If a surface $S$ admits two differentiable orthogonal families of geodesics $Rightarrow$ The Gaussian...












7












$begingroup$


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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















7












$begingroup$


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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















7












7








7


3



$begingroup$


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?










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 15:24







Matheus Manzatto

















asked Oct 24 '17 at 23:50









Matheus ManzattoMatheus Manzatto

1,4081523




1,4081523












  • $begingroup$
    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












  • $begingroup$
    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












0






active

oldest

votes











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2488432%2fif-a-surface-s-admits-two-differentiable-orthogonal-families-of-geodesics-ri%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























0






active

oldest

votes








0






active

oldest

votes









active

oldest

votes






active

oldest

votes
















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2488432%2fif-a-surface-s-admits-two-differentiable-orthogonal-families-of-geodesics-ri%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Ellipse (mathématiques)

Quarter-circle Tiles

Mont Emei