|
10
|
|
edited Oct 14 2011 at 1:44 |
I've got a letter from Idjad Sabitov which answer the question completely. Here is a short extract from it:
half-torus has rigidity of second order (Rembs' theorem, see Е. Rembs. Verbiegungen hoeherer Ordnung und ebene Flaechenrinnen. Math. Zeitschrift 36 (1932) or Ефимов, УМН, 1948, т.3, вып.2, стр. 135) Any second order rigid surface does not admit analytic deformation (i.e., the deformation $h_t(u,v)$ which is analytic on $t$). For the surfaces of revolution, the assumption of analyticity can be removed.
Below is the best part of my original post. It contains an idea which was not used by Rembs.
Let $h(u,v)$ be a small perturbation of the standard embedding;
$u\in (-\varepsilon,\varepsilon)$ and $v\in\mathbb S^1$.
Consider convex hull $K$ of $\mathop{\rm Im}h$
and look at the closed curve $\gamma_0$ which is formed by boundary of $\partial K\cap \mathop{\rm Im}h$.
I claim that $\gamma_0=h(0,{*})$ i.e. the Gauss curvature at points of $\gamma_0$ has to be $0$.
Indeed since $\gamma_0$ lies on convex part,
the Gauss curvature at the points of $\gamma_0$ has to be nonnegative.
On the other hand $\gamma_0$ bounds a flat disc in $\partial K$;
therefore its integral intrinsic curvature (in $\partial K$ and in the torus) has to be $2{\cdot}\pi$.
If the Gauss curvature is positive at some point of $\gamma_0$ then total intrinsic curvature of $\gamma_0$ has to be $<2{\cdot}\pi$, a contradiction.
Note that if the asymptotic direction goes transversally to $\gamma_0$ at $\gamma_0(v)$ then $\gamma_v$ can not lie on the $\partial K$.
I.e., $\gamma_0=h(0,{*})$ is an asymptotic curve.
WLOG we can assume that the length of $\gamma_0=h(0,{*})$ is $2{\cdot}\pi$ and its intrinsic curvature is $\equiv 1$.
In the space $\gamma_0$ has to be a curve with constant curvature $1$ and it should be closed --- the only such curve is a flat circle.
|
|
|
|
|
9
|
|
edited Oct 13 2011 at 17:20 |
I've got a letter from Idjad Sabitov which answer the question completely. Here is a short extract from it:
half-torus has rigidity of second order (Rembs' theorem, see Е. Rembs. Verbiegungen hoeherer Ordnung und ebene Flaechenrinnen. Math. Zeitschrift 36 (1932) or Ефимов, УМН, 1948, т.3, вып.2, стр. 135) Any second order rigid surface does not admit analytic deformation For the surfaces of revolution, the assumption of analyticity can be removed.
Below is the best part of my original post, it . It contains a proof in analytic casean idea which was not used by Rembs.
Let $h(u,v)$ be a small perturbation of the standard embedding;
$u\in (-\varepsilon,\varepsilon)$ and $v\in\mathbb S^1$.
Consider convex hull $K$ of $\mathop{\rm Im}h$
and look at the closed curve $\gamma_0$ which is formed by boundary of $\partial K\cap \mathop{\rm Im}h$.
I claim that $\gamma_0=h(0,{*})$ i.e. the Gauss curvature at points of $\gamma_0$ has to be $0$.
Indeed since $\gamma_0$ lies on convex part,
the Gauss curvature at the points of $\gamma_0$ has to be nonnegative.
On the other hand $\gamma_0$ bounds a flat disc in $\partial K$;
therefore its integral intrinsic curvature (in $\partial K$ and in the torus) has to be $2{\cdot}\pi$.
If the Gauss curvature is positive at some point of $\gamma_0$ then total intrinsic curvature of $\gamma_0$ has to be $<2{\cdot}\pi$, a contradiction.
Note that if the asymptotic direction goes transversally to $\gamma_0$ at $\gamma_0(v)$ then $\gamma_v$ can not lie on the $\partial K$.
I.e., $\gamma_0=h(0,{*})$ is an asymptotic curve.
WLOG we can assume that the length of $\gamma_0=h(0,{*})$ is $2{\cdot}\pi$ and its intrinsic curvature is $\equiv 1$.
In the space $\gamma_0$ has to be a curve with constant curvature $1$ and it should be closed --- the only such curve is a flat circle.
According to Rembs' theorem the convex part of torus is rigid. Therefor if $h$ is analytic then we are done.
|
|
|
|
|
8
|
|
edited Oct 13 2011 at 16:42 |
I've got a letter from Idjad Sabitov which answer the question completely. Here is a short extract from it:
half-torus has rigidity of second order (Rembs' theorem, see Ефимов, УМН, 1948, т.3, вып.2, стр. 135) Any second order rigid surface does not admit analytic deformation For the surfaces of revolution, the assumption of analyticity can be removed.
Below is the best part of my original post, I am proud to be able to make it contains a proof of partial result in such a short timeanalytic case
Let $h(u,v)$ be a small perturbation of the standard embedding;
$u\in (-\varepsilon,\varepsilon)$ and $v\in\mathbb S^1$.
Consider convex hull $K$ of $\mathop{\rm Im}h$
and look at the closed curve $\gamma_0$ which is formed by boundary of $\partial K\cap \mathop{\rm Im}h$.
I claim that $\gamma_0=h(0,{*})$ i.e. the Gauss curvature at points of $\gamma_0$ has to be $0$.
Indeed since $\gamma_0$ lies on convex part,
the Gauss curvature at the points of $\gamma_0$ has to be nonnegative.
On the other hand $\gamma_0$ bounds a flat disc in $\partial K$;
therefore its integral intrinsic curvature (in $\partial K$ and in the torus) has to be $2{\cdot}\pi$.
If the Gauss curvature is positive at some point of $\gamma_0$ then total intrinsic curvature of $\gamma_0$ has to be $<2{\cdot}\pi$, a contradiction.
Note that if the asymptotic direction goes transversally to $\gamma_0$ at $\gamma_0(v)$ then $\gamma_v$ can not lie on the $\partial K$.
I.e., $\gamma_0=h(0,{*})$ is an asymptotic curve.
WLOG we can assume that the length of $\gamma_0=h(0,{*})$ is $2{\cdot}\pi$ and its intrinsic curvature is $\equiv 1$.
In the space $\gamma_0$ has to be a curve with constant curvature $1$ and it should be closed --- the only such curve is a flat circle.
According to Rembs' theorem the convex part of torus is rigid. Therefor if $h$ is analytic then we are done.
|
|
|
|
7
|
|
edited Oct 13 2011 at 16:26 |
I've got a letter from Idjad Sabitov which answer the question completely. Here is a short extract from it: half-torus has rigidity of second order (Rembs' theorem, see Ефимов, УМН, 1948, т.3, вып.2, стр. 135) Any second order rigid surface does not admit analytic deformation For the surfaces of revolution, the assumption of analyticity can be removed. Below is the best part of my original post, I am proud to be able to make a proof of partial result in such a short time In the space $\gamma_0$ has to be a curve with constant curvature $1$ and it should be closed --- the only such curve is a flat circle.We already get a bit of rigidity; now let us show that the surface is a surface of revolution. Look at an other circle $\gamma_t=h(t,{*})$.If $n$ denotes the Gauss map for $h$, then we know the area bounded by $n(\gamma_t)$ in $\mathbb S^2$. We also know the intrinsic curvature and length of $\gamma_t$.These three things (area, length and curvature) force $\gamma_t$ to lie in a cone as a central circle. Denote by $p_t$ the vertex of this cone and let $R_t$ be the distance from $\gamma_t$.So the sphere $S(p_t,R_t)$ cuts the torus orthogonally along $\gamma_t$.Taking into account that $\gamma_t$ are equidistant in the torus plus $p_t$ can not stay on the same place, we get that all $\gamma_t$ are flat circles;it implies that the the half-torus is a surface of revolution. Hence the result.
|
|
|
|
|
6
|
|
edited Oct 12 2011 at 2:37 |
Let $h(u,v)$ be a small perturbation of the standard embedding;
$u\in (-\varepsilon,\varepsilon)$ and $v\in\mathbb S^1$.
Consider convex hull $K$ of $\mathop{\rm Im}h$
and look at the closed curve $\gamma_0$ which is formed by boundary of $\partial K\cap \mathop{\rm Im}h$.
I claim that $\gamma_0=h(0,{*})$ i.e. the Gauss curvature at points of $\gamma_0$ has to be $0$.
Indeed since $\gamma_0$ lies on convex part,
the Gauss curvature at the points of $\gamma_0$ has to be nonnegative.
On the other hand $\gamma_0$ bounds a flat disc in $\partial K$;
therefore its integral intrinsic curvature (in $\partial K$ and in the torus) has to be $2{\cdot}\pi$.
If the Gauss curvature is positive at some point of $\gamma_0$ then total intrinsic curvature of $\gamma_0$ has to be $<2{\cdot}\pi$, a contradiction.In particular
Note that if the asymptotic direction goes transversally to $\gamma_0$ at $\gamma_0(v)$ then $\gamma_v$ can not lie on the $\partial K$.
I.e., $\gamma_0=h(0,{*})$ is an asymptotic curve.
WLOG we can assume that the length of $\gamma_0=h(0,{*})$ is $2{\cdot}\pi$ and its intrinsic curvature is $\equiv 1$.
In the space $\gamma_0$ has to be a curve with constant curvature $1$ and it should be closed --- the only such curve is a flat circle.
We already get a bit of rigidity; now let us show that the surface is a surface of revolution.
Look at an other circle $\gamma_t=h(t,{*})$.
If $n$ denotes the Gauss map for $h$, then we know the area bounded by $n(\gamma_t)$ in $\mathbb S^2$.
We also know the intrinsic curvature and length of $\gamma_t$.
These three things (area, length and curvature) force $\gamma_t$ to lie in a cone as a central circle.
Denote by $p_t$ the vertex of this cone and let $R_t$ be the distance from $\gamma_t$.
So the sphere $S(p_t,R_t)$ cuts the torus orthogonally along $\gamma_t$.
Taking into account that $\gamma_t$ are equidistant in the torus plus $p_t$ can not stay on the same place, we get that all $\gamma_t$ are flat circles;
it implies that the the half-torus is a surface of revolution. Hence the result.
|
|
|
|
|
5
|
|
edited Oct 12 2011 at 2:09 |
Let $h(u,v)$ be a small perturbation of the standard embedding, ;
$u\in (-\varepsilon,\epsilon)$ -\varepsilon,\varepsilon)$ and $v\in\mathbb S^1$.
Consider convex hull $K$ of $\mathop{\rm Im}h$
and look at the closed curve $\gamma_0$ which is formed by boundary of $\partial K\cap \mathop{\rm Im}h$.
I claim that $\gamma_0=h(0,{*})$ i.e. the Gauss curvature at points of $\gamma_t$ \gamma_0$ has to be $0$.
Indeed since $\gamma_0$ lies on convex part,
the Gauss curvature at the points of $\gamma_0$ has to be nonnegativeand .
On the total other hand $\gamma_0$ bounds a flat disc in $\partial K$;
therefore its integral intrinsic curvature of (in $\gamma_0$ \partial K$ and in the torus) has to be $2{\cdot}\pi$.
If the Gauss curvature is positive at some point of $\gamma_0$ then total intrinsic curvature of $\gamma_0$ has to be $<2{\cdot}\pi$, a contradiction.
In particular $\gamma_0=h(0,{*})$ is an asymptotic curve.
WLOG we can assume that the length of $\gamma_0=h(0,{*})$ is $2{\cdot}\pi$ and its intrinsic curvature is $\equiv 1$.
In the space $\gamma_0$ has to be a curve with constant curvature $1$ and it should be closed --- the only such curve is a flat circle.
Now
We already get a bit of rigidity; now let us apply show that the same argument to surface is a surface of revolution.
Look at an other circle $\gamma_t$ which is parallel to $\gamma_0$.
\gamma_t=h(t,{*})$.
If $n$ denotes the normal vector Gauss map for $h$, then we know the area bounded by $n\circ\gamma_t$ n(\gamma_t)$ in $\mathbb S^2$.
We also know the intrinsic curvature and length of $\gamma_t$.
These three things will (area, length and curvature) force $h\circ\gamma_t$ \gamma_t$ to lie in a cone as a central circle.
Denote by $p_t$ the vertex of this cone and let $R_t$ be the distance from $\gamma_t$.
So the sphere $S(p_t,R_t)$ cuts the torus orthogonally along $\gamma_t$.
Taking into account that $\gamma_t$ are equidistant in the torus plus $p_t$ can not stay on the same place, we get that all $\gamma_t$ are flat circles;
it implies that the the half-torus is a surface of revolution. Hence the result.
|
|
|
| |
|
Post Undeleted by Anton Petrunin
|
occurred Oct 11 2011 at 17:53
|
|
|
|
|
|
|
4
|
|
edited Oct 11 2011 at 17:53 |
Look Let $h(u,v)$ a small perturbation of the standard embedding, $u\in (-\varepsilon,\epsilon)$ and $v\in\mathbb S^1$.
Consider convex hull $K$ of $\mathop{\rm Im}h$
and look at the circle closed curve $\gamma_0$ which divides negative is formed by boundary of $\partial K\cap \mathop{\rm Im}h$.
I claim that $\gamma_0=h(0,{*})$ i.e. the Gauss curvature at points of $\gamma_t$ has to be $0$.
Indeed the Gauss curvature at the points of $\gamma_0$ has to be nonnegative and the total intrinsic curvature of $\gamma_0$ has to be $2{\cdot}\pi$.
If the Gauss curvature is positive at some point of $\gamma_0$ then total intrinsic curvature of $\gamma_0$ has to be $<2{\cdot}\pi$, a contradiction.
Denote by In particular $h$ the embedding\gamma_0=h(0,{*})$ is an asymptotic curve.
WLOG we can assume that the length of $\gamma_0$ \gamma_0=h(0,{*})$ is $2{\cdot}\pi$ and its intrinsic curvature is $\equiv 1$.
In the space $h\circ\gamma_0$ \gamma_0$ has to be a curve with constant curvature $1$ and it should be closed --- the only such curve is a flat circle.
Now let us apply the same argument to an other circle $\gamma_t$ which is parallel to $\gamma_0$.
If $n$ denotes the normal vector then we know the area bounded by $n\circ\gamma_t$ in $\mathbb S^2$.
We also know the intrinsic curvature and length of $\gamma_t$.
These three things will force $h\circ\gamma_t$ to lie in a cone as a central circle.
Denote by $p_t$ the vertex of this cone and let $R_t$ be the distance from $\gamma_t$.
So the sphere $S(p_t,R_t)$ cuts the torus orthogonally along $\gamma_t$.
Taking into account that $\gamma_t$ are equidistant in the torus plus $p_t$ can not stay on the same place, we get that all $\gamma_t$ are flat circles;
it implies that the the half-torus is a surface of revolution. Hence the result.
|
|
|
| |
|
Post Deleted by Anton Petrunin
|
occurred Oct 11 2011 at 16:45
|
|
|
|
|
|
|
3
|
|
edited Oct 11 2011 at 4:16 |
Look at the circle $\gamma_0$ which divides negative and positive curvature.
Denote by $h$ the embedding.
WLOG we can assume that its the length of $\gamma_0$ is $2{\cdot}\pi$ and its intrinsic curvature is $\equiv 1$.
In the space $h\circ\gamma_0$ has to be a curve with constant curvature $1$ and it should be closed --- the only such curve is a flat circle.
It seems to be almost solution but I am not sure how to proceed...
One may try to
Now let us apply the same argument to an other circle $\gamma_t$ which is parallel to $\gamma_0$.
If $n$ denotes the normal vector then we know the area bounded by $n\circ\gamma_t$ in $\mathbb S^2$.
We also know the intrinsic curvature and length of $\gamma_t$.
These three things will force $h\circ\gamma_t$ to lie in a cone as a central circle.
Denote by $p_t$ the vertex of this cone .
and let $p_t$ is moving in such a way that R_t$ be the tangent lines to distance from $\gamma_t$.
So the sphere $S(p_t,R_t)$ cuts the torus passing through orthogonally along $p_t$ always touch \gamma_t$.
Taking into account that $\gamma_t$ are equidistant in the torus equidistantly plus $p_t$.
Can someone finish p_t$ can not stay on the same place, we get that all $\gamma_t$ are flat circles;
it ?implies that the the half-torus is a surface of revolution. Hence the result.
|
|
|
|
|
2
|
|
edited Oct 11 2011 at 2:21 |
Look at the circle $\gamma_0$ which divides negative and positive curvature.
Denote by $h$ the embedding.
WLOG we can assume that its length $2{\cdot}\pi$ and its intrinsic curvature is $\equiv 1$.
In the space it $h\circ\gamma_0$ has to be a curve with constant curvature $1$ and it should be closed --- the only such curve is a flat circle.
It seems to be almost solution but I think if Alexandrov's statement am not sure how to proceed...
One may try to apply the same argument to an other circle $\gamma_t$ which is true, it should follow easely from here..parallel to $\gamma_0$.
If $n$ denotes the normal vector then we know the area bounded by $n\circ\gamma_t$ in $\mathbb S^2$.
We also know the intrinsic curvature and length of $\gamma_t$.
These three things will force $h\circ\gamma_t$ to lie in a cone as a central circle.
Denote by $p_t$ the vertex of this cone.
$p_t$ is moving in such a way that the tangent lines to the torus passing through $p_t$ always touch the torus equidistantly $p_t$.
Can someone finish it?
|
|
|
|
|
1
|
|
answered Oct 11 2011 at 1:10 |
Look at the circle which divides negative and positive curvature.
WLOG we can assume that its length $2{\cdot}\pi$ and its intrinsic curvature is $\equiv 1$.
In the space it has to be a curve with constant curvature $1$ and it should be closed --- the only such curve is a flat circle.
I think if Alexandrov's statement is true, it should follow easely from here...
|
|
|
