If a surface $S$ admits two differentiable orthogonal families of geodesics $Rightarrow$ The Gaussian...
$begingroup$
This question was previously posted in Orthogonal differentiable family of curves . But I'm facing unsolved issues.
I'm still interested in solve the following exercise:
QUESTION: We say that a set of regular curves on a surface $S$ is a differentiable family of curves on $S$ if the tangent lines to the curves of the set make up a differentiable field of directions. Assume that a surface $S$ admits two differentiable orthogonal families of geodesics. Prove that the Gaussian curvature of $S$ is zero.
where a differentiable field of directions is:
Definition: A field of direction $r$ in a open $U cap S$ is a correspondence which assigns to each $p$ $in$ $Ucap S$ a line $r(p)$ in $T_pS$ passing through $p$. $r$ is said to be diferentiable at $p$ $in$ $U cap S$ if there exists a nonzero differentiable vector field $w$, defined in a neighborhood $Vcap S$ $subset$ $U cap S$ of $p$, such that for each $q$ $in$ $Vcap S$, $w(q) neq 0$ is a basis of $r(q)$; $r$ is diferentiable in $Ucap S$ if it is differentiable for every $p$ $in$ $U$.
At the end of the book, Manfredo give us the following hint
Manfredo's Hint: Parametrize a neighborhood of $pin S$ in such a way that the two families of geodesics are coordinate curves (Corollary 1, Sec. 3-4). Show that this implies that $F=0$, $E_v=0, $ $G_u = 0$. Make a change of parameters to obtain that $bar{F} = 0$, $bar{E} = bar{G} =1$.
where Corollary 1, Sec.3-4 is:
Corollary 1. (Sec.3-4): Given two fields of directions $r$ and $r'$ in an open set $U subset S$ such that at $p$ $in$ $U$, $r(p) neq r'(p)$, there exists a parametrization $x$ in a neighborhood of $p$ such that the coordinate curves of $x$ are the integral curves of $r$ and $r'$.
And now my problem appears.
Let $tau_1$ be the field of directions associated to the first family, and $tau_2$ be the field of directions associated to the second family (note that $tau_1$ and $tau_2$ are orthogonal).
I really don't understand why when we apply the Corollary 1 (Sec. 3-4), on the fields of directions $tau_1$ and $tau_2$, we end up getting a chart $x: U subset mathbb{R}^2 rightarrow W cap S$, in such a way that the coordinate curves, $x(u, c^{te})$ and $x(c^{te},v)$, are the two families of geodesics.
From what I know, when the coordinate curves are the integral curves of
fields of directions $tau_1$ and $tau_2$, we can only conclude that $x_u(u, c^{te})$ is l.d with the direction $tau_1(x(u, c^{te}))$ and $x_v(c^{te},v)$ is l.d. with the direction $tau_2(x(c^{te}, v))$.
GI can't see any reason for these coordinate curves to solve the geodesic's differential equation. In the best case scenario I was only able to conclude that the coordinate curves have the same path of a geodesic (which implies nothing).
Does anyone know how I outline this problem and solve the exercise?
geometry differential-geometry surfaces curvature geodesic
asked Oct 24 '17 at 23:50
Matheus ManzattoMatheus Manzatto
1,3771625
$endgroup$
|
show 3 more comments
$begingroup$
This question was previously posted in Orthogonal differentiable family of curves . But I'm facing unsolved issues.
I'm still interested in solve the following exercise:
QUESTION: We say that a set of regular curves on a surface $S$ is a differentiable family of curves on $S$ if the tangent lines to the curves of the set make up a differentiable field of directions. Assume that a surface $S$ admits two differentiable orthogonal families of geodesics. Prove that the Gaussian curvature of $S$ is zero.
where a differentiable field of directions is:
Definition: A field of direction $r$ in a open $U cap S$ is a correspondence which assigns to each $p$ $in$ $Ucap S$ a line $r(p)$ in $T_pS$ passing through $p$. $r$ is said to be diferentiable at $p$ $in$ $U cap S$ if there exists a nonzero differentiable vector field $w$, defined in a neighborhood $Vcap S$ $subset$ $U cap S$ of $p$, such that for each $q$ $in$ $Vcap S$, $w(q) neq 0$ is a basis of $r(q)$; $r$ is diferentiable in $Ucap S$ if it is differentiable for every $p$ $in$ $U$.
At the end of the book, Manfredo give us the following hint
Manfredo's Hint: Parametrize a neighborhood of $pin S$ in such a way that the two families of geodesics are coordinate curves (Corollary 1, Sec. 3-4). Show that this implies that $F=0$, $E_v=0, $ $G_u = 0$. Make a change of parameters to obtain that $bar{F} = 0$, $bar{E} = bar{G} =1$.
where Corollary 1, Sec.3-4 is:
Corollary 1. (Sec.3-4): Given two fields of directions $r$ and $r'$ in an open set $U subset S$ such that at $p$ $in$ $U$, $r(p) neq r'(p)$, there exists a parametrization $x$ in a neighborhood of $p$ such that the coordinate curves of $x$ are the integral curves of $r$ and $r'$.
And now my problem appears.
Let $tau_1$ be the field of directions associated to the first family, and $tau_2$ be the field of directions associated to the second family (note that $tau_1$ and $tau_2$ are orthogonal).
I really don't understand why when we apply the Corollary 1 (Sec. 3-4), on the fields of directions $tau_1$ and $tau_2$, we end up getting a chart $x: U subset mathbb{R}^2 rightarrow W cap S$, in such a way that the coordinate curves, $x(u, c^{te})$ and $x(c^{te},v)$, are the two families of geodesics.
From what I know, when the coordinate curves are the integral curves of
fields of directions $tau_1$ and $tau_2$, we can only conclude that $x_u(u, c^{te})$ is l.d with the direction $tau_1(x(u, c^{te}))$ and $x_v(c^{te},v)$ is l.d. with the direction $tau_2(x(c^{te}, v))$.
GI can't see any reason for these coordinate curves to solve the geodesic's differential equation. In the best case scenario I was only able to conclude that the coordinate curves have the same path of a geodesic (which implies nothing).
Does anyone know how I outline this problem and solve the exercise?
geometry differential-geometry surfaces curvature geodesic
asked Oct 24 '17 at 23:50
Matheus ManzattoMatheus Manzatto
1,3771625
$endgroup$
$begingroup$
First of all, I think that the first $W$ in the definition of field of direction should be a $V$, and that the line $r(p)$ should pass by $0$, not $p$. Then I find this notion of field of direction very misleading. It seems equivalent to the notion of non-vanishing vector fields. I think it's better to work with nonvanishing vector fields.
$endgroup$
– David Tewodrose
Oct 27 '17 at 16:18
$begingroup$
Then let me denote by $(gamma^{s})_s$ the first family of curves, and by $t$ the entry variable of $gamma^{s}$. Denote by $(tilde{gamma}^a)_a$ the second family of curves and by $b$ the entry variable of $gamma^{b}$. Couldn't you take for $tau_{1}$ the vector field $frac{mathrm{d}}{mathrm{d} b}gamma^{a}$?
$endgroup$
– David Tewodrose
Oct 27 '17 at 16:30
$begingroup$
About your first question, the deffinition is wrong I will fix it. I need change $W$ for $U$ (I copied Manfredo's definition). The line $r(p)$ pass by $p$ because Manfredo treats $T_p S$ as an affine space. But you can think in $T_p S$ using the usual definition, in this exercise doesn't change anything, but using the "usual" deffinition of Tangent Space, as you said, $r(p)$ should pass by $0$.
$endgroup$
– Matheus Manzatto
Oct 27 '17 at 21:01
1
$begingroup$
I recently answer a very similar question on this site. If my answer on "Orthogonal differentiable family of curves" posted by AguirreK is not helpfull for you, please let me know.
$endgroup$
– Upax
Dec 28 '17 at 18:01
2
$begingroup$
I know you cannot use Gauss-Bonnet at this point, but this is so worth noticing: if you have a geodesic square $T$ with 90 degrees on each vertex, then Gauss-Bonnet gives: $$intint_T kappa~ dA=0.$$ Now you have a contradiction, since if you have a point where the curvature is non-zero, you can take a very small square around the point so that the curvature keeps its sign inside the square.
$endgroup$
– Llohann
Dec 17 '18 at 9:13
|
show 3 more comments
$begingroup$
This question was previously posted in Orthogonal differentiable family of curves . But I'm facing unsolved issues.
I'm still interested in solve the following exercise:
QUESTION: We say that a set of regular curves on a surface $S$ is a differentiable family of curves on $S$ if the tangent lines to the curves of the set make up a differentiable field of directions. Assume that a surface $S$ admits two differentiable orthogonal families of geodesics. Prove that the Gaussian curvature of $S$ is zero.
where a differentiable field of directions is:
Definition: A field of direction $r$ in a open $U cap S$ is a correspondence which assigns to each $p$ $in$ $Ucap S$ a line $r(p)$ in $T_pS$ passing through $p$. $r$ is said to be diferentiable at $p$ $in$ $U cap S$ if there exists a nonzero differentiable vector field $w$, defined in a neighborhood $Vcap S$ $subset$ $U cap S$ of $p$, such that for each $q$ $in$ $Vcap S$, $w(q) neq 0$ is a basis of $r(q)$; $r$ is diferentiable in $Ucap S$ if it is differentiable for every $p$ $in$ $U$.
At the end of the book, Manfredo give us the following hint
Manfredo's Hint: Parametrize a neighborhood of $pin S$ in such a way that the two families of geodesics are coordinate curves (Corollary 1, Sec. 3-4). Show that this implies that $F=0$, $E_v=0, $ $G_u = 0$. Make a change of parameters to obtain that $bar{F} = 0$, $bar{E} = bar{G} =1$.
where Corollary 1, Sec.3-4 is:
Corollary 1. (Sec.3-4): Given two fields of directions $r$ and $r'$ in an open set $U subset S$ such that at $p$ $in$ $U$, $r(p) neq r'(p)$, there exists a parametrization $x$ in a neighborhood of $p$ such that the coordinate curves of $x$ are the integral curves of $r$ and $r'$.
And now my problem appears.
Let $tau_1$ be the field of directions associated to the first family, and $tau_2$ be the field of directions associated to the second family (note that $tau_1$ and $tau_2$ are orthogonal).
I really don't understand why when we apply the Corollary 1 (Sec. 3-4), on the fields of directions $tau_1$ and $tau_2$, we end up getting a chart $x: U subset mathbb{R}^2 rightarrow W cap S$, in such a way that the coordinate curves, $x(u, c^{te})$ and $x(c^{te},v)$, are the two families of geodesics.
From what I know, when the coordinate curves are the integral curves of
fields of directions $tau_1$ and $tau_2$, we can only conclude that $x_u(u, c^{te})$ is l.d with the direction $tau_1(x(u, c^{te}))$ and $x_v(c^{te},v)$ is l.d. with the direction $tau_2(x(c^{te}, v))$.
GI can't see any reason for these coordinate curves to solve the geodesic's differential equation. In the best case scenario I was only able to conclude that the coordinate curves have the same path of a geodesic (which implies nothing).
Does anyone know how I outline this problem and solve the exercise?
geometry differential-geometry surfaces curvature geodesic
asked Oct 24 '17 at 23:50
Matheus ManzattoMatheus Manzatto
1,3771625
$endgroup$
This question was previously posted in Orthogonal differentiable family of curves . But I'm facing unsolved issues.
I'm still interested in solve the following exercise:
QUESTION: We say that a set of regular curves on a surface $S$ is a differentiable family of curves on $S$ if the tangent lines to the curves of the set make up a differentiable field of directions. Assume that a surface $S$ admits two differentiable orthogonal families of geodesics. Prove that the Gaussian curvature of $S$ is zero.
where a differentiable field of directions is:
Definition: A field of direction $r$ in a open $U cap S$ is a correspondence which assigns to each $p$ $in$ $Ucap S$ a line $r(p)$ in $T_pS$ passing through $p$. $r$ is said to be diferentiable at $p$ $in$ $U cap S$ if there exists a nonzero differentiable vector field $w$, defined in a neighborhood $Vcap S$ $subset$ $U cap S$ of $p$, such that for each $q$ $in$ $Vcap S$, $w(q) neq 0$ is a basis of $r(q)$; $r$ is diferentiable in $Ucap S$ if it is differentiable for every $p$ $in$ $U$.
At the end of the book, Manfredo give us the following hint
Manfredo's Hint: Parametrize a neighborhood of $pin S$ in such a way that the two families of geodesics are coordinate curves (Corollary 1, Sec. 3-4). Show that this implies that $F=0$, $E_v=0, $ $G_u = 0$. Make a change of parameters to obtain that $bar{F} = 0$, $bar{E} = bar{G} =1$.
where Corollary 1, Sec.3-4 is:
Corollary 1. (Sec.3-4): Given two fields of directions $r$ and $r'$ in an open set $U subset S$ such that at $p$ $in$ $U$, $r(p) neq r'(p)$, there exists a parametrization $x$ in a neighborhood of $p$ such that the coordinate curves of $x$ are the integral curves of $r$ and $r'$.
And now my problem appears.
Let $tau_1$ be the field of directions associated to the first family, and $tau_2$ be the field of directions associated to the second family (note that $tau_1$ and $tau_2$ are orthogonal).
I really don't understand why when we apply the Corollary 1 (Sec. 3-4), on the fields of directions $tau_1$ and $tau_2$, we end up getting a chart $x: U subset mathbb{R}^2 rightarrow W cap S$, in such a way that the coordinate curves, $x(u, c^{te})$ and $x(c^{te},v)$, are the two families of geodesics.
From what I know, when the coordinate curves are the integral curves of
fields of directions $tau_1$ and $tau_2$, we can only conclude that $x_u(u, c^{te})$ is l.d with the direction $tau_1(x(u, c^{te}))$ and $x_v(c^{te},v)$ is l.d. with the direction $tau_2(x(c^{te}, v))$.
GI can't see any reason for these coordinate curves to solve the geodesic's differential equation. In the best case scenario I was only able to conclude that the coordinate curves have the same path of a geodesic (which implies nothing).
Does anyone know how I outline this problem and solve the exercise?
geometry differential-geometry surfaces curvature geodesic
geometry differential-geometry surfaces curvature geodesic
asked Oct 24 '17 at 23:50
Matheus ManzattoMatheus Manzatto
1,3771625
asked Oct 24 '17 at 23:50
Matheus ManzattoMatheus Manzatto
1,3771625
edited Dec 17 '18 at 15:24
Matheus Manzatto
asked Oct 24 '17 at 23:50
Matheus ManzattoMatheus Manzatto
1,3771625
asked Oct 24 '17 at 23:50
Matheus ManzattoMatheus Manzatto
1,3771625
asked Oct 24 '17 at 23:50
Matheus ManzattoMatheus Manzatto
1,3771625
1,3771625
$begingroup$
First of all, I think that the first $W$ in the definition of field of direction should be a $V$, and that the line $r(p)$ should pass by $0$, not $p$. Then I find this notion of field of direction very misleading. It seems equivalent to the notion of non-vanishing vector fields. I think it's better to work with nonvanishing vector fields.
$endgroup$
– David Tewodrose
Oct 27 '17 at 16:18
$begingroup$
Then let me denote by $(gamma^{s})_s$ the first family of curves, and by $t$ the entry variable of $gamma^{s}$. Denote by $(tilde{gamma}^a)_a$ the second family of curves and by $b$ the entry variable of $gamma^{b}$. Couldn't you take for $tau_{1}$ the vector field $frac{mathrm{d}}{mathrm{d} b}gamma^{a}$?
$endgroup$
– David Tewodrose
Oct 27 '17 at 16:30
$begingroup$
About your first question, the deffinition is wrong I will fix it. I need change $W$ for $U$ (I copied Manfredo's definition). The line $r(p)$ pass by $p$ because Manfredo treats $T_p S$ as an affine space. But you can think in $T_p S$ using the usual definition, in this exercise doesn't change anything, but using the "usual" deffinition of Tangent Space, as you said, $r(p)$ should pass by $0$.
$endgroup$
– Matheus Manzatto
Oct 27 '17 at 21:01
1
$begingroup$
I recently answer a very similar question on this site. If my answer on "Orthogonal differentiable family of curves" posted by AguirreK is not helpfull for you, please let me know.
$endgroup$
– Upax
Dec 28 '17 at 18:01
2
$begingroup$
I know you cannot use Gauss-Bonnet at this point, but this is so worth noticing: if you have a geodesic square $T$ with 90 degrees on each vertex, then Gauss-Bonnet gives: $$intint_T kappa~ dA=0.$$ Now you have a contradiction, since if you have a point where the curvature is non-zero, you can take a very small square around the point so that the curvature keeps its sign inside the square.
$endgroup$
– Llohann
Dec 17 '18 at 9:13
|
show 3 more comments
$begingroup$
First of all, I think that the first $W$ in the definition of field of direction should be a $V$, and that the line $r(p)$ should pass by $0$, not $p$. Then I find this notion of field of direction very misleading. It seems equivalent to the notion of non-vanishing vector fields. I think it's better to work with nonvanishing vector fields.
$endgroup$
– David Tewodrose
Oct 27 '17 at 16:18
$begingroup$
Then let me denote by $(gamma^{s})_s$ the first family of curves, and by $t$ the entry variable of $gamma^{s}$. Denote by $(tilde{gamma}^a)_a$ the second family of curves and by $b$ the entry variable of $gamma^{b}$. Couldn't you take for $tau_{1}$ the vector field $frac{mathrm{d}}{mathrm{d} b}gamma^{a}$?
$endgroup$
– David Tewodrose
Oct 27 '17 at 16:30
$begingroup$
About your first question, the deffinition is wrong I will fix it. I need change $W$ for $U$ (I copied Manfredo's definition). The line $r(p)$ pass by $p$ because Manfredo treats $T_p S$ as an affine space. But you can think in $T_p S$ using the usual definition, in this exercise doesn't change anything, but using the "usual" deffinition of Tangent Space, as you said, $r(p)$ should pass by $0$.
$endgroup$
– Matheus Manzatto
Oct 27 '17 at 21:01
1
$begingroup$
I recently answer a very similar question on this site. If my answer on "Orthogonal differentiable family of curves" posted by AguirreK is not helpfull for you, please let me know.
$endgroup$
– Upax
Dec 28 '17 at 18:01
2
$begingroup$
I know you cannot use Gauss-Bonnet at this point, but this is so worth noticing: if you have a geodesic square $T$ with 90 degrees on each vertex, then Gauss-Bonnet gives: $$intint_T kappa~ dA=0.$$ Now you have a contradiction, since if you have a point where the curvature is non-zero, you can take a very small square around the point so that the curvature keeps its sign inside the square.
$endgroup$
– Llohann
Dec 17 '18 at 9:13
$begingroup$
First of all, I think that the first $W$ in the definition of field of direction should be a $V$, and that the line $r(p)$ should pass by $0$, not $p$. Then I find this notion of field of direction very misleading. It seems equivalent to the notion of non-vanishing vector fields. I think it's better to work with nonvanishing vector fields.
$endgroup$
– David Tewodrose
Oct 27 '17 at 16:18
$begingroup$
First of all, I think that the first $W$ in the definition of field of direction should be a $V$, and that the line $r(p)$ should pass by $0$, not $p$. Then I find this notion of field of direction very misleading. It seems equivalent to the notion of non-vanishing vector fields. I think it's better to work with nonvanishing vector fields.
$endgroup$
– David Tewodrose
Oct 27 '17 at 16:18
$begingroup$
Then let me denote by $(gamma^{s})_s$ the first family of curves, and by $t$ the entry variable of $gamma^{s}$. Denote by $(tilde{gamma}^a)_a$ the second family of curves and by $b$ the entry variable of $gamma^{b}$. Couldn't you take for $tau_{1}$ the vector field $frac{mathrm{d}}{mathrm{d} b}gamma^{a}$?
$endgroup$
– David Tewodrose
Oct 27 '17 at 16:30
$begingroup$
Then let me denote by $(gamma^{s})_s$ the first family of curves, and by $t$ the entry variable of $gamma^{s}$. Denote by $(tilde{gamma}^a)_a$ the second family of curves and by $b$ the entry variable of $gamma^{b}$. Couldn't you take for $tau_{1}$ the vector field $frac{mathrm{d}}{mathrm{d} b}gamma^{a}$?
$endgroup$
– David Tewodrose
Oct 27 '17 at 16:30
$begingroup$
About your first question, the deffinition is wrong I will fix it. I need change $W$ for $U$ (I copied Manfredo's definition). The line $r(p)$ pass by $p$ because Manfredo treats $T_p S$ as an affine space. But you can think in $T_p S$ using the usual definition, in this exercise doesn't change anything, but using the "usual" deffinition of Tangent Space, as you said, $r(p)$ should pass by $0$.
$endgroup$
– Matheus Manzatto
Oct 27 '17 at 21:01
$begingroup$
About your first question, the deffinition is wrong I will fix it. I need change $W$ for $U$ (I copied Manfredo's definition). The line $r(p)$ pass by $p$ because Manfredo treats $T_p S$ as an affine space. But you can think in $T_p S$ using the usual definition, in this exercise doesn't change anything, but using the "usual" deffinition of Tangent Space, as you said, $r(p)$ should pass by $0$.
$endgroup$
– Matheus Manzatto
Oct 27 '17 at 21:01
1
1
$begingroup$
I recently answer a very similar question on this site. If my answer on "Orthogonal differentiable family of curves" posted by AguirreK is not helpfull for you, please let me know.
$endgroup$
– Upax
Dec 28 '17 at 18:01
$begingroup$
I recently answer a very similar question on this site. If my answer on "Orthogonal differentiable family of curves" posted by AguirreK is not helpfull for you, please let me know.
$endgroup$
– Upax
Dec 28 '17 at 18:01
2
2
$begingroup$
I know you cannot use Gauss-Bonnet at this point, but this is so worth noticing: if you have a geodesic square $T$ with 90 degrees on each vertex, then Gauss-Bonnet gives: $$intint_T kappa~ dA=0.$$ Now you have a contradiction, since if you have a point where the curvature is non-zero, you can take a very small square around the point so that the curvature keeps its sign inside the square.
$endgroup$
– Llohann
Dec 17 '18 at 9:13
$begingroup$
I know you cannot use Gauss-Bonnet at this point, but this is so worth noticing: if you have a geodesic square $T$ with 90 degrees on each vertex, then Gauss-Bonnet gives: $$intint_T kappa~ dA=0.$$ Now you have a contradiction, since if you have a point where the curvature is non-zero, you can take a very small square around the point so that the curvature keeps its sign inside the square.
$endgroup$
– Llohann
Dec 17 '18 at 9:13
|
show 3 more comments
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Required, but never shown
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First of all, I think that the first $W$ in the definition of field of direction should be a $V$, and that the line $r(p)$ should pass by $0$, not $p$. Then I find this notion of field of direction very misleading. It seems equivalent to the notion of non-vanishing vector fields. I think it's better to work with nonvanishing vector fields.
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– David Tewodrose
Oct 27 '17 at 16:18
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Then let me denote by $(gamma^{s})_s$ the first family of curves, and by $t$ the entry variable of $gamma^{s}$. Denote by $(tilde{gamma}^a)_a$ the second family of curves and by $b$ the entry variable of $gamma^{b}$. Couldn't you take for $tau_{1}$ the vector field $frac{mathrm{d}}{mathrm{d} b}gamma^{a}$?
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– David Tewodrose
Oct 27 '17 at 16:30
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About your first question, the deffinition is wrong I will fix it. I need change $W$ for $U$ (I copied Manfredo's definition). The line $r(p)$ pass by $p$ because Manfredo treats $T_p S$ as an affine space. But you can think in $T_p S$ using the usual definition, in this exercise doesn't change anything, but using the "usual" deffinition of Tangent Space, as you said, $r(p)$ should pass by $0$.
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– Matheus Manzatto
Oct 27 '17 at 21:01
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I recently answer a very similar question on this site. If my answer on "Orthogonal differentiable family of curves" posted by AguirreK is not helpfull for you, please let me know.
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– Upax
Dec 28 '17 at 18:01
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I know you cannot use Gauss-Bonnet at this point, but this is so worth noticing: if you have a geodesic square $T$ with 90 degrees on each vertex, then Gauss-Bonnet gives: $$intint_T kappa~ dA=0.$$ Now you have a contradiction, since if you have a point where the curvature is non-zero, you can take a very small square around the point so that the curvature keeps its sign inside the square.
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– Llohann
Dec 17 '18 at 9:13