Does a surface with given boundary in $mathbb{R}^3$ exist?












1












$begingroup$


Given a (smooth) simple closed curve $C subset mathbb{R}^3$, is there a (smooth) surface $S$ with $partial S = C$? I'm aware there is a variational problem to find among such surfaces the one with minimal area. Here I'm interested in the statement and proof of some existence theorem. Some cases where the existence is clear: if $C$ is planar, and more generally if $C$ is the curve of intersection of a a graph and a cylinder, $x_1=g(x_2,x_3)$ and $f(x_2,x_3)=0$



(This question came up when I read in a calculus book that $text{curl }mathbf{F}=mathbf{0}$ implies $mathbf{F}=nabla f$ for some scalar function $f(x,y,z)$. The proof was: $mathbf{F}=nabla f$ iff $mathbf{F}$ is conservative; to show $mathbf{F}$ is conservative, consider $int_C mathbf{F}cdot dmathbf{r}$ for any closed curve $C$. "THERE IS" a surface $S$ with $C$ as its boundary. By stokes theorem, and the fact that $text{curl }mathbf{F}=0$, $int_C mathbf{F}cdot dmathbf{r} = 0$. So, my question is why this surface even exists.)










share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Yes. These are called Seifert surfaces.
    $endgroup$
    – Cheerful Parsnip
    Dec 5 '18 at 20:17










  • $begingroup$
    @CheerfulParsnip I clarified my question in a comment below. Thanks for referring me to Seifert surfaces. I can say my question is about a step in the algorithm that is taken for granted.
    $endgroup$
    – user25959
    Dec 7 '18 at 2:34










  • $begingroup$
    One idea I had was: under some conditions on the curve $C$, there is a point $P$ far away enough that the segments $(PX]$ for $Xin C$ are all disjoint. So we call this cone $S$, smooth out the vertex, and get a surface that way.
    $endgroup$
    – user25959
    Dec 7 '18 at 2:38
















1












$begingroup$


Given a (smooth) simple closed curve $C subset mathbb{R}^3$, is there a (smooth) surface $S$ with $partial S = C$? I'm aware there is a variational problem to find among such surfaces the one with minimal area. Here I'm interested in the statement and proof of some existence theorem. Some cases where the existence is clear: if $C$ is planar, and more generally if $C$ is the curve of intersection of a a graph and a cylinder, $x_1=g(x_2,x_3)$ and $f(x_2,x_3)=0$



(This question came up when I read in a calculus book that $text{curl }mathbf{F}=mathbf{0}$ implies $mathbf{F}=nabla f$ for some scalar function $f(x,y,z)$. The proof was: $mathbf{F}=nabla f$ iff $mathbf{F}$ is conservative; to show $mathbf{F}$ is conservative, consider $int_C mathbf{F}cdot dmathbf{r}$ for any closed curve $C$. "THERE IS" a surface $S$ with $C$ as its boundary. By stokes theorem, and the fact that $text{curl }mathbf{F}=0$, $int_C mathbf{F}cdot dmathbf{r} = 0$. So, my question is why this surface even exists.)










share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Yes. These are called Seifert surfaces.
    $endgroup$
    – Cheerful Parsnip
    Dec 5 '18 at 20:17










  • $begingroup$
    @CheerfulParsnip I clarified my question in a comment below. Thanks for referring me to Seifert surfaces. I can say my question is about a step in the algorithm that is taken for granted.
    $endgroup$
    – user25959
    Dec 7 '18 at 2:34










  • $begingroup$
    One idea I had was: under some conditions on the curve $C$, there is a point $P$ far away enough that the segments $(PX]$ for $Xin C$ are all disjoint. So we call this cone $S$, smooth out the vertex, and get a surface that way.
    $endgroup$
    – user25959
    Dec 7 '18 at 2:38














1












1








1





$begingroup$


Given a (smooth) simple closed curve $C subset mathbb{R}^3$, is there a (smooth) surface $S$ with $partial S = C$? I'm aware there is a variational problem to find among such surfaces the one with minimal area. Here I'm interested in the statement and proof of some existence theorem. Some cases where the existence is clear: if $C$ is planar, and more generally if $C$ is the curve of intersection of a a graph and a cylinder, $x_1=g(x_2,x_3)$ and $f(x_2,x_3)=0$



(This question came up when I read in a calculus book that $text{curl }mathbf{F}=mathbf{0}$ implies $mathbf{F}=nabla f$ for some scalar function $f(x,y,z)$. The proof was: $mathbf{F}=nabla f$ iff $mathbf{F}$ is conservative; to show $mathbf{F}$ is conservative, consider $int_C mathbf{F}cdot dmathbf{r}$ for any closed curve $C$. "THERE IS" a surface $S$ with $C$ as its boundary. By stokes theorem, and the fact that $text{curl }mathbf{F}=0$, $int_C mathbf{F}cdot dmathbf{r} = 0$. So, my question is why this surface even exists.)










share|cite|improve this question









$endgroup$




Given a (smooth) simple closed curve $C subset mathbb{R}^3$, is there a (smooth) surface $S$ with $partial S = C$? I'm aware there is a variational problem to find among such surfaces the one with minimal area. Here I'm interested in the statement and proof of some existence theorem. Some cases where the existence is clear: if $C$ is planar, and more generally if $C$ is the curve of intersection of a a graph and a cylinder, $x_1=g(x_2,x_3)$ and $f(x_2,x_3)=0$



(This question came up when I read in a calculus book that $text{curl }mathbf{F}=mathbf{0}$ implies $mathbf{F}=nabla f$ for some scalar function $f(x,y,z)$. The proof was: $mathbf{F}=nabla f$ iff $mathbf{F}$ is conservative; to show $mathbf{F}$ is conservative, consider $int_C mathbf{F}cdot dmathbf{r}$ for any closed curve $C$. "THERE IS" a surface $S$ with $C$ as its boundary. By stokes theorem, and the fact that $text{curl }mathbf{F}=0$, $int_C mathbf{F}cdot dmathbf{r} = 0$. So, my question is why this surface even exists.)







multivariable-calculus differential-geometry calculus-of-variations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 5 '18 at 20:14









user25959user25959

1,573916




1,573916








  • 2




    $begingroup$
    Yes. These are called Seifert surfaces.
    $endgroup$
    – Cheerful Parsnip
    Dec 5 '18 at 20:17










  • $begingroup$
    @CheerfulParsnip I clarified my question in a comment below. Thanks for referring me to Seifert surfaces. I can say my question is about a step in the algorithm that is taken for granted.
    $endgroup$
    – user25959
    Dec 7 '18 at 2:34










  • $begingroup$
    One idea I had was: under some conditions on the curve $C$, there is a point $P$ far away enough that the segments $(PX]$ for $Xin C$ are all disjoint. So we call this cone $S$, smooth out the vertex, and get a surface that way.
    $endgroup$
    – user25959
    Dec 7 '18 at 2:38














  • 2




    $begingroup$
    Yes. These are called Seifert surfaces.
    $endgroup$
    – Cheerful Parsnip
    Dec 5 '18 at 20:17










  • $begingroup$
    @CheerfulParsnip I clarified my question in a comment below. Thanks for referring me to Seifert surfaces. I can say my question is about a step in the algorithm that is taken for granted.
    $endgroup$
    – user25959
    Dec 7 '18 at 2:34










  • $begingroup$
    One idea I had was: under some conditions on the curve $C$, there is a point $P$ far away enough that the segments $(PX]$ for $Xin C$ are all disjoint. So we call this cone $S$, smooth out the vertex, and get a surface that way.
    $endgroup$
    – user25959
    Dec 7 '18 at 2:38








2




2




$begingroup$
Yes. These are called Seifert surfaces.
$endgroup$
– Cheerful Parsnip
Dec 5 '18 at 20:17




$begingroup$
Yes. These are called Seifert surfaces.
$endgroup$
– Cheerful Parsnip
Dec 5 '18 at 20:17












$begingroup$
@CheerfulParsnip I clarified my question in a comment below. Thanks for referring me to Seifert surfaces. I can say my question is about a step in the algorithm that is taken for granted.
$endgroup$
– user25959
Dec 7 '18 at 2:34




$begingroup$
@CheerfulParsnip I clarified my question in a comment below. Thanks for referring me to Seifert surfaces. I can say my question is about a step in the algorithm that is taken for granted.
$endgroup$
– user25959
Dec 7 '18 at 2:34












$begingroup$
One idea I had was: under some conditions on the curve $C$, there is a point $P$ far away enough that the segments $(PX]$ for $Xin C$ are all disjoint. So we call this cone $S$, smooth out the vertex, and get a surface that way.
$endgroup$
– user25959
Dec 7 '18 at 2:38




$begingroup$
One idea I had was: under some conditions on the curve $C$, there is a point $P$ far away enough that the segments $(PX]$ for $Xin C$ are all disjoint. So we call this cone $S$, smooth out the vertex, and get a surface that way.
$endgroup$
– user25959
Dec 7 '18 at 2:38










1 Answer
1






active

oldest

votes


















3












$begingroup$

As already stated by Cheerful Parsnip, such surfaces exist.



In fact, you can choose them to be compact and oriented (then they are Seifert surfaces). You can find an elementar proof in [Saveliev - Lectures on the Topology of 3-Manifolds].



If you know a little bit more machinery from differential topology, this (much sexier) argument will do the trick.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you. I'm looking at the algorithm and see that it does a lot more than what I had in mind (I was only imagining unknots). The algorithm's early steps involve 'attaching disks' - the existence of this attachment is I guess what I'm asking about. Shouldn't there be an easy analytic proof of existence of an attached disk under some smoothness/"injectivity-type" conditions on the curve?
    $endgroup$
    – user25959
    Dec 7 '18 at 2:34











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%2f3027587%2fdoes-a-surface-with-given-boundary-in-mathbbr3-exist%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

As already stated by Cheerful Parsnip, such surfaces exist.



In fact, you can choose them to be compact and oriented (then they are Seifert surfaces). You can find an elementar proof in [Saveliev - Lectures on the Topology of 3-Manifolds].



If you know a little bit more machinery from differential topology, this (much sexier) argument will do the trick.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you. I'm looking at the algorithm and see that it does a lot more than what I had in mind (I was only imagining unknots). The algorithm's early steps involve 'attaching disks' - the existence of this attachment is I guess what I'm asking about. Shouldn't there be an easy analytic proof of existence of an attached disk under some smoothness/"injectivity-type" conditions on the curve?
    $endgroup$
    – user25959
    Dec 7 '18 at 2:34
















3












$begingroup$

As already stated by Cheerful Parsnip, such surfaces exist.



In fact, you can choose them to be compact and oriented (then they are Seifert surfaces). You can find an elementar proof in [Saveliev - Lectures on the Topology of 3-Manifolds].



If you know a little bit more machinery from differential topology, this (much sexier) argument will do the trick.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you. I'm looking at the algorithm and see that it does a lot more than what I had in mind (I was only imagining unknots). The algorithm's early steps involve 'attaching disks' - the existence of this attachment is I guess what I'm asking about. Shouldn't there be an easy analytic proof of existence of an attached disk under some smoothness/"injectivity-type" conditions on the curve?
    $endgroup$
    – user25959
    Dec 7 '18 at 2:34














3












3








3





$begingroup$

As already stated by Cheerful Parsnip, such surfaces exist.



In fact, you can choose them to be compact and oriented (then they are Seifert surfaces). You can find an elementar proof in [Saveliev - Lectures on the Topology of 3-Manifolds].



If you know a little bit more machinery from differential topology, this (much sexier) argument will do the trick.






share|cite|improve this answer











$endgroup$



As already stated by Cheerful Parsnip, such surfaces exist.



In fact, you can choose them to be compact and oriented (then they are Seifert surfaces). You can find an elementar proof in [Saveliev - Lectures on the Topology of 3-Manifolds].



If you know a little bit more machinery from differential topology, this (much sexier) argument will do the trick.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 21:03

























answered Dec 5 '18 at 20:52









Yong-HooYong-Hoo

312




312












  • $begingroup$
    Thank you. I'm looking at the algorithm and see that it does a lot more than what I had in mind (I was only imagining unknots). The algorithm's early steps involve 'attaching disks' - the existence of this attachment is I guess what I'm asking about. Shouldn't there be an easy analytic proof of existence of an attached disk under some smoothness/"injectivity-type" conditions on the curve?
    $endgroup$
    – user25959
    Dec 7 '18 at 2:34


















  • $begingroup$
    Thank you. I'm looking at the algorithm and see that it does a lot more than what I had in mind (I was only imagining unknots). The algorithm's early steps involve 'attaching disks' - the existence of this attachment is I guess what I'm asking about. Shouldn't there be an easy analytic proof of existence of an attached disk under some smoothness/"injectivity-type" conditions on the curve?
    $endgroup$
    – user25959
    Dec 7 '18 at 2:34
















$begingroup$
Thank you. I'm looking at the algorithm and see that it does a lot more than what I had in mind (I was only imagining unknots). The algorithm's early steps involve 'attaching disks' - the existence of this attachment is I guess what I'm asking about. Shouldn't there be an easy analytic proof of existence of an attached disk under some smoothness/"injectivity-type" conditions on the curve?
$endgroup$
– user25959
Dec 7 '18 at 2:34




$begingroup$
Thank you. I'm looking at the algorithm and see that it does a lot more than what I had in mind (I was only imagining unknots). The algorithm's early steps involve 'attaching disks' - the existence of this attachment is I guess what I'm asking about. Shouldn't there be an easy analytic proof of existence of an attached disk under some smoothness/"injectivity-type" conditions on the curve?
$endgroup$
– user25959
Dec 7 '18 at 2:34


















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%2f3027587%2fdoes-a-surface-with-given-boundary-in-mathbbr3-exist%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

How to change which sound is reproduced for terminal bell?

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents