If an immersion $X$ maps circles into planes then its image $X(mathbb{D})$ is homeomorphic to the cylinder.
$begingroup$
Let $X:left( u,vright)in mathbb{D}backslash left{ 0right}subseteqmathbb{R}^2 mathbb{%
longmapsto }left( xleft( u,vright) ,yleft( u,vright) ,zleft(
u,vright) right) in mathbb{mathbb{R}}^{3}$ an minimal immersion, where $mathbb{D=}left{ pin mathbb{R}^{2};leftVert prightVert <1right} $ is unitary open disc.
If $X$ maps circles into planes (coordinate function $z$ constant) then $X(mathbb{D})$ is homeomorphic to cylindric.
I do not know if all these assumptions are necessary, but that's what I have.
Can you help me prove that? I studied geometry and topology a long time ago.
general-topology geometry analysis differential-geometry differential-topology
edited Jan 22 at 18:22
Lord Shark the Unknown
104k1160132
asked Jan 7 at 23:32
TakashiTakashi
17917
$endgroup$
add a comment |
$begingroup$
Let $X:left( u,vright)in mathbb{D}backslash left{ 0right}subseteqmathbb{R}^2 mathbb{%
longmapsto }left( xleft( u,vright) ,yleft( u,vright) ,zleft(
u,vright) right) in mathbb{mathbb{R}}^{3}$ an minimal immersion, where $mathbb{D=}left{ pin mathbb{R}^{2};leftVert prightVert <1right} $ is unitary open disc.
If $X$ maps circles into planes (coordinate function $z$ constant) then $X(mathbb{D})$ is homeomorphic to cylindric.
I do not know if all these assumptions are necessary, but that's what I have.
Can you help me prove that? I studied geometry and topology a long time ago.
general-topology geometry analysis differential-geometry differential-topology
edited Jan 22 at 18:22
Lord Shark the Unknown
104k1160132
asked Jan 7 at 23:32
TakashiTakashi
17917
$endgroup$
$begingroup$
How can be $z=(u,v)$? z is supposed to belong to $mathbb{R}$
$endgroup$
– ecrin
Jan 12 at 20:29
$begingroup$
$z:mathbb{D}rightarrowmathbb{mathbb{R}}$ is coordinate function. I edited the post.
$endgroup$
– Takashi
Jan 13 at 0:44
$begingroup$
If by $mathbb{D}$ you mean the unit disk (en.wikipedia.org/wiki/Unit_disk) then $0in mathbb{D}$ and so what you have written is false: $mathbb{D}$ is not a subset of $mathbb {R}^2setminus{0}$.
$endgroup$
– pre-kidney
Jan 13 at 2:52
$begingroup$
@pre-kidney. I corrected, thanks.
$endgroup$
– Takashi
Jan 13 at 13:32
add a comment |
$begingroup$
Let $X:left( u,vright)in mathbb{D}backslash left{ 0right}subseteqmathbb{R}^2 mathbb{%
longmapsto }left( xleft( u,vright) ,yleft( u,vright) ,zleft(
u,vright) right) in mathbb{mathbb{R}}^{3}$ an minimal immersion, where $mathbb{D=}left{ pin mathbb{R}^{2};leftVert prightVert <1right} $ is unitary open disc.
If $X$ maps circles into planes (coordinate function $z$ constant) then $X(mathbb{D})$ is homeomorphic to cylindric.
I do not know if all these assumptions are necessary, but that's what I have.
Can you help me prove that? I studied geometry and topology a long time ago.
general-topology geometry analysis differential-geometry differential-topology
edited Jan 22 at 18:22
Lord Shark the Unknown
104k1160132
asked Jan 7 at 23:32
TakashiTakashi
17917
$endgroup$
Let $X:left( u,vright)in mathbb{D}backslash left{ 0right}subseteqmathbb{R}^2 mathbb{%
longmapsto }left( xleft( u,vright) ,yleft( u,vright) ,zleft(
u,vright) right) in mathbb{mathbb{R}}^{3}$ an minimal immersion, where $mathbb{D=}left{ pin mathbb{R}^{2};leftVert prightVert <1right} $ is unitary open disc.
If $X$ maps circles into planes (coordinate function $z$ constant) then $X(mathbb{D})$ is homeomorphic to cylindric.
I do not know if all these assumptions are necessary, but that's what I have.
Can you help me prove that? I studied geometry and topology a long time ago.
general-topology geometry analysis differential-geometry differential-topology
general-topology geometry analysis differential-geometry differential-topology
edited Jan 22 at 18:22
Lord Shark the Unknown
104k1160132
asked Jan 7 at 23:32
TakashiTakashi
17917
edited Jan 22 at 18:22
Lord Shark the Unknown
104k1160132
asked Jan 7 at 23:32
TakashiTakashi
17917
edited Jan 22 at 18:22
Lord Shark the Unknown
104k1160132
edited Jan 22 at 18:22
Lord Shark the Unknown
104k1160132
edited Jan 22 at 18:22
Lord Shark the Unknown
104k1160132
104k1160132
asked Jan 7 at 23:32
TakashiTakashi
17917
asked Jan 7 at 23:32
TakashiTakashi
17917
asked Jan 7 at 23:32
TakashiTakashi
17917
17917
$begingroup$
How can be $z=(u,v)$? z is supposed to belong to $mathbb{R}$
$endgroup$
– ecrin
Jan 12 at 20:29
$begingroup$
$z:mathbb{D}rightarrowmathbb{mathbb{R}}$ is coordinate function. I edited the post.
$endgroup$
– Takashi
Jan 13 at 0:44
$begingroup$
If by $mathbb{D}$ you mean the unit disk (en.wikipedia.org/wiki/Unit_disk) then $0in mathbb{D}$ and so what you have written is false: $mathbb{D}$ is not a subset of $mathbb {R}^2setminus{0}$.
$endgroup$
– pre-kidney
Jan 13 at 2:52
$begingroup$
@pre-kidney. I corrected, thanks.
$endgroup$
– Takashi
Jan 13 at 13:32
add a comment |
$begingroup$
How can be $z=(u,v)$? z is supposed to belong to $mathbb{R}$
$endgroup$
– ecrin
Jan 12 at 20:29
$begingroup$
$z:mathbb{D}rightarrowmathbb{mathbb{R}}$ is coordinate function. I edited the post.
$endgroup$
– Takashi
Jan 13 at 0:44
$begingroup$
If by $mathbb{D}$ you mean the unit disk (en.wikipedia.org/wiki/Unit_disk) then $0in mathbb{D}$ and so what you have written is false: $mathbb{D}$ is not a subset of $mathbb {R}^2setminus{0}$.
$endgroup$
– pre-kidney
Jan 13 at 2:52
$begingroup$
@pre-kidney. I corrected, thanks.
$endgroup$
– Takashi
Jan 13 at 13:32
$begingroup$
How can be $z=(u,v)$? z is supposed to belong to $mathbb{R}$
$endgroup$
– ecrin
Jan 12 at 20:29
$begingroup$
How can be $z=(u,v)$? z is supposed to belong to $mathbb{R}$
$endgroup$
– ecrin
Jan 12 at 20:29
$begingroup$
$z:mathbb{D}rightarrowmathbb{mathbb{R}}$ is coordinate function. I edited the post.
$endgroup$
– Takashi
Jan 13 at 0:44
$begingroup$
$z:mathbb{D}rightarrowmathbb{mathbb{R}}$ is coordinate function. I edited the post.
$endgroup$
– Takashi
Jan 13 at 0:44
$begingroup$
If by $mathbb{D}$ you mean the unit disk (en.wikipedia.org/wiki/Unit_disk) then $0in mathbb{D}$ and so what you have written is false: $mathbb{D}$ is not a subset of $mathbb {R}^2setminus{0}$.
$endgroup$
– pre-kidney
Jan 13 at 2:52
$begingroup$
If by $mathbb{D}$ you mean the unit disk (en.wikipedia.org/wiki/Unit_disk) then $0in mathbb{D}$ and so what you have written is false: $mathbb{D}$ is not a subset of $mathbb {R}^2setminus{0}$.
$endgroup$
– pre-kidney
Jan 13 at 2:52
$begingroup$
@pre-kidney. I corrected, thanks.
$endgroup$
– Takashi
Jan 13 at 13:32
$begingroup$
@pre-kidney. I corrected, thanks.
$endgroup$
– Takashi
Jan 13 at 13:32
add a comment |
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$begingroup$
How can be $z=(u,v)$? z is supposed to belong to $mathbb{R}$
$endgroup$
– ecrin
Jan 12 at 20:29
$begingroup$
$z:mathbb{D}rightarrowmathbb{mathbb{R}}$ is coordinate function. I edited the post.
$endgroup$
– Takashi
Jan 13 at 0:44
$begingroup$
If by $mathbb{D}$ you mean the unit disk (en.wikipedia.org/wiki/Unit_disk) then $0in mathbb{D}$ and so what you have written is false: $mathbb{D}$ is not a subset of $mathbb {R}^2setminus{0}$.
$endgroup$
– pre-kidney
Jan 13 at 2:52
$begingroup$
@pre-kidney. I corrected, thanks.
$endgroup$
– Takashi
Jan 13 at 13:32