Characterization of being a submersion
$begingroup$
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.
differential-geometry manifolds smooth-manifolds
asked Dec 29 '18 at 23:03
Sergi GalveSergi Galve
1816
$endgroup$
add a comment |
$begingroup$
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.
differential-geometry manifolds smooth-manifolds
asked Dec 29 '18 at 23:03
Sergi GalveSergi Galve
1816
$endgroup$
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
add a comment |
$begingroup$
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.
differential-geometry manifolds smooth-manifolds
asked Dec 29 '18 at 23:03
Sergi GalveSergi Galve
1816
$endgroup$
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
differential-geometry manifolds smooth-manifolds
asked Dec 29 '18 at 23:03
Sergi GalveSergi Galve
1816
asked Dec 29 '18 at 23:03
Sergi GalveSergi Galve
1816
edited Dec 29 '18 at 23:36
Sergi Galve
asked Dec 29 '18 at 23:03
Sergi GalveSergi Galve
1816
asked Dec 29 '18 at 23:03
Sergi GalveSergi Galve
1816
asked Dec 29 '18 at 23:03
Sergi GalveSergi Galve
1816
1816
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
add a comment |
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
add a comment |
1 Answer
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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.
answered Dec 30 '18 at 0:04
AmrAmr
14.4k43292
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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.
answered Dec 30 '18 at 0:04
AmrAmr
14.4k43292
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 30 '18 at 0:04
AmrAmr
14.4k43292
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 30 '18 at 0:04
AmrAmr
14.4k43292
$endgroup$
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.
answered Dec 30 '18 at 0:04
AmrAmr
14.4k43292
edited Dec 30 '18 at 0:29
answered Dec 30 '18 at 0:04
AmrAmr
14.4k43292
answered Dec 30 '18 at 0:04
AmrAmr
14.4k43292
answered Dec 30 '18 at 0:04
AmrAmr
14.4k43292
14.4k43292
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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