Adapted charts for smooth manifold












1














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?











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    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?











    share|cite|improve this question

























      1












      1








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      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?











      share|cite|improve this question













      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






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      asked Dec 28 '18 at 0:59









      Minato

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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.






          share|cite|improve this answer





















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            1 Answer
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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.






            share|cite|improve this answer


























              0
















              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.






              share|cite|improve this answer
























                0












                0








                0








                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.






                share|cite|improve this answer














                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.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 28 '18 at 3:06









                CL.

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