Involution action on $H^1(S^1times S^2)$












0












$begingroup$


I am studying about an action $I^*$ on a de Rham cohomology group $H^1(S^1times S^2)$ induced from an action $Icdot (z,x)=(overline{z},-x) $ where $S^1times S^2subset mathbb{C}times mathbb{R}^3$. Note that, by Kunneth formula, $$H^1(S^1times S^2)=mathbb{R}.$$



Thus, I want to find a nonzero element in $H^1(S^1times S^2)$ and want to see how $I^*$ acts to the element.



And my teacher taught me as below.



Let $dtheta in Omega^1(S^1)$ be a generator of $H^1(S^1)=mathbb{R}.$ And let $pi:S^1times S^2rightarrow S^1$ and let $omega=pi^*(dtheta)$. Then clearly, $domega=0$ so $[omega]$ is nonzero element in $H^1(S^1times S^2)$. If $iota : S^1times {text{north pole}}hookrightarrow S^1times S^2$ is an embedding, observe that
$$picirciota =Id implies iota^*pi^*=Id implies iota^*(omega)=dtheta implies iota^*[omega]=[dtheta] $$



Thus, $$I^*omega=I^*pi^*(dtheta)=pi^*(I_1^*dtheta)=pi^*(-dtheta)=-omega.$$



So we get $I^*=-Id$.




But I got stuck at why $I_1^*dtheta=-dtheta$.



If we see carefully, $I_1^*dtheta(z)=dtheta(overline{z})$. I don't know where is wrong and where I am missing.




I would very appreciate for any help and solution for this issue! Thank you!






group-actions de-rham-cohomology pullback







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I am studying about an action $I^*$ on a de Rham cohomology group $H^1(S^1times S^2)$ induced from an action $Icdot (z,x)=(overline{z},-x) $ where $S^1times S^2subset mathbb{C}times mathbb{R}^3$. Note that, by Kunneth formula, $$H^1(S^1times S^2)=mathbb{R}.$$



    Thus, I want to find a nonzero element in $H^1(S^1times S^2)$ and want to see how $I^*$ acts to the element.



    And my teacher taught me as below.



    Let $dtheta in Omega^1(S^1)$ be a generator of $H^1(S^1)=mathbb{R}.$ And let $pi:S^1times S^2rightarrow S^1$ and let $omega=pi^*(dtheta)$. Then clearly, $domega=0$ so $[omega]$ is nonzero element in $H^1(S^1times S^2)$. If $iota : S^1times {text{north pole}}hookrightarrow S^1times S^2$ is an embedding, observe that
    $$picirciota =Id implies iota^*pi^*=Id implies iota^*(omega)=dtheta implies iota^*[omega]=[dtheta] $$



    Thus, $$I^*omega=I^*pi^*(dtheta)=pi^*(I_1^*dtheta)=pi^*(-dtheta)=-omega.$$



    So we get $I^*=-Id$.




    But I got stuck at why $I_1^*dtheta=-dtheta$.



    If we see carefully, $I_1^*dtheta(z)=dtheta(overline{z})$. I don't know where is wrong and where I am missing.




    I would very appreciate for any help and solution for this issue! Thank you!






    group-actions de-rham-cohomology pullback







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I am studying about an action $I^*$ on a de Rham cohomology group $H^1(S^1times S^2)$ induced from an action $Icdot (z,x)=(overline{z},-x) $ where $S^1times S^2subset mathbb{C}times mathbb{R}^3$. Note that, by Kunneth formula, $$H^1(S^1times S^2)=mathbb{R}.$$



      Thus, I want to find a nonzero element in $H^1(S^1times S^2)$ and want to see how $I^*$ acts to the element.



      And my teacher taught me as below.



      Let $dtheta in Omega^1(S^1)$ be a generator of $H^1(S^1)=mathbb{R}.$ And let $pi:S^1times S^2rightarrow S^1$ and let $omega=pi^*(dtheta)$. Then clearly, $domega=0$ so $[omega]$ is nonzero element in $H^1(S^1times S^2)$. If $iota : S^1times {text{north pole}}hookrightarrow S^1times S^2$ is an embedding, observe that
      $$picirciota =Id implies iota^*pi^*=Id implies iota^*(omega)=dtheta implies iota^*[omega]=[dtheta] $$



      Thus, $$I^*omega=I^*pi^*(dtheta)=pi^*(I_1^*dtheta)=pi^*(-dtheta)=-omega.$$



      So we get $I^*=-Id$.




      But I got stuck at why $I_1^*dtheta=-dtheta$.



      If we see carefully, $I_1^*dtheta(z)=dtheta(overline{z})$. I don't know where is wrong and where I am missing.




      I would very appreciate for any help and solution for this issue! Thank you!






      group-actions de-rham-cohomology pullback







      share|cite|improve this question











      $endgroup$




      I am studying about an action $I^*$ on a de Rham cohomology group $H^1(S^1times S^2)$ induced from an action $Icdot (z,x)=(overline{z},-x) $ where $S^1times S^2subset mathbb{C}times mathbb{R}^3$. Note that, by Kunneth formula, $$H^1(S^1times S^2)=mathbb{R}.$$



      Thus, I want to find a nonzero element in $H^1(S^1times S^2)$ and want to see how $I^*$ acts to the element.



      And my teacher taught me as below.



      Let $dtheta in Omega^1(S^1)$ be a generator of $H^1(S^1)=mathbb{R}.$ And let $pi:S^1times S^2rightarrow S^1$ and let $omega=pi^*(dtheta)$. Then clearly, $domega=0$ so $[omega]$ is nonzero element in $H^1(S^1times S^2)$. If $iota : S^1times {text{north pole}}hookrightarrow S^1times S^2$ is an embedding, observe that
      $$picirciota =Id implies iota^*pi^*=Id implies iota^*(omega)=dtheta implies iota^*[omega]=[dtheta] $$



      Thus, $$I^*omega=I^*pi^*(dtheta)=pi^*(I_1^*dtheta)=pi^*(-dtheta)=-omega.$$



      So we get $I^*=-Id$.




      But I got stuck at why $I_1^*dtheta=-dtheta$.



      If we see carefully, $I_1^*dtheta(z)=dtheta(overline{z})$. I don't know where is wrong and where I am missing.




      I would very appreciate for any help and solution for this issue! Thank you!






      group-actions de-rham-cohomology pullback




      group-actions de-rham-cohomology pullback






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 24 '18 at 5:04









      Michael Albanese

      65k1599318




      65k1599318










      asked Dec 24 '18 at 4:56









      Lev BanLev Ban

      1,0771317




      1,0771317






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Note that $I_1^*dtheta = d(thetacirc I_1)$. As $(thetacirc I_1)(z) = theta(I_1(z)) = theta(bar{z}) = -theta(z)$, so $thetacirc I_1 = -theta$ and hence $I_1^*dtheta = d(thetacirc I_1) = d(-theta) = -dtheta$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for the answer! But could I ask why $theta(overline{z})=-theta(z)$?
            $endgroup$
            – Lev Ban
            Dec 24 '18 at 5:06










          • $begingroup$
            Recall that $theta(z)$ is the argument of $z$. So if $z = e^{ialpha}$, then $theta(z) = alpha$. On the other hand $bar{z} = e^{-ialpha}$, so $theta(bar{z}) = -alpha = -theta(z)$.
            $endgroup$
            – Michael Albanese
            Dec 24 '18 at 5:08










          • $begingroup$
            I guess that is that. But the reason why I got confused it that my teacher has not specify what was $theta$. He just said $dtheta$ is the generator. Should $theta$ be such a function in order for $dtheta$ be a generator?
            $endgroup$
            – Lev Ban
            Dec 24 '18 at 5:11












          • $begingroup$
            @LeB: Yes. It is common to denote the generator of $H^1_{text{dR}}(S^1)$ by $dtheta$ where $theta$ is the argument function.
            $endgroup$
            – Michael Albanese
            Dec 24 '18 at 5:16










          • $begingroup$
            Ah.. I got it.. Thank you!
            $endgroup$
            – Lev Ban
            Dec 24 '18 at 5:20












          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%2f3050952%2finvolution-action-on-h1s1-times-s2%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









          1












          $begingroup$

          Note that $I_1^*dtheta = d(thetacirc I_1)$. As $(thetacirc I_1)(z) = theta(I_1(z)) = theta(bar{z}) = -theta(z)$, so $thetacirc I_1 = -theta$ and hence $I_1^*dtheta = d(thetacirc I_1) = d(-theta) = -dtheta$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for the answer! But could I ask why $theta(overline{z})=-theta(z)$?
            $endgroup$
            – Lev Ban
            Dec 24 '18 at 5:06










          • $begingroup$
            Recall that $theta(z)$ is the argument of $z$. So if $z = e^{ialpha}$, then $theta(z) = alpha$. On the other hand $bar{z} = e^{-ialpha}$, so $theta(bar{z}) = -alpha = -theta(z)$.
            $endgroup$
            – Michael Albanese
            Dec 24 '18 at 5:08










          • $begingroup$
            I guess that is that. But the reason why I got confused it that my teacher has not specify what was $theta$. He just said $dtheta$ is the generator. Should $theta$ be such a function in order for $dtheta$ be a generator?
            $endgroup$
            – Lev Ban
            Dec 24 '18 at 5:11












          • $begingroup$
            @LeB: Yes. It is common to denote the generator of $H^1_{text{dR}}(S^1)$ by $dtheta$ where $theta$ is the argument function.
            $endgroup$
            – Michael Albanese
            Dec 24 '18 at 5:16










          • $begingroup$
            Ah.. I got it.. Thank you!
            $endgroup$
            – Lev Ban
            Dec 24 '18 at 5:20
















          1












          $begingroup$

          Note that $I_1^*dtheta = d(thetacirc I_1)$. As $(thetacirc I_1)(z) = theta(I_1(z)) = theta(bar{z}) = -theta(z)$, so $thetacirc I_1 = -theta$ and hence $I_1^*dtheta = d(thetacirc I_1) = d(-theta) = -dtheta$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you for the answer! But could I ask why $theta(overline{z})=-theta(z)$?
            $endgroup$
            – Lev Ban
            Dec 24 '18 at 5:06










          • $begingroup$
            Recall that $theta(z)$ is the argument of $z$. So if $z = e^{ialpha}$, then $theta(z) = alpha$. On the other hand $bar{z} = e^{-ialpha}$, so $theta(bar{z}) = -alpha = -theta(z)$.
            $endgroup$
            – Michael Albanese
            Dec 24 '18 at 5:08










          • $begingroup$
            I guess that is that. But the reason why I got confused it that my teacher has not specify what was $theta$. He just said $dtheta$ is the generator. Should $theta$ be such a function in order for $dtheta$ be a generator?
            $endgroup$
            – Lev Ban
            Dec 24 '18 at 5:11












          • $begingroup$
            @LeB: Yes. It is common to denote the generator of $H^1_{text{dR}}(S^1)$ by $dtheta$ where $theta$ is the argument function.
            $endgroup$
            – Michael Albanese
            Dec 24 '18 at 5:16










          • $begingroup$
            Ah.. I got it.. Thank you!
            $endgroup$
            – Lev Ban
            Dec 24 '18 at 5:20














          1












          1








          1





          $begingroup$

          Note that $I_1^*dtheta = d(thetacirc I_1)$. As $(thetacirc I_1)(z) = theta(I_1(z)) = theta(bar{z}) = -theta(z)$, so $thetacirc I_1 = -theta$ and hence $I_1^*dtheta = d(thetacirc I_1) = d(-theta) = -dtheta$.






          share|cite|improve this answer









          $endgroup$



          Note that $I_1^*dtheta = d(thetacirc I_1)$. As $(thetacirc I_1)(z) = theta(I_1(z)) = theta(bar{z}) = -theta(z)$, so $thetacirc I_1 = -theta$ and hence $I_1^*dtheta = d(thetacirc I_1) = d(-theta) = -dtheta$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 24 '18 at 5:01









          Michael AlbaneseMichael Albanese

          65k1599318




          65k1599318












          • $begingroup$
            Thank you for the answer! But could I ask why $theta(overline{z})=-theta(z)$?
            $endgroup$
            – Lev Ban
            Dec 24 '18 at 5:06










          • $begingroup$
            Recall that $theta(z)$ is the argument of $z$. So if $z = e^{ialpha}$, then $theta(z) = alpha$. On the other hand $bar{z} = e^{-ialpha}$, so $theta(bar{z}) = -alpha = -theta(z)$.
            $endgroup$
            – Michael Albanese
            Dec 24 '18 at 5:08










          • $begingroup$
            I guess that is that. But the reason why I got confused it that my teacher has not specify what was $theta$. He just said $dtheta$ is the generator. Should $theta$ be such a function in order for $dtheta$ be a generator?
            $endgroup$
            – Lev Ban
            Dec 24 '18 at 5:11












          • $begingroup$
            @LeB: Yes. It is common to denote the generator of $H^1_{text{dR}}(S^1)$ by $dtheta$ where $theta$ is the argument function.
            $endgroup$
            – Michael Albanese
            Dec 24 '18 at 5:16










          • $begingroup$
            Ah.. I got it.. Thank you!
            $endgroup$
            – Lev Ban
            Dec 24 '18 at 5:20


















          • $begingroup$
            Thank you for the answer! But could I ask why $theta(overline{z})=-theta(z)$?
            $endgroup$
            – Lev Ban
            Dec 24 '18 at 5:06










          • $begingroup$
            Recall that $theta(z)$ is the argument of $z$. So if $z = e^{ialpha}$, then $theta(z) = alpha$. On the other hand $bar{z} = e^{-ialpha}$, so $theta(bar{z}) = -alpha = -theta(z)$.
            $endgroup$
            – Michael Albanese
            Dec 24 '18 at 5:08










          • $begingroup$
            I guess that is that. But the reason why I got confused it that my teacher has not specify what was $theta$. He just said $dtheta$ is the generator. Should $theta$ be such a function in order for $dtheta$ be a generator?
            $endgroup$
            – Lev Ban
            Dec 24 '18 at 5:11












          • $begingroup$
            @LeB: Yes. It is common to denote the generator of $H^1_{text{dR}}(S^1)$ by $dtheta$ where $theta$ is the argument function.
            $endgroup$
            – Michael Albanese
            Dec 24 '18 at 5:16










          • $begingroup$
            Ah.. I got it.. Thank you!
            $endgroup$
            – Lev Ban
            Dec 24 '18 at 5:20
















          $begingroup$
          Thank you for the answer! But could I ask why $theta(overline{z})=-theta(z)$?
          $endgroup$
          – Lev Ban
          Dec 24 '18 at 5:06




          $begingroup$
          Thank you for the answer! But could I ask why $theta(overline{z})=-theta(z)$?
          $endgroup$
          – Lev Ban
          Dec 24 '18 at 5:06












          $begingroup$
          Recall that $theta(z)$ is the argument of $z$. So if $z = e^{ialpha}$, then $theta(z) = alpha$. On the other hand $bar{z} = e^{-ialpha}$, so $theta(bar{z}) = -alpha = -theta(z)$.
          $endgroup$
          – Michael Albanese
          Dec 24 '18 at 5:08




          $begingroup$
          Recall that $theta(z)$ is the argument of $z$. So if $z = e^{ialpha}$, then $theta(z) = alpha$. On the other hand $bar{z} = e^{-ialpha}$, so $theta(bar{z}) = -alpha = -theta(z)$.
          $endgroup$
          – Michael Albanese
          Dec 24 '18 at 5:08












          $begingroup$
          I guess that is that. But the reason why I got confused it that my teacher has not specify what was $theta$. He just said $dtheta$ is the generator. Should $theta$ be such a function in order for $dtheta$ be a generator?
          $endgroup$
          – Lev Ban
          Dec 24 '18 at 5:11






          $begingroup$
          I guess that is that. But the reason why I got confused it that my teacher has not specify what was $theta$. He just said $dtheta$ is the generator. Should $theta$ be such a function in order for $dtheta$ be a generator?
          $endgroup$
          – Lev Ban
          Dec 24 '18 at 5:11














          $begingroup$
          @LeB: Yes. It is common to denote the generator of $H^1_{text{dR}}(S^1)$ by $dtheta$ where $theta$ is the argument function.
          $endgroup$
          – Michael Albanese
          Dec 24 '18 at 5:16




          $begingroup$
          @LeB: Yes. It is common to denote the generator of $H^1_{text{dR}}(S^1)$ by $dtheta$ where $theta$ is the argument function.
          $endgroup$
          – Michael Albanese
          Dec 24 '18 at 5:16












          $begingroup$
          Ah.. I got it.. Thank you!
          $endgroup$
          – Lev Ban
          Dec 24 '18 at 5:20




          $begingroup$
          Ah.. I got it.. Thank you!
          $endgroup$
          – Lev Ban
          Dec 24 '18 at 5:20


















          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%2f3050952%2finvolution-action-on-h1s1-times-s2%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

          Plaza Victoria

          Image
          Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
          Read more

          Puebla de Zaragoza

          Image
          Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
          Read more

          Musa

          Image
          Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
          Read more