Characterization of being a submersion












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Let $mathcal{M}$ be a smooth $m$-dimensional manifold and let $Fcolon mathcal{M} longrightarrow mathbb{R}^{n}$ be any smooth map, with m $geq$ n . I want to prove that, if $p in mathcal{M}$:
$$ text{$F$ is a submersion at $p$} iff lbrace d_{p}F^{1},dots,d_{p}F^{n}rbrace text{ are l.i. in $(T_{p}mathcal{M})^{*}$} $$
Were $d_{p}f := lambda^{-1}_{f(p)} circ T_{p}f colon T_{p}mathcal{M} longrightarrow mathbb{R} in (T_{p}mathcal{M})^{*}$, $T_{p}fcolon T_{p}mathcal{M}longrightarrow T_{f(p)}mathcal{M}$ is the tangent map at $p$ and $lambda_{p}colon V longrightarrow T_{p}V$ is the canonical isomorphism between a finite dimensional vector space over $mathbb{R}$ and its tangent space. We can extend the definition of differential of a smooth map $Fcolon mathcal{M} longrightarrow V$ by putting $D_{p}F := lambda^{-1}_{F(p)} circ T_{p}F colon T_{p}mathcal{M} longrightarrow V$, where $V$ is a finite dimensional vector space over $mathbb{R}$.

This is trivial if you take $mathcal{M} = mathbb{R}^{m}$ since you can identify the tangent map with the differential from ordinary calculus, but I want to formalize the proof as much as possible in the general case using the definitions I gave.










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    $begingroup$
    This is a linear algebra question. You just want to know that a map $V to W$ of finite dimensional vector spaces is surjective iff its adjoint $W^* to V^*$ is injective.
    $endgroup$
    – user98602
    Dec 29 '18 at 23:41


















1












$begingroup$


Let $mathcal{M}$ be a smooth $m$-dimensional manifold and let $Fcolon mathcal{M} longrightarrow mathbb{R}^{n}$ be any smooth map, with m $geq$ n . I want to prove that, if $p in mathcal{M}$:
$$ text{$F$ is a submersion at $p$} iff lbrace d_{p}F^{1},dots,d_{p}F^{n}rbrace text{ are l.i. in $(T_{p}mathcal{M})^{*}$} $$
Were $d_{p}f := lambda^{-1}_{f(p)} circ T_{p}f colon T_{p}mathcal{M} longrightarrow mathbb{R} in (T_{p}mathcal{M})^{*}$, $T_{p}fcolon T_{p}mathcal{M}longrightarrow T_{f(p)}mathcal{M}$ is the tangent map at $p$ and $lambda_{p}colon V longrightarrow T_{p}V$ is the canonical isomorphism between a finite dimensional vector space over $mathbb{R}$ and its tangent space. We can extend the definition of differential of a smooth map $Fcolon mathcal{M} longrightarrow V$ by putting $D_{p}F := lambda^{-1}_{F(p)} circ T_{p}F colon T_{p}mathcal{M} longrightarrow V$, where $V$ is a finite dimensional vector space over $mathbb{R}$.

This is trivial if you take $mathcal{M} = mathbb{R}^{m}$ since you can identify the tangent map with the differential from ordinary calculus, but I want to formalize the proof as much as possible in the general case using the definitions I gave.










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    This is a linear algebra question. You just want to know that a map $V to W$ of finite dimensional vector spaces is surjective iff its adjoint $W^* to V^*$ is injective.
    $endgroup$
    – user98602
    Dec 29 '18 at 23:41
















1












1








1





$begingroup$


Let $mathcal{M}$ be a smooth $m$-dimensional manifold and let $Fcolon mathcal{M} longrightarrow mathbb{R}^{n}$ be any smooth map, with m $geq$ n . I want to prove that, if $p in mathcal{M}$:
$$ text{$F$ is a submersion at $p$} iff lbrace d_{p}F^{1},dots,d_{p}F^{n}rbrace text{ are l.i. in $(T_{p}mathcal{M})^{*}$} $$
Were $d_{p}f := lambda^{-1}_{f(p)} circ T_{p}f colon T_{p}mathcal{M} longrightarrow mathbb{R} in (T_{p}mathcal{M})^{*}$, $T_{p}fcolon T_{p}mathcal{M}longrightarrow T_{f(p)}mathcal{M}$ is the tangent map at $p$ and $lambda_{p}colon V longrightarrow T_{p}V$ is the canonical isomorphism between a finite dimensional vector space over $mathbb{R}$ and its tangent space. We can extend the definition of differential of a smooth map $Fcolon mathcal{M} longrightarrow V$ by putting $D_{p}F := lambda^{-1}_{F(p)} circ T_{p}F colon T_{p}mathcal{M} longrightarrow V$, where $V$ is a finite dimensional vector space over $mathbb{R}$.

This is trivial if you take $mathcal{M} = mathbb{R}^{m}$ since you can identify the tangent map with the differential from ordinary calculus, but I want to formalize the proof as much as possible in the general case using the definitions I gave.










share|cite|improve this question











$endgroup$




Let $mathcal{M}$ be a smooth $m$-dimensional manifold and let $Fcolon mathcal{M} longrightarrow mathbb{R}^{n}$ be any smooth map, with m $geq$ n . I want to prove that, if $p in mathcal{M}$:
$$ text{$F$ is a submersion at $p$} iff lbrace d_{p}F^{1},dots,d_{p}F^{n}rbrace text{ are l.i. in $(T_{p}mathcal{M})^{*}$} $$
Were $d_{p}f := lambda^{-1}_{f(p)} circ T_{p}f colon T_{p}mathcal{M} longrightarrow mathbb{R} in (T_{p}mathcal{M})^{*}$, $T_{p}fcolon T_{p}mathcal{M}longrightarrow T_{f(p)}mathcal{M}$ is the tangent map at $p$ and $lambda_{p}colon V longrightarrow T_{p}V$ is the canonical isomorphism between a finite dimensional vector space over $mathbb{R}$ and its tangent space. We can extend the definition of differential of a smooth map $Fcolon mathcal{M} longrightarrow V$ by putting $D_{p}F := lambda^{-1}_{F(p)} circ T_{p}F colon T_{p}mathcal{M} longrightarrow V$, where $V$ is a finite dimensional vector space over $mathbb{R}$.

This is trivial if you take $mathcal{M} = mathbb{R}^{m}$ since you can identify the tangent map with the differential from ordinary calculus, but I want to formalize the proof as much as possible in the general case using the definitions I gave.







differential-geometry manifolds smooth-manifolds






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edited Dec 29 '18 at 23:36







Sergi

















asked Dec 29 '18 at 23:03









SergiSergi

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  • 2




    $begingroup$
    This is a linear algebra question. You just want to know that a map $V to W$ of finite dimensional vector spaces is surjective iff its adjoint $W^* to V^*$ is injective.
    $endgroup$
    – user98602
    Dec 29 '18 at 23:41
















  • 2




    $begingroup$
    This is a linear algebra question. You just want to know that a map $V to W$ of finite dimensional vector spaces is surjective iff its adjoint $W^* to V^*$ is injective.
    $endgroup$
    – user98602
    Dec 29 '18 at 23:41










2




2




$begingroup$
This is a linear algebra question. You just want to know that a map $V to W$ of finite dimensional vector spaces is surjective iff its adjoint $W^* to V^*$ is injective.
$endgroup$
– user98602
Dec 29 '18 at 23:41






$begingroup$
This is a linear algebra question. You just want to know that a map $V to W$ of finite dimensional vector spaces is surjective iff its adjoint $W^* to V^*$ is injective.
$endgroup$
– user98602
Dec 29 '18 at 23:41












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Linear algebra Fact: Let $A:Vrightarrow mathbb{R}^n$ be a linear transformation of vector spaces, then $A$ is surjective iff $pi_1A,pi_2A,...,pi_nA$ are linearly independent members of $V^*$. (Where $pi_i:mathbb{R}^nrightarrow mathbb{R}$ is ith projection).



Proof: Equip $V$with any inner product, then there exists $v_1,...,v_nin V$ such that $pi_iA(x)=x circ v_i$ for all $xin V , iin [n]$. $dim(Im(A))=dim(V)-dim(Ker(A))=dim(V)-dim({v_1,...,v_n}^{perp})=dim(span({v_1,...,v_n }))$.



Thus $A$ is surjective iff $dim(Im A)=n$ iff $dim(span({v_1,...,v_n}))=n$ iff $v_1,...,v_n$ are L.I. iff $pi_1A,...,pi_nA$ are L.I. $square$



Identify the tangent space of any point of $mathbb{R}^n$ with $mathbb{R}^n$



Now let's go back to your setting, then for any $vin T_pM$: $dF^i_p(v)=<v,F^i>=<v,pi_i circ f>=<Df|_pv,pi_i>=pi_i(Df|_pv)$.



So $dF^i_p=pi_icirc Df|_p$. Now apply the previous fact on the linear transformation $Df|_p:T_pMrightarrow mathbb{R}^n$ and you will be done.






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    $begingroup$

    Linear algebra Fact: Let $A:Vrightarrow mathbb{R}^n$ be a linear transformation of vector spaces, then $A$ is surjective iff $pi_1A,pi_2A,...,pi_nA$ are linearly independent members of $V^*$. (Where $pi_i:mathbb{R}^nrightarrow mathbb{R}$ is ith projection).



    Proof: Equip $V$with any inner product, then there exists $v_1,...,v_nin V$ such that $pi_iA(x)=x circ v_i$ for all $xin V , iin [n]$. $dim(Im(A))=dim(V)-dim(Ker(A))=dim(V)-dim({v_1,...,v_n}^{perp})=dim(span({v_1,...,v_n }))$.



    Thus $A$ is surjective iff $dim(Im A)=n$ iff $dim(span({v_1,...,v_n}))=n$ iff $v_1,...,v_n$ are L.I. iff $pi_1A,...,pi_nA$ are L.I. $square$



    Identify the tangent space of any point of $mathbb{R}^n$ with $mathbb{R}^n$



    Now let's go back to your setting, then for any $vin T_pM$: $dF^i_p(v)=<v,F^i>=<v,pi_i circ f>=<Df|_pv,pi_i>=pi_i(Df|_pv)$.



    So $dF^i_p=pi_icirc Df|_p$. Now apply the previous fact on the linear transformation $Df|_p:T_pMrightarrow mathbb{R}^n$ and you will be done.






    share|cite|improve this answer











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      2












      $begingroup$

      Linear algebra Fact: Let $A:Vrightarrow mathbb{R}^n$ be a linear transformation of vector spaces, then $A$ is surjective iff $pi_1A,pi_2A,...,pi_nA$ are linearly independent members of $V^*$. (Where $pi_i:mathbb{R}^nrightarrow mathbb{R}$ is ith projection).



      Proof: Equip $V$with any inner product, then there exists $v_1,...,v_nin V$ such that $pi_iA(x)=x circ v_i$ for all $xin V , iin [n]$. $dim(Im(A))=dim(V)-dim(Ker(A))=dim(V)-dim({v_1,...,v_n}^{perp})=dim(span({v_1,...,v_n }))$.



      Thus $A$ is surjective iff $dim(Im A)=n$ iff $dim(span({v_1,...,v_n}))=n$ iff $v_1,...,v_n$ are L.I. iff $pi_1A,...,pi_nA$ are L.I. $square$



      Identify the tangent space of any point of $mathbb{R}^n$ with $mathbb{R}^n$



      Now let's go back to your setting, then for any $vin T_pM$: $dF^i_p(v)=<v,F^i>=<v,pi_i circ f>=<Df|_pv,pi_i>=pi_i(Df|_pv)$.



      So $dF^i_p=pi_icirc Df|_p$. Now apply the previous fact on the linear transformation $Df|_p:T_pMrightarrow mathbb{R}^n$ and you will be done.






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        Linear algebra Fact: Let $A:Vrightarrow mathbb{R}^n$ be a linear transformation of vector spaces, then $A$ is surjective iff $pi_1A,pi_2A,...,pi_nA$ are linearly independent members of $V^*$. (Where $pi_i:mathbb{R}^nrightarrow mathbb{R}$ is ith projection).



        Proof: Equip $V$with any inner product, then there exists $v_1,...,v_nin V$ such that $pi_iA(x)=x circ v_i$ for all $xin V , iin [n]$. $dim(Im(A))=dim(V)-dim(Ker(A))=dim(V)-dim({v_1,...,v_n}^{perp})=dim(span({v_1,...,v_n }))$.



        Thus $A$ is surjective iff $dim(Im A)=n$ iff $dim(span({v_1,...,v_n}))=n$ iff $v_1,...,v_n$ are L.I. iff $pi_1A,...,pi_nA$ are L.I. $square$



        Identify the tangent space of any point of $mathbb{R}^n$ with $mathbb{R}^n$



        Now let's go back to your setting, then for any $vin T_pM$: $dF^i_p(v)=<v,F^i>=<v,pi_i circ f>=<Df|_pv,pi_i>=pi_i(Df|_pv)$.



        So $dF^i_p=pi_icirc Df|_p$. Now apply the previous fact on the linear transformation $Df|_p:T_pMrightarrow mathbb{R}^n$ and you will be done.






        share|cite|improve this answer











        $endgroup$



        Linear algebra Fact: Let $A:Vrightarrow mathbb{R}^n$ be a linear transformation of vector spaces, then $A$ is surjective iff $pi_1A,pi_2A,...,pi_nA$ are linearly independent members of $V^*$. (Where $pi_i:mathbb{R}^nrightarrow mathbb{R}$ is ith projection).



        Proof: Equip $V$with any inner product, then there exists $v_1,...,v_nin V$ such that $pi_iA(x)=x circ v_i$ for all $xin V , iin [n]$. $dim(Im(A))=dim(V)-dim(Ker(A))=dim(V)-dim({v_1,...,v_n}^{perp})=dim(span({v_1,...,v_n }))$.



        Thus $A$ is surjective iff $dim(Im A)=n$ iff $dim(span({v_1,...,v_n}))=n$ iff $v_1,...,v_n$ are L.I. iff $pi_1A,...,pi_nA$ are L.I. $square$



        Identify the tangent space of any point of $mathbb{R}^n$ with $mathbb{R}^n$



        Now let's go back to your setting, then for any $vin T_pM$: $dF^i_p(v)=<v,F^i>=<v,pi_i circ f>=<Df|_pv,pi_i>=pi_i(Df|_pv)$.



        So $dF^i_p=pi_icirc Df|_p$. Now apply the previous fact on the linear transformation $Df|_p:T_pMrightarrow mathbb{R}^n$ and you will be done.







        share|cite|improve this answer














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        edited Dec 30 '18 at 0:29

























        answered Dec 30 '18 at 0:04









        AmrAmr

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