Smooth quotient bundle












0












$begingroup$


Let $E rightarrow M$ be a smooth vector bundle.



This link gives a construction of quotient bundle for a subbundle. $E' subseteq E$.



We define an equivalence relation $sim$ on $E$ by $v_x sim v_y$ if and only if $x=y$ and $v_x-v_y in E'_x$ and define $E/sim rightarrow M$ to be our quotient bundle.



My question is, how does this construction also give smooth bundle?





My attempt, the problem is thus to prove we can give a smooth structure to $E/sim$.



Transition function formulation: This is equivalent to finding an open cover with local trivializations
$$ {phi_U:E/sim |_U rightarrow U times Bbb R^k }$$ such that the transition functions of any two on $U cap V$ are smooth map from
$$ U cap V rightarrow GL_k(Bbb R)$$



Using the transition functions of $E$, $E'$.



Let us choose transition functions, form $E'$ and $E$, on open sets



$$ phi': U cap V rightarrow GL_m(Bbb R)$$
$$ phi: U cap V rightarrow GL_n(Bbb R)$$
where $E$ is bundle of rank $n$, $E'$ is of rank $m$. But how does this induce a well defined smooth map on the quotient bundle?










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$endgroup$












  • $begingroup$
    If you choose a Riemannian metric on your bundle this is easy: it's just the orthogonal complement.
    $endgroup$
    – user98602
    Jan 2 at 16:12
















0












$begingroup$


Let $E rightarrow M$ be a smooth vector bundle.



This link gives a construction of quotient bundle for a subbundle. $E' subseteq E$.



We define an equivalence relation $sim$ on $E$ by $v_x sim v_y$ if and only if $x=y$ and $v_x-v_y in E'_x$ and define $E/sim rightarrow M$ to be our quotient bundle.



My question is, how does this construction also give smooth bundle?





My attempt, the problem is thus to prove we can give a smooth structure to $E/sim$.



Transition function formulation: This is equivalent to finding an open cover with local trivializations
$$ {phi_U:E/sim |_U rightarrow U times Bbb R^k }$$ such that the transition functions of any two on $U cap V$ are smooth map from
$$ U cap V rightarrow GL_k(Bbb R)$$



Using the transition functions of $E$, $E'$.



Let us choose transition functions, form $E'$ and $E$, on open sets



$$ phi': U cap V rightarrow GL_m(Bbb R)$$
$$ phi: U cap V rightarrow GL_n(Bbb R)$$
where $E$ is bundle of rank $n$, $E'$ is of rank $m$. But how does this induce a well defined smooth map on the quotient bundle?










share|cite|improve this question









$endgroup$












  • $begingroup$
    If you choose a Riemannian metric on your bundle this is easy: it's just the orthogonal complement.
    $endgroup$
    – user98602
    Jan 2 at 16:12














0












0








0





$begingroup$


Let $E rightarrow M$ be a smooth vector bundle.



This link gives a construction of quotient bundle for a subbundle. $E' subseteq E$.



We define an equivalence relation $sim$ on $E$ by $v_x sim v_y$ if and only if $x=y$ and $v_x-v_y in E'_x$ and define $E/sim rightarrow M$ to be our quotient bundle.



My question is, how does this construction also give smooth bundle?





My attempt, the problem is thus to prove we can give a smooth structure to $E/sim$.



Transition function formulation: This is equivalent to finding an open cover with local trivializations
$$ {phi_U:E/sim |_U rightarrow U times Bbb R^k }$$ such that the transition functions of any two on $U cap V$ are smooth map from
$$ U cap V rightarrow GL_k(Bbb R)$$



Using the transition functions of $E$, $E'$.



Let us choose transition functions, form $E'$ and $E$, on open sets



$$ phi': U cap V rightarrow GL_m(Bbb R)$$
$$ phi: U cap V rightarrow GL_n(Bbb R)$$
where $E$ is bundle of rank $n$, $E'$ is of rank $m$. But how does this induce a well defined smooth map on the quotient bundle?










share|cite|improve this question









$endgroup$




Let $E rightarrow M$ be a smooth vector bundle.



This link gives a construction of quotient bundle for a subbundle. $E' subseteq E$.



We define an equivalence relation $sim$ on $E$ by $v_x sim v_y$ if and only if $x=y$ and $v_x-v_y in E'_x$ and define $E/sim rightarrow M$ to be our quotient bundle.



My question is, how does this construction also give smooth bundle?





My attempt, the problem is thus to prove we can give a smooth structure to $E/sim$.



Transition function formulation: This is equivalent to finding an open cover with local trivializations
$$ {phi_U:E/sim |_U rightarrow U times Bbb R^k }$$ such that the transition functions of any two on $U cap V$ are smooth map from
$$ U cap V rightarrow GL_k(Bbb R)$$



Using the transition functions of $E$, $E'$.



Let us choose transition functions, form $E'$ and $E$, on open sets



$$ phi': U cap V rightarrow GL_m(Bbb R)$$
$$ phi: U cap V rightarrow GL_n(Bbb R)$$
where $E$ is bundle of rank $n$, $E'$ is of rank $m$. But how does this induce a well defined smooth map on the quotient bundle?







differential-geometry algebraic-topology smooth-manifolds vector-bundles






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asked Jan 2 at 9:56









CL.CL.

2,4163925




2,4163925












  • $begingroup$
    If you choose a Riemannian metric on your bundle this is easy: it's just the orthogonal complement.
    $endgroup$
    – user98602
    Jan 2 at 16:12


















  • $begingroup$
    If you choose a Riemannian metric on your bundle this is easy: it's just the orthogonal complement.
    $endgroup$
    – user98602
    Jan 2 at 16:12
















$begingroup$
If you choose a Riemannian metric on your bundle this is easy: it's just the orthogonal complement.
$endgroup$
– user98602
Jan 2 at 16:12




$begingroup$
If you choose a Riemannian metric on your bundle this is easy: it's just the orthogonal complement.
$endgroup$
– user98602
Jan 2 at 16:12










2 Answers
2






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oldest

votes


















1












$begingroup$

We will appeal to Godemont's theorem, which states that if $X$ is a smooth manifold and $R$ is an equivalence relation on $X$, then $X/R$ has the structure of a smooth manifold such that $Xrightarrow X/R$ becomes a smooth submersion if and only if $Rsubseteq Xtimes X$ is a smoooth submanifold, and $pr_1:Rrightarrow X$ is a submersion. I can't find the exact reference I was looking for at the moment for this theorem, but it is a standard result. I know Bourbaki has a statement (its in French). A good general reference is section 4 of Lee's book Introduction to Smooth Manifolds.



To this end we consider the fibred product $Etimes_ME={(e_1,e_2)in Etimes Emid pi(e_1)=pi(e_2)}$, which is the pullback of $pi$ along itself. This space is a smooth submanifold of the product $Etimes E$. Likewise we consider $Etimes_ME'={(e,e')in Etimes E'mid pi(e)=pi'(e')}$, which, under the assumption that $E'subseteq E$ is a smooth submanifold, is itself a smooth submanifold $Etimes_ME'subseteq Etimes_ME$.



We however, will consider a non-standard embedding of this manifold. Since $E$ is a vector bundle it has a fibrewise sum $mu:Etimes_MErightarrow E$, which restricts to the standard vector space addition $(e_1,e_2)mapsto e_1+e_2$ on each fibre $E_ptimes E_p$, and we use this map to define



$$Theta:Etimes_ME'rightarrow Etimes_ME,qquad Theta(e,e')=(e,e+e').$$



It is clear that this map is a diffeomorphism onto its image, and that its image $Theta(Etimes_ME'):=mathcal{R}$ is precisely the graph of the relation $e_1sim e_2$ iff $pi(e_1)=pi(e_2)$ and $e_2-e_1in E'_{pi(e_1)}$. Then $mathcal{R}subseteq Etimes_MEsubseteq Etimes E$ is a smooth submanifold, the projection $pr_1:mathcal{R}rightarrow E$ is smooth, and we can appeal to Godemont's critereon to get a smooth structure on the quotient $E/E'$ such that the canonical map $q:Erightarrow E/E'$ becomes a smooth submersion. This then induces from the map $pi$ a submersive projection $rho:E/E'rightarrow M$ satisfying $pi=rhocirc q$. Submersions always have local sections, so $E/E'$ is a locally trivial fibre bundle over $M$. In fact, if $s$ is any local section of $pi$, then $qcirc s$ is a local section of $rho$.



Now from here it is not difficult to see that the fibrewise linear structure on $E$ transfers to $E/E'$ in such a way as to make $rho$ into a topological vector bundle, although ee have not yet shown that the induced operations of fiberwise addition and scalar multiplication are smooth. However this is not hard to do using the fact that $q$ is a submersion. In fact it is not difficult to see that $qtimes_Mq:Etimes_MErightarrow E'times_ME'$ is a submersion, and so induces the smooth fibrewise addition $mu':(E/E')times_M(E/E')rightarrow E/E'$ uniquely from $mu$. Similarly, the smoothness of the fibrewise scalar multiplicaiton $mathbb{K}times E/E'rightarrow E/E'$ follows.



To construct charts on $E/E'$ we use the previous remark on local sections to construct fibrewise linear local trivialisations $(U,varphi)$, $Usubseteq M$, of $rho$ in such a way that the restriction $q|_U$ becomes the quotient $Utimesmathbb{K}^srightarrow Utimesmathbb{K}^{s-r}$ locally modelling the vector space quotients $E_xrightarrow E_x/E'_x$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I checked out lee's text but couldn't find it, you know which page it is?
    $endgroup$
    – CL.
    Jan 4 at 6:44










  • $begingroup$
    As I said, the theorem is not actually discussed by Lee, although the material in his book is a good general reference for submersions in general. Check out Roger Godement's book Introduction to the Theory of Lie Groups, section 4.9, Theorem 5, page 133 (Springer-Verlag 2017 English Translation) for the exact statement.
    $endgroup$
    – Tyrone
    Jan 4 at 10:06



















0












$begingroup$

You have to construct compatible trivializations of the bundles $E'$ and $E$, which is slightly complicated to express in terms of transition functions. The easilest way to describe this in my opinion is in terms of local frames. Choose a local frame for $E'$ defined on $U$. Shrinking $U$, you can extend this to a local frame for $E$. (In a point $xin U$, extend the values of your frame which form a basis for $E'_x$ to a basis for $E_x$ and then extend the additional elements to local smooth sections of $E$.) Then you get a local trivialization $E|_Uto Utimesmathbb R^n$ which restricts to a trivialzation $E'|_Uto Utimesmathbb R^k$. From this, you get a trivialization from the quotient bundle to $mathbb R^n/mathbb R^kcongmathbb R^{n-k}$.



In terms of transition functions, you can use the above construction to show that you can arrange things in such a way that the transtition functions for $E$ have values in block matrices of the form $begin{pmatrix} A & B \ 0 & Cend{pmatrix}$ such that the $A$-component gives you transition functions for $E'$. In this setting, the $C$-block gives you the transition functions for $E/E'$.






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    2 Answers
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    2 Answers
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    $begingroup$

    We will appeal to Godemont's theorem, which states that if $X$ is a smooth manifold and $R$ is an equivalence relation on $X$, then $X/R$ has the structure of a smooth manifold such that $Xrightarrow X/R$ becomes a smooth submersion if and only if $Rsubseteq Xtimes X$ is a smoooth submanifold, and $pr_1:Rrightarrow X$ is a submersion. I can't find the exact reference I was looking for at the moment for this theorem, but it is a standard result. I know Bourbaki has a statement (its in French). A good general reference is section 4 of Lee's book Introduction to Smooth Manifolds.



    To this end we consider the fibred product $Etimes_ME={(e_1,e_2)in Etimes Emid pi(e_1)=pi(e_2)}$, which is the pullback of $pi$ along itself. This space is a smooth submanifold of the product $Etimes E$. Likewise we consider $Etimes_ME'={(e,e')in Etimes E'mid pi(e)=pi'(e')}$, which, under the assumption that $E'subseteq E$ is a smooth submanifold, is itself a smooth submanifold $Etimes_ME'subseteq Etimes_ME$.



    We however, will consider a non-standard embedding of this manifold. Since $E$ is a vector bundle it has a fibrewise sum $mu:Etimes_MErightarrow E$, which restricts to the standard vector space addition $(e_1,e_2)mapsto e_1+e_2$ on each fibre $E_ptimes E_p$, and we use this map to define



    $$Theta:Etimes_ME'rightarrow Etimes_ME,qquad Theta(e,e')=(e,e+e').$$



    It is clear that this map is a diffeomorphism onto its image, and that its image $Theta(Etimes_ME'):=mathcal{R}$ is precisely the graph of the relation $e_1sim e_2$ iff $pi(e_1)=pi(e_2)$ and $e_2-e_1in E'_{pi(e_1)}$. Then $mathcal{R}subseteq Etimes_MEsubseteq Etimes E$ is a smooth submanifold, the projection $pr_1:mathcal{R}rightarrow E$ is smooth, and we can appeal to Godemont's critereon to get a smooth structure on the quotient $E/E'$ such that the canonical map $q:Erightarrow E/E'$ becomes a smooth submersion. This then induces from the map $pi$ a submersive projection $rho:E/E'rightarrow M$ satisfying $pi=rhocirc q$. Submersions always have local sections, so $E/E'$ is a locally trivial fibre bundle over $M$. In fact, if $s$ is any local section of $pi$, then $qcirc s$ is a local section of $rho$.



    Now from here it is not difficult to see that the fibrewise linear structure on $E$ transfers to $E/E'$ in such a way as to make $rho$ into a topological vector bundle, although ee have not yet shown that the induced operations of fiberwise addition and scalar multiplication are smooth. However this is not hard to do using the fact that $q$ is a submersion. In fact it is not difficult to see that $qtimes_Mq:Etimes_MErightarrow E'times_ME'$ is a submersion, and so induces the smooth fibrewise addition $mu':(E/E')times_M(E/E')rightarrow E/E'$ uniquely from $mu$. Similarly, the smoothness of the fibrewise scalar multiplicaiton $mathbb{K}times E/E'rightarrow E/E'$ follows.



    To construct charts on $E/E'$ we use the previous remark on local sections to construct fibrewise linear local trivialisations $(U,varphi)$, $Usubseteq M$, of $rho$ in such a way that the restriction $q|_U$ becomes the quotient $Utimesmathbb{K}^srightarrow Utimesmathbb{K}^{s-r}$ locally modelling the vector space quotients $E_xrightarrow E_x/E'_x$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I checked out lee's text but couldn't find it, you know which page it is?
      $endgroup$
      – CL.
      Jan 4 at 6:44










    • $begingroup$
      As I said, the theorem is not actually discussed by Lee, although the material in his book is a good general reference for submersions in general. Check out Roger Godement's book Introduction to the Theory of Lie Groups, section 4.9, Theorem 5, page 133 (Springer-Verlag 2017 English Translation) for the exact statement.
      $endgroup$
      – Tyrone
      Jan 4 at 10:06
















    1












    $begingroup$

    We will appeal to Godemont's theorem, which states that if $X$ is a smooth manifold and $R$ is an equivalence relation on $X$, then $X/R$ has the structure of a smooth manifold such that $Xrightarrow X/R$ becomes a smooth submersion if and only if $Rsubseteq Xtimes X$ is a smoooth submanifold, and $pr_1:Rrightarrow X$ is a submersion. I can't find the exact reference I was looking for at the moment for this theorem, but it is a standard result. I know Bourbaki has a statement (its in French). A good general reference is section 4 of Lee's book Introduction to Smooth Manifolds.



    To this end we consider the fibred product $Etimes_ME={(e_1,e_2)in Etimes Emid pi(e_1)=pi(e_2)}$, which is the pullback of $pi$ along itself. This space is a smooth submanifold of the product $Etimes E$. Likewise we consider $Etimes_ME'={(e,e')in Etimes E'mid pi(e)=pi'(e')}$, which, under the assumption that $E'subseteq E$ is a smooth submanifold, is itself a smooth submanifold $Etimes_ME'subseteq Etimes_ME$.



    We however, will consider a non-standard embedding of this manifold. Since $E$ is a vector bundle it has a fibrewise sum $mu:Etimes_MErightarrow E$, which restricts to the standard vector space addition $(e_1,e_2)mapsto e_1+e_2$ on each fibre $E_ptimes E_p$, and we use this map to define



    $$Theta:Etimes_ME'rightarrow Etimes_ME,qquad Theta(e,e')=(e,e+e').$$



    It is clear that this map is a diffeomorphism onto its image, and that its image $Theta(Etimes_ME'):=mathcal{R}$ is precisely the graph of the relation $e_1sim e_2$ iff $pi(e_1)=pi(e_2)$ and $e_2-e_1in E'_{pi(e_1)}$. Then $mathcal{R}subseteq Etimes_MEsubseteq Etimes E$ is a smooth submanifold, the projection $pr_1:mathcal{R}rightarrow E$ is smooth, and we can appeal to Godemont's critereon to get a smooth structure on the quotient $E/E'$ such that the canonical map $q:Erightarrow E/E'$ becomes a smooth submersion. This then induces from the map $pi$ a submersive projection $rho:E/E'rightarrow M$ satisfying $pi=rhocirc q$. Submersions always have local sections, so $E/E'$ is a locally trivial fibre bundle over $M$. In fact, if $s$ is any local section of $pi$, then $qcirc s$ is a local section of $rho$.



    Now from here it is not difficult to see that the fibrewise linear structure on $E$ transfers to $E/E'$ in such a way as to make $rho$ into a topological vector bundle, although ee have not yet shown that the induced operations of fiberwise addition and scalar multiplication are smooth. However this is not hard to do using the fact that $q$ is a submersion. In fact it is not difficult to see that $qtimes_Mq:Etimes_MErightarrow E'times_ME'$ is a submersion, and so induces the smooth fibrewise addition $mu':(E/E')times_M(E/E')rightarrow E/E'$ uniquely from $mu$. Similarly, the smoothness of the fibrewise scalar multiplicaiton $mathbb{K}times E/E'rightarrow E/E'$ follows.



    To construct charts on $E/E'$ we use the previous remark on local sections to construct fibrewise linear local trivialisations $(U,varphi)$, $Usubseteq M$, of $rho$ in such a way that the restriction $q|_U$ becomes the quotient $Utimesmathbb{K}^srightarrow Utimesmathbb{K}^{s-r}$ locally modelling the vector space quotients $E_xrightarrow E_x/E'_x$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I checked out lee's text but couldn't find it, you know which page it is?
      $endgroup$
      – CL.
      Jan 4 at 6:44










    • $begingroup$
      As I said, the theorem is not actually discussed by Lee, although the material in his book is a good general reference for submersions in general. Check out Roger Godement's book Introduction to the Theory of Lie Groups, section 4.9, Theorem 5, page 133 (Springer-Verlag 2017 English Translation) for the exact statement.
      $endgroup$
      – Tyrone
      Jan 4 at 10:06














    1












    1








    1





    $begingroup$

    We will appeal to Godemont's theorem, which states that if $X$ is a smooth manifold and $R$ is an equivalence relation on $X$, then $X/R$ has the structure of a smooth manifold such that $Xrightarrow X/R$ becomes a smooth submersion if and only if $Rsubseteq Xtimes X$ is a smoooth submanifold, and $pr_1:Rrightarrow X$ is a submersion. I can't find the exact reference I was looking for at the moment for this theorem, but it is a standard result. I know Bourbaki has a statement (its in French). A good general reference is section 4 of Lee's book Introduction to Smooth Manifolds.



    To this end we consider the fibred product $Etimes_ME={(e_1,e_2)in Etimes Emid pi(e_1)=pi(e_2)}$, which is the pullback of $pi$ along itself. This space is a smooth submanifold of the product $Etimes E$. Likewise we consider $Etimes_ME'={(e,e')in Etimes E'mid pi(e)=pi'(e')}$, which, under the assumption that $E'subseteq E$ is a smooth submanifold, is itself a smooth submanifold $Etimes_ME'subseteq Etimes_ME$.



    We however, will consider a non-standard embedding of this manifold. Since $E$ is a vector bundle it has a fibrewise sum $mu:Etimes_MErightarrow E$, which restricts to the standard vector space addition $(e_1,e_2)mapsto e_1+e_2$ on each fibre $E_ptimes E_p$, and we use this map to define



    $$Theta:Etimes_ME'rightarrow Etimes_ME,qquad Theta(e,e')=(e,e+e').$$



    It is clear that this map is a diffeomorphism onto its image, and that its image $Theta(Etimes_ME'):=mathcal{R}$ is precisely the graph of the relation $e_1sim e_2$ iff $pi(e_1)=pi(e_2)$ and $e_2-e_1in E'_{pi(e_1)}$. Then $mathcal{R}subseteq Etimes_MEsubseteq Etimes E$ is a smooth submanifold, the projection $pr_1:mathcal{R}rightarrow E$ is smooth, and we can appeal to Godemont's critereon to get a smooth structure on the quotient $E/E'$ such that the canonical map $q:Erightarrow E/E'$ becomes a smooth submersion. This then induces from the map $pi$ a submersive projection $rho:E/E'rightarrow M$ satisfying $pi=rhocirc q$. Submersions always have local sections, so $E/E'$ is a locally trivial fibre bundle over $M$. In fact, if $s$ is any local section of $pi$, then $qcirc s$ is a local section of $rho$.



    Now from here it is not difficult to see that the fibrewise linear structure on $E$ transfers to $E/E'$ in such a way as to make $rho$ into a topological vector bundle, although ee have not yet shown that the induced operations of fiberwise addition and scalar multiplication are smooth. However this is not hard to do using the fact that $q$ is a submersion. In fact it is not difficult to see that $qtimes_Mq:Etimes_MErightarrow E'times_ME'$ is a submersion, and so induces the smooth fibrewise addition $mu':(E/E')times_M(E/E')rightarrow E/E'$ uniquely from $mu$. Similarly, the smoothness of the fibrewise scalar multiplicaiton $mathbb{K}times E/E'rightarrow E/E'$ follows.



    To construct charts on $E/E'$ we use the previous remark on local sections to construct fibrewise linear local trivialisations $(U,varphi)$, $Usubseteq M$, of $rho$ in such a way that the restriction $q|_U$ becomes the quotient $Utimesmathbb{K}^srightarrow Utimesmathbb{K}^{s-r}$ locally modelling the vector space quotients $E_xrightarrow E_x/E'_x$.






    share|cite|improve this answer









    $endgroup$



    We will appeal to Godemont's theorem, which states that if $X$ is a smooth manifold and $R$ is an equivalence relation on $X$, then $X/R$ has the structure of a smooth manifold such that $Xrightarrow X/R$ becomes a smooth submersion if and only if $Rsubseteq Xtimes X$ is a smoooth submanifold, and $pr_1:Rrightarrow X$ is a submersion. I can't find the exact reference I was looking for at the moment for this theorem, but it is a standard result. I know Bourbaki has a statement (its in French). A good general reference is section 4 of Lee's book Introduction to Smooth Manifolds.



    To this end we consider the fibred product $Etimes_ME={(e_1,e_2)in Etimes Emid pi(e_1)=pi(e_2)}$, which is the pullback of $pi$ along itself. This space is a smooth submanifold of the product $Etimes E$. Likewise we consider $Etimes_ME'={(e,e')in Etimes E'mid pi(e)=pi'(e')}$, which, under the assumption that $E'subseteq E$ is a smooth submanifold, is itself a smooth submanifold $Etimes_ME'subseteq Etimes_ME$.



    We however, will consider a non-standard embedding of this manifold. Since $E$ is a vector bundle it has a fibrewise sum $mu:Etimes_MErightarrow E$, which restricts to the standard vector space addition $(e_1,e_2)mapsto e_1+e_2$ on each fibre $E_ptimes E_p$, and we use this map to define



    $$Theta:Etimes_ME'rightarrow Etimes_ME,qquad Theta(e,e')=(e,e+e').$$



    It is clear that this map is a diffeomorphism onto its image, and that its image $Theta(Etimes_ME'):=mathcal{R}$ is precisely the graph of the relation $e_1sim e_2$ iff $pi(e_1)=pi(e_2)$ and $e_2-e_1in E'_{pi(e_1)}$. Then $mathcal{R}subseteq Etimes_MEsubseteq Etimes E$ is a smooth submanifold, the projection $pr_1:mathcal{R}rightarrow E$ is smooth, and we can appeal to Godemont's critereon to get a smooth structure on the quotient $E/E'$ such that the canonical map $q:Erightarrow E/E'$ becomes a smooth submersion. This then induces from the map $pi$ a submersive projection $rho:E/E'rightarrow M$ satisfying $pi=rhocirc q$. Submersions always have local sections, so $E/E'$ is a locally trivial fibre bundle over $M$. In fact, if $s$ is any local section of $pi$, then $qcirc s$ is a local section of $rho$.



    Now from here it is not difficult to see that the fibrewise linear structure on $E$ transfers to $E/E'$ in such a way as to make $rho$ into a topological vector bundle, although ee have not yet shown that the induced operations of fiberwise addition and scalar multiplication are smooth. However this is not hard to do using the fact that $q$ is a submersion. In fact it is not difficult to see that $qtimes_Mq:Etimes_MErightarrow E'times_ME'$ is a submersion, and so induces the smooth fibrewise addition $mu':(E/E')times_M(E/E')rightarrow E/E'$ uniquely from $mu$. Similarly, the smoothness of the fibrewise scalar multiplicaiton $mathbb{K}times E/E'rightarrow E/E'$ follows.



    To construct charts on $E/E'$ we use the previous remark on local sections to construct fibrewise linear local trivialisations $(U,varphi)$, $Usubseteq M$, of $rho$ in such a way that the restriction $q|_U$ becomes the quotient $Utimesmathbb{K}^srightarrow Utimesmathbb{K}^{s-r}$ locally modelling the vector space quotients $E_xrightarrow E_x/E'_x$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 2 at 12:17









    TyroneTyrone

    5,47711228




    5,47711228












    • $begingroup$
      I checked out lee's text but couldn't find it, you know which page it is?
      $endgroup$
      – CL.
      Jan 4 at 6:44










    • $begingroup$
      As I said, the theorem is not actually discussed by Lee, although the material in his book is a good general reference for submersions in general. Check out Roger Godement's book Introduction to the Theory of Lie Groups, section 4.9, Theorem 5, page 133 (Springer-Verlag 2017 English Translation) for the exact statement.
      $endgroup$
      – Tyrone
      Jan 4 at 10:06


















    • $begingroup$
      I checked out lee's text but couldn't find it, you know which page it is?
      $endgroup$
      – CL.
      Jan 4 at 6:44










    • $begingroup$
      As I said, the theorem is not actually discussed by Lee, although the material in his book is a good general reference for submersions in general. Check out Roger Godement's book Introduction to the Theory of Lie Groups, section 4.9, Theorem 5, page 133 (Springer-Verlag 2017 English Translation) for the exact statement.
      $endgroup$
      – Tyrone
      Jan 4 at 10:06
















    $begingroup$
    I checked out lee's text but couldn't find it, you know which page it is?
    $endgroup$
    – CL.
    Jan 4 at 6:44




    $begingroup$
    I checked out lee's text but couldn't find it, you know which page it is?
    $endgroup$
    – CL.
    Jan 4 at 6:44












    $begingroup$
    As I said, the theorem is not actually discussed by Lee, although the material in his book is a good general reference for submersions in general. Check out Roger Godement's book Introduction to the Theory of Lie Groups, section 4.9, Theorem 5, page 133 (Springer-Verlag 2017 English Translation) for the exact statement.
    $endgroup$
    – Tyrone
    Jan 4 at 10:06




    $begingroup$
    As I said, the theorem is not actually discussed by Lee, although the material in his book is a good general reference for submersions in general. Check out Roger Godement's book Introduction to the Theory of Lie Groups, section 4.9, Theorem 5, page 133 (Springer-Verlag 2017 English Translation) for the exact statement.
    $endgroup$
    – Tyrone
    Jan 4 at 10:06











    0












    $begingroup$

    You have to construct compatible trivializations of the bundles $E'$ and $E$, which is slightly complicated to express in terms of transition functions. The easilest way to describe this in my opinion is in terms of local frames. Choose a local frame for $E'$ defined on $U$. Shrinking $U$, you can extend this to a local frame for $E$. (In a point $xin U$, extend the values of your frame which form a basis for $E'_x$ to a basis for $E_x$ and then extend the additional elements to local smooth sections of $E$.) Then you get a local trivialization $E|_Uto Utimesmathbb R^n$ which restricts to a trivialzation $E'|_Uto Utimesmathbb R^k$. From this, you get a trivialization from the quotient bundle to $mathbb R^n/mathbb R^kcongmathbb R^{n-k}$.



    In terms of transition functions, you can use the above construction to show that you can arrange things in such a way that the transtition functions for $E$ have values in block matrices of the form $begin{pmatrix} A & B \ 0 & Cend{pmatrix}$ such that the $A$-component gives you transition functions for $E'$. In this setting, the $C$-block gives you the transition functions for $E/E'$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      You have to construct compatible trivializations of the bundles $E'$ and $E$, which is slightly complicated to express in terms of transition functions. The easilest way to describe this in my opinion is in terms of local frames. Choose a local frame for $E'$ defined on $U$. Shrinking $U$, you can extend this to a local frame for $E$. (In a point $xin U$, extend the values of your frame which form a basis for $E'_x$ to a basis for $E_x$ and then extend the additional elements to local smooth sections of $E$.) Then you get a local trivialization $E|_Uto Utimesmathbb R^n$ which restricts to a trivialzation $E'|_Uto Utimesmathbb R^k$. From this, you get a trivialization from the quotient bundle to $mathbb R^n/mathbb R^kcongmathbb R^{n-k}$.



      In terms of transition functions, you can use the above construction to show that you can arrange things in such a way that the transtition functions for $E$ have values in block matrices of the form $begin{pmatrix} A & B \ 0 & Cend{pmatrix}$ such that the $A$-component gives you transition functions for $E'$. In this setting, the $C$-block gives you the transition functions for $E/E'$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        You have to construct compatible trivializations of the bundles $E'$ and $E$, which is slightly complicated to express in terms of transition functions. The easilest way to describe this in my opinion is in terms of local frames. Choose a local frame for $E'$ defined on $U$. Shrinking $U$, you can extend this to a local frame for $E$. (In a point $xin U$, extend the values of your frame which form a basis for $E'_x$ to a basis for $E_x$ and then extend the additional elements to local smooth sections of $E$.) Then you get a local trivialization $E|_Uto Utimesmathbb R^n$ which restricts to a trivialzation $E'|_Uto Utimesmathbb R^k$. From this, you get a trivialization from the quotient bundle to $mathbb R^n/mathbb R^kcongmathbb R^{n-k}$.



        In terms of transition functions, you can use the above construction to show that you can arrange things in such a way that the transtition functions for $E$ have values in block matrices of the form $begin{pmatrix} A & B \ 0 & Cend{pmatrix}$ such that the $A$-component gives you transition functions for $E'$. In this setting, the $C$-block gives you the transition functions for $E/E'$.






        share|cite|improve this answer









        $endgroup$



        You have to construct compatible trivializations of the bundles $E'$ and $E$, which is slightly complicated to express in terms of transition functions. The easilest way to describe this in my opinion is in terms of local frames. Choose a local frame for $E'$ defined on $U$. Shrinking $U$, you can extend this to a local frame for $E$. (In a point $xin U$, extend the values of your frame which form a basis for $E'_x$ to a basis for $E_x$ and then extend the additional elements to local smooth sections of $E$.) Then you get a local trivialization $E|_Uto Utimesmathbb R^n$ which restricts to a trivialzation $E'|_Uto Utimesmathbb R^k$. From this, you get a trivialization from the quotient bundle to $mathbb R^n/mathbb R^kcongmathbb R^{n-k}$.



        In terms of transition functions, you can use the above construction to show that you can arrange things in such a way that the transtition functions for $E$ have values in block matrices of the form $begin{pmatrix} A & B \ 0 & Cend{pmatrix}$ such that the $A$-component gives you transition functions for $E'$. In this setting, the $C$-block gives you the transition functions for $E/E'$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 at 11:45









        Andreas CapAndreas Cap

        11.5k923




        11.5k923






























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