
sin alpha cos v & sin alpha sin v \
-u sin alpha sin v & usin alpha cos v
end{matrix} right| = u sin alpha neq 0
$$
($J(u,v)$ has an invertible $2 times 2$ submatrix, so its rank is $geq 2$). Since $2$ is the maximal rank $J(u,v)$ can achieve, "some version of the inverse function theorem" tells you that $F$ is a local diffeomorphism between $mathbb{R}^2$ and the cone.
Clearly, all of this makes sense if $sin alpha neq 0$, otherwise your map would be $F(u,v)=(0,0,u cos alpha)$, which gives you a line.
answered Apr 8 '15 at 10:08
CrostulCrostul
27.7k22352
Wait, doesn't a parametrization of a cone have the form of (x,y,$(x^2+y^2)^{1/2}$)? But if you follow the calculations you get z = $sin{(alpha)}$ rather than $cos{(alpha)}$ like the parametrization of F suggests.
– Adam Staples
Apr 8 '15 at 10:18
That Jacobian doesn't seem quite right. According to the wiki article en.wikipedia.org/wiki/Jacobian_matrix_and_determinant It should be a 3 by 2 instead of a 2 by 3 (basically just transpose what you have). Also you last entry should be 0, not u*$cos{(alpha)}$. Also even though you mentioned that the Jacobian has rank 2 (I still don't see why) but I also don't see how you reduced it to a 2 by 2. I know you can only take determinants of square matrices, unlike the 3 by 2 or 2 by 3 we have.
– Adam Staples
Apr 8 '15 at 10:38
And can you elaborate how $F(u,v) cdot (0,0,1) = ||F(u,v)|| ||(0,0,1)|| cos(alpha)$ proves the angle of the vertex is 2*alpha?
– Adam Staples
Apr 8 '15 at 10:48
This is the best I can do. I am sorry: I have no references for the theorem I put between "$cdot$". I know that it is true, but I cannot find any reference on the Internet.
– Crostul
Apr 8 '15 at 11:09
mathworld.wolfram.com/Cone.html The angle of the cone in this article is $phi$ which is twice of that of the angle given by $F(u,v) cdot (0,0,1)$ so that makes sense why the angle of the vertex is $2alpha$. As for that thm. you mentioned. It looks similar to the regularity condition found on pages 54 and 55 of De Carmo. It means that the 2 columns of the Jacobian are linearly independent (the fact that the det. of at-least one submatrix of the Jacobian is non-zero).
– Adam Staples
Apr 8 '15 at 11:21
add a comment |
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2 Answers
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1) The Jacobian $Jac(F)(u,v)$ has rank $2$ at every point $(u,v)in U$, so that $F$ is an immersion, but certainly not an embedding: see point 3) below.
The image $F(U)$ is exactly the subset $zgt0, x^2+y^2=z^2tan ^2(alpha)$ of $mathbb R^3$.
2) This subset is indeed the intersection $Csubset mathbb R^3$ of the upper half-space $zgt 0$ with the cone of half-angle $alpha$ with vertex at the origin and axis along the $z$-axis.
That (blunted !) half-cone $C$ is a locally closed submanifold of $mathbb R^3$.
3) The coinduced morphism of differential manifolds $F_0:Uto F(U)=C$ is the universal covering map of $C$, and its restriction to any vertical line ${c}times mathbb Rsubset U ;(cgt 0)$ is the universal covering map of the horizontal circle $Ccap{z=ccos alpha} $ of $C$ .
An opinion
All in all, this exercise could have been written in a slightly more explicit way ...
answered Apr 8 '15 at 12:51
Georges ElencwajgGeorges Elencwajg
118k7180329
add a comment |
1) The Jacobian $Jac(F)(u,v)$ has rank $2$ at every point $(u,v)in U$, so that $F$ is an immersion, but certainly not an embedding: see point 3) below.
The image $F(U)$ is exactly the subset $zgt0, x^2+y^2=z^2tan ^2(alpha)$ of $mathbb R^3$.
2) This subset is indeed the intersection $Csubset mathbb R^3$ of the upper half-space $zgt 0$ with the cone of half-angle $alpha$ with vertex at the origin and axis along the $z$-axis.
That (blunted !) half-cone $C$ is a locally closed submanifold of $mathbb R^3$.
3) The coinduced morphism of differential manifolds $F_0:Uto F(U)=C$ is the universal covering map of $C$, and its restriction to any vertical line ${c}times mathbb Rsubset U ;(cgt 0)$ is the universal covering map of the horizontal circle $Ccap{z=ccos alpha} $ of $C$ .
An opinion
All in all, this exercise could have been written in a slightly more explicit way ...
answered Apr 8 '15 at 12:51
Georges ElencwajgGeorges Elencwajg
118k7180329
add a comment |
1) The Jacobian $Jac(F)(u,v)$ has rank $2$ at every point $(u,v)in U$, so that $F$ is an immersion, but certainly not an embedding: see point 3) below.
The image $F(U)$ is exactly the subset $zgt0, x^2+y^2=z^2tan ^2(alpha)$ of $mathbb R^3$.
2) This subset is indeed the intersection $Csubset mathbb R^3$ of the upper half-space $zgt 0$ with the cone of half-angle $alpha$ with vertex at the origin and axis along the $z$-axis.
That (blunted !) half-cone $C$ is a locally closed submanifold of $mathbb R^3$.
3) The coinduced morphism of differential manifolds $F_0:Uto F(U)=C$ is the universal covering map of $C$, and its restriction to any vertical line ${c}times mathbb Rsubset U ;(cgt 0)$ is the universal covering map of the horizontal circle $Ccap{z=ccos alpha} $ of $C$ .
An opinion
All in all, this exercise could have been written in a slightly more explicit way ...
answered Apr 8 '15 at 12:51
Georges ElencwajgGeorges Elencwajg
118k7180329
1) The Jacobian $Jac(F)(u,v)$ has rank $2$ at every point $(u,v)in U$, so that $F$ is an immersion, but certainly not an embedding: see point 3) below.
The image $F(U)$ is exactly the subset $zgt0, x^2+y^2=z^2tan ^2(alpha)$ of $mathbb R^3$.
2) This subset is indeed the intersection $Csubset mathbb R^3$ of the upper half-space $zgt 0$ with the cone of half-angle $alpha$ with vertex at the origin and axis along the $z$-axis.
That (blunted !) half-cone $C$ is a locally closed submanifold of $mathbb R^3$.
3) The coinduced morphism of differential manifolds $F_0:Uto F(U)=C$ is the universal covering map of $C$, and its restriction to any vertical line ${c}times mathbb Rsubset U ;(cgt 0)$ is the universal covering map of the horizontal circle $Ccap{z=ccos alpha} $ of $C$ .
An opinion
All in all, this exercise could have been written in a slightly more explicit way ...
answered Apr 8 '15 at 12:51
Georges ElencwajgGeorges Elencwajg
118k7180329
edited Aug 14 '18 at 21:20
answered Apr 8 '15 at 12:51
Georges ElencwajgGeorges Elencwajg
118k7180329
answered Apr 8 '15 at 12:51
Georges ElencwajgGeorges Elencwajg
118k7180329
answered Apr 8 '15 at 12:51
Georges ElencwajgGeorges Elencwajg
118k7180329
118k7180329
add a comment |
add a comment |
You have to do two things:
- Show that the image of $F$ is the required cone
- Show that $F$ is a local diffeomorphism
As for point 1, you can see that $F(u,v)=u (sin alpha cos v , sin alpha sin v, cos alpha)$, so it is clearly a cone (if you fix the variable $v$, and let $u in mathbb{R}$ you get a straight line through $0$, while if you fix the variable $u$ you get circles). To see that the angle of the vertex is $2 alpha$, you can use the fact that
$$F(u,v) cdot (0,0,1) = ||F(u,v)|| ||(0,0,1)|| cos( mbox{angle between $F(u,v)$ and $(0,0,1)$ } ) $$
where $cdot$ denotes the scalar product. LHS gives you $u cos alpha$, while RHS gives you $u cos( mbox{angle between $F(u,v)$ and $(0,0,1)$ } )$, so (supposing that $alpha in left( 0,pi right)$)
$$mbox{angle between $F(u,v)$ and $(0,0,1)$ } = alpha$$
As for point 2, compute the Jacobian
$$J(u,v) = left( begin{matrix}
sin alpha cos v & sin alpha sin v & cos alpha \
-u sin alpha sin v & usin alpha cos v & 0
end{matrix} right)
$$
Since $u >0$ (this hypothesis is important otherwise you would have the second row vanishing), you can see that this Jacobian has always rank $2$, because
$$
left| begin{matrix}
sin alpha cos v & sin alpha sin v \
-u sin alpha sin v & usin alpha cos v
end{matrix} right| = u sin alpha neq 0
$$
($J(u,v)$ has an invertible $2 times 2$ submatrix, so its rank is $geq 2$). Since $2$ is the maximal rank $J(u,v)$ can achieve, "some version of the inverse function theorem" tells you that $F$ is a local diffeomorphism between $mathbb{R}^2$ and the cone.
Clearly, all of this makes sense if $sin alpha neq 0$, otherwise your map would be $F(u,v)=(0,0,u cos alpha)$, which gives you a line.
answered Apr 8 '15 at 10:08
CrostulCrostul
27.7k22352
Wait, doesn't a parametrization of a cone have the form of (x,y,$(x^2+y^2)^{1/2}$)? But if you follow the calculations you get z = $sin{(alpha)}$ rather than $cos{(alpha)}$ like the parametrization of F suggests.
– Adam Staples
Apr 8 '15 at 10:18
That Jacobian doesn't seem quite right. According to the wiki article en.wikipedia.org/wiki/Jacobian_matrix_and_determinant It should be a 3 by 2 instead of a 2 by 3 (basically just transpose what you have). Also you last entry should be 0, not u*$cos{(alpha)}$. Also even though you mentioned that the Jacobian has rank 2 (I still don't see why) but I also don't see how you reduced it to a 2 by 2. I know you can only take determinants of square matrices, unlike the 3 by 2 or 2 by 3 we have.
– Adam Staples
Apr 8 '15 at 10:38
And can you elaborate how $F(u,v) cdot (0,0,1) = ||F(u,v)|| ||(0,0,1)|| cos(alpha)$ proves the angle of the vertex is 2*alpha?
– Adam Staples
Apr 8 '15 at 10:48
This is the best I can do. I am sorry: I have no references for the theorem I put between "$cdot$". I know that it is true, but I cannot find any reference on the Internet.
– Crostul
Apr 8 '15 at 11:09
mathworld.wolfram.com/Cone.html The angle of the cone in this article is $phi$ which is twice of that of the angle given by $F(u,v) cdot (0,0,1)$ so that makes sense why the angle of the vertex is $2alpha$. As for that thm. you mentioned. It looks similar to the regularity condition found on pages 54 and 55 of De Carmo. It means that the 2 columns of the Jacobian are linearly independent (the fact that the det. of at-least one submatrix of the Jacobian is non-zero).
– Adam Staples
Apr 8 '15 at 11:21
add a comment |
You have to do two things:
- Show that the image of $F$ is the required cone
- Show that $F$ is a local diffeomorphism
As for point 1, you can see that $F(u,v)=u (sin alpha cos v , sin alpha sin v, cos alpha)$, so it is clearly a cone (if you fix the variable $v$, and let $u in mathbb{R}$ you get a straight line through $0$, while if you fix the variable $u$ you get circles). To see that the angle of the vertex is $2 alpha$, you can use the fact that
$$F(u,v) cdot (0,0,1) = ||F(u,v)|| ||(0,0,1)|| cos( mbox{angle between $F(u,v)$ and $(0,0,1)$ } ) $$
where $cdot$ denotes the scalar product. LHS gives you $u cos alpha$, while RHS gives you $u cos( mbox{angle between $F(u,v)$ and $(0,0,1)$ } )$, so (supposing that $alpha in left( 0,pi right)$)
$$mbox{angle between $F(u,v)$ and $(0,0,1)$ } = alpha$$
As for point 2, compute the Jacobian
$$J(u,v) = left( begin{matrix}
sin alpha cos v & sin alpha sin v & cos alpha \
-u sin alpha sin v & usin alpha cos v & 0
end{matrix} right)
$$
Since $u >0$ (this hypothesis is important otherwise you would have the second row vanishing), you can see that this Jacobian has always rank $2$, because
$$
left| begin{matrix}
sin alpha cos v & sin alpha sin v \
-u sin alpha sin v & usin alpha cos v
end{matrix} right| = u sin alpha neq 0
$$
($J(u,v)$ has an invertible $2 times 2$ submatrix, so its rank is $geq 2$). Since $2$ is the maximal rank $J(u,v)$ can achieve, "some version of the inverse function theorem" tells you that $F$ is a local diffeomorphism between $mathbb{R}^2$ and the cone.
Clearly, all of this makes sense if $sin alpha neq 0$, otherwise your map would be $F(u,v)=(0,0,u cos alpha)$, which gives you a line.
answered Apr 8 '15 at 10:08
CrostulCrostul
27.7k22352
Wait, doesn't a parametrization of a cone have the form of (x,y,$(x^2+y^2)^{1/2}$)? But if you follow the calculations you get z = $sin{(alpha)}$ rather than $cos{(alpha)}$ like the parametrization of F suggests.
– Adam Staples
Apr 8 '15 at 10:18
That Jacobian doesn't seem quite right. According to the wiki article en.wikipedia.org/wiki/Jacobian_matrix_and_determinant It should be a 3 by 2 instead of a 2 by 3 (basically just transpose what you have). Also you last entry should be 0, not u*$cos{(alpha)}$. Also even though you mentioned that the Jacobian has rank 2 (I still don't see why) but I also don't see how you reduced it to a 2 by 2. I know you can only take determinants of square matrices, unlike the 3 by 2 or 2 by 3 we have.
– Adam Staples
Apr 8 '15 at 10:38
And can you elaborate how $F(u,v) cdot (0,0,1) = ||F(u,v)|| ||(0,0,1)|| cos(alpha)$ proves the angle of the vertex is 2*alpha?
– Adam Staples
Apr 8 '15 at 10:48
This is the best I can do. I am sorry: I have no references for the theorem I put between "$cdot$". I know that it is true, but I cannot find any reference on the Internet.
– Crostul
Apr 8 '15 at 11:09
mathworld.wolfram.com/Cone.html The angle of the cone in this article is $phi$ which is twice of that of the angle given by $F(u,v) cdot (0,0,1)$ so that makes sense why the angle of the vertex is $2alpha$. As for that thm. you mentioned. It looks similar to the regularity condition found on pages 54 and 55 of De Carmo. It means that the 2 columns of the Jacobian are linearly independent (the fact that the det. of at-least one submatrix of the Jacobian is non-zero).
– Adam Staples
Apr 8 '15 at 11:21
add a comment |
You have to do two things:
- Show that the image of $F$ is the required cone
- Show that $F$ is a local diffeomorphism
As for point 1, you can see that $F(u,v)=u (sin alpha cos v , sin alpha sin v, cos alpha)$, so it is clearly a cone (if you fix the variable $v$, and let $u in mathbb{R}$ you get a straight line through $0$, while if you fix the variable $u$ you get circles). To see that the angle of the vertex is $2 alpha$, you can use the fact that
$$F(u,v) cdot (0,0,1) = ||F(u,v)|| ||(0,0,1)|| cos( mbox{angle between $F(u,v)$ and $(0,0,1)$ } ) $$
where $cdot$ denotes the scalar product. LHS gives you $u cos alpha$, while RHS gives you $u cos( mbox{angle between $F(u,v)$ and $(0,0,1)$ } )$, so (supposing that $alpha in left( 0,pi right)$)
$$mbox{angle between $F(u,v)$ and $(0,0,1)$ } = alpha$$
As for point 2, compute the Jacobian
$$J(u,v) = left( begin{matrix}
sin alpha cos v & sin alpha sin v & cos alpha \
-u sin alpha sin v & usin alpha cos v & 0
end{matrix} right)
$$
Since $u >0$ (this hypothesis is important otherwise you would have the second row vanishing), you can see that this Jacobian has always rank $2$, because
$$
left| begin{matrix}
sin alpha cos v & sin alpha sin v \
-u sin alpha sin v & usin alpha cos v
end{matrix} right| = u sin alpha neq 0
$$
($J(u,v)$ has an invertible $2 times 2$ submatrix, so its rank is $geq 2$). Since $2$ is the maximal rank $J(u,v)$ can achieve, "some version of the inverse function theorem" tells you that $F$ is a local diffeomorphism between $mathbb{R}^2$ and the cone.
Clearly, all of this makes sense if $sin alpha neq 0$, otherwise your map would be $F(u,v)=(0,0,u cos alpha)$, which gives you a line.
answered Apr 8 '15 at 10:08
CrostulCrostul
27.7k22352
You have to do two things:
- Show that the image of $F$ is the required cone
- Show that $F$ is a local diffeomorphism
As for point 1, you can see that $F(u,v)=u (sin alpha cos v , sin alpha sin v, cos alpha)$, so it is clearly a cone (if you fix the variable $v$, and let $u in mathbb{R}$ you get a straight line through $0$, while if you fix the variable $u$ you get circles). To see that the angle of the vertex is $2 alpha$, you can use the fact that
$$F(u,v) cdot (0,0,1) = ||F(u,v)|| ||(0,0,1)|| cos( mbox{angle between $F(u,v)$ and $(0,0,1)$ } ) $$
where $cdot$ denotes the scalar product. LHS gives you $u cos alpha$, while RHS gives you $u cos( mbox{angle between $F(u,v)$ and $(0,0,1)$ } )$, so (supposing that $alpha in left( 0,pi right)$)
$$mbox{angle between $F(u,v)$ and $(0,0,1)$ } = alpha$$
As for point 2, compute the Jacobian
$$J(u,v) = left( begin{matrix}
sin alpha cos v & sin alpha sin v & cos alpha \
-u sin alpha sin v & usin alpha cos v & 0
end{matrix} right)
$$
Since $u >0$ (this hypothesis is important otherwise you would have the second row vanishing), you can see that this Jacobian has always rank $2$, because
$$
left| begin{matrix}
sin alpha cos v & sin alpha sin v \
-u sin alpha sin v & usin alpha cos v
end{matrix} right| = u sin alpha neq 0
$$
($J(u,v)$ has an invertible $2 times 2$ submatrix, so its rank is $geq 2$). Since $2$ is the maximal rank $J(u,v)$ can achieve, "some version of the inverse function theorem" tells you that $F$ is a local diffeomorphism between $mathbb{R}^2$ and the cone.
Clearly, all of this makes sense if $sin alpha neq 0$, otherwise your map would be $F(u,v)=(0,0,u cos alpha)$, which gives you a line.
answered Apr 8 '15 at 10:08
CrostulCrostul
27.7k22352
edited Apr 8 '15 at 11:04
answered Apr 8 '15 at 10:08
CrostulCrostul
27.7k22352
answered Apr 8 '15 at 10:08
CrostulCrostul
27.7k22352
answered Apr 8 '15 at 10:08
CrostulCrostul
27.7k22352
27.7k22352
Wait, doesn't a parametrization of a cone have the form of (x,y,$(x^2+y^2)^{1/2}$)? But if you follow the calculations you get z = $sin{(alpha)}$ rather than $cos{(alpha)}$ like the parametrization of F suggests.
– Adam Staples
Apr 8 '15 at 10:18
That Jacobian doesn't seem quite right. According to the wiki article en.wikipedia.org/wiki/Jacobian_matrix_and_determinant It should be a 3 by 2 instead of a 2 by 3 (basically just transpose what you have). Also you last entry should be 0, not u*$cos{(alpha)}$. Also even though you mentioned that the Jacobian has rank 2 (I still don't see why) but I also don't see how you reduced it to a 2 by 2. I know you can only take determinants of square matrices, unlike the 3 by 2 or 2 by 3 we have.
– Adam Staples
Apr 8 '15 at 10:38
And can you elaborate how $F(u,v) cdot (0,0,1) = ||F(u,v)|| ||(0,0,1)|| cos(alpha)$ proves the angle of the vertex is 2*alpha?
– Adam Staples
Apr 8 '15 at 10:48
This is the best I can do. I am sorry: I have no references for the theorem I put between "$cdot$". I know that it is true, but I cannot find any reference on the Internet.
– Crostul
Apr 8 '15 at 11:09
mathworld.wolfram.com/Cone.html The angle of the cone in this article is $phi$ which is twice of that of the angle given by $F(u,v) cdot (0,0,1)$ so that makes sense why the angle of the vertex is $2alpha$. As for that thm. you mentioned. It looks similar to the regularity condition found on pages 54 and 55 of De Carmo. It means that the 2 columns of the Jacobian are linearly independent (the fact that the det. of at-least one submatrix of the Jacobian is non-zero).
– Adam Staples
Apr 8 '15 at 11:21
add a comment |
Wait, doesn't a parametrization of a cone have the form of (x,y,$(x^2+y^2)^{1/2}$)? But if you follow the calculations you get z = $sin{(alpha)}$ rather than $cos{(alpha)}$ like the parametrization of F suggests.
– Adam Staples
Apr 8 '15 at 10:18
That Jacobian doesn't seem quite right. According to the wiki article en.wikipedia.org/wiki/Jacobian_matrix_and_determinant It should be a 3 by 2 instead of a 2 by 3 (basically just transpose what you have). Also you last entry should be 0, not u*$cos{(alpha)}$. Also even though you mentioned that the Jacobian has rank 2 (I still don't see why) but I also don't see how you reduced it to a 2 by 2. I know you can only take determinants of square matrices, unlike the 3 by 2 or 2 by 3 we have.
– Adam Staples
Apr 8 '15 at 10:38
And can you elaborate how $F(u,v) cdot (0,0,1) = ||F(u,v)|| ||(0,0,1)|| cos(alpha)$ proves the angle of the vertex is 2*alpha?
– Adam Staples
Apr 8 '15 at 10:48
This is the best I can do. I am sorry: I have no references for the theorem I put between "$cdot$". I know that it is true, but I cannot find any reference on the Internet.
– Crostul
Apr 8 '15 at 11:09
mathworld.wolfram.com/Cone.html The angle of the cone in this article is $phi$ which is twice of that of the angle given by $F(u,v) cdot (0,0,1)$ so that makes sense why the angle of the vertex is $2alpha$. As for that thm. you mentioned. It looks similar to the regularity condition found on pages 54 and 55 of De Carmo. It means that the 2 columns of the Jacobian are linearly independent (the fact that the det. of at-least one submatrix of the Jacobian is non-zero).
– Adam Staples
Apr 8 '15 at 11:21
Wait, doesn't a parametrization of a cone have the form of (x,y,$(x^2+y^2)^{1/2}$)? But if you follow the calculations you get z = $sin{(alpha)}$ rather than $cos{(alpha)}$ like the parametrization of F suggests.
– Adam Staples
Apr 8 '15 at 10:18
Wait, doesn't a parametrization of a cone have the form of (x,y,$(x^2+y^2)^{1/2}$)? But if you follow the calculations you get z = $sin{(alpha)}$ rather than $cos{(alpha)}$ like the parametrization of F suggests.
– Adam Staples
Apr 8 '15 at 10:18
That Jacobian doesn't seem quite right. According to the wiki article en.wikipedia.org/wiki/Jacobian_matrix_and_determinant It should be a 3 by 2 instead of a 2 by 3 (basically just transpose what you have). Also you last entry should be 0, not u*$cos{(alpha)}$. Also even though you mentioned that the Jacobian has rank 2 (I still don't see why) but I also don't see how you reduced it to a 2 by 2. I know you can only take determinants of square matrices, unlike the 3 by 2 or 2 by 3 we have.
– Adam Staples
Apr 8 '15 at 10:38
That Jacobian doesn't seem quite right. According to the wiki article en.wikipedia.org/wiki/Jacobian_matrix_and_determinant It should be a 3 by 2 instead of a 2 by 3 (basically just transpose what you have). Also you last entry should be 0, not u*$cos{(alpha)}$. Also even though you mentioned that the Jacobian has rank 2 (I still don't see why) but I also don't see how you reduced it to a 2 by 2. I know you can only take determinants of square matrices, unlike the 3 by 2 or 2 by 3 we have.
– Adam Staples
Apr 8 '15 at 10:38
And can you elaborate how $F(u,v) cdot (0,0,1) = ||F(u,v)|| ||(0,0,1)|| cos(alpha)$ proves the angle of the vertex is 2*alpha?
– Adam Staples
Apr 8 '15 at 10:48
And can you elaborate how $F(u,v) cdot (0,0,1) = ||F(u,v)|| ||(0,0,1)|| cos(alpha)$ proves the angle of the vertex is 2*alpha?
– Adam Staples
Apr 8 '15 at 10:48
This is the best I can do. I am sorry: I have no references for the theorem I put between "$cdot$". I know that it is true, but I cannot find any reference on the Internet.
– Crostul
Apr 8 '15 at 11:09
This is the best I can do. I am sorry: I have no references for the theorem I put between "$cdot$". I know that it is true, but I cannot find any reference on the Internet.
– Crostul
Apr 8 '15 at 11:09
mathworld.wolfram.com/Cone.html The angle of the cone in this article is $phi$ which is twice of that of the angle given by $F(u,v) cdot (0,0,1)$ so that makes sense why the angle of the vertex is $2alpha$. As for that thm. you mentioned. It looks similar to the regularity condition found on pages 54 and 55 of De Carmo. It means that the 2 columns of the Jacobian are linearly independent (the fact that the det. of at-least one submatrix of the Jacobian is non-zero).
– Adam Staples
Apr 8 '15 at 11:21
mathworld.wolfram.com/Cone.html The angle of the cone in this article is $phi$ which is twice of that of the angle given by $F(u,v) cdot (0,0,1)$ so that makes sense why the angle of the vertex is $2alpha$. As for that thm. you mentioned. It looks similar to the regularity condition found on pages 54 and 55 of De Carmo. It means that the 2 columns of the Jacobian are linearly independent (the fact that the det. of at-least one submatrix of the Jacobian is non-zero).
– Adam Staples
Apr 8 '15 at 11:21
add a comment |
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Also I'm guessing F is not a local isometry since it does not include the corner point (0,0)?
– Adam Staples
Apr 8 '15 at 10:53
The corner point is not in the domain.
– Crostul
Apr 8 '15 at 11:06
yes, Do Carmo. I'll fix it now.
– Adam Staples
Apr 8 '15 at 11:41