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Submanifold of $S^1 times S^1 times mathbbR^2$. [closed]
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Let's $f,g : S^1 longrightarrow mathbbR^2$ be embeding's. Consider $F: S^1 times S^1 times mathbbR^2 longrightarrow mathbbR^2 $, $F(x,y,v)= f(x) - g(y) -v$. Show that $N= F^-1(0)$ is submanifold of $S^1 times S^1 times mathbbR^2$.
geometry manifolds submanifold
asked Jul 26 at 16:41
Pampas
336110
closed as off-topic by Mike Miller, amWhy, Isaac Browne, Shailesh, Xander Henderson Jul 27 at 12:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Mike Miller, amWhy, Isaac Browne, Shailesh, Xander Henderson
add a comment |Â
up vote
-2
down vote
favorite
Let's $f,g : S^1 longrightarrow mathbbR^2$ be embeding's. Consider $F: S^1 times S^1 times mathbbR^2 longrightarrow mathbbR^2 $, $F(x,y,v)= f(x) - g(y) -v$. Show that $N= F^-1(0)$ is submanifold of $S^1 times S^1 times mathbbR^2$.
geometry manifolds submanifold
asked Jul 26 at 16:41
Pampas
336110
closed as off-topic by Mike Miller, amWhy, Isaac Browne, Shailesh, Xander Henderson Jul 27 at 12:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Mike Miller, amWhy, Isaac Browne, Shailesh, Xander Henderson
Prove it is a regular value
– Anubhav Mukherjee
Jul 26 at 17:34
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Let's $f,g : S^1 longrightarrow mathbbR^2$ be embeding's. Consider $F: S^1 times S^1 times mathbbR^2 longrightarrow mathbbR^2 $, $F(x,y,v)= f(x) - g(y) -v$. Show that $N= F^-1(0)$ is submanifold of $S^1 times S^1 times mathbbR^2$.
geometry manifolds submanifold
asked Jul 26 at 16:41
Pampas
336110
Let's $f,g : S^1 longrightarrow mathbbR^2$ be embeding's. Consider $F: S^1 times S^1 times mathbbR^2 longrightarrow mathbbR^2 $, $F(x,y,v)= f(x) - g(y) -v$. Show that $N= F^-1(0)$ is submanifold of $S^1 times S^1 times mathbbR^2$.
geometry manifolds submanifold
asked Jul 26 at 16:41
Pampas
336110
edited Jul 26 at 17:16
asked Jul 26 at 16:41
Pampas
336110
asked Jul 26 at 16:41
Pampas
336110
asked Jul 26 at 16:41
Pampas
336110
336110
closed as off-topic by Mike Miller, amWhy, Isaac Browne, Shailesh, Xander Henderson Jul 27 at 12:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Mike Miller, amWhy, Isaac Browne, Shailesh, Xander Henderson
closed as off-topic by Mike Miller, amWhy, Isaac Browne, Shailesh, Xander Henderson Jul 27 at 12:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Mike Miller, amWhy, Isaac Browne, Shailesh, Xander Henderson
Prove it is a regular value
– Anubhav Mukherjee
Jul 26 at 17:34
add a comment |Â
Prove it is a regular value
– Anubhav Mukherjee
Jul 26 at 17:34
Prove it is a regular value
– Anubhav Mukherjee
Jul 26 at 17:34
Prove it is a regular value
– Anubhav Mukherjee
Jul 26 at 17:34
add a comment |Â
1 Answer
1
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1
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You need only check that $F$ has rank $2$ at every point $(x,y,v) in S^1 times S^1 times mathbbR^2$, i.e. $dF$ on tangent spaces is a surjection everywhere it's defined. In fact, one only needs to check that $dF$ is a surjection on $F^-1(0)$, but we'll prove the stronger version, so that $F^-1(r)$ is an embedded submanifold of $S^1 times S^1 times mathbbR^2$ for all $r in mathbbR^2$.
Consider the map $h: mathbbR^2 to S^1 times S^1 times mathbbR^2$ given by $h(t) = (x,y,t)$. Then $F circ h(t) = f(x) - g(y) -t$, so $F circ h$ is a diffeomorphism from $mathbbR^2$ to itself. Therefore the tangent map $d(F circ h)$ is an isomorphism, but $d(F circ h) = dF circ dh$, so that $dF$ is in fact a surjection $T_(x,y,v)(S^1 times S^1 times mathbbR^2) to T_-v(mathbbR^2)$.
answered Jul 26 at 17:47
Point
6114
Why $F o h$ is a diffeomorphism from $mathbbR^2$?
– Pampas
Jul 26 at 20:53
1
@Pampas: Because $F circ h: mathbbR^2 to mathbbR^2$ has inverse. In fact, $(F circ h)^2 = (F circ h) circ (F circ h) = id_mathbbR^2$.
– Point
Jul 26 at 22:58
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You need only check that $F$ has rank $2$ at every point $(x,y,v) in S^1 times S^1 times mathbbR^2$, i.e. $dF$ on tangent spaces is a surjection everywhere it's defined. In fact, one only needs to check that $dF$ is a surjection on $F^-1(0)$, but we'll prove the stronger version, so that $F^-1(r)$ is an embedded submanifold of $S^1 times S^1 times mathbbR^2$ for all $r in mathbbR^2$.
Consider the map $h: mathbbR^2 to S^1 times S^1 times mathbbR^2$ given by $h(t) = (x,y,t)$. Then $F circ h(t) = f(x) - g(y) -t$, so $F circ h$ is a diffeomorphism from $mathbbR^2$ to itself. Therefore the tangent map $d(F circ h)$ is an isomorphism, but $d(F circ h) = dF circ dh$, so that $dF$ is in fact a surjection $T_(x,y,v)(S^1 times S^1 times mathbbR^2) to T_-v(mathbbR^2)$.
answered Jul 26 at 17:47
Point
6114
Why $F o h$ is a diffeomorphism from $mathbbR^2$?
– Pampas
Jul 26 at 20:53
1
@Pampas: Because $F circ h: mathbbR^2 to mathbbR^2$ has inverse. In fact, $(F circ h)^2 = (F circ h) circ (F circ h) = id_mathbbR^2$.
– Point
Jul 26 at 22:58
add a comment |Â
up vote
1
down vote
accepted
You need only check that $F$ has rank $2$ at every point $(x,y,v) in S^1 times S^1 times mathbbR^2$, i.e. $dF$ on tangent spaces is a surjection everywhere it's defined. In fact, one only needs to check that $dF$ is a surjection on $F^-1(0)$, but we'll prove the stronger version, so that $F^-1(r)$ is an embedded submanifold of $S^1 times S^1 times mathbbR^2$ for all $r in mathbbR^2$.
Consider the map $h: mathbbR^2 to S^1 times S^1 times mathbbR^2$ given by $h(t) = (x,y,t)$. Then $F circ h(t) = f(x) - g(y) -t$, so $F circ h$ is a diffeomorphism from $mathbbR^2$ to itself. Therefore the tangent map $d(F circ h)$ is an isomorphism, but $d(F circ h) = dF circ dh$, so that $dF$ is in fact a surjection $T_(x,y,v)(S^1 times S^1 times mathbbR^2) to T_-v(mathbbR^2)$.
answered Jul 26 at 17:47
Point
6114
Why $F o h$ is a diffeomorphism from $mathbbR^2$?
– Pampas
Jul 26 at 20:53
1
@Pampas: Because $F circ h: mathbbR^2 to mathbbR^2$ has inverse. In fact, $(F circ h)^2 = (F circ h) circ (F circ h) = id_mathbbR^2$.
– Point
Jul 26 at 22:58
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You need only check that $F$ has rank $2$ at every point $(x,y,v) in S^1 times S^1 times mathbbR^2$, i.e. $dF$ on tangent spaces is a surjection everywhere it's defined. In fact, one only needs to check that $dF$ is a surjection on $F^-1(0)$, but we'll prove the stronger version, so that $F^-1(r)$ is an embedded submanifold of $S^1 times S^1 times mathbbR^2$ for all $r in mathbbR^2$.
Consider the map $h: mathbbR^2 to S^1 times S^1 times mathbbR^2$ given by $h(t) = (x,y,t)$. Then $F circ h(t) = f(x) - g(y) -t$, so $F circ h$ is a diffeomorphism from $mathbbR^2$ to itself. Therefore the tangent map $d(F circ h)$ is an isomorphism, but $d(F circ h) = dF circ dh$, so that $dF$ is in fact a surjection $T_(x,y,v)(S^1 times S^1 times mathbbR^2) to T_-v(mathbbR^2)$.
answered Jul 26 at 17:47
Point
6114
You need only check that $F$ has rank $2$ at every point $(x,y,v) in S^1 times S^1 times mathbbR^2$, i.e. $dF$ on tangent spaces is a surjection everywhere it's defined. In fact, one only needs to check that $dF$ is a surjection on $F^-1(0)$, but we'll prove the stronger version, so that $F^-1(r)$ is an embedded submanifold of $S^1 times S^1 times mathbbR^2$ for all $r in mathbbR^2$.
Consider the map $h: mathbbR^2 to S^1 times S^1 times mathbbR^2$ given by $h(t) = (x,y,t)$. Then $F circ h(t) = f(x) - g(y) -t$, so $F circ h$ is a diffeomorphism from $mathbbR^2$ to itself. Therefore the tangent map $d(F circ h)$ is an isomorphism, but $d(F circ h) = dF circ dh$, so that $dF$ is in fact a surjection $T_(x,y,v)(S^1 times S^1 times mathbbR^2) to T_-v(mathbbR^2)$.
answered Jul 26 at 17:47
Point
6114
edited Jul 26 at 17:56
answered Jul 26 at 17:47
Point
6114
answered Jul 26 at 17:47
Point
6114
answered Jul 26 at 17:47
Point
6114
6114
Why $F o h$ is a diffeomorphism from $mathbbR^2$?
– Pampas
Jul 26 at 20:53
1
@Pampas: Because $F circ h: mathbbR^2 to mathbbR^2$ has inverse. In fact, $(F circ h)^2 = (F circ h) circ (F circ h) = id_mathbbR^2$.
– Point
Jul 26 at 22:58
add a comment |Â
Why $F o h$ is a diffeomorphism from $mathbbR^2$?
– Pampas
Jul 26 at 20:53
1
@Pampas: Because $F circ h: mathbbR^2 to mathbbR^2$ has inverse. In fact, $(F circ h)^2 = (F circ h) circ (F circ h) = id_mathbbR^2$.
– Point
Jul 26 at 22:58
Why $F o h$ is a diffeomorphism from $mathbbR^2$?
– Pampas
Jul 26 at 20:53
Why $F o h$ is a diffeomorphism from $mathbbR^2$?
– Pampas
Jul 26 at 20:53
1
1
@Pampas: Because $F circ h: mathbbR^2 to mathbbR^2$ has inverse. In fact, $(F circ h)^2 = (F circ h) circ (F circ h) = id_mathbbR^2$.
– Point
Jul 26 at 22:58
@Pampas: Because $F circ h: mathbbR^2 to mathbbR^2$ has inverse. In fact, $(F circ h)^2 = (F circ h) circ (F circ h) = id_mathbbR^2$.
– Point
Jul 26 at 22:58
add a comment |Â
Prove it is a regular value
– Anubhav Mukherjee
Jul 26 at 17:34