On the regularity of any reparametrezation of a regular curve The Next CEO of Stack...
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On the regularity of any reparametrezation of a regular curve
The Next CEO of Stack OverflowImmersed curve in $mathbb{R}^2$ and regular curveProof check for critical point definition with mean curvatureAny reparametrisation of a regular curve is regularHow can the vectors be non-zero and linearly independent?Reparametrization ConfusionPath, Curve, Contour in Complex AnalysisChecking the composition reparametrizations of a curve is a reparametrizationShowing that a regular curve is a pregeodesicChain Rule for Surface DifferentialsRealizing any vector in tangent space as tangent vector
$begingroup$
According to Differential Geometry by Pressley:
$mathsf{Proposition}:mathsf{1.3}$
Any reparametrisation of a regular curve is regular.
Proof $it{1.3}$
Suppose that $gamma$ and $tilde{gamma}$ are related as in Definition 1.5, let $t=phi(tilde{t})$, and let $psi=phi^{-1}$ so that $tilde{t}=psi(t)$. Differentiating both sides of the equation $phi(psi(t))=t$ with respect to $t$ and using the chain rule gives $${{raise{0.3pt}{dphi}}above{0.5pt}{dtilde{t}}} frac{dpsi}{lower{1pt}{dt}}=1.$$ This shows that $dphi/dtilde{t}$ is never zero. Since $tilde{gamma}(tilde{t})=gamma(phi(tilde{t}))$, another application of the chain rule gives $${{raise{0.3pt}{dtilde{gamma}}}above{0.5pt}{dtilde{t}}}=frac{dgamma}{lower{1pt}{dt}}{{raise{0.3pt}{dphi}}above{0.5pt}{dtilde{t}}},$$ which shows that $dtilde{gamma}/dtilde{t}$ is never zero if $dgamma/dt$ is never zero. $hspace{6.2cm}square$
Now suppose the parabola $y=x^2$ and the parametrazation $gamma(t) = (t, t^2)$ and also reparametrazation $delta(t)=(t^3, t^6)$. Obviously the function $phi(t)=t^3$ is strictly monotone and continuously differentiable. Nonetheless, $gamma(t)$ is regular but $delta(t)$ is not! What did I do wrong? Or is the proof incorrect?
Maybe the following definition from the same book is wrong; that is, it must include that $dphi/dt $ is never zero in all its domain?
$mathsf{Definition}:mathsf{1.5}$
A parametrised curve $tilde{gamma}:(tilde{alpha},tilde{beta})tomathbf{R}^n$ is a reparametrisation of a parametrised curve $gamma:(alpha,beta)tomathbf{R}^n$ if there is a smooth bijective map $phi:(tilde{alpha},tilde{beta})to(alpha,beta)$ (the reparametrisation map) such that the inverse map $phi^{-1}:(alpha,beta)to(tilde{alpha},tilde{beta})$ is also smooth and $$tilde{gamma}(tilde{t})=gamma(phi(tilde{t}))::text{for all}::tilde{t}in(tilde{alpha},tilde{beta}).$$
differential-geometry examples-counterexamples curves parametrization
edited Mar 17 at 6:27
Robert Howard
2,2933935
$endgroup$
add a comment |
$begingroup$
According to Differential Geometry by Pressley:
$mathsf{Proposition}:mathsf{1.3}$
Any reparametrisation of a regular curve is regular.
Proof $it{1.3}$
Suppose that $gamma$ and $tilde{gamma}$ are related as in Definition 1.5, let $t=phi(tilde{t})$, and let $psi=phi^{-1}$ so that $tilde{t}=psi(t)$. Differentiating both sides of the equation $phi(psi(t))=t$ with respect to $t$ and using the chain rule gives $${{raise{0.3pt}{dphi}}above{0.5pt}{dtilde{t}}} frac{dpsi}{lower{1pt}{dt}}=1.$$ This shows that $dphi/dtilde{t}$ is never zero. Since $tilde{gamma}(tilde{t})=gamma(phi(tilde{t}))$, another application of the chain rule gives $${{raise{0.3pt}{dtilde{gamma}}}above{0.5pt}{dtilde{t}}}=frac{dgamma}{lower{1pt}{dt}}{{raise{0.3pt}{dphi}}above{0.5pt}{dtilde{t}}},$$ which shows that $dtilde{gamma}/dtilde{t}$ is never zero if $dgamma/dt$ is never zero. $hspace{6.2cm}square$
Now suppose the parabola $y=x^2$ and the parametrazation $gamma(t) = (t, t^2)$ and also reparametrazation $delta(t)=(t^3, t^6)$. Obviously the function $phi(t)=t^3$ is strictly monotone and continuously differentiable. Nonetheless, $gamma(t)$ is regular but $delta(t)$ is not! What did I do wrong? Or is the proof incorrect?
Maybe the following definition from the same book is wrong; that is, it must include that $dphi/dt $ is never zero in all its domain?
$mathsf{Definition}:mathsf{1.5}$
A parametrised curve $tilde{gamma}:(tilde{alpha},tilde{beta})tomathbf{R}^n$ is a reparametrisation of a parametrised curve $gamma:(alpha,beta)tomathbf{R}^n$ if there is a smooth bijective map $phi:(tilde{alpha},tilde{beta})to(alpha,beta)$ (the reparametrisation map) such that the inverse map $phi^{-1}:(alpha,beta)to(tilde{alpha},tilde{beta})$ is also smooth and $$tilde{gamma}(tilde{t})=gamma(phi(tilde{t}))::text{for all}::tilde{t}in(tilde{alpha},tilde{beta}).$$
differential-geometry examples-counterexamples curves parametrization
edited Mar 17 at 6:27
Robert Howard
2,2933935
$endgroup$
add a comment |
$begingroup$
According to Differential Geometry by Pressley:
$mathsf{Proposition}:mathsf{1.3}$
Any reparametrisation of a regular curve is regular.
Proof $it{1.3}$
Suppose that $gamma$ and $tilde{gamma}$ are related as in Definition 1.5, let $t=phi(tilde{t})$, and let $psi=phi^{-1}$ so that $tilde{t}=psi(t)$. Differentiating both sides of the equation $phi(psi(t))=t$ with respect to $t$ and using the chain rule gives $${{raise{0.3pt}{dphi}}above{0.5pt}{dtilde{t}}} frac{dpsi}{lower{1pt}{dt}}=1.$$ This shows that $dphi/dtilde{t}$ is never zero. Since $tilde{gamma}(tilde{t})=gamma(phi(tilde{t}))$, another application of the chain rule gives $${{raise{0.3pt}{dtilde{gamma}}}above{0.5pt}{dtilde{t}}}=frac{dgamma}{lower{1pt}{dt}}{{raise{0.3pt}{dphi}}above{0.5pt}{dtilde{t}}},$$ which shows that $dtilde{gamma}/dtilde{t}$ is never zero if $dgamma/dt$ is never zero. $hspace{6.2cm}square$
Now suppose the parabola $y=x^2$ and the parametrazation $gamma(t) = (t, t^2)$ and also reparametrazation $delta(t)=(t^3, t^6)$. Obviously the function $phi(t)=t^3$ is strictly monotone and continuously differentiable. Nonetheless, $gamma(t)$ is regular but $delta(t)$ is not! What did I do wrong? Or is the proof incorrect?
Maybe the following definition from the same book is wrong; that is, it must include that $dphi/dt $ is never zero in all its domain?
$mathsf{Definition}:mathsf{1.5}$
A parametrised curve $tilde{gamma}:(tilde{alpha},tilde{beta})tomathbf{R}^n$ is a reparametrisation of a parametrised curve $gamma:(alpha,beta)tomathbf{R}^n$ if there is a smooth bijective map $phi:(tilde{alpha},tilde{beta})to(alpha,beta)$ (the reparametrisation map) such that the inverse map $phi^{-1}:(alpha,beta)to(tilde{alpha},tilde{beta})$ is also smooth and $$tilde{gamma}(tilde{t})=gamma(phi(tilde{t}))::text{for all}::tilde{t}in(tilde{alpha},tilde{beta}).$$
differential-geometry examples-counterexamples curves parametrization
edited Mar 17 at 6:27
Robert Howard
2,2933935
$endgroup$
According to Differential Geometry by Pressley:
$mathsf{Proposition}:mathsf{1.3}$
Any reparametrisation of a regular curve is regular.
Proof $it{1.3}$
Suppose that $gamma$ and $tilde{gamma}$ are related as in Definition 1.5, let $t=phi(tilde{t})$, and let $psi=phi^{-1}$ so that $tilde{t}=psi(t)$. Differentiating both sides of the equation $phi(psi(t))=t$ with respect to $t$ and using the chain rule gives $${{raise{0.3pt}{dphi}}above{0.5pt}{dtilde{t}}} frac{dpsi}{lower{1pt}{dt}}=1.$$ This shows that $dphi/dtilde{t}$ is never zero. Since $tilde{gamma}(tilde{t})=gamma(phi(tilde{t}))$, another application of the chain rule gives $${{raise{0.3pt}{dtilde{gamma}}}above{0.5pt}{dtilde{t}}}=frac{dgamma}{lower{1pt}{dt}}{{raise{0.3pt}{dphi}}above{0.5pt}{dtilde{t}}},$$ which shows that $dtilde{gamma}/dtilde{t}$ is never zero if $dgamma/dt$ is never zero. $hspace{6.2cm}square$
Now suppose the parabola $y=x^2$ and the parametrazation $gamma(t) = (t, t^2)$ and also reparametrazation $delta(t)=(t^3, t^6)$. Obviously the function $phi(t)=t^3$ is strictly monotone and continuously differentiable. Nonetheless, $gamma(t)$ is regular but $delta(t)$ is not! What did I do wrong? Or is the proof incorrect?
Maybe the following definition from the same book is wrong; that is, it must include that $dphi/dt $ is never zero in all its domain?
$mathsf{Definition}:mathsf{1.5}$
A parametrised curve $tilde{gamma}:(tilde{alpha},tilde{beta})tomathbf{R}^n$ is a reparametrisation of a parametrised curve $gamma:(alpha,beta)tomathbf{R}^n$ if there is a smooth bijective map $phi:(tilde{alpha},tilde{beta})to(alpha,beta)$ (the reparametrisation map) such that the inverse map $phi^{-1}:(alpha,beta)to(tilde{alpha},tilde{beta})$ is also smooth and $$tilde{gamma}(tilde{t})=gamma(phi(tilde{t}))::text{for all}::tilde{t}in(tilde{alpha},tilde{beta}).$$
differential-geometry examples-counterexamples curves parametrization
differential-geometry examples-counterexamples curves parametrization
edited Mar 17 at 6:27
Robert Howard
2,2933935
edited Mar 17 at 6:27
Robert Howard
2,2933935
edited Mar 17 at 6:27
Robert Howard
2,2933935
edited Mar 17 at 6:27
Robert Howard
2,2933935
edited Mar 17 at 6:27
Robert Howard
2,2933935
2,2933935
asked Sep 10 '18 at 10:54
user231343
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, if $phi$ is a diffeomorphism (a smooth bijection with a smooth inverse) then $frac{dphi}{dt} ne 0$.
This is because $phi circ phi^{-1} = text{id}$ so $$1 = frac{d(text{id})}{dt} = d(phi^{-1} circ phi) = frac{dphi^{-1}}{dphi} frac{dphi}{dt}$$
In particular we see that $frac{dphi}{dt} ne 0$.
answered Sep 10 '18 at 11:03
mechanodroidmechanodroid
29k62648
$endgroup$
$begingroup$
Am I right to say that 1) if derivative of a contentiously differentiable function is nonzero at some point its inverse exists at that point and it is contentiously differentiable for some interval about that point? 2) The converse of the (1) holds?
$endgroup$
– user231343
Sep 10 '18 at 11:10
1
$begingroup$
@Edi Yes, the first statement is a consequence of the inverse mapping theorem. The converse is also true: if $phi$ is a smooth bijection around $t_0$ with smooth inverse then $dotphi(t_0) ne 0$, as the above discussion shows.
$endgroup$
– mechanodroid
Sep 10 '18 at 11:14
add a comment |
$begingroup$
See definition 1.3.1 in Pressley's textbook:
For him, a reparametrization is a smooth bijective map whose inverse is also smooth. That does not occur with $tmapsto t^3$.
answered Sep 10 '18 at 11:00
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
there is no "definition 1.3.1" in the book (springer.com/gp/book/9781848828902)! But the same definition the Pressley gives and I don't understand why $tmapsto t^3$ is not bijective or smooth
$endgroup$
– user231343
Sep 10 '18 at 11:03
1
$begingroup$
@Edi Yes, there is. I have it before my eyes. And $tmapstosqrt[3]t$ is not smooth.
$endgroup$
– José Carlos Santos
Sep 10 '18 at 11:09
$begingroup$
Haha it's Def. 1.5 in the edition that I am reading (same definition)!
$endgroup$
– user231343
Sep 10 '18 at 11:12
$begingroup$
Following the comment below @mechanodroid's answer, I think the inverse of of $tmapsto t^3$ is not smooth because d/dt (t^3) is not never-zero?
$endgroup$
– user231343
Sep 10 '18 at 11:15
1
$begingroup$
@Edi Yes, that would be correct.
$endgroup$
– José Carlos Santos
Sep 10 '18 at 11:16
add a comment |
Your Answer
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, if $phi$ is a diffeomorphism (a smooth bijection with a smooth inverse) then $frac{dphi}{dt} ne 0$.
This is because $phi circ phi^{-1} = text{id}$ so $$1 = frac{d(text{id})}{dt} = d(phi^{-1} circ phi) = frac{dphi^{-1}}{dphi} frac{dphi}{dt}$$
In particular we see that $frac{dphi}{dt} ne 0$.
answered Sep 10 '18 at 11:03
mechanodroidmechanodroid
29k62648
$endgroup$
$begingroup$
Am I right to say that 1) if derivative of a contentiously differentiable function is nonzero at some point its inverse exists at that point and it is contentiously differentiable for some interval about that point? 2) The converse of the (1) holds?
$endgroup$
– user231343
Sep 10 '18 at 11:10
1
$begingroup$
@Edi Yes, the first statement is a consequence of the inverse mapping theorem. The converse is also true: if $phi$ is a smooth bijection around $t_0$ with smooth inverse then $dotphi(t_0) ne 0$, as the above discussion shows.
$endgroup$
– mechanodroid
Sep 10 '18 at 11:14
add a comment |
$begingroup$
Yes, if $phi$ is a diffeomorphism (a smooth bijection with a smooth inverse) then $frac{dphi}{dt} ne 0$.
This is because $phi circ phi^{-1} = text{id}$ so $$1 = frac{d(text{id})}{dt} = d(phi^{-1} circ phi) = frac{dphi^{-1}}{dphi} frac{dphi}{dt}$$
In particular we see that $frac{dphi}{dt} ne 0$.
answered Sep 10 '18 at 11:03
mechanodroidmechanodroid
29k62648
$endgroup$
$begingroup$
Am I right to say that 1) if derivative of a contentiously differentiable function is nonzero at some point its inverse exists at that point and it is contentiously differentiable for some interval about that point? 2) The converse of the (1) holds?
$endgroup$
– user231343
Sep 10 '18 at 11:10
1
$begingroup$
@Edi Yes, the first statement is a consequence of the inverse mapping theorem. The converse is also true: if $phi$ is a smooth bijection around $t_0$ with smooth inverse then $dotphi(t_0) ne 0$, as the above discussion shows.
$endgroup$
– mechanodroid
Sep 10 '18 at 11:14
add a comment |
$begingroup$
Yes, if $phi$ is a diffeomorphism (a smooth bijection with a smooth inverse) then $frac{dphi}{dt} ne 0$.
This is because $phi circ phi^{-1} = text{id}$ so $$1 = frac{d(text{id})}{dt} = d(phi^{-1} circ phi) = frac{dphi^{-1}}{dphi} frac{dphi}{dt}$$
In particular we see that $frac{dphi}{dt} ne 0$.
answered Sep 10 '18 at 11:03
mechanodroidmechanodroid
29k62648
$endgroup$
Yes, if $phi$ is a diffeomorphism (a smooth bijection with a smooth inverse) then $frac{dphi}{dt} ne 0$.
This is because $phi circ phi^{-1} = text{id}$ so $$1 = frac{d(text{id})}{dt} = d(phi^{-1} circ phi) = frac{dphi^{-1}}{dphi} frac{dphi}{dt}$$
In particular we see that $frac{dphi}{dt} ne 0$.
answered Sep 10 '18 at 11:03
mechanodroidmechanodroid
29k62648
answered Sep 10 '18 at 11:03
mechanodroidmechanodroid
29k62648
answered Sep 10 '18 at 11:03
mechanodroidmechanodroid
29k62648
answered Sep 10 '18 at 11:03
mechanodroidmechanodroid
29k62648
29k62648
$begingroup$
Am I right to say that 1) if derivative of a contentiously differentiable function is nonzero at some point its inverse exists at that point and it is contentiously differentiable for some interval about that point? 2) The converse of the (1) holds?
$endgroup$
– user231343
Sep 10 '18 at 11:10
1
$begingroup$
@Edi Yes, the first statement is a consequence of the inverse mapping theorem. The converse is also true: if $phi$ is a smooth bijection around $t_0$ with smooth inverse then $dotphi(t_0) ne 0$, as the above discussion shows.
$endgroup$
– mechanodroid
Sep 10 '18 at 11:14
add a comment |
$begingroup$
Am I right to say that 1) if derivative of a contentiously differentiable function is nonzero at some point its inverse exists at that point and it is contentiously differentiable for some interval about that point? 2) The converse of the (1) holds?
$endgroup$
– user231343
Sep 10 '18 at 11:10
1
$begingroup$
@Edi Yes, the first statement is a consequence of the inverse mapping theorem. The converse is also true: if $phi$ is a smooth bijection around $t_0$ with smooth inverse then $dotphi(t_0) ne 0$, as the above discussion shows.
$endgroup$
– mechanodroid
Sep 10 '18 at 11:14
$begingroup$
Am I right to say that 1) if derivative of a contentiously differentiable function is nonzero at some point its inverse exists at that point and it is contentiously differentiable for some interval about that point? 2) The converse of the (1) holds?
$endgroup$
– user231343
Sep 10 '18 at 11:10
$begingroup$
Am I right to say that 1) if derivative of a contentiously differentiable function is nonzero at some point its inverse exists at that point and it is contentiously differentiable for some interval about that point? 2) The converse of the (1) holds?
$endgroup$
– user231343
Sep 10 '18 at 11:10
1
1
$begingroup$
@Edi Yes, the first statement is a consequence of the inverse mapping theorem. The converse is also true: if $phi$ is a smooth bijection around $t_0$ with smooth inverse then $dotphi(t_0) ne 0$, as the above discussion shows.
$endgroup$
– mechanodroid
Sep 10 '18 at 11:14
$begingroup$
@Edi Yes, the first statement is a consequence of the inverse mapping theorem. The converse is also true: if $phi$ is a smooth bijection around $t_0$ with smooth inverse then $dotphi(t_0) ne 0$, as the above discussion shows.
$endgroup$
– mechanodroid
Sep 10 '18 at 11:14
add a comment |
$begingroup$
See definition 1.3.1 in Pressley's textbook:
For him, a reparametrization is a smooth bijective map whose inverse is also smooth. That does not occur with $tmapsto t^3$.
answered Sep 10 '18 at 11:00
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
there is no "definition 1.3.1" in the book (springer.com/gp/book/9781848828902)! But the same definition the Pressley gives and I don't understand why $tmapsto t^3$ is not bijective or smooth
$endgroup$
– user231343
Sep 10 '18 at 11:03
1
$begingroup$
@Edi Yes, there is. I have it before my eyes. And $tmapstosqrt[3]t$ is not smooth.
$endgroup$
– José Carlos Santos
Sep 10 '18 at 11:09
$begingroup$
Haha it's Def. 1.5 in the edition that I am reading (same definition)!
$endgroup$
– user231343
Sep 10 '18 at 11:12
$begingroup$
Following the comment below @mechanodroid's answer, I think the inverse of of $tmapsto t^3$ is not smooth because d/dt (t^3) is not never-zero?
$endgroup$
– user231343
Sep 10 '18 at 11:15
1
$begingroup$
@Edi Yes, that would be correct.
$endgroup$
– José Carlos Santos
Sep 10 '18 at 11:16
add a comment |
$begingroup$
See definition 1.3.1 in Pressley's textbook:
For him, a reparametrization is a smooth bijective map whose inverse is also smooth. That does not occur with $tmapsto t^3$.
answered Sep 10 '18 at 11:00
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
there is no "definition 1.3.1" in the book (springer.com/gp/book/9781848828902)! But the same definition the Pressley gives and I don't understand why $tmapsto t^3$ is not bijective or smooth
$endgroup$
– user231343
Sep 10 '18 at 11:03
1
$begingroup$
@Edi Yes, there is. I have it before my eyes. And $tmapstosqrt[3]t$ is not smooth.
$endgroup$
– José Carlos Santos
Sep 10 '18 at 11:09
$begingroup$
Haha it's Def. 1.5 in the edition that I am reading (same definition)!
$endgroup$
– user231343
Sep 10 '18 at 11:12
$begingroup$
Following the comment below @mechanodroid's answer, I think the inverse of of $tmapsto t^3$ is not smooth because d/dt (t^3) is not never-zero?
$endgroup$
– user231343
Sep 10 '18 at 11:15
1
$begingroup$
@Edi Yes, that would be correct.
$endgroup$
– José Carlos Santos
Sep 10 '18 at 11:16
add a comment |
$begingroup$
See definition 1.3.1 in Pressley's textbook:
For him, a reparametrization is a smooth bijective map whose inverse is also smooth. That does not occur with $tmapsto t^3$.
answered Sep 10 '18 at 11:00
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
See definition 1.3.1 in Pressley's textbook:
For him, a reparametrization is a smooth bijective map whose inverse is also smooth. That does not occur with $tmapsto t^3$.
answered Sep 10 '18 at 11:00
José Carlos SantosJosé Carlos Santos
171k23132240
edited Sep 10 '18 at 11:08
answered Sep 10 '18 at 11:00
José Carlos SantosJosé Carlos Santos
171k23132240
answered Sep 10 '18 at 11:00
José Carlos SantosJosé Carlos Santos
171k23132240
answered Sep 10 '18 at 11:00
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
$begingroup$
there is no "definition 1.3.1" in the book (springer.com/gp/book/9781848828902)! But the same definition the Pressley gives and I don't understand why $tmapsto t^3$ is not bijective or smooth
$endgroup$
– user231343
Sep 10 '18 at 11:03
1
$begingroup$
@Edi Yes, there is. I have it before my eyes. And $tmapstosqrt[3]t$ is not smooth.
$endgroup$
– José Carlos Santos
Sep 10 '18 at 11:09
$begingroup$
Haha it's Def. 1.5 in the edition that I am reading (same definition)!
$endgroup$
– user231343
Sep 10 '18 at 11:12
$begingroup$
Following the comment below @mechanodroid's answer, I think the inverse of of $tmapsto t^3$ is not smooth because d/dt (t^3) is not never-zero?
$endgroup$
– user231343
Sep 10 '18 at 11:15
1
$begingroup$
@Edi Yes, that would be correct.
$endgroup$
– José Carlos Santos
Sep 10 '18 at 11:16
add a comment |
$begingroup$
there is no "definition 1.3.1" in the book (springer.com/gp/book/9781848828902)! But the same definition the Pressley gives and I don't understand why $tmapsto t^3$ is not bijective or smooth
$endgroup$
– user231343
Sep 10 '18 at 11:03
1
$begingroup$
@Edi Yes, there is. I have it before my eyes. And $tmapstosqrt[3]t$ is not smooth.
$endgroup$
– José Carlos Santos
Sep 10 '18 at 11:09
$begingroup$
Haha it's Def. 1.5 in the edition that I am reading (same definition)!
$endgroup$
– user231343
Sep 10 '18 at 11:12
$begingroup$
Following the comment below @mechanodroid's answer, I think the inverse of of $tmapsto t^3$ is not smooth because d/dt (t^3) is not never-zero?
$endgroup$
– user231343
Sep 10 '18 at 11:15
1
$begingroup$
@Edi Yes, that would be correct.
$endgroup$
– José Carlos Santos
Sep 10 '18 at 11:16
$begingroup$
there is no "definition 1.3.1" in the book (springer.com/gp/book/9781848828902)! But the same definition the Pressley gives and I don't understand why $tmapsto t^3$ is not bijective or smooth
$endgroup$
– user231343
Sep 10 '18 at 11:03
$begingroup$
there is no "definition 1.3.1" in the book (springer.com/gp/book/9781848828902)! But the same definition the Pressley gives and I don't understand why $tmapsto t^3$ is not bijective or smooth
$endgroup$
– user231343
Sep 10 '18 at 11:03
1
1
$begingroup$
@Edi Yes, there is. I have it before my eyes. And $tmapstosqrt[3]t$ is not smooth.
$endgroup$
– José Carlos Santos
Sep 10 '18 at 11:09
$begingroup$
@Edi Yes, there is. I have it before my eyes. And $tmapstosqrt[3]t$ is not smooth.
$endgroup$
– José Carlos Santos
Sep 10 '18 at 11:09
$begingroup$
Haha it's Def. 1.5 in the edition that I am reading (same definition)!
$endgroup$
– user231343
Sep 10 '18 at 11:12
$begingroup$
Haha it's Def. 1.5 in the edition that I am reading (same definition)!
$endgroup$
– user231343
Sep 10 '18 at 11:12
$begingroup$
Following the comment below @mechanodroid's answer, I think the inverse of of $tmapsto t^3$ is not smooth because d/dt (t^3) is not never-zero?
$endgroup$
– user231343
Sep 10 '18 at 11:15
$begingroup$
Following the comment below @mechanodroid's answer, I think the inverse of of $tmapsto t^3$ is not smooth because d/dt (t^3) is not never-zero?
$endgroup$
– user231343
Sep 10 '18 at 11:15
1
1
$begingroup$
@Edi Yes, that would be correct.
$endgroup$
– José Carlos Santos
Sep 10 '18 at 11:16
$begingroup$
@Edi Yes, that would be correct.
$endgroup$
– José Carlos Santos
Sep 10 '18 at 11:16
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