$f:Mrightarrow M$ is an isometry and $ker(Id - d_{gamma(t)}f) = mathbb{R} dot{gamma}(t)$, show that $gamma$...












1












$begingroup$


Let M be a Riemannian manifold, $gamma: I rightarrow M$ be a curve which is parameterized with constant speed and $f:Mrightarrow M$ be an isometry with $f circ gamma = gamma$. Furthermore $ker(Id - d_{gamma(t)}f) = mathbb{R} dot{gamma}(t), forall t$. Show that $gamma$ is a geodesic then.



I need some help with this. I don't know how f plays into this. We know that $d_{gamma(t)}f(adot{gamma}(t)) = adot{gamma}(t)$ for any $a in mathbb{R}$ , then for $ mathbb{R} ni c = langle dot{gamma(t)}, dot{gamma(t)} rangle = langle d_{gamma(t)}f(dot{gamma}(t)), d_{gamma(t)}f(dot{gamma}(t)) rangle$.



I don't see how any of this is supposed to be related to $nabla_{dot{gamma}}dot{gamma}$.



Why is it important that f is an isometry?










share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Let M be a Riemannian manifold, $gamma: I rightarrow M$ be a curve which is parameterized with constant speed and $f:Mrightarrow M$ be an isometry with $f circ gamma = gamma$. Furthermore $ker(Id - d_{gamma(t)}f) = mathbb{R} dot{gamma}(t), forall t$. Show that $gamma$ is a geodesic then.



    I need some help with this. I don't know how f plays into this. We know that $d_{gamma(t)}f(adot{gamma}(t)) = adot{gamma}(t)$ for any $a in mathbb{R}$ , then for $ mathbb{R} ni c = langle dot{gamma(t)}, dot{gamma(t)} rangle = langle d_{gamma(t)}f(dot{gamma}(t)), d_{gamma(t)}f(dot{gamma}(t)) rangle$.



    I don't see how any of this is supposed to be related to $nabla_{dot{gamma}}dot{gamma}$.



    Why is it important that f is an isometry?










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let M be a Riemannian manifold, $gamma: I rightarrow M$ be a curve which is parameterized with constant speed and $f:Mrightarrow M$ be an isometry with $f circ gamma = gamma$. Furthermore $ker(Id - d_{gamma(t)}f) = mathbb{R} dot{gamma}(t), forall t$. Show that $gamma$ is a geodesic then.



      I need some help with this. I don't know how f plays into this. We know that $d_{gamma(t)}f(adot{gamma}(t)) = adot{gamma}(t)$ for any $a in mathbb{R}$ , then for $ mathbb{R} ni c = langle dot{gamma(t)}, dot{gamma(t)} rangle = langle d_{gamma(t)}f(dot{gamma}(t)), d_{gamma(t)}f(dot{gamma}(t)) rangle$.



      I don't see how any of this is supposed to be related to $nabla_{dot{gamma}}dot{gamma}$.



      Why is it important that f is an isometry?










      share|cite|improve this question









      $endgroup$




      Let M be a Riemannian manifold, $gamma: I rightarrow M$ be a curve which is parameterized with constant speed and $f:Mrightarrow M$ be an isometry with $f circ gamma = gamma$. Furthermore $ker(Id - d_{gamma(t)}f) = mathbb{R} dot{gamma}(t), forall t$. Show that $gamma$ is a geodesic then.



      I need some help with this. I don't know how f plays into this. We know that $d_{gamma(t)}f(adot{gamma}(t)) = adot{gamma}(t)$ for any $a in mathbb{R}$ , then for $ mathbb{R} ni c = langle dot{gamma(t)}, dot{gamma(t)} rangle = langle d_{gamma(t)}f(dot{gamma}(t)), d_{gamma(t)}f(dot{gamma}(t)) rangle$.



      I don't see how any of this is supposed to be related to $nabla_{dot{gamma}}dot{gamma}$.



      Why is it important that f is an isometry?







      differential-geometry






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 7 at 13:26









      eager2learneager2learn

      1,23711430




      1,23711430






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          This is a consequence of the fact that, under your given assumptions, $gamma$ constitutes a local parameterization of an immersed submanifold of $M$ which is fixed by an isometry. But such immersed submanifolds are totally-geodesic. In this instance, the immersed submanifold is of dimension 1 and an immersed, totally-geodesic submanifold of dimension 1 is just a geodesic.






          share|cite|improve this answer









          $endgroup$













            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: "69"
            };
            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',
            autoActivateHeartbeat: false,
            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%2fmath.stackexchange.com%2fquestions%2f3065003%2ffm-rightarrow-m-is-an-isometry-and-kerid-d-gammatf-mathbbr-do%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









            0












            $begingroup$

            This is a consequence of the fact that, under your given assumptions, $gamma$ constitutes a local parameterization of an immersed submanifold of $M$ which is fixed by an isometry. But such immersed submanifolds are totally-geodesic. In this instance, the immersed submanifold is of dimension 1 and an immersed, totally-geodesic submanifold of dimension 1 is just a geodesic.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              This is a consequence of the fact that, under your given assumptions, $gamma$ constitutes a local parameterization of an immersed submanifold of $M$ which is fixed by an isometry. But such immersed submanifolds are totally-geodesic. In this instance, the immersed submanifold is of dimension 1 and an immersed, totally-geodesic submanifold of dimension 1 is just a geodesic.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                This is a consequence of the fact that, under your given assumptions, $gamma$ constitutes a local parameterization of an immersed submanifold of $M$ which is fixed by an isometry. But such immersed submanifolds are totally-geodesic. In this instance, the immersed submanifold is of dimension 1 and an immersed, totally-geodesic submanifold of dimension 1 is just a geodesic.






                share|cite|improve this answer









                $endgroup$



                This is a consequence of the fact that, under your given assumptions, $gamma$ constitutes a local parameterization of an immersed submanifold of $M$ which is fixed by an isometry. But such immersed submanifolds are totally-geodesic. In this instance, the immersed submanifold is of dimension 1 and an immersed, totally-geodesic submanifold of dimension 1 is just a geodesic.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                Or EisenbergOr Eisenberg

                1346




                1346






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • 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.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065003%2ffm-rightarrow-m-is-an-isometry-and-kerid-d-gammatf-mathbbr-do%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

                    Mario Kart Wii

                    The Binding of Isaac: Rebirth/Afterbirth

                    What does “Dominus providebit” mean?