Mixing
Understanding Embedded submanifolds and immersions
$begingroup$
I am trying to understand the concept of embedded submanifolds and have the following understanding:
Suppose we have a smooth manifold $M$ and $N$ of dimensions $m$ and $n$ such that $mgt n$ .Let $Fcolon Nto M$ be a smooth map and a topological embedding onto its image $F(N)$ under subspace topology that $F(N)$ inherits from $M$. So, there is a smooth structure on $F(N)$ such that $Fcolon N to F(N)$ is a diffeomorphism (the smooth coordinate maps are just maps of the form $fcirc F^{-1}$, where $f$ is a smooth coordinate map for $N$). Therefore, it gives us a smooth manifold $F(N)$ of dimension similar as $N$ which sits inside $M$.
Now, the definition of embedded submanifolds as given in the text of boothby is:
image of a topological embedding+immersion is an embedded submanifold
My problem is:
If just taking $Fcolon N to F(N)$ as a topological embedding gives $F(N)$ diffeomorphic to $N$. So, it gives a smooth manifold $F(N)$ with same dimension as $N$, why do we even need to consider immersions?
Thanks in advance
differential-geometry manifolds differential-topology smooth-manifolds
edited Jan 10 at 18:44
Christoph
12k1642
asked Jan 10 at 18:39
Abhishek ShrivastavaAbhishek Shrivastava
414413
$endgroup$
add a comment |
$begingroup$
I am trying to understand the concept of embedded submanifolds and have the following understanding:
Suppose we have a smooth manifold $M$ and $N$ of dimensions $m$ and $n$ such that $mgt n$ .Let $Fcolon Nto M$ be a smooth map and a topological embedding onto its image $F(N)$ under subspace topology that $F(N)$ inherits from $M$. So, there is a smooth structure on $F(N)$ such that $Fcolon N to F(N)$ is a diffeomorphism (the smooth coordinate maps are just maps of the form $fcirc F^{-1}$, where $f$ is a smooth coordinate map for $N$). Therefore, it gives us a smooth manifold $F(N)$ of dimension similar as $N$ which sits inside $M$.
Now, the definition of embedded submanifolds as given in the text of boothby is:
image of a topological embedding+immersion is an embedded submanifold
My problem is:
If just taking $Fcolon N to F(N)$ as a topological embedding gives $F(N)$ diffeomorphic to $N$. So, it gives a smooth manifold $F(N)$ with same dimension as $N$, why do we even need to consider immersions?
Thanks in advance
differential-geometry manifolds differential-topology smooth-manifolds
edited Jan 10 at 18:44
Christoph
12k1642
asked Jan 10 at 18:39
Abhishek ShrivastavaAbhishek Shrivastava
414413
$endgroup$
$begingroup$
What is the definition of topological embedding here?
$endgroup$
– Anubhav Mukherjee
Jan 10 at 18:48
$begingroup$
@Anubhav Mukherjee:it is a homeomorphism from N to f(N),where f(N) has subspace topology
$endgroup$
– Abhishek Shrivastava
Jan 10 at 18:50
add a comment |
$begingroup$
I am trying to understand the concept of embedded submanifolds and have the following understanding:
Suppose we have a smooth manifold $M$ and $N$ of dimensions $m$ and $n$ such that $mgt n$ .Let $Fcolon Nto M$ be a smooth map and a topological embedding onto its image $F(N)$ under subspace topology that $F(N)$ inherits from $M$. So, there is a smooth structure on $F(N)$ such that $Fcolon N to F(N)$ is a diffeomorphism (the smooth coordinate maps are just maps of the form $fcirc F^{-1}$, where $f$ is a smooth coordinate map for $N$). Therefore, it gives us a smooth manifold $F(N)$ of dimension similar as $N$ which sits inside $M$.
Now, the definition of embedded submanifolds as given in the text of boothby is:
image of a topological embedding+immersion is an embedded submanifold
My problem is:
If just taking $Fcolon N to F(N)$ as a topological embedding gives $F(N)$ diffeomorphic to $N$. So, it gives a smooth manifold $F(N)$ with same dimension as $N$, why do we even need to consider immersions?
Thanks in advance
differential-geometry manifolds differential-topology smooth-manifolds
edited Jan 10 at 18:44
Christoph
12k1642
asked Jan 10 at 18:39
Abhishek ShrivastavaAbhishek Shrivastava
414413
$endgroup$
I am trying to understand the concept of embedded submanifolds and have the following understanding:
Suppose we have a smooth manifold $M$ and $N$ of dimensions $m$ and $n$ such that $mgt n$ .Let $Fcolon Nto M$ be a smooth map and a topological embedding onto its image $F(N)$ under subspace topology that $F(N)$ inherits from $M$. So, there is a smooth structure on $F(N)$ such that $Fcolon N to F(N)$ is a diffeomorphism (the smooth coordinate maps are just maps of the form $fcirc F^{-1}$, where $f$ is a smooth coordinate map for $N$). Therefore, it gives us a smooth manifold $F(N)$ of dimension similar as $N$ which sits inside $M$.
Now, the definition of embedded submanifolds as given in the text of boothby is:
image of a topological embedding+immersion is an embedded submanifold
My problem is:
If just taking $Fcolon N to F(N)$ as a topological embedding gives $F(N)$ diffeomorphic to $N$. So, it gives a smooth manifold $F(N)$ with same dimension as $N$, why do we even need to consider immersions?
Thanks in advance
differential-geometry manifolds differential-topology smooth-manifolds
differential-geometry manifolds differential-topology smooth-manifolds
edited Jan 10 at 18:44
Christoph
12k1642
asked Jan 10 at 18:39
Abhishek ShrivastavaAbhishek Shrivastava
414413
edited Jan 10 at 18:44
Christoph
12k1642
asked Jan 10 at 18:39
Abhishek ShrivastavaAbhishek Shrivastava
414413
edited Jan 10 at 18:44
Christoph
12k1642
edited Jan 10 at 18:44
Christoph
12k1642
edited Jan 10 at 18:44
Christoph
12k1642
12k1642
asked Jan 10 at 18:39
Abhishek ShrivastavaAbhishek Shrivastava
414413
asked Jan 10 at 18:39
Abhishek ShrivastavaAbhishek Shrivastava
414413
asked Jan 10 at 18:39
Abhishek ShrivastavaAbhishek Shrivastava
414413
414413
$begingroup$
What is the definition of topological embedding here?
$endgroup$
– Anubhav Mukherjee
Jan 10 at 18:48
$begingroup$
@Anubhav Mukherjee:it is a homeomorphism from N to f(N),where f(N) has subspace topology
$endgroup$
– Abhishek Shrivastava
Jan 10 at 18:50
add a comment |
$begingroup$
What is the definition of topological embedding here?
$endgroup$
– Anubhav Mukherjee
Jan 10 at 18:48
$begingroup$
@Anubhav Mukherjee:it is a homeomorphism from N to f(N),where f(N) has subspace topology
$endgroup$
– Abhishek Shrivastava
Jan 10 at 18:50
$begingroup$
What is the definition of topological embedding here?
$endgroup$
– Anubhav Mukherjee
Jan 10 at 18:48
$begingroup$
What is the definition of topological embedding here?
$endgroup$
– Anubhav Mukherjee
Jan 10 at 18:48
$begingroup$
@Anubhav Mukherjee:it is a homeomorphism from N to f(N),where f(N) has subspace topology
$endgroup$
– Abhishek Shrivastava
Jan 10 at 18:50
$begingroup$
@Anubhav Mukherjee:it is a homeomorphism from N to f(N),where f(N) has subspace topology
$endgroup$
– Abhishek Shrivastava
Jan 10 at 18:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $N=Bbb R$ and $M=Bbb R^2$. The map $Fcolon Nto M$, $xmapsto (x^3,0)$ is a topological embedding of $Bbb R$ into $Bbb R^2$. However, the image $F(N)=Bbb Rtimes 0subset Bbb R^2$ does not inherit the same smooth structure as a subspace of $M$ as it does by pushing forward the smooth structure from $N$. (The two smooth structures on $F(N)$ yield diffeomorphic manifolds, but are not equivalent.) To avoid this, you want $F$ to be an immersion, so that both smooth structures (the one pushed forward via $F$ and the one inherited from $M$) are the same.
From $M$ the subspace $Bbb Rtimes 0$ inherits a smooth structure with atlas defined by the global chart $(x,0)mapsto x$. Pushed forward via $F$, we get a smooth structure on $Bbb Rtimes 0$ with atlas defined by the global chart $(x,0)mapsto x^{1/3}$. Now check that with respect to these two atlases the identity map $Bbb Rtimes 0 to Bbb Rtimes 0$ is not a diffeomorphism, so the two smooth structures are different. They still define diffeomorphic manifolds, since $(x,0)mapsto (x^3,0)$ is a diffeomorphism with respect to these two atlases.
answered Jan 10 at 19:01
ChristophChristoph
12k1642
$endgroup$
$begingroup$
:thanks for the answer.could you elaborate:"However, the image $F(N)=Bbb Rtimes 0subset Bbb R^2$ does not inherit the same smooth structure as a subspace of $M$ as it does by pushing forward the smooth structure from $N$"
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:05
$begingroup$
I added some details, does this help?
$endgroup$
– Christoph
Jan 10 at 19:14
$begingroup$
:understood....thanks a lot
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:16
$begingroup$
:just to ask one more thing:
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:16
$begingroup$
:just to ask one more thing:as you told,i think i can have f(N) as a differentiable manifold just by considering a topological embedding as it gives a diffeomorphism but the differential structure so obtained on f(N) might not be related at all to the differential structure of manifold M in which f(N) sits.to avoid this,we define them as immersion which identifies tangent space at point p of $pmb N$ with a n dimensional subspace of tangent space at f(p) of $pmb M$"is this correct??
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:25
|
show 2 more comments
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1 Answer
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1 Answer
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$begingroup$
Let $N=Bbb R$ and $M=Bbb R^2$. The map $Fcolon Nto M$, $xmapsto (x^3,0)$ is a topological embedding of $Bbb R$ into $Bbb R^2$. However, the image $F(N)=Bbb Rtimes 0subset Bbb R^2$ does not inherit the same smooth structure as a subspace of $M$ as it does by pushing forward the smooth structure from $N$. (The two smooth structures on $F(N)$ yield diffeomorphic manifolds, but are not equivalent.) To avoid this, you want $F$ to be an immersion, so that both smooth structures (the one pushed forward via $F$ and the one inherited from $M$) are the same.
From $M$ the subspace $Bbb Rtimes 0$ inherits a smooth structure with atlas defined by the global chart $(x,0)mapsto x$. Pushed forward via $F$, we get a smooth structure on $Bbb Rtimes 0$ with atlas defined by the global chart $(x,0)mapsto x^{1/3}$. Now check that with respect to these two atlases the identity map $Bbb Rtimes 0 to Bbb Rtimes 0$ is not a diffeomorphism, so the two smooth structures are different. They still define diffeomorphic manifolds, since $(x,0)mapsto (x^3,0)$ is a diffeomorphism with respect to these two atlases.
answered Jan 10 at 19:01
ChristophChristoph
12k1642
$endgroup$
$begingroup$
:thanks for the answer.could you elaborate:"However, the image $F(N)=Bbb Rtimes 0subset Bbb R^2$ does not inherit the same smooth structure as a subspace of $M$ as it does by pushing forward the smooth structure from $N$"
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:05
$begingroup$
I added some details, does this help?
$endgroup$
– Christoph
Jan 10 at 19:14
$begingroup$
:understood....thanks a lot
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:16
$begingroup$
:just to ask one more thing:
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:16
$begingroup$
:just to ask one more thing:as you told,i think i can have f(N) as a differentiable manifold just by considering a topological embedding as it gives a diffeomorphism but the differential structure so obtained on f(N) might not be related at all to the differential structure of manifold M in which f(N) sits.to avoid this,we define them as immersion which identifies tangent space at point p of $pmb N$ with a n dimensional subspace of tangent space at f(p) of $pmb M$"is this correct??
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:25
|
show 2 more comments
$begingroup$
Let $N=Bbb R$ and $M=Bbb R^2$. The map $Fcolon Nto M$, $xmapsto (x^3,0)$ is a topological embedding of $Bbb R$ into $Bbb R^2$. However, the image $F(N)=Bbb Rtimes 0subset Bbb R^2$ does not inherit the same smooth structure as a subspace of $M$ as it does by pushing forward the smooth structure from $N$. (The two smooth structures on $F(N)$ yield diffeomorphic manifolds, but are not equivalent.) To avoid this, you want $F$ to be an immersion, so that both smooth structures (the one pushed forward via $F$ and the one inherited from $M$) are the same.
From $M$ the subspace $Bbb Rtimes 0$ inherits a smooth structure with atlas defined by the global chart $(x,0)mapsto x$. Pushed forward via $F$, we get a smooth structure on $Bbb Rtimes 0$ with atlas defined by the global chart $(x,0)mapsto x^{1/3}$. Now check that with respect to these two atlases the identity map $Bbb Rtimes 0 to Bbb Rtimes 0$ is not a diffeomorphism, so the two smooth structures are different. They still define diffeomorphic manifolds, since $(x,0)mapsto (x^3,0)$ is a diffeomorphism with respect to these two atlases.
answered Jan 10 at 19:01
ChristophChristoph
12k1642
$endgroup$
$begingroup$
:thanks for the answer.could you elaborate:"However, the image $F(N)=Bbb Rtimes 0subset Bbb R^2$ does not inherit the same smooth structure as a subspace of $M$ as it does by pushing forward the smooth structure from $N$"
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:05
$begingroup$
I added some details, does this help?
$endgroup$
– Christoph
Jan 10 at 19:14
$begingroup$
:understood....thanks a lot
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:16
$begingroup$
:just to ask one more thing:
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:16
$begingroup$
:just to ask one more thing:as you told,i think i can have f(N) as a differentiable manifold just by considering a topological embedding as it gives a diffeomorphism but the differential structure so obtained on f(N) might not be related at all to the differential structure of manifold M in which f(N) sits.to avoid this,we define them as immersion which identifies tangent space at point p of $pmb N$ with a n dimensional subspace of tangent space at f(p) of $pmb M$"is this correct??
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:25
|
show 2 more comments
$begingroup$
Let $N=Bbb R$ and $M=Bbb R^2$. The map $Fcolon Nto M$, $xmapsto (x^3,0)$ is a topological embedding of $Bbb R$ into $Bbb R^2$. However, the image $F(N)=Bbb Rtimes 0subset Bbb R^2$ does not inherit the same smooth structure as a subspace of $M$ as it does by pushing forward the smooth structure from $N$. (The two smooth structures on $F(N)$ yield diffeomorphic manifolds, but are not equivalent.) To avoid this, you want $F$ to be an immersion, so that both smooth structures (the one pushed forward via $F$ and the one inherited from $M$) are the same.
From $M$ the subspace $Bbb Rtimes 0$ inherits a smooth structure with atlas defined by the global chart $(x,0)mapsto x$. Pushed forward via $F$, we get a smooth structure on $Bbb Rtimes 0$ with atlas defined by the global chart $(x,0)mapsto x^{1/3}$. Now check that with respect to these two atlases the identity map $Bbb Rtimes 0 to Bbb Rtimes 0$ is not a diffeomorphism, so the two smooth structures are different. They still define diffeomorphic manifolds, since $(x,0)mapsto (x^3,0)$ is a diffeomorphism with respect to these two atlases.
answered Jan 10 at 19:01
ChristophChristoph
12k1642
$endgroup$
Let $N=Bbb R$ and $M=Bbb R^2$. The map $Fcolon Nto M$, $xmapsto (x^3,0)$ is a topological embedding of $Bbb R$ into $Bbb R^2$. However, the image $F(N)=Bbb Rtimes 0subset Bbb R^2$ does not inherit the same smooth structure as a subspace of $M$ as it does by pushing forward the smooth structure from $N$. (The two smooth structures on $F(N)$ yield diffeomorphic manifolds, but are not equivalent.) To avoid this, you want $F$ to be an immersion, so that both smooth structures (the one pushed forward via $F$ and the one inherited from $M$) are the same.
From $M$ the subspace $Bbb Rtimes 0$ inherits a smooth structure with atlas defined by the global chart $(x,0)mapsto x$. Pushed forward via $F$, we get a smooth structure on $Bbb Rtimes 0$ with atlas defined by the global chart $(x,0)mapsto x^{1/3}$. Now check that with respect to these two atlases the identity map $Bbb Rtimes 0 to Bbb Rtimes 0$ is not a diffeomorphism, so the two smooth structures are different. They still define diffeomorphic manifolds, since $(x,0)mapsto (x^3,0)$ is a diffeomorphism with respect to these two atlases.
answered Jan 10 at 19:01
ChristophChristoph
12k1642
edited Jan 10 at 19:13
answered Jan 10 at 19:01
ChristophChristoph
12k1642
answered Jan 10 at 19:01
ChristophChristoph
12k1642
answered Jan 10 at 19:01
ChristophChristoph
12k1642
12k1642
$begingroup$
:thanks for the answer.could you elaborate:"However, the image $F(N)=Bbb Rtimes 0subset Bbb R^2$ does not inherit the same smooth structure as a subspace of $M$ as it does by pushing forward the smooth structure from $N$"
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:05
$begingroup$
I added some details, does this help?
$endgroup$
– Christoph
Jan 10 at 19:14
$begingroup$
:understood....thanks a lot
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:16
$begingroup$
:just to ask one more thing:
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:16
$begingroup$
:just to ask one more thing:as you told,i think i can have f(N) as a differentiable manifold just by considering a topological embedding as it gives a diffeomorphism but the differential structure so obtained on f(N) might not be related at all to the differential structure of manifold M in which f(N) sits.to avoid this,we define them as immersion which identifies tangent space at point p of $pmb N$ with a n dimensional subspace of tangent space at f(p) of $pmb M$"is this correct??
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:25
|
show 2 more comments
$begingroup$
:thanks for the answer.could you elaborate:"However, the image $F(N)=Bbb Rtimes 0subset Bbb R^2$ does not inherit the same smooth structure as a subspace of $M$ as it does by pushing forward the smooth structure from $N$"
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:05
$begingroup$
I added some details, does this help?
$endgroup$
– Christoph
Jan 10 at 19:14
$begingroup$
:understood....thanks a lot
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:16
$begingroup$
:just to ask one more thing:
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:16
$begingroup$
:just to ask one more thing:as you told,i think i can have f(N) as a differentiable manifold just by considering a topological embedding as it gives a diffeomorphism but the differential structure so obtained on f(N) might not be related at all to the differential structure of manifold M in which f(N) sits.to avoid this,we define them as immersion which identifies tangent space at point p of $pmb N$ with a n dimensional subspace of tangent space at f(p) of $pmb M$"is this correct??
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:25
$begingroup$
:thanks for the answer.could you elaborate:"However, the image $F(N)=Bbb Rtimes 0subset Bbb R^2$ does not inherit the same smooth structure as a subspace of $M$ as it does by pushing forward the smooth structure from $N$"
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:05
$begingroup$
:thanks for the answer.could you elaborate:"However, the image $F(N)=Bbb Rtimes 0subset Bbb R^2$ does not inherit the same smooth structure as a subspace of $M$ as it does by pushing forward the smooth structure from $N$"
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:05
$begingroup$
I added some details, does this help?
$endgroup$
– Christoph
Jan 10 at 19:14
$begingroup$
I added some details, does this help?
$endgroup$
– Christoph
Jan 10 at 19:14
$begingroup$
:understood....thanks a lot
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:16
$begingroup$
:understood....thanks a lot
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:16
$begingroup$
:just to ask one more thing:
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:16
$begingroup$
:just to ask one more thing:
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:16
$begingroup$
:just to ask one more thing:as you told,i think i can have f(N) as a differentiable manifold just by considering a topological embedding as it gives a diffeomorphism but the differential structure so obtained on f(N) might not be related at all to the differential structure of manifold M in which f(N) sits.to avoid this,we define them as immersion which identifies tangent space at point p of $pmb N$ with a n dimensional subspace of tangent space at f(p) of $pmb M$"is this correct??
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:25
$begingroup$
:just to ask one more thing:as you told,i think i can have f(N) as a differentiable manifold just by considering a topological embedding as it gives a diffeomorphism but the differential structure so obtained on f(N) might not be related at all to the differential structure of manifold M in which f(N) sits.to avoid this,we define them as immersion which identifies tangent space at point p of $pmb N$ with a n dimensional subspace of tangent space at f(p) of $pmb M$"is this correct??
$endgroup$
– Abhishek Shrivastava
Jan 10 at 19:25
|
show 2 more comments
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$begingroup$
What is the definition of topological embedding here?
$endgroup$
– Anubhav Mukherjee
Jan 10 at 18:48
$begingroup$
@Anubhav Mukherjee:it is a homeomorphism from N to f(N),where f(N) has subspace topology
$endgroup$
– Abhishek Shrivastava
Jan 10 at 18:50