Lower bound on length of closed curve












1














Let $Sigma = { (x,y,z)in mathbb{R}^3 colon -z^2+x^2+y^2 =1 }$ be the one-sheeted hyperboloid, so there is a unique closed geodesic, which is of length $2pi$, given by the intersection of $Sigma$ with the plane ${z=0}$. Fix a point $p=(x_0,y_0,z_0)in Sigma$ with $z_0>0$. Does there exist an $epsilon>0$ such that every smooth, homotopically non-trivial, closed curve $gamma colon [0,2pi] to Sigma$ with $gamma(0)=p$ has length $l(gamma)geq 2pi+epsilon$?



This seems to me to be intuitively the case, but I do not know how to prove it. Can anyone offer a proof or any suggestions towards a proof? Many thanks!






differential-geometry







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    Let $Sigma = { (x,y,z)in mathbb{R}^3 colon -z^2+x^2+y^2 =1 }$ be the one-sheeted hyperboloid, so there is a unique closed geodesic, which is of length $2pi$, given by the intersection of $Sigma$ with the plane ${z=0}$. Fix a point $p=(x_0,y_0,z_0)in Sigma$ with $z_0>0$. Does there exist an $epsilon>0$ such that every smooth, homotopically non-trivial, closed curve $gamma colon [0,2pi] to Sigma$ with $gamma(0)=p$ has length $l(gamma)geq 2pi+epsilon$?



    This seems to me to be intuitively the case, but I do not know how to prove it. Can anyone offer a proof or any suggestions towards a proof? Many thanks!






    differential-geometry







    share|cite|improve this question

























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      1







      Let $Sigma = { (x,y,z)in mathbb{R}^3 colon -z^2+x^2+y^2 =1 }$ be the one-sheeted hyperboloid, so there is a unique closed geodesic, which is of length $2pi$, given by the intersection of $Sigma$ with the plane ${z=0}$. Fix a point $p=(x_0,y_0,z_0)in Sigma$ with $z_0>0$. Does there exist an $epsilon>0$ such that every smooth, homotopically non-trivial, closed curve $gamma colon [0,2pi] to Sigma$ with $gamma(0)=p$ has length $l(gamma)geq 2pi+epsilon$?



      This seems to me to be intuitively the case, but I do not know how to prove it. Can anyone offer a proof or any suggestions towards a proof? Many thanks!






      differential-geometry







      share|cite|improve this question













      Let $Sigma = { (x,y,z)in mathbb{R}^3 colon -z^2+x^2+y^2 =1 }$ be the one-sheeted hyperboloid, so there is a unique closed geodesic, which is of length $2pi$, given by the intersection of $Sigma$ with the plane ${z=0}$. Fix a point $p=(x_0,y_0,z_0)in Sigma$ with $z_0>0$. Does there exist an $epsilon>0$ such that every smooth, homotopically non-trivial, closed curve $gamma colon [0,2pi] to Sigma$ with $gamma(0)=p$ has length $l(gamma)geq 2pi+epsilon$?



      This seems to me to be intuitively the case, but I do not know how to prove it. Can anyone offer a proof or any suggestions towards a proof? Many thanks!






      differential-geometry




      differential-geometry






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      asked Dec 5 '18 at 16:40









      Aerinmund FagelsonAerinmund Fagelson

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          I would use following arguments:




          • The curve $gamma$ lies outside the right cylinder of axis $Oz$ and radius equal to one.

          • The length of $gamma$ is larger or equal to the one of its projection $delta$ on the $Oxy$ plane. This is a general result: the length of a curve is larger or equal to the one of its orthogonal projection on any plane.


          • $delta$ is included in the region of the plane $Oxy$ minus the disk $D$ centered on $O$ with radius $1$ (consequence of the first argument).

          • Using the Isoperimetric inequality you get that the length of $delta$ is greater of equal to $2pi$. This is therefore also the case for the length of $gamma$.


          You have to use an extra argument to get that $l(gamma) ge 2pi +epsilon$ with $epsilon >0$ by using the isoperimetric inequality and the fact that the projection of $p$ on $Oxy$ is outside the disk $D$






          share|cite|improve this answer























          • Nice idea! Do you have a specific improved version of the isoperimetric inequality in mind? If $hat{p}inmathbb{R}^2$ is the projection of $p$ to the plane, then $|hat{p}|>1$, and the standard isoperimetric inequality (with jordan curve lemma) will tell me that any simple closed curve $xcolon [0,1]to mathbb{R}^2$ which has non-trivial winding number around the origin $(0,0)$ and satisfies $x(0)=hat{p}$ and $|x(t)|geq1$ must have length $>2pi$, but it does not give me the uniform bound $>2pi + epsilon$
            – Aerinmund Fagelson
            Dec 6 '18 at 19:31












          • You can use a small disk centered on $|hat(p)|$ not intersecting the disk centered O with radius one.
            – mathcounterexamples.net
            Dec 6 '18 at 19:52










          • Sorry i'm still confused, how do you make use of the little disk?
            – Aerinmund Fagelson
            Dec 6 '18 at 21:03











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          active

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          1














          I would use following arguments:




          • The curve $gamma$ lies outside the right cylinder of axis $Oz$ and radius equal to one.

          • The length of $gamma$ is larger or equal to the one of its projection $delta$ on the $Oxy$ plane. This is a general result: the length of a curve is larger or equal to the one of its orthogonal projection on any plane.


          • $delta$ is included in the region of the plane $Oxy$ minus the disk $D$ centered on $O$ with radius $1$ (consequence of the first argument).

          • Using the Isoperimetric inequality you get that the length of $delta$ is greater of equal to $2pi$. This is therefore also the case for the length of $gamma$.


          You have to use an extra argument to get that $l(gamma) ge 2pi +epsilon$ with $epsilon >0$ by using the isoperimetric inequality and the fact that the projection of $p$ on $Oxy$ is outside the disk $D$






          share|cite|improve this answer























          • Nice idea! Do you have a specific improved version of the isoperimetric inequality in mind? If $hat{p}inmathbb{R}^2$ is the projection of $p$ to the plane, then $|hat{p}|>1$, and the standard isoperimetric inequality (with jordan curve lemma) will tell me that any simple closed curve $xcolon [0,1]to mathbb{R}^2$ which has non-trivial winding number around the origin $(0,0)$ and satisfies $x(0)=hat{p}$ and $|x(t)|geq1$ must have length $>2pi$, but it does not give me the uniform bound $>2pi + epsilon$
            – Aerinmund Fagelson
            Dec 6 '18 at 19:31












          • You can use a small disk centered on $|hat(p)|$ not intersecting the disk centered O with radius one.
            – mathcounterexamples.net
            Dec 6 '18 at 19:52










          • Sorry i'm still confused, how do you make use of the little disk?
            – Aerinmund Fagelson
            Dec 6 '18 at 21:03
















          1














          I would use following arguments:




          • The curve $gamma$ lies outside the right cylinder of axis $Oz$ and radius equal to one.

          • The length of $gamma$ is larger or equal to the one of its projection $delta$ on the $Oxy$ plane. This is a general result: the length of a curve is larger or equal to the one of its orthogonal projection on any plane.


          • $delta$ is included in the region of the plane $Oxy$ minus the disk $D$ centered on $O$ with radius $1$ (consequence of the first argument).

          • Using the Isoperimetric inequality you get that the length of $delta$ is greater of equal to $2pi$. This is therefore also the case for the length of $gamma$.


          You have to use an extra argument to get that $l(gamma) ge 2pi +epsilon$ with $epsilon >0$ by using the isoperimetric inequality and the fact that the projection of $p$ on $Oxy$ is outside the disk $D$






          share|cite|improve this answer























          • Nice idea! Do you have a specific improved version of the isoperimetric inequality in mind? If $hat{p}inmathbb{R}^2$ is the projection of $p$ to the plane, then $|hat{p}|>1$, and the standard isoperimetric inequality (with jordan curve lemma) will tell me that any simple closed curve $xcolon [0,1]to mathbb{R}^2$ which has non-trivial winding number around the origin $(0,0)$ and satisfies $x(0)=hat{p}$ and $|x(t)|geq1$ must have length $>2pi$, but it does not give me the uniform bound $>2pi + epsilon$
            – Aerinmund Fagelson
            Dec 6 '18 at 19:31












          • You can use a small disk centered on $|hat(p)|$ not intersecting the disk centered O with radius one.
            – mathcounterexamples.net
            Dec 6 '18 at 19:52










          • Sorry i'm still confused, how do you make use of the little disk?
            – Aerinmund Fagelson
            Dec 6 '18 at 21:03














          1












          1








          1






          I would use following arguments:




          • The curve $gamma$ lies outside the right cylinder of axis $Oz$ and radius equal to one.

          • The length of $gamma$ is larger or equal to the one of its projection $delta$ on the $Oxy$ plane. This is a general result: the length of a curve is larger or equal to the one of its orthogonal projection on any plane.


          • $delta$ is included in the region of the plane $Oxy$ minus the disk $D$ centered on $O$ with radius $1$ (consequence of the first argument).

          • Using the Isoperimetric inequality you get that the length of $delta$ is greater of equal to $2pi$. This is therefore also the case for the length of $gamma$.


          You have to use an extra argument to get that $l(gamma) ge 2pi +epsilon$ with $epsilon >0$ by using the isoperimetric inequality and the fact that the projection of $p$ on $Oxy$ is outside the disk $D$






          share|cite|improve this answer














          I would use following arguments:




          • The curve $gamma$ lies outside the right cylinder of axis $Oz$ and radius equal to one.

          • The length of $gamma$ is larger or equal to the one of its projection $delta$ on the $Oxy$ plane. This is a general result: the length of a curve is larger or equal to the one of its orthogonal projection on any plane.


          • $delta$ is included in the region of the plane $Oxy$ minus the disk $D$ centered on $O$ with radius $1$ (consequence of the first argument).

          • Using the Isoperimetric inequality you get that the length of $delta$ is greater of equal to $2pi$. This is therefore also the case for the length of $gamma$.


          You have to use an extra argument to get that $l(gamma) ge 2pi +epsilon$ with $epsilon >0$ by using the isoperimetric inequality and the fact that the projection of $p$ on $Oxy$ is outside the disk $D$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 6 '18 at 8:45

























          answered Dec 5 '18 at 18:41









          mathcounterexamples.netmathcounterexamples.net

          25.3k21953




          25.3k21953












          • Nice idea! Do you have a specific improved version of the isoperimetric inequality in mind? If $hat{p}inmathbb{R}^2$ is the projection of $p$ to the plane, then $|hat{p}|>1$, and the standard isoperimetric inequality (with jordan curve lemma) will tell me that any simple closed curve $xcolon [0,1]to mathbb{R}^2$ which has non-trivial winding number around the origin $(0,0)$ and satisfies $x(0)=hat{p}$ and $|x(t)|geq1$ must have length $>2pi$, but it does not give me the uniform bound $>2pi + epsilon$
            – Aerinmund Fagelson
            Dec 6 '18 at 19:31












          • You can use a small disk centered on $|hat(p)|$ not intersecting the disk centered O with radius one.
            – mathcounterexamples.net
            Dec 6 '18 at 19:52










          • Sorry i'm still confused, how do you make use of the little disk?
            – Aerinmund Fagelson
            Dec 6 '18 at 21:03


















          • Nice idea! Do you have a specific improved version of the isoperimetric inequality in mind? If $hat{p}inmathbb{R}^2$ is the projection of $p$ to the plane, then $|hat{p}|>1$, and the standard isoperimetric inequality (with jordan curve lemma) will tell me that any simple closed curve $xcolon [0,1]to mathbb{R}^2$ which has non-trivial winding number around the origin $(0,0)$ and satisfies $x(0)=hat{p}$ and $|x(t)|geq1$ must have length $>2pi$, but it does not give me the uniform bound $>2pi + epsilon$
            – Aerinmund Fagelson
            Dec 6 '18 at 19:31












          • You can use a small disk centered on $|hat(p)|$ not intersecting the disk centered O with radius one.
            – mathcounterexamples.net
            Dec 6 '18 at 19:52










          • Sorry i'm still confused, how do you make use of the little disk?
            – Aerinmund Fagelson
            Dec 6 '18 at 21:03
















          Nice idea! Do you have a specific improved version of the isoperimetric inequality in mind? If $hat{p}inmathbb{R}^2$ is the projection of $p$ to the plane, then $|hat{p}|>1$, and the standard isoperimetric inequality (with jordan curve lemma) will tell me that any simple closed curve $xcolon [0,1]to mathbb{R}^2$ which has non-trivial winding number around the origin $(0,0)$ and satisfies $x(0)=hat{p}$ and $|x(t)|geq1$ must have length $>2pi$, but it does not give me the uniform bound $>2pi + epsilon$
          – Aerinmund Fagelson
          Dec 6 '18 at 19:31






          Nice idea! Do you have a specific improved version of the isoperimetric inequality in mind? If $hat{p}inmathbb{R}^2$ is the projection of $p$ to the plane, then $|hat{p}|>1$, and the standard isoperimetric inequality (with jordan curve lemma) will tell me that any simple closed curve $xcolon [0,1]to mathbb{R}^2$ which has non-trivial winding number around the origin $(0,0)$ and satisfies $x(0)=hat{p}$ and $|x(t)|geq1$ must have length $>2pi$, but it does not give me the uniform bound $>2pi + epsilon$
          – Aerinmund Fagelson
          Dec 6 '18 at 19:31














          You can use a small disk centered on $|hat(p)|$ not intersecting the disk centered O with radius one.
          – mathcounterexamples.net
          Dec 6 '18 at 19:52




          You can use a small disk centered on $|hat(p)|$ not intersecting the disk centered O with radius one.
          – mathcounterexamples.net
          Dec 6 '18 at 19:52












          Sorry i'm still confused, how do you make use of the little disk?
          – Aerinmund Fagelson
          Dec 6 '18 at 21:03




          Sorry i'm still confused, how do you make use of the little disk?
          – Aerinmund Fagelson
          Dec 6 '18 at 21:03


















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