Construct a nontrivial symplectomorphism of cotangent bundle












2














I have tried to prove that exercise 4 on the page 20, Lectures on Symplectic Geometry, Ana Cannas da Silva, which is available on professor's website: https://people.math.ethz.ch/~acannas/Papers/lsg.pdf



Let $X$ be an arbitrary $n$-manifold, and let $M=T^{*}X$ its cotangent bundle. Let h be a smooth function on $X$. Define $tau_h : M to M$ by setting $$tau_h(x,xi)=(x,xi+dh_x).$$ Prove that $$tau_h^{*} alpha= alpha + pi^*dh$$ where $pi$ is the projection map $pi: M to X$ defined by $(x,xi) to x$. Deduce that $$tau_h^{*} omega= omega,$$ i.e., $tau_h$ is a symplectomorphism.



I haven't discovered a possible pattern of proof yet. Any hints or suggestions?






symplectic-geometry







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    2














    I have tried to prove that exercise 4 on the page 20, Lectures on Symplectic Geometry, Ana Cannas da Silva, which is available on professor's website: https://people.math.ethz.ch/~acannas/Papers/lsg.pdf



    Let $X$ be an arbitrary $n$-manifold, and let $M=T^{*}X$ its cotangent bundle. Let h be a smooth function on $X$. Define $tau_h : M to M$ by setting $$tau_h(x,xi)=(x,xi+dh_x).$$ Prove that $$tau_h^{*} alpha= alpha + pi^*dh$$ where $pi$ is the projection map $pi: M to X$ defined by $(x,xi) to x$. Deduce that $$tau_h^{*} omega= omega,$$ i.e., $tau_h$ is a symplectomorphism.



    I haven't discovered a possible pattern of proof yet. Any hints or suggestions?






    symplectic-geometry







    share|cite|improve this question

























      2












      2








      2


      3





      I have tried to prove that exercise 4 on the page 20, Lectures on Symplectic Geometry, Ana Cannas da Silva, which is available on professor's website: https://people.math.ethz.ch/~acannas/Papers/lsg.pdf



      Let $X$ be an arbitrary $n$-manifold, and let $M=T^{*}X$ its cotangent bundle. Let h be a smooth function on $X$. Define $tau_h : M to M$ by setting $$tau_h(x,xi)=(x,xi+dh_x).$$ Prove that $$tau_h^{*} alpha= alpha + pi^*dh$$ where $pi$ is the projection map $pi: M to X$ defined by $(x,xi) to x$. Deduce that $$tau_h^{*} omega= omega,$$ i.e., $tau_h$ is a symplectomorphism.



      I haven't discovered a possible pattern of proof yet. Any hints or suggestions?






      symplectic-geometry







      share|cite|improve this question













      I have tried to prove that exercise 4 on the page 20, Lectures on Symplectic Geometry, Ana Cannas da Silva, which is available on professor's website: https://people.math.ethz.ch/~acannas/Papers/lsg.pdf



      Let $X$ be an arbitrary $n$-manifold, and let $M=T^{*}X$ its cotangent bundle. Let h be a smooth function on $X$. Define $tau_h : M to M$ by setting $$tau_h(x,xi)=(x,xi+dh_x).$$ Prove that $$tau_h^{*} alpha= alpha + pi^*dh$$ where $pi$ is the projection map $pi: M to X$ defined by $(x,xi) to x$. Deduce that $$tau_h^{*} omega= omega,$$ i.e., $tau_h$ is a symplectomorphism.



      I haven't discovered a possible pattern of proof yet. Any hints or suggestions?






      symplectic-geometry




      symplectic-geometry






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      asked Mar 6 '18 at 19:10







      user530422





























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          $alpha_{(x,xi)}(u,v)=xi(u)$, $d(tau_h)_{(x,xi)}(u,v)=(u,v+d^2h_x(u))$, this implies that $tau_h^*alpha_{(x,xi)}(u,v)=alpha_{(x,xi+dh_x)}(u,v+dh^2_x.u)=(xi+dh_x)(u)=xi(u)+dh_x(u)$



          $=alpha_{(x,xi)}(u)+ pi^*dh_x(u)$.



          We deduce that $tau_h^*alpha=alpha+pi^*dh$.



          $tau_h^*alpha=alpha+pi^*dh$ implies that $(dtau_h^*alpha)=dalpha+d(pi^*dh)$



          This is equivalent to say that:
          $tau_h^*(dalpha)=dalpha+pi^*(d(d(h))$, since $d^2=0$, and $-dalpha=omega$, we deduce that $tau_h^*omega=omega$.






          share|cite|improve this answer























          • Why $tau_h^*alpha_{(x,xi)}(u,v)=xi(u)+dh_x(u)$ implies that $tau_h^*alpha_{(x,xi)}=xi+dh_x$? Is it independent from $v$?
            – user530422
            Mar 7 '18 at 10:25










          • yes, it is independent of $v$, $alpha_{(x,xi)}(u,v)=xi(u)$ where $(u,v)in T_{(x,xi)}T^*X$.
            – Tsemo Aristide
            Mar 7 '18 at 12:22



















          0














          For any $p = (x ,xi) in M$ and $u in T_pM$



          $$begin{aligned} (tau_h alpha)^*_p (u) &= alpha_{tau_h (p)} ((dtau_h)_p(u))\&= alpha_{(x, xi + dh_x)} ((dtau_h)_p(u)) \&= (xi + dh_x)(dpi_p circ (dtau_h)_p(u))\&= (xi + dh_x) (underbrace{d(pi circ tau)_p}_{dpi_p} (u))\&=xi (dpi_p (u)) + dh_{pi(p)} (dpi_p (u)) \&= alpha_p (u) + (pi^*dh)_p (u)
          end{aligned}$$
          as required.



          Finally, we see that



          $$tau_h^* omega = tau_h^* (- dalpha) = - d(tau^*alpha) = -d (alpha + pi^*dh) = underbrace{-dalpha}_{omega} - pi^*underbrace{d^2h}_0= omega$$






          share|cite|improve this answer





















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            2 Answers
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            $alpha_{(x,xi)}(u,v)=xi(u)$, $d(tau_h)_{(x,xi)}(u,v)=(u,v+d^2h_x(u))$, this implies that $tau_h^*alpha_{(x,xi)}(u,v)=alpha_{(x,xi+dh_x)}(u,v+dh^2_x.u)=(xi+dh_x)(u)=xi(u)+dh_x(u)$



            $=alpha_{(x,xi)}(u)+ pi^*dh_x(u)$.



            We deduce that $tau_h^*alpha=alpha+pi^*dh$.



            $tau_h^*alpha=alpha+pi^*dh$ implies that $(dtau_h^*alpha)=dalpha+d(pi^*dh)$



            This is equivalent to say that:
            $tau_h^*(dalpha)=dalpha+pi^*(d(d(h))$, since $d^2=0$, and $-dalpha=omega$, we deduce that $tau_h^*omega=omega$.






            share|cite|improve this answer























            • Why $tau_h^*alpha_{(x,xi)}(u,v)=xi(u)+dh_x(u)$ implies that $tau_h^*alpha_{(x,xi)}=xi+dh_x$? Is it independent from $v$?
              – user530422
              Mar 7 '18 at 10:25










            • yes, it is independent of $v$, $alpha_{(x,xi)}(u,v)=xi(u)$ where $(u,v)in T_{(x,xi)}T^*X$.
              – Tsemo Aristide
              Mar 7 '18 at 12:22
















            0














            $alpha_{(x,xi)}(u,v)=xi(u)$, $d(tau_h)_{(x,xi)}(u,v)=(u,v+d^2h_x(u))$, this implies that $tau_h^*alpha_{(x,xi)}(u,v)=alpha_{(x,xi+dh_x)}(u,v+dh^2_x.u)=(xi+dh_x)(u)=xi(u)+dh_x(u)$



            $=alpha_{(x,xi)}(u)+ pi^*dh_x(u)$.



            We deduce that $tau_h^*alpha=alpha+pi^*dh$.



            $tau_h^*alpha=alpha+pi^*dh$ implies that $(dtau_h^*alpha)=dalpha+d(pi^*dh)$



            This is equivalent to say that:
            $tau_h^*(dalpha)=dalpha+pi^*(d(d(h))$, since $d^2=0$, and $-dalpha=omega$, we deduce that $tau_h^*omega=omega$.






            share|cite|improve this answer























            • Why $tau_h^*alpha_{(x,xi)}(u,v)=xi(u)+dh_x(u)$ implies that $tau_h^*alpha_{(x,xi)}=xi+dh_x$? Is it independent from $v$?
              – user530422
              Mar 7 '18 at 10:25










            • yes, it is independent of $v$, $alpha_{(x,xi)}(u,v)=xi(u)$ where $(u,v)in T_{(x,xi)}T^*X$.
              – Tsemo Aristide
              Mar 7 '18 at 12:22














            0












            0








            0






            $alpha_{(x,xi)}(u,v)=xi(u)$, $d(tau_h)_{(x,xi)}(u,v)=(u,v+d^2h_x(u))$, this implies that $tau_h^*alpha_{(x,xi)}(u,v)=alpha_{(x,xi+dh_x)}(u,v+dh^2_x.u)=(xi+dh_x)(u)=xi(u)+dh_x(u)$



            $=alpha_{(x,xi)}(u)+ pi^*dh_x(u)$.



            We deduce that $tau_h^*alpha=alpha+pi^*dh$.



            $tau_h^*alpha=alpha+pi^*dh$ implies that $(dtau_h^*alpha)=dalpha+d(pi^*dh)$



            This is equivalent to say that:
            $tau_h^*(dalpha)=dalpha+pi^*(d(d(h))$, since $d^2=0$, and $-dalpha=omega$, we deduce that $tau_h^*omega=omega$.






            share|cite|improve this answer














            $alpha_{(x,xi)}(u,v)=xi(u)$, $d(tau_h)_{(x,xi)}(u,v)=(u,v+d^2h_x(u))$, this implies that $tau_h^*alpha_{(x,xi)}(u,v)=alpha_{(x,xi+dh_x)}(u,v+dh^2_x.u)=(xi+dh_x)(u)=xi(u)+dh_x(u)$



            $=alpha_{(x,xi)}(u)+ pi^*dh_x(u)$.



            We deduce that $tau_h^*alpha=alpha+pi^*dh$.



            $tau_h^*alpha=alpha+pi^*dh$ implies that $(dtau_h^*alpha)=dalpha+d(pi^*dh)$



            This is equivalent to say that:
            $tau_h^*(dalpha)=dalpha+pi^*(d(d(h))$, since $d^2=0$, and $-dalpha=omega$, we deduce that $tau_h^*omega=omega$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 8 '18 at 12:38







            user530422

















            answered Mar 6 '18 at 19:20









            Tsemo Aristide

            55.7k11444




            55.7k11444












            • Why $tau_h^*alpha_{(x,xi)}(u,v)=xi(u)+dh_x(u)$ implies that $tau_h^*alpha_{(x,xi)}=xi+dh_x$? Is it independent from $v$?
              – user530422
              Mar 7 '18 at 10:25










            • yes, it is independent of $v$, $alpha_{(x,xi)}(u,v)=xi(u)$ where $(u,v)in T_{(x,xi)}T^*X$.
              – Tsemo Aristide
              Mar 7 '18 at 12:22


















            • Why $tau_h^*alpha_{(x,xi)}(u,v)=xi(u)+dh_x(u)$ implies that $tau_h^*alpha_{(x,xi)}=xi+dh_x$? Is it independent from $v$?
              – user530422
              Mar 7 '18 at 10:25










            • yes, it is independent of $v$, $alpha_{(x,xi)}(u,v)=xi(u)$ where $(u,v)in T_{(x,xi)}T^*X$.
              – Tsemo Aristide
              Mar 7 '18 at 12:22
















            Why $tau_h^*alpha_{(x,xi)}(u,v)=xi(u)+dh_x(u)$ implies that $tau_h^*alpha_{(x,xi)}=xi+dh_x$? Is it independent from $v$?
            – user530422
            Mar 7 '18 at 10:25




            Why $tau_h^*alpha_{(x,xi)}(u,v)=xi(u)+dh_x(u)$ implies that $tau_h^*alpha_{(x,xi)}=xi+dh_x$? Is it independent from $v$?
            – user530422
            Mar 7 '18 at 10:25












            yes, it is independent of $v$, $alpha_{(x,xi)}(u,v)=xi(u)$ where $(u,v)in T_{(x,xi)}T^*X$.
            – Tsemo Aristide
            Mar 7 '18 at 12:22




            yes, it is independent of $v$, $alpha_{(x,xi)}(u,v)=xi(u)$ where $(u,v)in T_{(x,xi)}T^*X$.
            – Tsemo Aristide
            Mar 7 '18 at 12:22











            0














            For any $p = (x ,xi) in M$ and $u in T_pM$



            $$begin{aligned} (tau_h alpha)^*_p (u) &= alpha_{tau_h (p)} ((dtau_h)_p(u))\&= alpha_{(x, xi + dh_x)} ((dtau_h)_p(u)) \&= (xi + dh_x)(dpi_p circ (dtau_h)_p(u))\&= (xi + dh_x) (underbrace{d(pi circ tau)_p}_{dpi_p} (u))\&=xi (dpi_p (u)) + dh_{pi(p)} (dpi_p (u)) \&= alpha_p (u) + (pi^*dh)_p (u)
            end{aligned}$$
            as required.



            Finally, we see that



            $$tau_h^* omega = tau_h^* (- dalpha) = - d(tau^*alpha) = -d (alpha + pi^*dh) = underbrace{-dalpha}_{omega} - pi^*underbrace{d^2h}_0= omega$$






            share|cite|improve this answer


























              0














              For any $p = (x ,xi) in M$ and $u in T_pM$



              $$begin{aligned} (tau_h alpha)^*_p (u) &= alpha_{tau_h (p)} ((dtau_h)_p(u))\&= alpha_{(x, xi + dh_x)} ((dtau_h)_p(u)) \&= (xi + dh_x)(dpi_p circ (dtau_h)_p(u))\&= (xi + dh_x) (underbrace{d(pi circ tau)_p}_{dpi_p} (u))\&=xi (dpi_p (u)) + dh_{pi(p)} (dpi_p (u)) \&= alpha_p (u) + (pi^*dh)_p (u)
              end{aligned}$$
              as required.



              Finally, we see that



              $$tau_h^* omega = tau_h^* (- dalpha) = - d(tau^*alpha) = -d (alpha + pi^*dh) = underbrace{-dalpha}_{omega} - pi^*underbrace{d^2h}_0= omega$$






              share|cite|improve this answer
























                0












                0








                0






                For any $p = (x ,xi) in M$ and $u in T_pM$



                $$begin{aligned} (tau_h alpha)^*_p (u) &= alpha_{tau_h (p)} ((dtau_h)_p(u))\&= alpha_{(x, xi + dh_x)} ((dtau_h)_p(u)) \&= (xi + dh_x)(dpi_p circ (dtau_h)_p(u))\&= (xi + dh_x) (underbrace{d(pi circ tau)_p}_{dpi_p} (u))\&=xi (dpi_p (u)) + dh_{pi(p)} (dpi_p (u)) \&= alpha_p (u) + (pi^*dh)_p (u)
                end{aligned}$$
                as required.



                Finally, we see that



                $$tau_h^* omega = tau_h^* (- dalpha) = - d(tau^*alpha) = -d (alpha + pi^*dh) = underbrace{-dalpha}_{omega} - pi^*underbrace{d^2h}_0= omega$$






                share|cite|improve this answer












                For any $p = (x ,xi) in M$ and $u in T_pM$



                $$begin{aligned} (tau_h alpha)^*_p (u) &= alpha_{tau_h (p)} ((dtau_h)_p(u))\&= alpha_{(x, xi + dh_x)} ((dtau_h)_p(u)) \&= (xi + dh_x)(dpi_p circ (dtau_h)_p(u))\&= (xi + dh_x) (underbrace{d(pi circ tau)_p}_{dpi_p} (u))\&=xi (dpi_p (u)) + dh_{pi(p)} (dpi_p (u)) \&= alpha_p (u) + (pi^*dh)_p (u)
                end{aligned}$$
                as required.



                Finally, we see that



                $$tau_h^* omega = tau_h^* (- dalpha) = - d(tau^*alpha) = -d (alpha + pi^*dh) = underbrace{-dalpha}_{omega} - pi^*underbrace{d^2h}_0= omega$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 27 '18 at 1:08









                Aaron Maroja

                15.5k51445




                15.5k51445






























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