Adapted charts for smooth manifold
$begingroup$
Let $M$ be a smooth manifold of dimension $n$ and let $S$ be an embedded submanifold of $M$, i.e. a subset of $M$ which is given a structure of smooth manifold of dimension $kleq n$ such that the inclusion map $iota:S to M$ is a smooth embedding.
By the constant rank theorem, if $p$ is a point of $S$, then there exists smooth charts $(W,chi)$ for $S$ centered at $p$ and $(V,psi)$ for $M$ centered at $p$ with $W subseteq V$ such that the map $psicircchi^{-1}:chi(W)topsi (V)$ has the expression $(x^1,dots,x^k)mapsto (x^1,dots,x^k,0,dots,0) $.
In particular the map $psicircchi^{-1}:chi(W)topsi (W)$ is a bijection and thus $$psi(W)=chi(W)times{ underline{0} }$$
Since $W$ is open in $S$, we have $W=Scap A$ with $A$ open subset of $M$.
Let be $U=V cap A$. Then $U$ is open in $V$ and thus $psi(U)$ is open in $mathbb{R}^n$. We also have $Ucap S=W$. So we have $$psi(Ucap S)=chi(W)times{ underline{0} }quad [1]$$
I want to say that $psi(Ucap S)={ xinpsi(U) : x^{k+1}=dots=x^n=0}$.
My notes say that I can suppose $psi(U)=B^n_varepsilon(0)$ and $chi(W)=B^k_varepsilon(0)$ be open balls in $mathbb{R}^n$ and $mathbb{R}^k$ respectively with the same radius but I can't see why.
I know that I can WLOG assume $chi(W)=B^k_varepsilon(0)$ for a certain $varepsilon>0$ but then $U$ is determinated by $W$! (See the definition of $U$) So I can't modify $U$ and then still pretend that $[1]$ holds.
Can you tell me why the conclusion of my notes is true (if it is true!) or provide another strategy to conclude the argument?
general-topology differential-geometry smooth-manifolds smooth-functions
$endgroup$
add a comment |
$begingroup$
Let $M$ be a smooth manifold of dimension $n$ and let $S$ be an embedded submanifold of $M$, i.e. a subset of $M$ which is given a structure of smooth manifold of dimension $kleq n$ such that the inclusion map $iota:S to M$ is a smooth embedding.
By the constant rank theorem, if $p$ is a point of $S$, then there exists smooth charts $(W,chi)$ for $S$ centered at $p$ and $(V,psi)$ for $M$ centered at $p$ with $W subseteq V$ such that the map $psicircchi^{-1}:chi(W)topsi (V)$ has the expression $(x^1,dots,x^k)mapsto (x^1,dots,x^k,0,dots,0) $.
In particular the map $psicircchi^{-1}:chi(W)topsi (W)$ is a bijection and thus $$psi(W)=chi(W)times{ underline{0} }$$
Since $W$ is open in $S$, we have $W=Scap A$ with $A$ open subset of $M$.
Let be $U=V cap A$. Then $U$ is open in $V$ and thus $psi(U)$ is open in $mathbb{R}^n$. We also have $Ucap S=W$. So we have $$psi(Ucap S)=chi(W)times{ underline{0} }quad [1]$$
I want to say that $psi(Ucap S)={ xinpsi(U) : x^{k+1}=dots=x^n=0}$.
My notes say that I can suppose $psi(U)=B^n_varepsilon(0)$ and $chi(W)=B^k_varepsilon(0)$ be open balls in $mathbb{R}^n$ and $mathbb{R}^k$ respectively with the same radius but I can't see why.
I know that I can WLOG assume $chi(W)=B^k_varepsilon(0)$ for a certain $varepsilon>0$ but then $U$ is determinated by $W$! (See the definition of $U$) So I can't modify $U$ and then still pretend that $[1]$ holds.
Can you tell me why the conclusion of my notes is true (if it is true!) or provide another strategy to conclude the argument?
general-topology differential-geometry smooth-manifolds smooth-functions
$endgroup$
add a comment |
$begingroup$
Let $M$ be a smooth manifold of dimension $n$ and let $S$ be an embedded submanifold of $M$, i.e. a subset of $M$ which is given a structure of smooth manifold of dimension $kleq n$ such that the inclusion map $iota:S to M$ is a smooth embedding.
By the constant rank theorem, if $p$ is a point of $S$, then there exists smooth charts $(W,chi)$ for $S$ centered at $p$ and $(V,psi)$ for $M$ centered at $p$ with $W subseteq V$ such that the map $psicircchi^{-1}:chi(W)topsi (V)$ has the expression $(x^1,dots,x^k)mapsto (x^1,dots,x^k,0,dots,0) $.
In particular the map $psicircchi^{-1}:chi(W)topsi (W)$ is a bijection and thus $$psi(W)=chi(W)times{ underline{0} }$$
Since $W$ is open in $S$, we have $W=Scap A$ with $A$ open subset of $M$.
Let be $U=V cap A$. Then $U$ is open in $V$ and thus $psi(U)$ is open in $mathbb{R}^n$. We also have $Ucap S=W$. So we have $$psi(Ucap S)=chi(W)times{ underline{0} }quad [1]$$
I want to say that $psi(Ucap S)={ xinpsi(U) : x^{k+1}=dots=x^n=0}$.
My notes say that I can suppose $psi(U)=B^n_varepsilon(0)$ and $chi(W)=B^k_varepsilon(0)$ be open balls in $mathbb{R}^n$ and $mathbb{R}^k$ respectively with the same radius but I can't see why.
I know that I can WLOG assume $chi(W)=B^k_varepsilon(0)$ for a certain $varepsilon>0$ but then $U$ is determinated by $W$! (See the definition of $U$) So I can't modify $U$ and then still pretend that $[1]$ holds.
Can you tell me why the conclusion of my notes is true (if it is true!) or provide another strategy to conclude the argument?
general-topology differential-geometry smooth-manifolds smooth-functions
$endgroup$
Let $M$ be a smooth manifold of dimension $n$ and let $S$ be an embedded submanifold of $M$, i.e. a subset of $M$ which is given a structure of smooth manifold of dimension $kleq n$ such that the inclusion map $iota:S to M$ is a smooth embedding.
By the constant rank theorem, if $p$ is a point of $S$, then there exists smooth charts $(W,chi)$ for $S$ centered at $p$ and $(V,psi)$ for $M$ centered at $p$ with $W subseteq V$ such that the map $psicircchi^{-1}:chi(W)topsi (V)$ has the expression $(x^1,dots,x^k)mapsto (x^1,dots,x^k,0,dots,0) $.
In particular the map $psicircchi^{-1}:chi(W)topsi (W)$ is a bijection and thus $$psi(W)=chi(W)times{ underline{0} }$$
Since $W$ is open in $S$, we have $W=Scap A$ with $A$ open subset of $M$.
Let be $U=V cap A$. Then $U$ is open in $V$ and thus $psi(U)$ is open in $mathbb{R}^n$. We also have $Ucap S=W$. So we have $$psi(Ucap S)=chi(W)times{ underline{0} }quad [1]$$
I want to say that $psi(Ucap S)={ xinpsi(U) : x^{k+1}=dots=x^n=0}$.
My notes say that I can suppose $psi(U)=B^n_varepsilon(0)$ and $chi(W)=B^k_varepsilon(0)$ be open balls in $mathbb{R}^n$ and $mathbb{R}^k$ respectively with the same radius but I can't see why.
I know that I can WLOG assume $chi(W)=B^k_varepsilon(0)$ for a certain $varepsilon>0$ but then $U$ is determinated by $W$! (See the definition of $U$) So I can't modify $U$ and then still pretend that $[1]$ holds.
Can you tell me why the conclusion of my notes is true (if it is true!) or provide another strategy to conclude the argument?
general-topology differential-geometry smooth-manifolds smooth-functions
general-topology differential-geometry smooth-manifolds smooth-functions
asked Dec 28 '18 at 0:59
MinatoMinato
492313
492313
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Let us understand what the claim is doing here.
Definition: Let $S subseteq M$, $S$ satisfies the $k$-slice condition if each point of $S$ lies in smooth chart $(U, varphi)$ for $M$ such that $S cap U$ is a single $k$-slice in $U$, i.e. for a constant $c$, in coordinates,
$$ S = { x in U , : , x^{i}=c^i, , k+1 le i le n}$$
So what we want to show is thatany embedded manifold satisfies the condition.
Your set equality is by definition: but let me explain the steps.
Making the radius smaller is used in first changing your $W$ and $V$. Choose open balls $B' subseteq V$, $B subseteq W$, such that the image of $B$ is precisely a $k$-sliceof $B'$. This is possible by making the radius of the balls sufficiently small. We may take these as our $W$ and $V$.
One might think we are done here. That $V$ is our desired $k$-slice. i.e. $V cap S=W$. Nonetheless, we have:
$$ W subseteq V cap S$$
The remedy is to introduce your open set $A$.
As you wrote, $W = S cap A$ for an open set $A$ in $M$. Then $U= A cap V$ is our desired $k$-slice. This is a set chase. One direction is already proven. Now $$ U cap S = (A cap V) cap S = (A cap S) cap V = V cap W = W$$
In words, this means,
$U cap S$ is precisely the $k$-slice of $U$
At this point it is clear, in coordinates: if $(x,0) in U cap S$, then it lies both in $S$ and is in the $k$-slice of $U$. But above shows this implies it must lie in $W$. Hence $U cap S subseteq W$. The converse is also clear, we know $W$ is represented as the $k$-slice in $U cap S$. Thus your equality.
In fact more is true. If a subset satisfies the condition, then it can be given a (unique) manifold structure such that it is an embedded $k$-manifold.
$endgroup$
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$begingroup$
Let us understand what the claim is doing here.
Definition: Let $S subseteq M$, $S$ satisfies the $k$-slice condition if each point of $S$ lies in smooth chart $(U, varphi)$ for $M$ such that $S cap U$ is a single $k$-slice in $U$, i.e. for a constant $c$, in coordinates,
$$ S = { x in U , : , x^{i}=c^i, , k+1 le i le n}$$
So what we want to show is thatany embedded manifold satisfies the condition.
Your set equality is by definition: but let me explain the steps.
Making the radius smaller is used in first changing your $W$ and $V$. Choose open balls $B' subseteq V$, $B subseteq W$, such that the image of $B$ is precisely a $k$-sliceof $B'$. This is possible by making the radius of the balls sufficiently small. We may take these as our $W$ and $V$.
One might think we are done here. That $V$ is our desired $k$-slice. i.e. $V cap S=W$. Nonetheless, we have:
$$ W subseteq V cap S$$
The remedy is to introduce your open set $A$.
As you wrote, $W = S cap A$ for an open set $A$ in $M$. Then $U= A cap V$ is our desired $k$-slice. This is a set chase. One direction is already proven. Now $$ U cap S = (A cap V) cap S = (A cap S) cap V = V cap W = W$$
In words, this means,
$U cap S$ is precisely the $k$-slice of $U$
At this point it is clear, in coordinates: if $(x,0) in U cap S$, then it lies both in $S$ and is in the $k$-slice of $U$. But above shows this implies it must lie in $W$. Hence $U cap S subseteq W$. The converse is also clear, we know $W$ is represented as the $k$-slice in $U cap S$. Thus your equality.
In fact more is true. If a subset satisfies the condition, then it can be given a (unique) manifold structure such that it is an embedded $k$-manifold.
$endgroup$
add a comment |
$begingroup$
Let us understand what the claim is doing here.
Definition: Let $S subseteq M$, $S$ satisfies the $k$-slice condition if each point of $S$ lies in smooth chart $(U, varphi)$ for $M$ such that $S cap U$ is a single $k$-slice in $U$, i.e. for a constant $c$, in coordinates,
$$ S = { x in U , : , x^{i}=c^i, , k+1 le i le n}$$
So what we want to show is thatany embedded manifold satisfies the condition.
Your set equality is by definition: but let me explain the steps.
Making the radius smaller is used in first changing your $W$ and $V$. Choose open balls $B' subseteq V$, $B subseteq W$, such that the image of $B$ is precisely a $k$-sliceof $B'$. This is possible by making the radius of the balls sufficiently small. We may take these as our $W$ and $V$.
One might think we are done here. That $V$ is our desired $k$-slice. i.e. $V cap S=W$. Nonetheless, we have:
$$ W subseteq V cap S$$
The remedy is to introduce your open set $A$.
As you wrote, $W = S cap A$ for an open set $A$ in $M$. Then $U= A cap V$ is our desired $k$-slice. This is a set chase. One direction is already proven. Now $$ U cap S = (A cap V) cap S = (A cap S) cap V = V cap W = W$$
In words, this means,
$U cap S$ is precisely the $k$-slice of $U$
At this point it is clear, in coordinates: if $(x,0) in U cap S$, then it lies both in $S$ and is in the $k$-slice of $U$. But above shows this implies it must lie in $W$. Hence $U cap S subseteq W$. The converse is also clear, we know $W$ is represented as the $k$-slice in $U cap S$. Thus your equality.
In fact more is true. If a subset satisfies the condition, then it can be given a (unique) manifold structure such that it is an embedded $k$-manifold.
$endgroup$
add a comment |
$begingroup$
Let us understand what the claim is doing here.
Definition: Let $S subseteq M$, $S$ satisfies the $k$-slice condition if each point of $S$ lies in smooth chart $(U, varphi)$ for $M$ such that $S cap U$ is a single $k$-slice in $U$, i.e. for a constant $c$, in coordinates,
$$ S = { x in U , : , x^{i}=c^i, , k+1 le i le n}$$
So what we want to show is thatany embedded manifold satisfies the condition.
Your set equality is by definition: but let me explain the steps.
Making the radius smaller is used in first changing your $W$ and $V$. Choose open balls $B' subseteq V$, $B subseteq W$, such that the image of $B$ is precisely a $k$-sliceof $B'$. This is possible by making the radius of the balls sufficiently small. We may take these as our $W$ and $V$.
One might think we are done here. That $V$ is our desired $k$-slice. i.e. $V cap S=W$. Nonetheless, we have:
$$ W subseteq V cap S$$
The remedy is to introduce your open set $A$.
As you wrote, $W = S cap A$ for an open set $A$ in $M$. Then $U= A cap V$ is our desired $k$-slice. This is a set chase. One direction is already proven. Now $$ U cap S = (A cap V) cap S = (A cap S) cap V = V cap W = W$$
In words, this means,
$U cap S$ is precisely the $k$-slice of $U$
At this point it is clear, in coordinates: if $(x,0) in U cap S$, then it lies both in $S$ and is in the $k$-slice of $U$. But above shows this implies it must lie in $W$. Hence $U cap S subseteq W$. The converse is also clear, we know $W$ is represented as the $k$-slice in $U cap S$. Thus your equality.
In fact more is true. If a subset satisfies the condition, then it can be given a (unique) manifold structure such that it is an embedded $k$-manifold.
$endgroup$
Let us understand what the claim is doing here.
Definition: Let $S subseteq M$, $S$ satisfies the $k$-slice condition if each point of $S$ lies in smooth chart $(U, varphi)$ for $M$ such that $S cap U$ is a single $k$-slice in $U$, i.e. for a constant $c$, in coordinates,
$$ S = { x in U , : , x^{i}=c^i, , k+1 le i le n}$$
So what we want to show is thatany embedded manifold satisfies the condition.
Your set equality is by definition: but let me explain the steps.
Making the radius smaller is used in first changing your $W$ and $V$. Choose open balls $B' subseteq V$, $B subseteq W$, such that the image of $B$ is precisely a $k$-sliceof $B'$. This is possible by making the radius of the balls sufficiently small. We may take these as our $W$ and $V$.
One might think we are done here. That $V$ is our desired $k$-slice. i.e. $V cap S=W$. Nonetheless, we have:
$$ W subseteq V cap S$$
The remedy is to introduce your open set $A$.
As you wrote, $W = S cap A$ for an open set $A$ in $M$. Then $U= A cap V$ is our desired $k$-slice. This is a set chase. One direction is already proven. Now $$ U cap S = (A cap V) cap S = (A cap S) cap V = V cap W = W$$
In words, this means,
$U cap S$ is precisely the $k$-slice of $U$
At this point it is clear, in coordinates: if $(x,0) in U cap S$, then it lies both in $S$ and is in the $k$-slice of $U$. But above shows this implies it must lie in $W$. Hence $U cap S subseteq W$. The converse is also clear, we know $W$ is represented as the $k$-slice in $U cap S$. Thus your equality.
In fact more is true. If a subset satisfies the condition, then it can be given a (unique) manifold structure such that it is an embedded $k$-manifold.
answered Dec 28 '18 at 3:06
CL.CL.
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2,2402925
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