Does a surface with given boundary in $mathbb{R}^3$ exist?
$begingroup$
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.)
multivariable-calculus differential-geometry calculus-of-variations
asked Dec 5 '18 at 20:14
user25959user25959
1,573816
$endgroup$
add a comment |
$begingroup$
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.)
multivariable-calculus differential-geometry calculus-of-variations
asked Dec 5 '18 at 20:14
user25959user25959
1,573816
$endgroup$
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
add a comment |
$begingroup$
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.)
multivariable-calculus differential-geometry calculus-of-variations
asked Dec 5 '18 at 20:14
user25959user25959
1,573816
$endgroup$
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
multivariable-calculus differential-geometry calculus-of-variations
asked Dec 5 '18 at 20:14
user25959user25959
1,573816
asked Dec 5 '18 at 20:14
user25959user25959
1,573816
asked Dec 5 '18 at 20:14
user25959user25959
1,573816
asked Dec 5 '18 at 20:14
user25959user25959
1,573816
asked Dec 5 '18 at 20:14
user25959user25959
1,573816
1,573816
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
add a comment |
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
add a comment |
1 Answer
1
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oldest
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$begingroup$
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.
answered Dec 5 '18 at 20:52
Yong-HooYong-Hoo
312
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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$begingroup$
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.
answered Dec 5 '18 at 20:52
Yong-HooYong-Hoo
312
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 5 '18 at 20:52
Yong-HooYong-Hoo
312
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 5 '18 at 20:52
Yong-HooYong-Hoo
312
$endgroup$
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.
answered Dec 5 '18 at 20:52
Yong-HooYong-Hoo
312
edited Dec 5 '18 at 21:03
answered Dec 5 '18 at 20:52
Yong-HooYong-Hoo
312
answered Dec 5 '18 at 20:52
Yong-HooYong-Hoo
312
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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