Cogeodesic flow on compact manifold has compact leaves and manifold












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Let $g_{ij}$ be positive definite metric on a compact manifold $M$. Consider hamiltonian $H=sum_{ij}frac{1}{2}g_{ij}(x) p_ip_j$ with $p_iinGamma(T^star M)$. Then define $E_lambda={(x,p)in T^star Mvert H(x,p)=lambda}$



$textbf{Q:}$ $E_lambda$ is compact. How do I see this obviously? I did the following. Take any open covering of $E_lambda$ and refine it if necessary s.t. one has trivialization of $T^star M$. Then project down to $M$ and then lift up to $T^star M$ and this selects some of the open covering. Since $M$ is compact, $g_{ij}(x)$ has minimal value $g$. Then use $sum_{ij}g p_ip_jleqlambda$ to bound $p_i$ in closed set. Hence, $E_lambda$ is compact by selecting the finite open covering.



Ref: Riemannian Geometry and Geometric Analysis Jost. Chpt 2 Lie Groups and Vector Bunbldes Pg 70






geometry differential-equations differential-geometry riemannian-geometry







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    Let $g_{ij}$ be positive definite metric on a compact manifold $M$. Consider hamiltonian $H=sum_{ij}frac{1}{2}g_{ij}(x) p_ip_j$ with $p_iinGamma(T^star M)$. Then define $E_lambda={(x,p)in T^star Mvert H(x,p)=lambda}$



    $textbf{Q:}$ $E_lambda$ is compact. How do I see this obviously? I did the following. Take any open covering of $E_lambda$ and refine it if necessary s.t. one has trivialization of $T^star M$. Then project down to $M$ and then lift up to $T^star M$ and this selects some of the open covering. Since $M$ is compact, $g_{ij}(x)$ has minimal value $g$. Then use $sum_{ij}g p_ip_jleqlambda$ to bound $p_i$ in closed set. Hence, $E_lambda$ is compact by selecting the finite open covering.



    Ref: Riemannian Geometry and Geometric Analysis Jost. Chpt 2 Lie Groups and Vector Bunbldes Pg 70






    geometry differential-equations differential-geometry riemannian-geometry







    share|cite|improve this question



























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      Let $g_{ij}$ be positive definite metric on a compact manifold $M$. Consider hamiltonian $H=sum_{ij}frac{1}{2}g_{ij}(x) p_ip_j$ with $p_iinGamma(T^star M)$. Then define $E_lambda={(x,p)in T^star Mvert H(x,p)=lambda}$



      $textbf{Q:}$ $E_lambda$ is compact. How do I see this obviously? I did the following. Take any open covering of $E_lambda$ and refine it if necessary s.t. one has trivialization of $T^star M$. Then project down to $M$ and then lift up to $T^star M$ and this selects some of the open covering. Since $M$ is compact, $g_{ij}(x)$ has minimal value $g$. Then use $sum_{ij}g p_ip_jleqlambda$ to bound $p_i$ in closed set. Hence, $E_lambda$ is compact by selecting the finite open covering.



      Ref: Riemannian Geometry and Geometric Analysis Jost. Chpt 2 Lie Groups and Vector Bunbldes Pg 70






      geometry differential-equations differential-geometry riemannian-geometry







      share|cite|improve this question















      Let $g_{ij}$ be positive definite metric on a compact manifold $M$. Consider hamiltonian $H=sum_{ij}frac{1}{2}g_{ij}(x) p_ip_j$ with $p_iinGamma(T^star M)$. Then define $E_lambda={(x,p)in T^star Mvert H(x,p)=lambda}$



      $textbf{Q:}$ $E_lambda$ is compact. How do I see this obviously? I did the following. Take any open covering of $E_lambda$ and refine it if necessary s.t. one has trivialization of $T^star M$. Then project down to $M$ and then lift up to $T^star M$ and this selects some of the open covering. Since $M$ is compact, $g_{ij}(x)$ has minimal value $g$. Then use $sum_{ij}g p_ip_jleqlambda$ to bound $p_i$ in closed set. Hence, $E_lambda$ is compact by selecting the finite open covering.



      Ref: Riemannian Geometry and Geometric Analysis Jost. Chpt 2 Lie Groups and Vector Bunbldes Pg 70






      geometry differential-equations differential-geometry riemannian-geometry




      geometry differential-equations differential-geometry riemannian-geometry






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      edited Nov 26 at 2:29

























      asked Nov 26 at 1:59









      user45765

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          This is equivalent to showing the sphere bundle $pi:SMto M$ is compact for compact manifolds $M$, and your case is just looking at the co-sphere bundle of radius $sqrt{2lambda}$, which are equivalent via the bundle isomorphism $F:TMto T^*M$, $F(v)=2lambda g(v,cdot).$



          Let ${x_j,v_j}subset SM$ be any sequence. Since $M$ is compact, by possibly passing to a subsequence there exists $xin M$ such that $x_jto x$. Let $Usubseteq M$ be a trivializing neighborhood of $x$, i.e., $pi^{-1}(U)cong Utimes S^{n-1}$. Since $x_jto xin U$, for infinitely many $j$, we have that $(x_j,v_j)in Utimes S^{n-1}$. In particular, we have infinitely many $v_j$in the compact set $S^{n-1}$. Again by possibly passing down to a further subsequence there exists some $vin S^{n-1}$ such that $v_jto v$. That is, the arbitrary sequence ${x_j,v_j}subset SM$ has an accumulation point in $SM$, thus showing $SM$ is compact.






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            This is equivalent to showing the sphere bundle $pi:SMto M$ is compact for compact manifolds $M$, and your case is just looking at the co-sphere bundle of radius $sqrt{2lambda}$, which are equivalent via the bundle isomorphism $F:TMto T^*M$, $F(v)=2lambda g(v,cdot).$



            Let ${x_j,v_j}subset SM$ be any sequence. Since $M$ is compact, by possibly passing to a subsequence there exists $xin M$ such that $x_jto x$. Let $Usubseteq M$ be a trivializing neighborhood of $x$, i.e., $pi^{-1}(U)cong Utimes S^{n-1}$. Since $x_jto xin U$, for infinitely many $j$, we have that $(x_j,v_j)in Utimes S^{n-1}$. In particular, we have infinitely many $v_j$in the compact set $S^{n-1}$. Again by possibly passing down to a further subsequence there exists some $vin S^{n-1}$ such that $v_jto v$. That is, the arbitrary sequence ${x_j,v_j}subset SM$ has an accumulation point in $SM$, thus showing $SM$ is compact.






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              This is equivalent to showing the sphere bundle $pi:SMto M$ is compact for compact manifolds $M$, and your case is just looking at the co-sphere bundle of radius $sqrt{2lambda}$, which are equivalent via the bundle isomorphism $F:TMto T^*M$, $F(v)=2lambda g(v,cdot).$



              Let ${x_j,v_j}subset SM$ be any sequence. Since $M$ is compact, by possibly passing to a subsequence there exists $xin M$ such that $x_jto x$. Let $Usubseteq M$ be a trivializing neighborhood of $x$, i.e., $pi^{-1}(U)cong Utimes S^{n-1}$. Since $x_jto xin U$, for infinitely many $j$, we have that $(x_j,v_j)in Utimes S^{n-1}$. In particular, we have infinitely many $v_j$in the compact set $S^{n-1}$. Again by possibly passing down to a further subsequence there exists some $vin S^{n-1}$ such that $v_jto v$. That is, the arbitrary sequence ${x_j,v_j}subset SM$ has an accumulation point in $SM$, thus showing $SM$ is compact.






              share|cite|improve this answer
























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                This is equivalent to showing the sphere bundle $pi:SMto M$ is compact for compact manifolds $M$, and your case is just looking at the co-sphere bundle of radius $sqrt{2lambda}$, which are equivalent via the bundle isomorphism $F:TMto T^*M$, $F(v)=2lambda g(v,cdot).$



                Let ${x_j,v_j}subset SM$ be any sequence. Since $M$ is compact, by possibly passing to a subsequence there exists $xin M$ such that $x_jto x$. Let $Usubseteq M$ be a trivializing neighborhood of $x$, i.e., $pi^{-1}(U)cong Utimes S^{n-1}$. Since $x_jto xin U$, for infinitely many $j$, we have that $(x_j,v_j)in Utimes S^{n-1}$. In particular, we have infinitely many $v_j$in the compact set $S^{n-1}$. Again by possibly passing down to a further subsequence there exists some $vin S^{n-1}$ such that $v_jto v$. That is, the arbitrary sequence ${x_j,v_j}subset SM$ has an accumulation point in $SM$, thus showing $SM$ is compact.






                share|cite|improve this answer












                This is equivalent to showing the sphere bundle $pi:SMto M$ is compact for compact manifolds $M$, and your case is just looking at the co-sphere bundle of radius $sqrt{2lambda}$, which are equivalent via the bundle isomorphism $F:TMto T^*M$, $F(v)=2lambda g(v,cdot).$



                Let ${x_j,v_j}subset SM$ be any sequence. Since $M$ is compact, by possibly passing to a subsequence there exists $xin M$ such that $x_jto x$. Let $Usubseteq M$ be a trivializing neighborhood of $x$, i.e., $pi^{-1}(U)cong Utimes S^{n-1}$. Since $x_jto xin U$, for infinitely many $j$, we have that $(x_j,v_j)in Utimes S^{n-1}$. In particular, we have infinitely many $v_j$in the compact set $S^{n-1}$. Again by possibly passing down to a further subsequence there exists some $vin S^{n-1}$ such that $v_jto v$. That is, the arbitrary sequence ${x_j,v_j}subset SM$ has an accumulation point in $SM$, thus showing $SM$ is compact.







                share|cite|improve this answer












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                answered Nov 30 at 13:10









                Matt

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