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







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$endgroup$

















    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






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      share|cite|improve this question













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      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
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          $begingroup$

          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












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          $begingroup$

          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


















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