Is the metric completion of a Riemannian manifold always a geodesic space?












16












$begingroup$


A length space is a metric space $X$, where the distance between two points is the infimum of the lengths of curves joining them. The length of a curve $c: [0,1] rightarrow X$ is the sup of
$$ d(c(0), c(t_1)) + d(c(t_1), d(t_2)) + cdots + d(c(t_{N-1}), c(1)) $$
over all $0 < t_1 < t_2cdots < t_{N-1} < 1$ and $N > 0$.



A geodesic space is a length space, where for each $x,y in X$, there is a curve $c$ connecting $x$ to $y$ whose length is equal to $d(x,y)$.



A Riemannian manifold $M$ and its metric completion $overline{M}$ are length spaces. If the Riemannian manifold is complete, then it is a geodesic space.



But is $overline{M}$ necessarily a geodesic space? If not, what is a counterexample?



This was motivated by my flawed answer to Minimizing geodesics in incomplete Riemannian manifolds



Also, note that if $overline{M}$ is locally compact, then it is a geodesic space by the usual proof. One example of $M$, where $overline{M}$ is not locally compact is the universal cover of the punctured flat plane. However, this is still a geodesic space.






dg.differential-geometry mg.metric-geometry riemannian-geometry







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$endgroup$

















    16












    $begingroup$


    A length space is a metric space $X$, where the distance between two points is the infimum of the lengths of curves joining them. The length of a curve $c: [0,1] rightarrow X$ is the sup of
    $$ d(c(0), c(t_1)) + d(c(t_1), d(t_2)) + cdots + d(c(t_{N-1}), c(1)) $$
    over all $0 < t_1 < t_2cdots < t_{N-1} < 1$ and $N > 0$.



    A geodesic space is a length space, where for each $x,y in X$, there is a curve $c$ connecting $x$ to $y$ whose length is equal to $d(x,y)$.



    A Riemannian manifold $M$ and its metric completion $overline{M}$ are length spaces. If the Riemannian manifold is complete, then it is a geodesic space.



    But is $overline{M}$ necessarily a geodesic space? If not, what is a counterexample?



    This was motivated by my flawed answer to Minimizing geodesics in incomplete Riemannian manifolds



    Also, note that if $overline{M}$ is locally compact, then it is a geodesic space by the usual proof. One example of $M$, where $overline{M}$ is not locally compact is the universal cover of the punctured flat plane. However, this is still a geodesic space.






    dg.differential-geometry mg.metric-geometry riemannian-geometry







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    $endgroup$















      16












      16








      16


      2



      $begingroup$


      A length space is a metric space $X$, where the distance between two points is the infimum of the lengths of curves joining them. The length of a curve $c: [0,1] rightarrow X$ is the sup of
      $$ d(c(0), c(t_1)) + d(c(t_1), d(t_2)) + cdots + d(c(t_{N-1}), c(1)) $$
      over all $0 < t_1 < t_2cdots < t_{N-1} < 1$ and $N > 0$.



      A geodesic space is a length space, where for each $x,y in X$, there is a curve $c$ connecting $x$ to $y$ whose length is equal to $d(x,y)$.



      A Riemannian manifold $M$ and its metric completion $overline{M}$ are length spaces. If the Riemannian manifold is complete, then it is a geodesic space.



      But is $overline{M}$ necessarily a geodesic space? If not, what is a counterexample?



      This was motivated by my flawed answer to Minimizing geodesics in incomplete Riemannian manifolds



      Also, note that if $overline{M}$ is locally compact, then it is a geodesic space by the usual proof. One example of $M$, where $overline{M}$ is not locally compact is the universal cover of the punctured flat plane. However, this is still a geodesic space.






      dg.differential-geometry mg.metric-geometry riemannian-geometry







      share|cite|improve this question









      $endgroup$




      A length space is a metric space $X$, where the distance between two points is the infimum of the lengths of curves joining them. The length of a curve $c: [0,1] rightarrow X$ is the sup of
      $$ d(c(0), c(t_1)) + d(c(t_1), d(t_2)) + cdots + d(c(t_{N-1}), c(1)) $$
      over all $0 < t_1 < t_2cdots < t_{N-1} < 1$ and $N > 0$.



      A geodesic space is a length space, where for each $x,y in X$, there is a curve $c$ connecting $x$ to $y$ whose length is equal to $d(x,y)$.



      A Riemannian manifold $M$ and its metric completion $overline{M}$ are length spaces. If the Riemannian manifold is complete, then it is a geodesic space.



      But is $overline{M}$ necessarily a geodesic space? If not, what is a counterexample?



      This was motivated by my flawed answer to Minimizing geodesics in incomplete Riemannian manifolds



      Also, note that if $overline{M}$ is locally compact, then it is a geodesic space by the usual proof. One example of $M$, where $overline{M}$ is not locally compact is the universal cover of the punctured flat plane. However, this is still a geodesic space.






      dg.differential-geometry mg.metric-geometry riemannian-geometry




      dg.differential-geometry mg.metric-geometry riemannian-geometry






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      asked 9 hours ago









      Deane YangDeane Yang

      20.1k562141




      20.1k562141






















          1 Answer
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          10












          $begingroup$

          I have been thinking about this since Deane and I discussed it this morning, and I came up with the following idea. Let $Sigma:={1,tfrac{1}{2},tfrac{1}{3},ldots}cup {-1,-tfrac{1}{2},-tfrac{1}{3},ldots}$. The set $Sigmacup {0}$ is closed in $mathbb{R}$.



          Let $(M,g)$ be the complement of $[0,1]times (Sigmacup{0})$ in the Euclidean plane. Offhand, it seems to me that the metric completion $overline{M}$ of $(M,g)$ contains the following "extra points":




          • ${0,1}times (Sigmacup{0})$


          • for each $(t,s)in(0,1)timesSigma$, two points $(t,s)_pm$, coming from the (two different) directional limits $lim_{yto s^pm}(t,y)$.



          Importantly, as far as I can tell, there is nothing in $overline{M}$ corresponding to the points in the segment $(0,1)times{0}$.



          If that's so, then the distance between the points $(0,0)$ and $(1,0)$ is 1, but there is no curve of distance 1 in $overline{M}$ connecting them.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Yes, this does work. It is similar to Ballmann's example mentioned in Benoit's answer here: mathoverflow.net/questions/15592/…
            $endgroup$
            – Misha
            6 hours ago






          • 1




            $begingroup$
            Oh, that's interesting. Yes, this seems to be a "fattening" of the Ballman example which makes it a Riemannian manifold.
            $endgroup$
            – macbeth
            6 hours ago






          • 1




            $begingroup$
            Right, topologically speaking, you are taking a neighborhood of Ballmann's example as embedded in $R^2$.
            $endgroup$
            – Misha
            6 hours ago










          • $begingroup$
            @macbeth, thanks!
            $endgroup$
            – Deane Yang
            4 hours ago











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          1 Answer
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          1 Answer
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          active

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          10












          $begingroup$

          I have been thinking about this since Deane and I discussed it this morning, and I came up with the following idea. Let $Sigma:={1,tfrac{1}{2},tfrac{1}{3},ldots}cup {-1,-tfrac{1}{2},-tfrac{1}{3},ldots}$. The set $Sigmacup {0}$ is closed in $mathbb{R}$.



          Let $(M,g)$ be the complement of $[0,1]times (Sigmacup{0})$ in the Euclidean plane. Offhand, it seems to me that the metric completion $overline{M}$ of $(M,g)$ contains the following "extra points":




          • ${0,1}times (Sigmacup{0})$


          • for each $(t,s)in(0,1)timesSigma$, two points $(t,s)_pm$, coming from the (two different) directional limits $lim_{yto s^pm}(t,y)$.



          Importantly, as far as I can tell, there is nothing in $overline{M}$ corresponding to the points in the segment $(0,1)times{0}$.



          If that's so, then the distance between the points $(0,0)$ and $(1,0)$ is 1, but there is no curve of distance 1 in $overline{M}$ connecting them.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Yes, this does work. It is similar to Ballmann's example mentioned in Benoit's answer here: mathoverflow.net/questions/15592/…
            $endgroup$
            – Misha
            6 hours ago






          • 1




            $begingroup$
            Oh, that's interesting. Yes, this seems to be a "fattening" of the Ballman example which makes it a Riemannian manifold.
            $endgroup$
            – macbeth
            6 hours ago






          • 1




            $begingroup$
            Right, topologically speaking, you are taking a neighborhood of Ballmann's example as embedded in $R^2$.
            $endgroup$
            – Misha
            6 hours ago










          • $begingroup$
            @macbeth, thanks!
            $endgroup$
            – Deane Yang
            4 hours ago
















          10












          $begingroup$

          I have been thinking about this since Deane and I discussed it this morning, and I came up with the following idea. Let $Sigma:={1,tfrac{1}{2},tfrac{1}{3},ldots}cup {-1,-tfrac{1}{2},-tfrac{1}{3},ldots}$. The set $Sigmacup {0}$ is closed in $mathbb{R}$.



          Let $(M,g)$ be the complement of $[0,1]times (Sigmacup{0})$ in the Euclidean plane. Offhand, it seems to me that the metric completion $overline{M}$ of $(M,g)$ contains the following "extra points":




          • ${0,1}times (Sigmacup{0})$


          • for each $(t,s)in(0,1)timesSigma$, two points $(t,s)_pm$, coming from the (two different) directional limits $lim_{yto s^pm}(t,y)$.



          Importantly, as far as I can tell, there is nothing in $overline{M}$ corresponding to the points in the segment $(0,1)times{0}$.



          If that's so, then the distance between the points $(0,0)$ and $(1,0)$ is 1, but there is no curve of distance 1 in $overline{M}$ connecting them.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Yes, this does work. It is similar to Ballmann's example mentioned in Benoit's answer here: mathoverflow.net/questions/15592/…
            $endgroup$
            – Misha
            6 hours ago






          • 1




            $begingroup$
            Oh, that's interesting. Yes, this seems to be a "fattening" of the Ballman example which makes it a Riemannian manifold.
            $endgroup$
            – macbeth
            6 hours ago






          • 1




            $begingroup$
            Right, topologically speaking, you are taking a neighborhood of Ballmann's example as embedded in $R^2$.
            $endgroup$
            – Misha
            6 hours ago










          • $begingroup$
            @macbeth, thanks!
            $endgroup$
            – Deane Yang
            4 hours ago














          10












          10








          10





          $begingroup$

          I have been thinking about this since Deane and I discussed it this morning, and I came up with the following idea. Let $Sigma:={1,tfrac{1}{2},tfrac{1}{3},ldots}cup {-1,-tfrac{1}{2},-tfrac{1}{3},ldots}$. The set $Sigmacup {0}$ is closed in $mathbb{R}$.



          Let $(M,g)$ be the complement of $[0,1]times (Sigmacup{0})$ in the Euclidean plane. Offhand, it seems to me that the metric completion $overline{M}$ of $(M,g)$ contains the following "extra points":




          • ${0,1}times (Sigmacup{0})$


          • for each $(t,s)in(0,1)timesSigma$, two points $(t,s)_pm$, coming from the (two different) directional limits $lim_{yto s^pm}(t,y)$.



          Importantly, as far as I can tell, there is nothing in $overline{M}$ corresponding to the points in the segment $(0,1)times{0}$.



          If that's so, then the distance between the points $(0,0)$ and $(1,0)$ is 1, but there is no curve of distance 1 in $overline{M}$ connecting them.






          share|cite|improve this answer









          $endgroup$



          I have been thinking about this since Deane and I discussed it this morning, and I came up with the following idea. Let $Sigma:={1,tfrac{1}{2},tfrac{1}{3},ldots}cup {-1,-tfrac{1}{2},-tfrac{1}{3},ldots}$. The set $Sigmacup {0}$ is closed in $mathbb{R}$.



          Let $(M,g)$ be the complement of $[0,1]times (Sigmacup{0})$ in the Euclidean plane. Offhand, it seems to me that the metric completion $overline{M}$ of $(M,g)$ contains the following "extra points":




          • ${0,1}times (Sigmacup{0})$


          • for each $(t,s)in(0,1)timesSigma$, two points $(t,s)_pm$, coming from the (two different) directional limits $lim_{yto s^pm}(t,y)$.



          Importantly, as far as I can tell, there is nothing in $overline{M}$ corresponding to the points in the segment $(0,1)times{0}$.



          If that's so, then the distance between the points $(0,0)$ and $(1,0)$ is 1, but there is no curve of distance 1 in $overline{M}$ connecting them.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 7 hours ago









          macbethmacbeth

          1,7781427




          1,7781427








          • 1




            $begingroup$
            Yes, this does work. It is similar to Ballmann's example mentioned in Benoit's answer here: mathoverflow.net/questions/15592/…
            $endgroup$
            – Misha
            6 hours ago






          • 1




            $begingroup$
            Oh, that's interesting. Yes, this seems to be a "fattening" of the Ballman example which makes it a Riemannian manifold.
            $endgroup$
            – macbeth
            6 hours ago






          • 1




            $begingroup$
            Right, topologically speaking, you are taking a neighborhood of Ballmann's example as embedded in $R^2$.
            $endgroup$
            – Misha
            6 hours ago










          • $begingroup$
            @macbeth, thanks!
            $endgroup$
            – Deane Yang
            4 hours ago














          • 1




            $begingroup$
            Yes, this does work. It is similar to Ballmann's example mentioned in Benoit's answer here: mathoverflow.net/questions/15592/…
            $endgroup$
            – Misha
            6 hours ago






          • 1




            $begingroup$
            Oh, that's interesting. Yes, this seems to be a "fattening" of the Ballman example which makes it a Riemannian manifold.
            $endgroup$
            – macbeth
            6 hours ago






          • 1




            $begingroup$
            Right, topologically speaking, you are taking a neighborhood of Ballmann's example as embedded in $R^2$.
            $endgroup$
            – Misha
            6 hours ago










          • $begingroup$
            @macbeth, thanks!
            $endgroup$
            – Deane Yang
            4 hours ago








          1




          1




          $begingroup$
          Yes, this does work. It is similar to Ballmann's example mentioned in Benoit's answer here: mathoverflow.net/questions/15592/…
          $endgroup$
          – Misha
          6 hours ago




          $begingroup$
          Yes, this does work. It is similar to Ballmann's example mentioned in Benoit's answer here: mathoverflow.net/questions/15592/…
          $endgroup$
          – Misha
          6 hours ago




          1




          1




          $begingroup$
          Oh, that's interesting. Yes, this seems to be a "fattening" of the Ballman example which makes it a Riemannian manifold.
          $endgroup$
          – macbeth
          6 hours ago




          $begingroup$
          Oh, that's interesting. Yes, this seems to be a "fattening" of the Ballman example which makes it a Riemannian manifold.
          $endgroup$
          – macbeth
          6 hours ago




          1




          1




          $begingroup$
          Right, topologically speaking, you are taking a neighborhood of Ballmann's example as embedded in $R^2$.
          $endgroup$
          – Misha
          6 hours ago




          $begingroup$
          Right, topologically speaking, you are taking a neighborhood of Ballmann's example as embedded in $R^2$.
          $endgroup$
          – Misha
          6 hours ago












          $begingroup$
          @macbeth, thanks!
          $endgroup$
          – Deane Yang
          4 hours ago




          $begingroup$
          @macbeth, thanks!
          $endgroup$
          – Deane Yang
          4 hours ago


















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