The automorphism group of a symplectic symmetric space











up vote
1
down vote

favorite












Why is the automorphism group of a sympelctic symmetric space a Lie group?



$\$



A symplectic symmetric space is a triple $(M, omega, s)$, where $(M, omega)$ is a symplectic manifold and $ s ; colon M times M to M $, $(x, y) mapsto s_x(y)$, is such that $s_x$ is an involutive symplectic diffeomorphism with an isolated fixed point at $x$ and $s_xs_ys_x = s_{s_x(y)} ; forall ; x, y in M; $ $big($this can be read as $s_x s = ss_x ; forall x in M$$big)$.



An interesting fact about a symplectic symmetric space is the one that once $x$ is an isolated fixed point of $s_x$ and $s_x^2 = Id$, $ds_{x (x)} = -Id_{T_xM}$. Furthermore, a symplectic symmetric space admits a unique affine connection $nabla$ on it such that $s_x$ is an affinity for every $x in M$ and such that $nabla omega = 0$. This connection has no torsion and is thus a symplectic connection.



The automorphism group $Aut(M, omega, s)$ of $(M, omega, s)$ is the group of symplectic automorphisms $varphi$ of $(M, omega)$ that satisfy $varphi circ s_x = s_{varphi(x)} circ varphi ; forall ; x in M$ $big($ or, in other words, $varphi circ s = scirc varphi$ $big)$. It's simple to check that this group is the intersection of the symplectic automorphism group of $(M, omega)$ and the affine group of $(M, nabla)$.



My question then is: why is $Aut(M, omega, s)$ a Lie group?










share|cite|improve this question


























    up vote
    1
    down vote

    favorite












    Why is the automorphism group of a sympelctic symmetric space a Lie group?



    $\$



    A symplectic symmetric space is a triple $(M, omega, s)$, where $(M, omega)$ is a symplectic manifold and $ s ; colon M times M to M $, $(x, y) mapsto s_x(y)$, is such that $s_x$ is an involutive symplectic diffeomorphism with an isolated fixed point at $x$ and $s_xs_ys_x = s_{s_x(y)} ; forall ; x, y in M; $ $big($this can be read as $s_x s = ss_x ; forall x in M$$big)$.



    An interesting fact about a symplectic symmetric space is the one that once $x$ is an isolated fixed point of $s_x$ and $s_x^2 = Id$, $ds_{x (x)} = -Id_{T_xM}$. Furthermore, a symplectic symmetric space admits a unique affine connection $nabla$ on it such that $s_x$ is an affinity for every $x in M$ and such that $nabla omega = 0$. This connection has no torsion and is thus a symplectic connection.



    The automorphism group $Aut(M, omega, s)$ of $(M, omega, s)$ is the group of symplectic automorphisms $varphi$ of $(M, omega)$ that satisfy $varphi circ s_x = s_{varphi(x)} circ varphi ; forall ; x in M$ $big($ or, in other words, $varphi circ s = scirc varphi$ $big)$. It's simple to check that this group is the intersection of the symplectic automorphism group of $(M, omega)$ and the affine group of $(M, nabla)$.



    My question then is: why is $Aut(M, omega, s)$ a Lie group?










    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Why is the automorphism group of a sympelctic symmetric space a Lie group?



      $\$



      A symplectic symmetric space is a triple $(M, omega, s)$, where $(M, omega)$ is a symplectic manifold and $ s ; colon M times M to M $, $(x, y) mapsto s_x(y)$, is such that $s_x$ is an involutive symplectic diffeomorphism with an isolated fixed point at $x$ and $s_xs_ys_x = s_{s_x(y)} ; forall ; x, y in M; $ $big($this can be read as $s_x s = ss_x ; forall x in M$$big)$.



      An interesting fact about a symplectic symmetric space is the one that once $x$ is an isolated fixed point of $s_x$ and $s_x^2 = Id$, $ds_{x (x)} = -Id_{T_xM}$. Furthermore, a symplectic symmetric space admits a unique affine connection $nabla$ on it such that $s_x$ is an affinity for every $x in M$ and such that $nabla omega = 0$. This connection has no torsion and is thus a symplectic connection.



      The automorphism group $Aut(M, omega, s)$ of $(M, omega, s)$ is the group of symplectic automorphisms $varphi$ of $(M, omega)$ that satisfy $varphi circ s_x = s_{varphi(x)} circ varphi ; forall ; x in M$ $big($ or, in other words, $varphi circ s = scirc varphi$ $big)$. It's simple to check that this group is the intersection of the symplectic automorphism group of $(M, omega)$ and the affine group of $(M, nabla)$.



      My question then is: why is $Aut(M, omega, s)$ a Lie group?










      share|cite|improve this question













      Why is the automorphism group of a sympelctic symmetric space a Lie group?



      $\$



      A symplectic symmetric space is a triple $(M, omega, s)$, where $(M, omega)$ is a symplectic manifold and $ s ; colon M times M to M $, $(x, y) mapsto s_x(y)$, is such that $s_x$ is an involutive symplectic diffeomorphism with an isolated fixed point at $x$ and $s_xs_ys_x = s_{s_x(y)} ; forall ; x, y in M; $ $big($this can be read as $s_x s = ss_x ; forall x in M$$big)$.



      An interesting fact about a symplectic symmetric space is the one that once $x$ is an isolated fixed point of $s_x$ and $s_x^2 = Id$, $ds_{x (x)} = -Id_{T_xM}$. Furthermore, a symplectic symmetric space admits a unique affine connection $nabla$ on it such that $s_x$ is an affinity for every $x in M$ and such that $nabla omega = 0$. This connection has no torsion and is thus a symplectic connection.



      The automorphism group $Aut(M, omega, s)$ of $(M, omega, s)$ is the group of symplectic automorphisms $varphi$ of $(M, omega)$ that satisfy $varphi circ s_x = s_{varphi(x)} circ varphi ; forall ; x in M$ $big($ or, in other words, $varphi circ s = scirc varphi$ $big)$. It's simple to check that this group is the intersection of the symplectic automorphism group of $(M, omega)$ and the affine group of $(M, nabla)$.



      My question then is: why is $Aut(M, omega, s)$ a Lie group?







      sg.symplectic-geometry connections symmetric-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 3 hours ago









      Valentino

      1055




      1055






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          4
          down vote



          accepted










          The affine group of $(M,nabla)$ is a Lie group $G$ by Kobayashi's theorem that shows that the automorphism group of any affine connection is a Lie group (see Kobayashi and Nomizu's Foundations of Differential Geometry). The dimension of $G$ is at most $n+n^2$ (where $n=dim M$).



          The subgroup $H$ of $G$ consisting of those elements of $G$ that preserve the symplectic structure on $M$ is clearly a closed subgroup, and since every closed subgroup of a Lie group is a Lie group (see, for example, Helgason's proof in his Differential Geometry, Lie Groups, and Symmetric Spaces), it follows that $H$ is a Lie group as well.






          share|cite|improve this answer





















            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "504"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f317497%2fthe-automorphism-group-of-a-symplectic-symmetric-space%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            4
            down vote



            accepted










            The affine group of $(M,nabla)$ is a Lie group $G$ by Kobayashi's theorem that shows that the automorphism group of any affine connection is a Lie group (see Kobayashi and Nomizu's Foundations of Differential Geometry). The dimension of $G$ is at most $n+n^2$ (where $n=dim M$).



            The subgroup $H$ of $G$ consisting of those elements of $G$ that preserve the symplectic structure on $M$ is clearly a closed subgroup, and since every closed subgroup of a Lie group is a Lie group (see, for example, Helgason's proof in his Differential Geometry, Lie Groups, and Symmetric Spaces), it follows that $H$ is a Lie group as well.






            share|cite|improve this answer

























              up vote
              4
              down vote



              accepted










              The affine group of $(M,nabla)$ is a Lie group $G$ by Kobayashi's theorem that shows that the automorphism group of any affine connection is a Lie group (see Kobayashi and Nomizu's Foundations of Differential Geometry). The dimension of $G$ is at most $n+n^2$ (where $n=dim M$).



              The subgroup $H$ of $G$ consisting of those elements of $G$ that preserve the symplectic structure on $M$ is clearly a closed subgroup, and since every closed subgroup of a Lie group is a Lie group (see, for example, Helgason's proof in his Differential Geometry, Lie Groups, and Symmetric Spaces), it follows that $H$ is a Lie group as well.






              share|cite|improve this answer























                up vote
                4
                down vote



                accepted







                up vote
                4
                down vote



                accepted






                The affine group of $(M,nabla)$ is a Lie group $G$ by Kobayashi's theorem that shows that the automorphism group of any affine connection is a Lie group (see Kobayashi and Nomizu's Foundations of Differential Geometry). The dimension of $G$ is at most $n+n^2$ (where $n=dim M$).



                The subgroup $H$ of $G$ consisting of those elements of $G$ that preserve the symplectic structure on $M$ is clearly a closed subgroup, and since every closed subgroup of a Lie group is a Lie group (see, for example, Helgason's proof in his Differential Geometry, Lie Groups, and Symmetric Spaces), it follows that $H$ is a Lie group as well.






                share|cite|improve this answer












                The affine group of $(M,nabla)$ is a Lie group $G$ by Kobayashi's theorem that shows that the automorphism group of any affine connection is a Lie group (see Kobayashi and Nomizu's Foundations of Differential Geometry). The dimension of $G$ is at most $n+n^2$ (where $n=dim M$).



                The subgroup $H$ of $G$ consisting of those elements of $G$ that preserve the symplectic structure on $M$ is clearly a closed subgroup, and since every closed subgroup of a Lie group is a Lie group (see, for example, Helgason's proof in his Differential Geometry, Lie Groups, and Symmetric Spaces), it follows that $H$ is a Lie group as well.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 3 hours ago









                Robert Bryant

                72.5k5213313




                72.5k5213313






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to MathOverflow!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f317497%2fthe-automorphism-group-of-a-symplectic-symmetric-space%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    What visual should I use to simply compare current year value vs last year in Power BI desktop

                    Alexandru Averescu

                    Trompette piccolo