Embedding a compact manifold in $mathbb{R}^N$












2












$begingroup$


I have an attempt at solving the following problem. This is not so much a question asking for a solution in general, but more on how to complete my own.




Let $M^n$ be a compact smooth manifold. Show that there exists an embedding of $M$ into $mathbb{R}^N$ for some $N$.




(Recall that an embedding of smooth manifolds is a topological embedding such that each differential is injective.)



My attempt:



Since $M$ is compact, we can cover it by finitely many coordinate neighbourhoods $U_i$ for $i=1,dotsc,k$, where $varphi_i : U_i to mathbb{R}^n$ are the corresponding charts. Choose a subordinate (smooth) partition of unity $psi_i : M to mathbb{R}$. Then the functions
$$ f_i := psi_i cdot varphi_i$$
are smooth, where we interpret $varphi$ as being zero outside of the support of $psi$. Now define
$$ F : M to mathbb{R}^{ncdot k + k} : xmapsto (f_1(x),dotsc,f_k(x),psi_1(x),dotsc,psi_k(x)).$$
My hope was that $F$ is an embedding of smooth manifolds. For this, we verify:





  • Injectivity. This is why the $psi$'s were stuck at the end of the map. If all the $psi(x)$'s are the same, then all the $varphi(x)$'s are the same, but these are local diffeomorphisms, in particular bijections.


  • Smoothness. Trivial.


  • Topological embedding. Immediate since $M$ is compact and $F$ is injective and continuous.


  • Injective differentials. This is my issue.


Are all the differentials injective for $F$ as defined above? Or would we need more assumptions on the covering or of the partition of unity for this to work (or for this to work more easily)?






differential-geometry smooth-manifolds compact-manifolds







share|cite|improve this question









$endgroup$












  • $begingroup$
    This is whitney lemma
    $endgroup$
    – Tsemo Aristide
    Dec 12 '18 at 18:26












  • $begingroup$
    @TsemoAristide Which lemma are you referring to? If it's just about embedding a compact manifold in $mathbb{R}^N$, then like I stated, that's not the point of me raising this question.
    $endgroup$
    – SvanN
    Dec 12 '18 at 18:27










  • $begingroup$
    en.m.wikipedia.org/wiki/Whitney_embedding_theorem
    $endgroup$
    – Tsemo Aristide
    Dec 12 '18 at 18:29










  • $begingroup$
    @TsemoAristide Thanks for the reference, but again, I am asking this question about my own attempted solution, not a general proof.
    $endgroup$
    – SvanN
    Dec 12 '18 at 18:30










  • $begingroup$
    Read the proof of Whitney's theorem, it starts like your idea, then by using sard lemma it can be shown that $M$ can be embedded in $mathbb{R} ^{2n+1}$ where $n$ is the dimension of $M$.
    $endgroup$
    – Tsemo Aristide
    Dec 12 '18 at 18:43
















2












$begingroup$


I have an attempt at solving the following problem. This is not so much a question asking for a solution in general, but more on how to complete my own.




Let $M^n$ be a compact smooth manifold. Show that there exists an embedding of $M$ into $mathbb{R}^N$ for some $N$.




(Recall that an embedding of smooth manifolds is a topological embedding such that each differential is injective.)



My attempt:



Since $M$ is compact, we can cover it by finitely many coordinate neighbourhoods $U_i$ for $i=1,dotsc,k$, where $varphi_i : U_i to mathbb{R}^n$ are the corresponding charts. Choose a subordinate (smooth) partition of unity $psi_i : M to mathbb{R}$. Then the functions
$$ f_i := psi_i cdot varphi_i$$
are smooth, where we interpret $varphi$ as being zero outside of the support of $psi$. Now define
$$ F : M to mathbb{R}^{ncdot k + k} : xmapsto (f_1(x),dotsc,f_k(x),psi_1(x),dotsc,psi_k(x)).$$
My hope was that $F$ is an embedding of smooth manifolds. For this, we verify:





  • Injectivity. This is why the $psi$'s were stuck at the end of the map. If all the $psi(x)$'s are the same, then all the $varphi(x)$'s are the same, but these are local diffeomorphisms, in particular bijections.


  • Smoothness. Trivial.


  • Topological embedding. Immediate since $M$ is compact and $F$ is injective and continuous.


  • Injective differentials. This is my issue.


Are all the differentials injective for $F$ as defined above? Or would we need more assumptions on the covering or of the partition of unity for this to work (or for this to work more easily)?






differential-geometry smooth-manifolds compact-manifolds







share|cite|improve this question









$endgroup$












  • $begingroup$
    This is whitney lemma
    $endgroup$
    – Tsemo Aristide
    Dec 12 '18 at 18:26












  • $begingroup$
    @TsemoAristide Which lemma are you referring to? If it's just about embedding a compact manifold in $mathbb{R}^N$, then like I stated, that's not the point of me raising this question.
    $endgroup$
    – SvanN
    Dec 12 '18 at 18:27










  • $begingroup$
    en.m.wikipedia.org/wiki/Whitney_embedding_theorem
    $endgroup$
    – Tsemo Aristide
    Dec 12 '18 at 18:29










  • $begingroup$
    @TsemoAristide Thanks for the reference, but again, I am asking this question about my own attempted solution, not a general proof.
    $endgroup$
    – SvanN
    Dec 12 '18 at 18:30










  • $begingroup$
    Read the proof of Whitney's theorem, it starts like your idea, then by using sard lemma it can be shown that $M$ can be embedded in $mathbb{R} ^{2n+1}$ where $n$ is the dimension of $M$.
    $endgroup$
    – Tsemo Aristide
    Dec 12 '18 at 18:43














2












2








2





$begingroup$


I have an attempt at solving the following problem. This is not so much a question asking for a solution in general, but more on how to complete my own.




Let $M^n$ be a compact smooth manifold. Show that there exists an embedding of $M$ into $mathbb{R}^N$ for some $N$.




(Recall that an embedding of smooth manifolds is a topological embedding such that each differential is injective.)



My attempt:



Since $M$ is compact, we can cover it by finitely many coordinate neighbourhoods $U_i$ for $i=1,dotsc,k$, where $varphi_i : U_i to mathbb{R}^n$ are the corresponding charts. Choose a subordinate (smooth) partition of unity $psi_i : M to mathbb{R}$. Then the functions
$$ f_i := psi_i cdot varphi_i$$
are smooth, where we interpret $varphi$ as being zero outside of the support of $psi$. Now define
$$ F : M to mathbb{R}^{ncdot k + k} : xmapsto (f_1(x),dotsc,f_k(x),psi_1(x),dotsc,psi_k(x)).$$
My hope was that $F$ is an embedding of smooth manifolds. For this, we verify:





  • Injectivity. This is why the $psi$'s were stuck at the end of the map. If all the $psi(x)$'s are the same, then all the $varphi(x)$'s are the same, but these are local diffeomorphisms, in particular bijections.


  • Smoothness. Trivial.


  • Topological embedding. Immediate since $M$ is compact and $F$ is injective and continuous.


  • Injective differentials. This is my issue.


Are all the differentials injective for $F$ as defined above? Or would we need more assumptions on the covering or of the partition of unity for this to work (or for this to work more easily)?






differential-geometry smooth-manifolds compact-manifolds







share|cite|improve this question









$endgroup$




I have an attempt at solving the following problem. This is not so much a question asking for a solution in general, but more on how to complete my own.




Let $M^n$ be a compact smooth manifold. Show that there exists an embedding of $M$ into $mathbb{R}^N$ for some $N$.




(Recall that an embedding of smooth manifolds is a topological embedding such that each differential is injective.)



My attempt:



Since $M$ is compact, we can cover it by finitely many coordinate neighbourhoods $U_i$ for $i=1,dotsc,k$, where $varphi_i : U_i to mathbb{R}^n$ are the corresponding charts. Choose a subordinate (smooth) partition of unity $psi_i : M to mathbb{R}$. Then the functions
$$ f_i := psi_i cdot varphi_i$$
are smooth, where we interpret $varphi$ as being zero outside of the support of $psi$. Now define
$$ F : M to mathbb{R}^{ncdot k + k} : xmapsto (f_1(x),dotsc,f_k(x),psi_1(x),dotsc,psi_k(x)).$$
My hope was that $F$ is an embedding of smooth manifolds. For this, we verify:





  • Injectivity. This is why the $psi$'s were stuck at the end of the map. If all the $psi(x)$'s are the same, then all the $varphi(x)$'s are the same, but these are local diffeomorphisms, in particular bijections.


  • Smoothness. Trivial.


  • Topological embedding. Immediate since $M$ is compact and $F$ is injective and continuous.


  • Injective differentials. This is my issue.


Are all the differentials injective for $F$ as defined above? Or would we need more assumptions on the covering or of the partition of unity for this to work (or for this to work more easily)?






differential-geometry smooth-manifolds compact-manifolds




differential-geometry smooth-manifolds compact-manifolds






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 12 '18 at 18:25









SvanNSvanN

2,0661422




2,0661422












  • $begingroup$
    This is whitney lemma
    $endgroup$
    – Tsemo Aristide
    Dec 12 '18 at 18:26












  • $begingroup$
    @TsemoAristide Which lemma are you referring to? If it's just about embedding a compact manifold in $mathbb{R}^N$, then like I stated, that's not the point of me raising this question.
    $endgroup$
    – SvanN
    Dec 12 '18 at 18:27










  • $begingroup$
    en.m.wikipedia.org/wiki/Whitney_embedding_theorem
    $endgroup$
    – Tsemo Aristide
    Dec 12 '18 at 18:29










  • $begingroup$
    @TsemoAristide Thanks for the reference, but again, I am asking this question about my own attempted solution, not a general proof.
    $endgroup$
    – SvanN
    Dec 12 '18 at 18:30










  • $begingroup$
    Read the proof of Whitney's theorem, it starts like your idea, then by using sard lemma it can be shown that $M$ can be embedded in $mathbb{R} ^{2n+1}$ where $n$ is the dimension of $M$.
    $endgroup$
    – Tsemo Aristide
    Dec 12 '18 at 18:43


















  • $begingroup$
    This is whitney lemma
    $endgroup$
    – Tsemo Aristide
    Dec 12 '18 at 18:26












  • $begingroup$
    @TsemoAristide Which lemma are you referring to? If it's just about embedding a compact manifold in $mathbb{R}^N$, then like I stated, that's not the point of me raising this question.
    $endgroup$
    – SvanN
    Dec 12 '18 at 18:27










  • $begingroup$
    en.m.wikipedia.org/wiki/Whitney_embedding_theorem
    $endgroup$
    – Tsemo Aristide
    Dec 12 '18 at 18:29










  • $begingroup$
    @TsemoAristide Thanks for the reference, but again, I am asking this question about my own attempted solution, not a general proof.
    $endgroup$
    – SvanN
    Dec 12 '18 at 18:30










  • $begingroup$
    Read the proof of Whitney's theorem, it starts like your idea, then by using sard lemma it can be shown that $M$ can be embedded in $mathbb{R} ^{2n+1}$ where $n$ is the dimension of $M$.
    $endgroup$
    – Tsemo Aristide
    Dec 12 '18 at 18:43
















$begingroup$
This is whitney lemma
$endgroup$
– Tsemo Aristide
Dec 12 '18 at 18:26






$begingroup$
This is whitney lemma
$endgroup$
– Tsemo Aristide
Dec 12 '18 at 18:26














$begingroup$
@TsemoAristide Which lemma are you referring to? If it's just about embedding a compact manifold in $mathbb{R}^N$, then like I stated, that's not the point of me raising this question.
$endgroup$
– SvanN
Dec 12 '18 at 18:27




$begingroup$
@TsemoAristide Which lemma are you referring to? If it's just about embedding a compact manifold in $mathbb{R}^N$, then like I stated, that's not the point of me raising this question.
$endgroup$
– SvanN
Dec 12 '18 at 18:27












$begingroup$
en.m.wikipedia.org/wiki/Whitney_embedding_theorem
$endgroup$
– Tsemo Aristide
Dec 12 '18 at 18:29




$begingroup$
en.m.wikipedia.org/wiki/Whitney_embedding_theorem
$endgroup$
– Tsemo Aristide
Dec 12 '18 at 18:29












$begingroup$
@TsemoAristide Thanks for the reference, but again, I am asking this question about my own attempted solution, not a general proof.
$endgroup$
– SvanN
Dec 12 '18 at 18:30




$begingroup$
@TsemoAristide Thanks for the reference, but again, I am asking this question about my own attempted solution, not a general proof.
$endgroup$
– SvanN
Dec 12 '18 at 18:30












$begingroup$
Read the proof of Whitney's theorem, it starts like your idea, then by using sard lemma it can be shown that $M$ can be embedded in $mathbb{R} ^{2n+1}$ where $n$ is the dimension of $M$.
$endgroup$
– Tsemo Aristide
Dec 12 '18 at 18:43




$begingroup$
Read the proof of Whitney's theorem, it starts like your idea, then by using sard lemma it can be shown that $M$ can be embedded in $mathbb{R} ^{2n+1}$ where $n$ is the dimension of $M$.
$endgroup$
– Tsemo Aristide
Dec 12 '18 at 18:43










1 Answer
1






active

oldest

votes


















0












$begingroup$

Let $x in M$ be such that $T_x F$ is not injective. Then there exists some nonzero $X in T_x M$ such that $T_xF(X)=0$.



Let $i$ be such that $psi_i(x)>0$, then $d_x psi_i(X)=0$ (It is a component of $F$) and $T_x(psi_iphi_i)(X)=0$.
Now, since $d_xpsi_i(X)=0$, $0=T_x(psi_iphi_i)(X) = psi_i(x)T_xphi_i(X)$.



Thus, $T_xphi_i(X)=0$, a contradiction.
So your differentials are injective.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How does it follow that $nabla_x psi_i =0$? (And I’m presuming that by $nabla psi$ you mean the differential of $psi$?)
    $endgroup$
    – SvanN
    Dec 12 '18 at 20:24










  • $begingroup$
    Edited, I hope I answered your concern.
    $endgroup$
    – Mindlack
    Dec 12 '18 at 20:31










  • $begingroup$
    You did, yes :) Thank you!
    $endgroup$
    – SvanN
    Dec 12 '18 at 20:49











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Let $x in M$ be such that $T_x F$ is not injective. Then there exists some nonzero $X in T_x M$ such that $T_xF(X)=0$.



Let $i$ be such that $psi_i(x)>0$, then $d_x psi_i(X)=0$ (It is a component of $F$) and $T_x(psi_iphi_i)(X)=0$.
Now, since $d_xpsi_i(X)=0$, $0=T_x(psi_iphi_i)(X) = psi_i(x)T_xphi_i(X)$.



Thus, $T_xphi_i(X)=0$, a contradiction.
So your differentials are injective.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How does it follow that $nabla_x psi_i =0$? (And I’m presuming that by $nabla psi$ you mean the differential of $psi$?)
    $endgroup$
    – SvanN
    Dec 12 '18 at 20:24










  • $begingroup$
    Edited, I hope I answered your concern.
    $endgroup$
    – Mindlack
    Dec 12 '18 at 20:31










  • $begingroup$
    You did, yes :) Thank you!
    $endgroup$
    – SvanN
    Dec 12 '18 at 20:49
















0












$begingroup$

Let $x in M$ be such that $T_x F$ is not injective. Then there exists some nonzero $X in T_x M$ such that $T_xF(X)=0$.



Let $i$ be such that $psi_i(x)>0$, then $d_x psi_i(X)=0$ (It is a component of $F$) and $T_x(psi_iphi_i)(X)=0$.
Now, since $d_xpsi_i(X)=0$, $0=T_x(psi_iphi_i)(X) = psi_i(x)T_xphi_i(X)$.



Thus, $T_xphi_i(X)=0$, a contradiction.
So your differentials are injective.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How does it follow that $nabla_x psi_i =0$? (And I’m presuming that by $nabla psi$ you mean the differential of $psi$?)
    $endgroup$
    – SvanN
    Dec 12 '18 at 20:24










  • $begingroup$
    Edited, I hope I answered your concern.
    $endgroup$
    – Mindlack
    Dec 12 '18 at 20:31










  • $begingroup$
    You did, yes :) Thank you!
    $endgroup$
    – SvanN
    Dec 12 '18 at 20:49














0












0








0





$begingroup$

Let $x in M$ be such that $T_x F$ is not injective. Then there exists some nonzero $X in T_x M$ such that $T_xF(X)=0$.



Let $i$ be such that $psi_i(x)>0$, then $d_x psi_i(X)=0$ (It is a component of $F$) and $T_x(psi_iphi_i)(X)=0$.
Now, since $d_xpsi_i(X)=0$, $0=T_x(psi_iphi_i)(X) = psi_i(x)T_xphi_i(X)$.



Thus, $T_xphi_i(X)=0$, a contradiction.
So your differentials are injective.






share|cite|improve this answer











$endgroup$



Let $x in M$ be such that $T_x F$ is not injective. Then there exists some nonzero $X in T_x M$ such that $T_xF(X)=0$.



Let $i$ be such that $psi_i(x)>0$, then $d_x psi_i(X)=0$ (It is a component of $F$) and $T_x(psi_iphi_i)(X)=0$.
Now, since $d_xpsi_i(X)=0$, $0=T_x(psi_iphi_i)(X) = psi_i(x)T_xphi_i(X)$.



Thus, $T_xphi_i(X)=0$, a contradiction.
So your differentials are injective.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 12 '18 at 20:30

























answered Dec 12 '18 at 20:21









MindlackMindlack

4,760210




4,760210












  • $begingroup$
    How does it follow that $nabla_x psi_i =0$? (And I’m presuming that by $nabla psi$ you mean the differential of $psi$?)
    $endgroup$
    – SvanN
    Dec 12 '18 at 20:24










  • $begingroup$
    Edited, I hope I answered your concern.
    $endgroup$
    – Mindlack
    Dec 12 '18 at 20:31










  • $begingroup$
    You did, yes :) Thank you!
    $endgroup$
    – SvanN
    Dec 12 '18 at 20:49


















  • $begingroup$
    How does it follow that $nabla_x psi_i =0$? (And I’m presuming that by $nabla psi$ you mean the differential of $psi$?)
    $endgroup$
    – SvanN
    Dec 12 '18 at 20:24










  • $begingroup$
    Edited, I hope I answered your concern.
    $endgroup$
    – Mindlack
    Dec 12 '18 at 20:31










  • $begingroup$
    You did, yes :) Thank you!
    $endgroup$
    – SvanN
    Dec 12 '18 at 20:49
















$begingroup$
How does it follow that $nabla_x psi_i =0$? (And I’m presuming that by $nabla psi$ you mean the differential of $psi$?)
$endgroup$
– SvanN
Dec 12 '18 at 20:24




$begingroup$
How does it follow that $nabla_x psi_i =0$? (And I’m presuming that by $nabla psi$ you mean the differential of $psi$?)
$endgroup$
– SvanN
Dec 12 '18 at 20:24












$begingroup$
Edited, I hope I answered your concern.
$endgroup$
– Mindlack
Dec 12 '18 at 20:31




$begingroup$
Edited, I hope I answered your concern.
$endgroup$
– Mindlack
Dec 12 '18 at 20:31












$begingroup$
You did, yes :) Thank you!
$endgroup$
– SvanN
Dec 12 '18 at 20:49




$begingroup$
You did, yes :) Thank you!
$endgroup$
– SvanN
Dec 12 '18 at 20:49


















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Cáceres

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Para otros usos de este término, véase Cáceres (desambiguación). Cáceres Municipio y ciudad Bandera Escudo Collage de Cáceres Cáceres Ubicación de Cáceres en España. Cáceres Ubicación de Cáceres en la provincia de Cáceres. País  España • Com. autónoma  Extremadura • Provincia  Cáceres • Partido judicial Cáceres Ubicación 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 Coordenadas: 39°28′23″N 6°22′16″O  /  39.473055555556, -6.3711111111111 • Altitud 457 [ 1 ] ​ msnm (mín.:211 [ 2 ] ​, máx.:708 [ 2 ] ​) Superficie 1750,33 km² Núcleos de población Cáceres Estación Arroyo-Malpartida Rincón de Ballesteros Valdesalor Fundación 34 a. C. Población 96 068 hab. (2018) • Densidad 54,8 hab./km² Gentilicio Cacereño/a [ 3 ] ​ Código postal 10001-10005, 10195 Alcaldesa (2015) María Elena Nevado del Campo [ 4 ] ​ Presupuesto 69.045.004€ [ 5 ] ​  (año ...
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