Prove $Phi$ is a diffeomorphism
Let $d in mathbb N$.
Define $Phi:]0, infty[ times { x in mathbb R^{d-1}: |x|<1}to {y in mathbb R^d:y_{1}>0}, (r,x) mapsto r(sqrt{1-|x|^2},x)$
Show that $Phi$ is a diffeomorphism.
Ideas:
$1.$ Bijectivity:
$1.1$ Injectivity:
Let $x,y in mathbb R^d$ and $r,s in ]0, infty[$ whereby $x neq y$ or $r neq s$
if $x neq y$ is clear that $Phi(r,x)neq Phi(r,y)Rightarrow$ injectivity
So let $s neq r$ and $Phi(s,x)=s(sqrt{1-|x|^2},x) neqPhi(r,x)=r(sqrt{1-|x|^2},x) Rightarrow$ injectivity
$1.2$ Surjectivity: Let $z in {y in mathbb R^d:y_{1}>0}$ it is clear that there is an $r in ]0, infty[$ so that $r sqrt{1-|x|^2}=z_{1}Rightarrow$ surjectivity
$Rightarrow$ Bijectivity
On the issue of Differentiability of $Phi$ as well as Differentiability $Phi^{-1}$ I am lost, as this is the first time doing this...
Any ideas, corrections, tips?
real-analysis derivatives diffeomorphism
asked 16 hours ago
SABOY
568311
|
show 2 more comments
Let $d in mathbb N$.
Define $Phi:]0, infty[ times { x in mathbb R^{d-1}: |x|<1}to {y in mathbb R^d:y_{1}>0}, (r,x) mapsto r(sqrt{1-|x|^2},x)$
Show that $Phi$ is a diffeomorphism.
Ideas:
$1.$ Bijectivity:
$1.1$ Injectivity:
Let $x,y in mathbb R^d$ and $r,s in ]0, infty[$ whereby $x neq y$ or $r neq s$
if $x neq y$ is clear that $Phi(r,x)neq Phi(r,y)Rightarrow$ injectivity
So let $s neq r$ and $Phi(s,x)=s(sqrt{1-|x|^2},x) neqPhi(r,x)=r(sqrt{1-|x|^2},x) Rightarrow$ injectivity
$1.2$ Surjectivity: Let $z in {y in mathbb R^d:y_{1}>0}$ it is clear that there is an $r in ]0, infty[$ so that $r sqrt{1-|x|^2}=z_{1}Rightarrow$ surjectivity
$Rightarrow$ Bijectivity
On the issue of Differentiability of $Phi$ as well as Differentiability $Phi^{-1}$ I am lost, as this is the first time doing this...
Any ideas, corrections, tips?
real-analysis derivatives diffeomorphism
asked 16 hours ago
SABOY
568311
Your injectivity proof does not work. You also have to consider $rneq s$ and $xneq y$ while $Phi(r,x)= Phi(s,y)$. For the differentability you can use the fact that a function is differentiable iff each coordinate is differentiable. An easy way to check differentiabilty of the inverse function is to use the inverse function theorem. Otherwise you have to find the inverse and check by hand that it is differentiable.
– Severin Schraven
10 hours ago
So $Phi$ is not injective?
– SABOY
9 hours ago
It is, but your proof is not complete.
– Severin Schraven
4 hours ago
Also your proof of surjectivity is not clear at all. If you change $r$, you also change the second coordinate.
– Severin Schraven
4 hours ago
On the issue of injectivity where $ r neq s$ and $x neq y$ do you mean we generate a contradiction after assuming $Phi(r,x) = Phi(s,y)$? Because assuming $r(sqrt{1-|x|^2},x)=s(sqrt{1-|y|^2},y)$ that would mean $r(sqrt{1-|x|^2})=s(sqrt{1-|y|^2})$ and that $rx_{i}=sy_{i}, forall i in {1,...,n}$. I have no idea how to generate a contradiction from this
– SABOY
3 hours ago
|
show 2 more comments
Let $d in mathbb N$.
Define $Phi:]0, infty[ times { x in mathbb R^{d-1}: |x|<1}to {y in mathbb R^d:y_{1}>0}, (r,x) mapsto r(sqrt{1-|x|^2},x)$
Show that $Phi$ is a diffeomorphism.
Ideas:
$1.$ Bijectivity:
$1.1$ Injectivity:
Let $x,y in mathbb R^d$ and $r,s in ]0, infty[$ whereby $x neq y$ or $r neq s$
if $x neq y$ is clear that $Phi(r,x)neq Phi(r,y)Rightarrow$ injectivity
So let $s neq r$ and $Phi(s,x)=s(sqrt{1-|x|^2},x) neqPhi(r,x)=r(sqrt{1-|x|^2},x) Rightarrow$ injectivity
$1.2$ Surjectivity: Let $z in {y in mathbb R^d:y_{1}>0}$ it is clear that there is an $r in ]0, infty[$ so that $r sqrt{1-|x|^2}=z_{1}Rightarrow$ surjectivity
$Rightarrow$ Bijectivity
On the issue of Differentiability of $Phi$ as well as Differentiability $Phi^{-1}$ I am lost, as this is the first time doing this...
Any ideas, corrections, tips?
real-analysis derivatives diffeomorphism
asked 16 hours ago
SABOY
568311
Let $d in mathbb N$.
Define $Phi:]0, infty[ times { x in mathbb R^{d-1}: |x|<1}to {y in mathbb R^d:y_{1}>0}, (r,x) mapsto r(sqrt{1-|x|^2},x)$
Show that $Phi$ is a diffeomorphism.
Ideas:
$1.$ Bijectivity:
$1.1$ Injectivity:
Let $x,y in mathbb R^d$ and $r,s in ]0, infty[$ whereby $x neq y$ or $r neq s$
if $x neq y$ is clear that $Phi(r,x)neq Phi(r,y)Rightarrow$ injectivity
So let $s neq r$ and $Phi(s,x)=s(sqrt{1-|x|^2},x) neqPhi(r,x)=r(sqrt{1-|x|^2},x) Rightarrow$ injectivity
$1.2$ Surjectivity: Let $z in {y in mathbb R^d:y_{1}>0}$ it is clear that there is an $r in ]0, infty[$ so that $r sqrt{1-|x|^2}=z_{1}Rightarrow$ surjectivity
$Rightarrow$ Bijectivity
On the issue of Differentiability of $Phi$ as well as Differentiability $Phi^{-1}$ I am lost, as this is the first time doing this...
Any ideas, corrections, tips?
real-analysis derivatives diffeomorphism
real-analysis derivatives diffeomorphism
asked 16 hours ago
SABOY
568311
asked 16 hours ago
SABOY
568311
asked 16 hours ago
SABOY
568311
asked 16 hours ago
SABOY
568311
asked 16 hours ago
SABOY
568311
568311
Your injectivity proof does not work. You also have to consider $rneq s$ and $xneq y$ while $Phi(r,x)= Phi(s,y)$. For the differentability you can use the fact that a function is differentiable iff each coordinate is differentiable. An easy way to check differentiabilty of the inverse function is to use the inverse function theorem. Otherwise you have to find the inverse and check by hand that it is differentiable.
– Severin Schraven
10 hours ago
So $Phi$ is not injective?
– SABOY
9 hours ago
It is, but your proof is not complete.
– Severin Schraven
4 hours ago
Also your proof of surjectivity is not clear at all. If you change $r$, you also change the second coordinate.
– Severin Schraven
4 hours ago
On the issue of injectivity where $ r neq s$ and $x neq y$ do you mean we generate a contradiction after assuming $Phi(r,x) = Phi(s,y)$? Because assuming $r(sqrt{1-|x|^2},x)=s(sqrt{1-|y|^2},y)$ that would mean $r(sqrt{1-|x|^2})=s(sqrt{1-|y|^2})$ and that $rx_{i}=sy_{i}, forall i in {1,...,n}$. I have no idea how to generate a contradiction from this
– SABOY
3 hours ago
|
show 2 more comments
Your injectivity proof does not work. You also have to consider $rneq s$ and $xneq y$ while $Phi(r,x)= Phi(s,y)$. For the differentability you can use the fact that a function is differentiable iff each coordinate is differentiable. An easy way to check differentiabilty of the inverse function is to use the inverse function theorem. Otherwise you have to find the inverse and check by hand that it is differentiable.
– Severin Schraven
10 hours ago
So $Phi$ is not injective?
– SABOY
9 hours ago
It is, but your proof is not complete.
– Severin Schraven
4 hours ago
Also your proof of surjectivity is not clear at all. If you change $r$, you also change the second coordinate.
– Severin Schraven
4 hours ago
On the issue of injectivity where $ r neq s$ and $x neq y$ do you mean we generate a contradiction after assuming $Phi(r,x) = Phi(s,y)$? Because assuming $r(sqrt{1-|x|^2},x)=s(sqrt{1-|y|^2},y)$ that would mean $r(sqrt{1-|x|^2})=s(sqrt{1-|y|^2})$ and that $rx_{i}=sy_{i}, forall i in {1,...,n}$. I have no idea how to generate a contradiction from this
– SABOY
3 hours ago
Your injectivity proof does not work. You also have to consider $rneq s$ and $xneq y$ while $Phi(r,x)= Phi(s,y)$. For the differentability you can use the fact that a function is differentiable iff each coordinate is differentiable. An easy way to check differentiabilty of the inverse function is to use the inverse function theorem. Otherwise you have to find the inverse and check by hand that it is differentiable.
– Severin Schraven
10 hours ago
Your injectivity proof does not work. You also have to consider $rneq s$ and $xneq y$ while $Phi(r,x)= Phi(s,y)$. For the differentability you can use the fact that a function is differentiable iff each coordinate is differentiable. An easy way to check differentiabilty of the inverse function is to use the inverse function theorem. Otherwise you have to find the inverse and check by hand that it is differentiable.
– Severin Schraven
10 hours ago
So $Phi$ is not injective?
– SABOY
9 hours ago
So $Phi$ is not injective?
– SABOY
9 hours ago
It is, but your proof is not complete.
– Severin Schraven
4 hours ago
It is, but your proof is not complete.
– Severin Schraven
4 hours ago
Also your proof of surjectivity is not clear at all. If you change $r$, you also change the second coordinate.
– Severin Schraven
4 hours ago
Also your proof of surjectivity is not clear at all. If you change $r$, you also change the second coordinate.
– Severin Schraven
4 hours ago
On the issue of injectivity where $ r neq s$ and $x neq y$ do you mean we generate a contradiction after assuming $Phi(r,x) = Phi(s,y)$? Because assuming $r(sqrt{1-|x|^2},x)=s(sqrt{1-|y|^2},y)$ that would mean $r(sqrt{1-|x|^2})=s(sqrt{1-|y|^2})$ and that $rx_{i}=sy_{i}, forall i in {1,...,n}$. I have no idea how to generate a contradiction from this
– SABOY
3 hours ago
On the issue of injectivity where $ r neq s$ and $x neq y$ do you mean we generate a contradiction after assuming $Phi(r,x) = Phi(s,y)$? Because assuming $r(sqrt{1-|x|^2},x)=s(sqrt{1-|y|^2},y)$ that would mean $r(sqrt{1-|x|^2})=s(sqrt{1-|y|^2})$ and that $rx_{i}=sy_{i}, forall i in {1,...,n}$. I have no idea how to generate a contradiction from this
– SABOY
3 hours ago
|
show 2 more comments
1 Answer
1
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oldest
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Hint: The easiest way is to find the inverse function. For this we compute
$$ vert Phi(r,x) vert = r $$
as
$$ vert Phi(r,x) vert = vert r vert sqrt{(1-vert x vert^2)+ x_1^2 + dots + x_{d_1}^2} = vert r vert sqrt{1- vert x vert^2 + vert x vert^2} = vert r vert = r. $$
Hence, the inverse function is (for $y=(y_1, dots, y_d)$)
$$ Psi (y_1, dots, y_d) = left(vert y vert, frac{1}{vert y vert} y_2, dots, frac{1}{vert y vert} y_d right).$$
answered 2 hours ago
Severin Schraven
5,8281934
add a comment |
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Hint: The easiest way is to find the inverse function. For this we compute
$$ vert Phi(r,x) vert = r $$
as
$$ vert Phi(r,x) vert = vert r vert sqrt{(1-vert x vert^2)+ x_1^2 + dots + x_{d_1}^2} = vert r vert sqrt{1- vert x vert^2 + vert x vert^2} = vert r vert = r. $$
Hence, the inverse function is (for $y=(y_1, dots, y_d)$)
$$ Psi (y_1, dots, y_d) = left(vert y vert, frac{1}{vert y vert} y_2, dots, frac{1}{vert y vert} y_d right).$$
answered 2 hours ago
Severin Schraven
5,8281934
add a comment |
Hint: The easiest way is to find the inverse function. For this we compute
$$ vert Phi(r,x) vert = r $$
as
$$ vert Phi(r,x) vert = vert r vert sqrt{(1-vert x vert^2)+ x_1^2 + dots + x_{d_1}^2} = vert r vert sqrt{1- vert x vert^2 + vert x vert^2} = vert r vert = r. $$
Hence, the inverse function is (for $y=(y_1, dots, y_d)$)
$$ Psi (y_1, dots, y_d) = left(vert y vert, frac{1}{vert y vert} y_2, dots, frac{1}{vert y vert} y_d right).$$
answered 2 hours ago
Severin Schraven
5,8281934
add a comment |
Hint: The easiest way is to find the inverse function. For this we compute
$$ vert Phi(r,x) vert = r $$
as
$$ vert Phi(r,x) vert = vert r vert sqrt{(1-vert x vert^2)+ x_1^2 + dots + x_{d_1}^2} = vert r vert sqrt{1- vert x vert^2 + vert x vert^2} = vert r vert = r. $$
Hence, the inverse function is (for $y=(y_1, dots, y_d)$)
$$ Psi (y_1, dots, y_d) = left(vert y vert, frac{1}{vert y vert} y_2, dots, frac{1}{vert y vert} y_d right).$$
answered 2 hours ago
Severin Schraven
5,8281934
Hint: The easiest way is to find the inverse function. For this we compute
$$ vert Phi(r,x) vert = r $$
as
$$ vert Phi(r,x) vert = vert r vert sqrt{(1-vert x vert^2)+ x_1^2 + dots + x_{d_1}^2} = vert r vert sqrt{1- vert x vert^2 + vert x vert^2} = vert r vert = r. $$
Hence, the inverse function is (for $y=(y_1, dots, y_d)$)
$$ Psi (y_1, dots, y_d) = left(vert y vert, frac{1}{vert y vert} y_2, dots, frac{1}{vert y vert} y_d right).$$
answered 2 hours ago
Severin Schraven
5,8281934
answered 2 hours ago
Severin Schraven
5,8281934
answered 2 hours ago
Severin Schraven
5,8281934
answered 2 hours ago
Severin Schraven
5,8281934
5,8281934
add a comment |
add a comment |
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Your injectivity proof does not work. You also have to consider $rneq s$ and $xneq y$ while $Phi(r,x)= Phi(s,y)$. For the differentability you can use the fact that a function is differentiable iff each coordinate is differentiable. An easy way to check differentiabilty of the inverse function is to use the inverse function theorem. Otherwise you have to find the inverse and check by hand that it is differentiable.
– Severin Schraven
10 hours ago
So $Phi$ is not injective?
– SABOY
9 hours ago
It is, but your proof is not complete.
– Severin Schraven
4 hours ago
Also your proof of surjectivity is not clear at all. If you change $r$, you also change the second coordinate.
– Severin Schraven
4 hours ago
On the issue of injectivity where $ r neq s$ and $x neq y$ do you mean we generate a contradiction after assuming $Phi(r,x) = Phi(s,y)$? Because assuming $r(sqrt{1-|x|^2},x)=s(sqrt{1-|y|^2},y)$ that would mean $r(sqrt{1-|x|^2})=s(sqrt{1-|y|^2})$ and that $rx_{i}=sy_{i}, forall i in {1,...,n}$. I have no idea how to generate a contradiction from this
– SABOY
3 hours ago