Mixing
Prove $Phi$ is a diffeomorphism
$begingroup$
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 Jan 4 at 11:57
SABOYSABOY
567311
$endgroup$
|
show 2 more comments
$begingroup$
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 Jan 4 at 11:57
SABOYSABOY
567311
$endgroup$
$begingroup$
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.
$endgroup$
– Severin Schraven
Jan 4 at 17:28
$begingroup$
So $Phi$ is not injective?
$endgroup$
– SABOY
Jan 4 at 18:30
$begingroup$
It is, but your proof is not complete.
$endgroup$
– Severin Schraven
Jan 4 at 23:15
$begingroup$
Also your proof of surjectivity is not clear at all. If you change $r$, you also change the second coordinate.
$endgroup$
– Severin Schraven
Jan 4 at 23:25
$begingroup$
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
$endgroup$
– SABOY
Jan 4 at 23:59
|
show 2 more comments
$begingroup$
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 Jan 4 at 11:57
SABOYSABOY
567311
$endgroup$
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 Jan 4 at 11:57
SABOYSABOY
567311
asked Jan 4 at 11:57
SABOYSABOY
567311
asked Jan 4 at 11:57
SABOYSABOY
567311
asked Jan 4 at 11:57
SABOYSABOY
567311
asked Jan 4 at 11:57
SABOYSABOY
567311
567311
$begingroup$
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.
$endgroup$
– Severin Schraven
Jan 4 at 17:28
$begingroup$
So $Phi$ is not injective?
$endgroup$
– SABOY
Jan 4 at 18:30
$begingroup$
It is, but your proof is not complete.
$endgroup$
– Severin Schraven
Jan 4 at 23:15
$begingroup$
Also your proof of surjectivity is not clear at all. If you change $r$, you also change the second coordinate.
$endgroup$
– Severin Schraven
Jan 4 at 23:25
$begingroup$
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
$endgroup$
– SABOY
Jan 4 at 23:59
|
show 2 more comments
$begingroup$
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.
$endgroup$
– Severin Schraven
Jan 4 at 17:28
$begingroup$
So $Phi$ is not injective?
$endgroup$
– SABOY
Jan 4 at 18:30
$begingroup$
It is, but your proof is not complete.
$endgroup$
– Severin Schraven
Jan 4 at 23:15
$begingroup$
Also your proof of surjectivity is not clear at all. If you change $r$, you also change the second coordinate.
$endgroup$
– Severin Schraven
Jan 4 at 23:25
$begingroup$
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
$endgroup$
– SABOY
Jan 4 at 23:59
$begingroup$
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.
$endgroup$
– Severin Schraven
Jan 4 at 17:28
$begingroup$
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.
$endgroup$
– Severin Schraven
Jan 4 at 17:28
$begingroup$
So $Phi$ is not injective?
$endgroup$
– SABOY
Jan 4 at 18:30
$begingroup$
So $Phi$ is not injective?
$endgroup$
– SABOY
Jan 4 at 18:30
$begingroup$
It is, but your proof is not complete.
$endgroup$
– Severin Schraven
Jan 4 at 23:15
$begingroup$
It is, but your proof is not complete.
$endgroup$
– Severin Schraven
Jan 4 at 23:15
$begingroup$
Also your proof of surjectivity is not clear at all. If you change $r$, you also change the second coordinate.
$endgroup$
– Severin Schraven
Jan 4 at 23:25
$begingroup$
Also your proof of surjectivity is not clear at all. If you change $r$, you also change the second coordinate.
$endgroup$
– Severin Schraven
Jan 4 at 23:25
$begingroup$
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
$endgroup$
– SABOY
Jan 4 at 23:59
$begingroup$
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
$endgroup$
– SABOY
Jan 4 at 23:59
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
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 Jan 5 at 0:59
Severin SchravenSeverin Schraven
6,0231934
$endgroup$
add a comment |
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$begingroup$
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 Jan 5 at 0:59
Severin SchravenSeverin Schraven
6,0231934
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 0:59
Severin SchravenSeverin Schraven
6,0231934
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 0:59
Severin SchravenSeverin Schraven
6,0231934
$endgroup$
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 Jan 5 at 0:59
Severin SchravenSeverin Schraven
6,0231934
answered Jan 5 at 0:59
Severin SchravenSeverin Schraven
6,0231934
answered Jan 5 at 0:59
Severin SchravenSeverin Schraven
6,0231934
answered Jan 5 at 0:59
Severin SchravenSeverin Schraven
6,0231934
6,0231934
add a comment |
add a comment |
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$begingroup$
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.
$endgroup$
– Severin Schraven
Jan 4 at 17:28
$begingroup$
So $Phi$ is not injective?
$endgroup$
– SABOY
Jan 4 at 18:30
$begingroup$
It is, but your proof is not complete.
$endgroup$
– Severin Schraven
Jan 4 at 23:15
$begingroup$
Also your proof of surjectivity is not clear at all. If you change $r$, you also change the second coordinate.
$endgroup$
– Severin Schraven
Jan 4 at 23:25
$begingroup$
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
$endgroup$
– SABOY
Jan 4 at 23:59