Loomis and Sternberg: Tangent Space to a manifold, using equivalence classes; help justifying one step of an...












1












$begingroup$


I am currently reading through the section in Loomis and Sternberg's Advanced Calculus on Tangent Spaces, but I'm having trouble justifying one step of the argument (shown below).



Here's the definitions and notations used by them. Let $M$ be a differentiable manifold (here they model their manifolds on a banach space $V$ rather than some $mathbb{R}^n$). Let $x in M$, and let $varphi : I to M$ be a differentiable map where $I$ is an interval in $mathbb{R}$ containing $0$, and $varphi(0) = x.$ Then, they define an operator $D_{varphi}: C^{infty}(M) to mathbb{R}$ by $D_{varphi}(f) = (f circ varphi)'(0)$. Next, they define an equivalence relation on all the curves passing through $x$ by $varphi sim psi$ if and only if $D_{varphi} = D_{psi}$, and call an equivalence class of curves, $xi$ to be a tangent vector at $x$.



So far so good. Next, they go to a chart $(W, alpha)$, and the underlined section below is what I don't fully understand. I get that $varphi sim psi$ if and only if for every $f in C^{infty}(M)$,
$d(f circ alpha^{-1})_{alpha(x)}((alpha circ varphi)'(0)) =
d(f circ alpha^{-1})_{alpha(x)}((alpha circ psi)'(0))$.

But I don't see how to conclude from here that the two vectors
$(alpha circ varphi)'(0)$ and
$(alpha circ psi)'(0)$ are equal.



I'm guessing it has something to do with the fact that the two derivatives are equal for every $f$; if we somehow choose an $f$ such that the differential $d(f circ alpha^{-1})_{alpha(x)} : V to mathbb{R}$ is injective, then its kernel is ${0}$, and thus equality follows. However I doubt this is always possible, since in general $dim(V) > 1$, so a linear map from $V$ to $mathbb{R}$ cannot be injective.



Any help justifying this step is much appreciated.



enter image description here






differential-topology smooth-manifolds differential tangent-spaces







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I am currently reading through the section in Loomis and Sternberg's Advanced Calculus on Tangent Spaces, but I'm having trouble justifying one step of the argument (shown below).



    Here's the definitions and notations used by them. Let $M$ be a differentiable manifold (here they model their manifolds on a banach space $V$ rather than some $mathbb{R}^n$). Let $x in M$, and let $varphi : I to M$ be a differentiable map where $I$ is an interval in $mathbb{R}$ containing $0$, and $varphi(0) = x.$ Then, they define an operator $D_{varphi}: C^{infty}(M) to mathbb{R}$ by $D_{varphi}(f) = (f circ varphi)'(0)$. Next, they define an equivalence relation on all the curves passing through $x$ by $varphi sim psi$ if and only if $D_{varphi} = D_{psi}$, and call an equivalence class of curves, $xi$ to be a tangent vector at $x$.



    So far so good. Next, they go to a chart $(W, alpha)$, and the underlined section below is what I don't fully understand. I get that $varphi sim psi$ if and only if for every $f in C^{infty}(M)$,
    $d(f circ alpha^{-1})_{alpha(x)}((alpha circ varphi)'(0)) =
    d(f circ alpha^{-1})_{alpha(x)}((alpha circ psi)'(0))$    .

    But I don't see how to conclude from here that the two vectors
    $(alpha circ varphi)'(0)$ and
    $(alpha circ psi)'(0)$ are equal.



    I'm guessing it has something to do with the fact that the two derivatives are equal for every $f$; if we somehow choose an $f$ such that the differential $d(f circ alpha^{-1})_{alpha(x)} : V to mathbb{R}$ is injective, then its kernel is ${0}$, and thus equality follows. However I doubt this is always possible, since in general $dim(V) > 1$, so a linear map from $V$ to $mathbb{R}$ cannot be injective.



    Any help justifying this step is much appreciated.



    enter image description here






    differential-topology smooth-manifolds differential tangent-spaces







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am currently reading through the section in Loomis and Sternberg's Advanced Calculus on Tangent Spaces, but I'm having trouble justifying one step of the argument (shown below).



      Here's the definitions and notations used by them. Let $M$ be a differentiable manifold (here they model their manifolds on a banach space $V$ rather than some $mathbb{R}^n$). Let $x in M$, and let $varphi : I to M$ be a differentiable map where $I$ is an interval in $mathbb{R}$ containing $0$, and $varphi(0) = x.$ Then, they define an operator $D_{varphi}: C^{infty}(M) to mathbb{R}$ by $D_{varphi}(f) = (f circ varphi)'(0)$. Next, they define an equivalence relation on all the curves passing through $x$ by $varphi sim psi$ if and only if $D_{varphi} = D_{psi}$, and call an equivalence class of curves, $xi$ to be a tangent vector at $x$.



      So far so good. Next, they go to a chart $(W, alpha)$, and the underlined section below is what I don't fully understand. I get that $varphi sim psi$ if and only if for every $f in C^{infty}(M)$,
      $d(f circ alpha^{-1})_{alpha(x)}((alpha circ varphi)'(0)) =
      d(f circ alpha^{-1})_{alpha(x)}((alpha circ psi)'(0))$      .

      But I don't see how to conclude from here that the two vectors
      $(alpha circ varphi)'(0)$ and
      $(alpha circ psi)'(0)$ are equal.



      I'm guessing it has something to do with the fact that the two derivatives are equal for every $f$; if we somehow choose an $f$ such that the differential $d(f circ alpha^{-1})_{alpha(x)} : V to mathbb{R}$ is injective, then its kernel is ${0}$, and thus equality follows. However I doubt this is always possible, since in general $dim(V) > 1$, so a linear map from $V$ to $mathbb{R}$ cannot be injective.



      Any help justifying this step is much appreciated.



      enter image description here






      differential-topology smooth-manifolds differential tangent-spaces







      share|cite|improve this question









      $endgroup$




      I am currently reading through the section in Loomis and Sternberg's Advanced Calculus on Tangent Spaces, but I'm having trouble justifying one step of the argument (shown below).



      Here's the definitions and notations used by them. Let $M$ be a differentiable manifold (here they model their manifolds on a banach space $V$ rather than some $mathbb{R}^n$). Let $x in M$, and let $varphi : I to M$ be a differentiable map where $I$ is an interval in $mathbb{R}$ containing $0$, and $varphi(0) = x.$ Then, they define an operator $D_{varphi}: C^{infty}(M) to mathbb{R}$ by $D_{varphi}(f) = (f circ varphi)'(0)$. Next, they define an equivalence relation on all the curves passing through $x$ by $varphi sim psi$ if and only if $D_{varphi} = D_{psi}$, and call an equivalence class of curves, $xi$ to be a tangent vector at $x$.



      So far so good. Next, they go to a chart $(W, alpha)$, and the underlined section below is what I don't fully understand. I get that $varphi sim psi$ if and only if for every $f in C^{infty}(M)$,
      $d(f circ alpha^{-1})_{alpha(x)}((alpha circ varphi)'(0)) =
      d(f circ alpha^{-1})_{alpha(x)}((alpha circ psi)'(0))$      .

      But I don't see how to conclude from here that the two vectors
      $(alpha circ varphi)'(0)$ and
      $(alpha circ psi)'(0)$ are equal.



      I'm guessing it has something to do with the fact that the two derivatives are equal for every $f$; if we somehow choose an $f$ such that the differential $d(f circ alpha^{-1})_{alpha(x)} : V to mathbb{R}$ is injective, then its kernel is ${0}$, and thus equality follows. However I doubt this is always possible, since in general $dim(V) > 1$, so a linear map from $V$ to $mathbb{R}$ cannot be injective.



      Any help justifying this step is much appreciated.



      enter image description here






      differential-topology smooth-manifolds differential tangent-spaces




      differential-topology smooth-manifolds differential tangent-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 26 '18 at 20:39









      peek-a-boopeek-a-boo

      555




      555






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          The differential does not have to be injective. Note that for any vector space $W$, $phi(v) = 0$ for any linear functional $phi$ on $W$ implies that $v = 0$ necessarily. Therefore, since $mathrm{d}F$ is always a linear functional regardless of the choice of $f$, you simply need to show that you can obtain any linear functional on $V$ with an appropriate choice of $f$.






          share|cite|improve this answer









          $endgroup$














            Your Answer








            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%2f3053298%2floomis-and-sternberg-tangent-space-to-a-manifold-using-equivalence-classes-he%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












            $begingroup$

            The differential does not have to be injective. Note that for any vector space $W$, $phi(v) = 0$ for any linear functional $phi$ on $W$ implies that $v = 0$ necessarily. Therefore, since $mathrm{d}F$ is always a linear functional regardless of the choice of $f$, you simply need to show that you can obtain any linear functional on $V$ with an appropriate choice of $f$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              The differential does not have to be injective. Note that for any vector space $W$, $phi(v) = 0$ for any linear functional $phi$ on $W$ implies that $v = 0$ necessarily. Therefore, since $mathrm{d}F$ is always a linear functional regardless of the choice of $f$, you simply need to show that you can obtain any linear functional on $V$ with an appropriate choice of $f$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                The differential does not have to be injective. Note that for any vector space $W$, $phi(v) = 0$ for any linear functional $phi$ on $W$ implies that $v = 0$ necessarily. Therefore, since $mathrm{d}F$ is always a linear functional regardless of the choice of $f$, you simply need to show that you can obtain any linear functional on $V$ with an appropriate choice of $f$.






                share|cite|improve this answer









                $endgroup$



                The differential does not have to be injective. Note that for any vector space $W$, $phi(v) = 0$ for any linear functional $phi$ on $W$ implies that $v = 0$ necessarily. Therefore, since $mathrm{d}F$ is always a linear functional regardless of the choice of $f$, you simply need to show that you can obtain any linear functional on $V$ with an appropriate choice of $f$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 15 at 9:18









                Kurtland ChuaKurtland Chua

                3541212




                3541212






























                    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%2f3053298%2floomis-and-sternberg-tangent-space-to-a-manifold-using-equivalence-classes-he%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

                    Willebadessen

                    Image
                    Wappen Deutschlandkarte 51.633055555556 9.0330555555556 295 Koordinaten: 51° 38′  N , 9° 2′  O Basisdaten Bundesland: Nordrhein-Westfalen Regierungsbezirk: Detmold Kreis: Höxter Höhe: 295 m ü. NHN Fläche: 128,41 km 2 Einwohner: 8227 (31. Dez. 2017) [1] Bevölkerungsdichte: 64 Einwohner je km 2 Postleitzahlen: 34435, 34437, 34439 Vorwahlen: 05644, 05646, 05642 Kfz-Kennzeichen: HX Gemeindeschlüssel: 05 7 62 040 Stadtgliederung: 13 Stadtteile Adresse der Stadtverwaltung: Abdinghofweg 1 34439 Willebadessen Website: www.willebadessen.de Bürgermeister: Hans Hermann Bluhm (CDU) Lage der Stadt Willebadessen im Kreis Höxter Willebadessen ist eine Kleinstadt in Nordrhein-Westfalen und gehört zum Kreis Höxter im Regierungsbezirk Detmold. Die Stadt liegt im Osten des Regierungsbezirkes mitten im Naturpark Teutoburger Wald/Eggegebirge. Für den heu...
                    Read more

                    Ida-Boy-Ed-Garten

                    Image
                    Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
                    Read more

                    Residenzschloss Arolsen

                    Image
                    Residenzschloss Arolsen Residenzschloss Arolsen Entstehungszeit: 1710–1728 Erhaltungszustand: Vollständig Ständische Stellung: Grafen Ort: Bad Arolsen Geographische Lage 51° 22′ 51″  N , 9° 1′ 19″  O 51.380833333333 9.0219444444444 293 Koordinaten: 51° 22′ 51″  N , 9° 1′ 19″  O Höhe: 293  m ü. NHN Das Residenzschloss Arolsen ist ein barockes Schloss in Bad Arolsen im Kreis Waldeck-Frankenberg in Nordhessen (Deutschland). Das Schloss wurde als dreiflügelige Anlage gebaut, an die sich ein englischer Garten anschließt. Zentrales landschaftsarchitektonisches Gestaltungselement ist ein ausgedehntes Rondell. Inhaltsverzeichnis 1 Geschichte 1.1 Vorgängerbau 1.2 Bau und Geschichte des neuen Schlosses 1.3 Bibliothek 1.4 Finanzielle und staatsrechtliche Konsequenzen 1.5 Heutige Nutzung 2 Innenausstattung 3 Umgebung 3.1 Park 3.2 Wirtschaftsgebäude 4 Literatur 5 Weblinks 6...
                    Read more