existence of a non-negative smooth function on a neighborhood of point on a boundary of a smooth manifold.












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Let $X$ be an n-dimensional manifold with boundary and let $x in partial X$. Show that there exists a smooth non-negative function $f$ on some open neighborhood $U$ of $x$, such that $f(z)=0$ iff $z in partial U$, with $0$ be the regular value of $f$.




Could someone help me in solving this problem? thank you.






multivariable-calculus differential-geometry differential-topology smooth-manifolds







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  • 2




    $begingroup$
    The most elementary solution is to let $U$ be a coordinate chart for the point $xinpartial X$ and use the definition of a manifold with boundary.
    $endgroup$
    – Ted Shifrin
    Nov 30 '18 at 20:32
















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$begingroup$


Let $X$ be an n-dimensional manifold with boundary and let $x in partial X$. Show that there exists a smooth non-negative function $f$ on some open neighborhood $U$ of $x$, such that $f(z)=0$ iff $z in partial U$, with $0$ be the regular value of $f$.




Could someone help me in solving this problem? thank you.






multivariable-calculus differential-geometry differential-topology smooth-manifolds







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The most elementary solution is to let $U$ be a coordinate chart for the point $xinpartial X$ and use the definition of a manifold with boundary.
    $endgroup$
    – Ted Shifrin
    Nov 30 '18 at 20:32














0












0








0





$begingroup$


Let $X$ be an n-dimensional manifold with boundary and let $x in partial X$. Show that there exists a smooth non-negative function $f$ on some open neighborhood $U$ of $x$, such that $f(z)=0$ iff $z in partial U$, with $0$ be the regular value of $f$.




Could someone help me in solving this problem? thank you.






multivariable-calculus differential-geometry differential-topology smooth-manifolds







share|cite|improve this question











$endgroup$




Let $X$ be an n-dimensional manifold with boundary and let $x in partial X$. Show that there exists a smooth non-negative function $f$ on some open neighborhood $U$ of $x$, such that $f(z)=0$ iff $z in partial U$, with $0$ be the regular value of $f$.




Could someone help me in solving this problem? thank you.






multivariable-calculus differential-geometry differential-topology smooth-manifolds




multivariable-calculus differential-geometry differential-topology smooth-manifolds






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edited Nov 30 '18 at 16:49







Amirhossein

















asked Nov 30 '18 at 16:15









AmirhosseinAmirhossein

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  • 2




    $begingroup$
    The most elementary solution is to let $U$ be a coordinate chart for the point $xinpartial X$ and use the definition of a manifold with boundary.
    $endgroup$
    – Ted Shifrin
    Nov 30 '18 at 20:32














  • 2




    $begingroup$
    The most elementary solution is to let $U$ be a coordinate chart for the point $xinpartial X$ and use the definition of a manifold with boundary.
    $endgroup$
    – Ted Shifrin
    Nov 30 '18 at 20:32








2




2




$begingroup$
The most elementary solution is to let $U$ be a coordinate chart for the point $xinpartial X$ and use the definition of a manifold with boundary.
$endgroup$
– Ted Shifrin
Nov 30 '18 at 20:32




$begingroup$
The most elementary solution is to let $U$ be a coordinate chart for the point $xinpartial X$ and use the definition of a manifold with boundary.
$endgroup$
– Ted Shifrin
Nov 30 '18 at 20:32










1 Answer
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$begingroup$

I'll denote the manifold by $M$. I assume you want to take $x$ as a boundary point (?).



Denote by $n$ the interior unit normal along the boundary (on a neighbourhood of some point, if you want to do this locally)



If you know that the exponentional map
$$partial M times [0,varepsilon ): (x,t)mapsto exp_x(t n(x))$$
is a diffeormorphism onto a neighbourhood of $partial M$ in $M$ (for sufficiently small $varepsilon >0$, if $partial M$ is compact) then you know that (on that neighbourhood)
$$f(exp_x(t n(x))) := t$$
is well defined. It has the properties you asked for.



If $partial M$ is not compact you need to modify this a bit or do this locally.



If you do not know about the exponential map you may work in local coordinates (which will result in a local result, which you could patch together to a global one using a partition of unity):



Consider a chart $$varphi:mathbb{R}^{n-1}times [0,infty) rightarrow M$$
at the boundary and define
$$h(x, t) = t$$
on
$mathbb{R}^{n-1}times [0,infty)$
and let $f(p) = h(varphi^{-1}(p))$






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    $begingroup$

    I'll denote the manifold by $M$. I assume you want to take $x$ as a boundary point (?).



    Denote by $n$ the interior unit normal along the boundary (on a neighbourhood of some point, if you want to do this locally)



    If you know that the exponentional map
    $$partial M times [0,varepsilon ): (x,t)mapsto exp_x(t n(x))$$
    is a diffeormorphism onto a neighbourhood of $partial M$ in $M$ (for sufficiently small $varepsilon >0$, if $partial M$ is compact) then you know that (on that neighbourhood)
    $$f(exp_x(t n(x))) := t$$
    is well defined. It has the properties you asked for.



    If $partial M$ is not compact you need to modify this a bit or do this locally.



    If you do not know about the exponential map you may work in local coordinates (which will result in a local result, which you could patch together to a global one using a partition of unity):



    Consider a chart $$varphi:mathbb{R}^{n-1}times [0,infty) rightarrow M$$
    at the boundary and define
    $$h(x, t) = t$$
    on
    $mathbb{R}^{n-1}times [0,infty)$
    and let $f(p) = h(varphi^{-1}(p))$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      I'll denote the manifold by $M$. I assume you want to take $x$ as a boundary point (?).



      Denote by $n$ the interior unit normal along the boundary (on a neighbourhood of some point, if you want to do this locally)



      If you know that the exponentional map
      $$partial M times [0,varepsilon ): (x,t)mapsto exp_x(t n(x))$$
      is a diffeormorphism onto a neighbourhood of $partial M$ in $M$ (for sufficiently small $varepsilon >0$, if $partial M$ is compact) then you know that (on that neighbourhood)
      $$f(exp_x(t n(x))) := t$$
      is well defined. It has the properties you asked for.



      If $partial M$ is not compact you need to modify this a bit or do this locally.



      If you do not know about the exponential map you may work in local coordinates (which will result in a local result, which you could patch together to a global one using a partition of unity):



      Consider a chart $$varphi:mathbb{R}^{n-1}times [0,infty) rightarrow M$$
      at the boundary and define
      $$h(x, t) = t$$
      on
      $mathbb{R}^{n-1}times [0,infty)$
      and let $f(p) = h(varphi^{-1}(p))$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        I'll denote the manifold by $M$. I assume you want to take $x$ as a boundary point (?).



        Denote by $n$ the interior unit normal along the boundary (on a neighbourhood of some point, if you want to do this locally)



        If you know that the exponentional map
        $$partial M times [0,varepsilon ): (x,t)mapsto exp_x(t n(x))$$
        is a diffeormorphism onto a neighbourhood of $partial M$ in $M$ (for sufficiently small $varepsilon >0$, if $partial M$ is compact) then you know that (on that neighbourhood)
        $$f(exp_x(t n(x))) := t$$
        is well defined. It has the properties you asked for.



        If $partial M$ is not compact you need to modify this a bit or do this locally.



        If you do not know about the exponential map you may work in local coordinates (which will result in a local result, which you could patch together to a global one using a partition of unity):



        Consider a chart $$varphi:mathbb{R}^{n-1}times [0,infty) rightarrow M$$
        at the boundary and define
        $$h(x, t) = t$$
        on
        $mathbb{R}^{n-1}times [0,infty)$
        and let $f(p) = h(varphi^{-1}(p))$






        share|cite|improve this answer











        $endgroup$



        I'll denote the manifold by $M$. I assume you want to take $x$ as a boundary point (?).



        Denote by $n$ the interior unit normal along the boundary (on a neighbourhood of some point, if you want to do this locally)



        If you know that the exponentional map
        $$partial M times [0,varepsilon ): (x,t)mapsto exp_x(t n(x))$$
        is a diffeormorphism onto a neighbourhood of $partial M$ in $M$ (for sufficiently small $varepsilon >0$, if $partial M$ is compact) then you know that (on that neighbourhood)
        $$f(exp_x(t n(x))) := t$$
        is well defined. It has the properties you asked for.



        If $partial M$ is not compact you need to modify this a bit or do this locally.



        If you do not know about the exponential map you may work in local coordinates (which will result in a local result, which you could patch together to a global one using a partition of unity):



        Consider a chart $$varphi:mathbb{R}^{n-1}times [0,infty) rightarrow M$$
        at the boundary and define
        $$h(x, t) = t$$
        on
        $mathbb{R}^{n-1}times [0,infty)$
        and let $f(p) = h(varphi^{-1}(p))$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 30 '18 at 16:54

























        answered Nov 30 '18 at 16:47









        ThomasThomas

        16.7k21631




        16.7k21631






























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