Involution action on $H^3(S^1times S^2)$












1












$begingroup$


I am studying about involution action $I^*$ on de Rham cohomology group $H^3(S^1times S^2)$ 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, $$H^3(S^1times S^2)=mathbb{R}.$$



I figured out the action of $I^*$ on $H^2(S^1times S^2)$ in a similar way with my previous question
Involution action on $H^1(S^1times S^2)$.




However, I am stuck at identifying the action on $H^3(S^1times S^2)$.



I was trying to pick $dtheta in Omega^1(S^1)$ and $dsin Omega^2(S^2)$ where $theta(overline{z})=-theta(z)$ and $s(-x)=-s(x)$. And it seems leading me to create $omega=dthetawedge ds$ as a generator of $H^3(S^1times S^2)$. But I am not sure if I am going in right direction since the $omega$ seems not the one for some anticlimatic reason.




Any help would be appreciated! Thank you in advance!






group-actions de-rham-cohomology involutions







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I am studying about involution action $I^*$ on de Rham cohomology group $H^3(S^1times S^2)$ 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, $$H^3(S^1times S^2)=mathbb{R}.$$



    I figured out the action of $I^*$ on $H^2(S^1times S^2)$ in a similar way with my previous question
    Involution action on $H^1(S^1times S^2)$.




    However, I am stuck at identifying the action on $H^3(S^1times S^2)$.



    I was trying to pick $dtheta in Omega^1(S^1)$ and $dsin Omega^2(S^2)$ where $theta(overline{z})=-theta(z)$ and $s(-x)=-s(x)$. And it seems leading me to create $omega=dthetawedge ds$ as a generator of $H^3(S^1times S^2)$. But I am not sure if I am going in right direction since the $omega$ seems not the one for some anticlimatic reason.




    Any help would be appreciated! Thank you in advance!






    group-actions de-rham-cohomology involutions







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am studying about involution action $I^*$ on de Rham cohomology group $H^3(S^1times S^2)$ 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, $$H^3(S^1times S^2)=mathbb{R}.$$



      I figured out the action of $I^*$ on $H^2(S^1times S^2)$ in a similar way with my previous question
      Involution action on $H^1(S^1times S^2)$.




      However, I am stuck at identifying the action on $H^3(S^1times S^2)$.



      I was trying to pick $dtheta in Omega^1(S^1)$ and $dsin Omega^2(S^2)$ where $theta(overline{z})=-theta(z)$ and $s(-x)=-s(x)$. And it seems leading me to create $omega=dthetawedge ds$ as a generator of $H^3(S^1times S^2)$. But I am not sure if I am going in right direction since the $omega$ seems not the one for some anticlimatic reason.




      Any help would be appreciated! Thank you in advance!






      group-actions de-rham-cohomology involutions







      share|cite|improve this question









      $endgroup$




      I am studying about involution action $I^*$ on de Rham cohomology group $H^3(S^1times S^2)$ 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, $$H^3(S^1times S^2)=mathbb{R}.$$



      I figured out the action of $I^*$ on $H^2(S^1times S^2)$ in a similar way with my previous question
      Involution action on $H^1(S^1times S^2)$.




      However, I am stuck at identifying the action on $H^3(S^1times S^2)$.



      I was trying to pick $dtheta in Omega^1(S^1)$ and $dsin Omega^2(S^2)$ where $theta(overline{z})=-theta(z)$ and $s(-x)=-s(x)$. And it seems leading me to create $omega=dthetawedge ds$ as a generator of $H^3(S^1times S^2)$. But I am not sure if I am going in right direction since the $omega$ seems not the one for some anticlimatic reason.




      Any help would be appreciated! Thank you in advance!






      group-actions de-rham-cohomology involutions




      group-actions de-rham-cohomology involutions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 24 '18 at 6:11









      Lev BanLev Ban

      1,0721317




      1,0721317






















          1 Answer
          1






          active

          oldest

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          2












          $begingroup$

          As you have defined it, $omega$ doesn't make sense because you can't take the wedge product of a form on $S^1$ with a form on $S^2$. Instead, let $pi_i : S^1times S^2 to S^i$ be the natural projections. Then $pi_1^*dtheta$ and $pi_2^*ds$ are both forms on $S^1times S^2$, so their wedge product $omega = (pi_1^*dtheta)wedge(pi_2^*ds)$ is defined.



          Let $I_1 : S^1to S^1$ be given by $I_1(z) = bar{z}$ and $I_2 : S^2to S^2$ be given by $I_2(x) = -x$, then $I(z, x) = (I_1(z), I_2(x))$. Said another way, $pi_1circ I = I_1circpi_1$ and $pi_2circ I = I_2circ pi_2$. Therefore,



          $$I^*pi_1^*dtheta = (pi_1circ I)^*dtheta = (I_1circ pi_1)^*dtheta = pi_1^*(I_1^*dtheta) = pi_1^*(-dtheta) = -pi_1^*dtheta$$



          and



          $$I^*pi_2^*ds = (pi_2circ I)^*ds = (I_2circ pi_2)^*ds = pi_2^*(I_2^*ds) = pi_2^*(-ds) = -pi_2^*ds.$$



          Therefore $I^*omega = (I^*pi_1^*dtheta)wedge(I^*pi_2^*ds) = (-dtheta)wedge(-ds) = dthetawedge ds = omega$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            More generally, there's a Kunneth theorem for deRham cohomology: $H^*(X)otimes H^*(Y)longrightarrow H^*(Xtimes Y)$ given by the formula of this post is an isomorphism of algebras.
            $endgroup$
            – Pedro Tamaroff♦
            Dec 24 '18 at 22:18










          • $begingroup$
            @Michael Albanese Come to think of it, $ds$ seems not $2$ -form. Is it still vaild argument?
            $endgroup$
            – Lev Ban
            Dec 26 '18 at 19:48










          • $begingroup$
            @LeB: What do you mean by $ds$? I thought you meant a two-form on $S^2$ (not the exterior derivative of a function, which is not a two-form).
            $endgroup$
            – Michael Albanese
            Dec 26 '18 at 21:29










          • $begingroup$
            @MichaelAlbanese I am sorry! I got a bit confused. If it is not too much, could you let me know which kind of 2 form I can use of for this problem?
            $endgroup$
            – Lev Ban
            Dec 26 '18 at 23:38








          • 1




            $begingroup$
            The two form $omega = x,dywedge dz - y,dxwedge dz + z,dxwedge dy$ on $mathbb{R}^3$ restricts to a closed form on $S^2$ which generates $H^2_{text{dR}}(S^2)$, and satisfies $I_2^*omega = -omega$.
            $endgroup$
            – Michael Albanese
            Dec 27 '18 at 1:19













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          $begingroup$

          As you have defined it, $omega$ doesn't make sense because you can't take the wedge product of a form on $S^1$ with a form on $S^2$. Instead, let $pi_i : S^1times S^2 to S^i$ be the natural projections. Then $pi_1^*dtheta$ and $pi_2^*ds$ are both forms on $S^1times S^2$, so their wedge product $omega = (pi_1^*dtheta)wedge(pi_2^*ds)$ is defined.



          Let $I_1 : S^1to S^1$ be given by $I_1(z) = bar{z}$ and $I_2 : S^2to S^2$ be given by $I_2(x) = -x$, then $I(z, x) = (I_1(z), I_2(x))$. Said another way, $pi_1circ I = I_1circpi_1$ and $pi_2circ I = I_2circ pi_2$. Therefore,



          $$I^*pi_1^*dtheta = (pi_1circ I)^*dtheta = (I_1circ pi_1)^*dtheta = pi_1^*(I_1^*dtheta) = pi_1^*(-dtheta) = -pi_1^*dtheta$$



          and



          $$I^*pi_2^*ds = (pi_2circ I)^*ds = (I_2circ pi_2)^*ds = pi_2^*(I_2^*ds) = pi_2^*(-ds) = -pi_2^*ds.$$



          Therefore $I^*omega = (I^*pi_1^*dtheta)wedge(I^*pi_2^*ds) = (-dtheta)wedge(-ds) = dthetawedge ds = omega$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            More generally, there's a Kunneth theorem for deRham cohomology: $H^*(X)otimes H^*(Y)longrightarrow H^*(Xtimes Y)$ given by the formula of this post is an isomorphism of algebras.
            $endgroup$
            – Pedro Tamaroff♦
            Dec 24 '18 at 22:18










          • $begingroup$
            @Michael Albanese Come to think of it, $ds$ seems not $2$ -form. Is it still vaild argument?
            $endgroup$
            – Lev Ban
            Dec 26 '18 at 19:48










          • $begingroup$
            @LeB: What do you mean by $ds$? I thought you meant a two-form on $S^2$ (not the exterior derivative of a function, which is not a two-form).
            $endgroup$
            – Michael Albanese
            Dec 26 '18 at 21:29










          • $begingroup$
            @MichaelAlbanese I am sorry! I got a bit confused. If it is not too much, could you let me know which kind of 2 form I can use of for this problem?
            $endgroup$
            – Lev Ban
            Dec 26 '18 at 23:38








          • 1




            $begingroup$
            The two form $omega = x,dywedge dz - y,dxwedge dz + z,dxwedge dy$ on $mathbb{R}^3$ restricts to a closed form on $S^2$ which generates $H^2_{text{dR}}(S^2)$, and satisfies $I_2^*omega = -omega$.
            $endgroup$
            – Michael Albanese
            Dec 27 '18 at 1:19


















          2












          $begingroup$

          As you have defined it, $omega$ doesn't make sense because you can't take the wedge product of a form on $S^1$ with a form on $S^2$. Instead, let $pi_i : S^1times S^2 to S^i$ be the natural projections. Then $pi_1^*dtheta$ and $pi_2^*ds$ are both forms on $S^1times S^2$, so their wedge product $omega = (pi_1^*dtheta)wedge(pi_2^*ds)$ is defined.



          Let $I_1 : S^1to S^1$ be given by $I_1(z) = bar{z}$ and $I_2 : S^2to S^2$ be given by $I_2(x) = -x$, then $I(z, x) = (I_1(z), I_2(x))$. Said another way, $pi_1circ I = I_1circpi_1$ and $pi_2circ I = I_2circ pi_2$. Therefore,



          $$I^*pi_1^*dtheta = (pi_1circ I)^*dtheta = (I_1circ pi_1)^*dtheta = pi_1^*(I_1^*dtheta) = pi_1^*(-dtheta) = -pi_1^*dtheta$$



          and



          $$I^*pi_2^*ds = (pi_2circ I)^*ds = (I_2circ pi_2)^*ds = pi_2^*(I_2^*ds) = pi_2^*(-ds) = -pi_2^*ds.$$



          Therefore $I^*omega = (I^*pi_1^*dtheta)wedge(I^*pi_2^*ds) = (-dtheta)wedge(-ds) = dthetawedge ds = omega$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            More generally, there's a Kunneth theorem for deRham cohomology: $H^*(X)otimes H^*(Y)longrightarrow H^*(Xtimes Y)$ given by the formula of this post is an isomorphism of algebras.
            $endgroup$
            – Pedro Tamaroff♦
            Dec 24 '18 at 22:18










          • $begingroup$
            @Michael Albanese Come to think of it, $ds$ seems not $2$ -form. Is it still vaild argument?
            $endgroup$
            – Lev Ban
            Dec 26 '18 at 19:48










          • $begingroup$
            @LeB: What do you mean by $ds$? I thought you meant a two-form on $S^2$ (not the exterior derivative of a function, which is not a two-form).
            $endgroup$
            – Michael Albanese
            Dec 26 '18 at 21:29










          • $begingroup$
            @MichaelAlbanese I am sorry! I got a bit confused. If it is not too much, could you let me know which kind of 2 form I can use of for this problem?
            $endgroup$
            – Lev Ban
            Dec 26 '18 at 23:38








          • 1




            $begingroup$
            The two form $omega = x,dywedge dz - y,dxwedge dz + z,dxwedge dy$ on $mathbb{R}^3$ restricts to a closed form on $S^2$ which generates $H^2_{text{dR}}(S^2)$, and satisfies $I_2^*omega = -omega$.
            $endgroup$
            – Michael Albanese
            Dec 27 '18 at 1:19
















          2












          2








          2





          $begingroup$

          As you have defined it, $omega$ doesn't make sense because you can't take the wedge product of a form on $S^1$ with a form on $S^2$. Instead, let $pi_i : S^1times S^2 to S^i$ be the natural projections. Then $pi_1^*dtheta$ and $pi_2^*ds$ are both forms on $S^1times S^2$, so their wedge product $omega = (pi_1^*dtheta)wedge(pi_2^*ds)$ is defined.



          Let $I_1 : S^1to S^1$ be given by $I_1(z) = bar{z}$ and $I_2 : S^2to S^2$ be given by $I_2(x) = -x$, then $I(z, x) = (I_1(z), I_2(x))$. Said another way, $pi_1circ I = I_1circpi_1$ and $pi_2circ I = I_2circ pi_2$. Therefore,



          $$I^*pi_1^*dtheta = (pi_1circ I)^*dtheta = (I_1circ pi_1)^*dtheta = pi_1^*(I_1^*dtheta) = pi_1^*(-dtheta) = -pi_1^*dtheta$$



          and



          $$I^*pi_2^*ds = (pi_2circ I)^*ds = (I_2circ pi_2)^*ds = pi_2^*(I_2^*ds) = pi_2^*(-ds) = -pi_2^*ds.$$



          Therefore $I^*omega = (I^*pi_1^*dtheta)wedge(I^*pi_2^*ds) = (-dtheta)wedge(-ds) = dthetawedge ds = omega$.






          share|cite|improve this answer









          $endgroup$



          As you have defined it, $omega$ doesn't make sense because you can't take the wedge product of a form on $S^1$ with a form on $S^2$. Instead, let $pi_i : S^1times S^2 to S^i$ be the natural projections. Then $pi_1^*dtheta$ and $pi_2^*ds$ are both forms on $S^1times S^2$, so their wedge product $omega = (pi_1^*dtheta)wedge(pi_2^*ds)$ is defined.



          Let $I_1 : S^1to S^1$ be given by $I_1(z) = bar{z}$ and $I_2 : S^2to S^2$ be given by $I_2(x) = -x$, then $I(z, x) = (I_1(z), I_2(x))$. Said another way, $pi_1circ I = I_1circpi_1$ and $pi_2circ I = I_2circ pi_2$. Therefore,



          $$I^*pi_1^*dtheta = (pi_1circ I)^*dtheta = (I_1circ pi_1)^*dtheta = pi_1^*(I_1^*dtheta) = pi_1^*(-dtheta) = -pi_1^*dtheta$$



          and



          $$I^*pi_2^*ds = (pi_2circ I)^*ds = (I_2circ pi_2)^*ds = pi_2^*(I_2^*ds) = pi_2^*(-ds) = -pi_2^*ds.$$



          Therefore $I^*omega = (I^*pi_1^*dtheta)wedge(I^*pi_2^*ds) = (-dtheta)wedge(-ds) = dthetawedge ds = omega$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 24 '18 at 22:02









          Michael AlbaneseMichael Albanese

          63.8k1599309




          63.8k1599309








          • 1




            $begingroup$
            More generally, there's a Kunneth theorem for deRham cohomology: $H^*(X)otimes H^*(Y)longrightarrow H^*(Xtimes Y)$ given by the formula of this post is an isomorphism of algebras.
            $endgroup$
            – Pedro Tamaroff♦
            Dec 24 '18 at 22:18










          • $begingroup$
            @Michael Albanese Come to think of it, $ds$ seems not $2$ -form. Is it still vaild argument?
            $endgroup$
            – Lev Ban
            Dec 26 '18 at 19:48










          • $begingroup$
            @LeB: What do you mean by $ds$? I thought you meant a two-form on $S^2$ (not the exterior derivative of a function, which is not a two-form).
            $endgroup$
            – Michael Albanese
            Dec 26 '18 at 21:29










          • $begingroup$
            @MichaelAlbanese I am sorry! I got a bit confused. If it is not too much, could you let me know which kind of 2 form I can use of for this problem?
            $endgroup$
            – Lev Ban
            Dec 26 '18 at 23:38








          • 1




            $begingroup$
            The two form $omega = x,dywedge dz - y,dxwedge dz + z,dxwedge dy$ on $mathbb{R}^3$ restricts to a closed form on $S^2$ which generates $H^2_{text{dR}}(S^2)$, and satisfies $I_2^*omega = -omega$.
            $endgroup$
            – Michael Albanese
            Dec 27 '18 at 1:19
















          • 1




            $begingroup$
            More generally, there's a Kunneth theorem for deRham cohomology: $H^*(X)otimes H^*(Y)longrightarrow H^*(Xtimes Y)$ given by the formula of this post is an isomorphism of algebras.
            $endgroup$
            – Pedro Tamaroff♦
            Dec 24 '18 at 22:18










          • $begingroup$
            @Michael Albanese Come to think of it, $ds$ seems not $2$ -form. Is it still vaild argument?
            $endgroup$
            – Lev Ban
            Dec 26 '18 at 19:48










          • $begingroup$
            @LeB: What do you mean by $ds$? I thought you meant a two-form on $S^2$ (not the exterior derivative of a function, which is not a two-form).
            $endgroup$
            – Michael Albanese
            Dec 26 '18 at 21:29










          • $begingroup$
            @MichaelAlbanese I am sorry! I got a bit confused. If it is not too much, could you let me know which kind of 2 form I can use of for this problem?
            $endgroup$
            – Lev Ban
            Dec 26 '18 at 23:38








          • 1




            $begingroup$
            The two form $omega = x,dywedge dz - y,dxwedge dz + z,dxwedge dy$ on $mathbb{R}^3$ restricts to a closed form on $S^2$ which generates $H^2_{text{dR}}(S^2)$, and satisfies $I_2^*omega = -omega$.
            $endgroup$
            – Michael Albanese
            Dec 27 '18 at 1:19










          1




          1




          $begingroup$
          More generally, there's a Kunneth theorem for deRham cohomology: $H^*(X)otimes H^*(Y)longrightarrow H^*(Xtimes Y)$ given by the formula of this post is an isomorphism of algebras.
          $endgroup$
          – Pedro Tamaroff♦
          Dec 24 '18 at 22:18




          $begingroup$
          More generally, there's a Kunneth theorem for deRham cohomology: $H^*(X)otimes H^*(Y)longrightarrow H^*(Xtimes Y)$ given by the formula of this post is an isomorphism of algebras.
          $endgroup$
          – Pedro Tamaroff♦
          Dec 24 '18 at 22:18












          $begingroup$
          @Michael Albanese Come to think of it, $ds$ seems not $2$ -form. Is it still vaild argument?
          $endgroup$
          – Lev Ban
          Dec 26 '18 at 19:48




          $begingroup$
          @Michael Albanese Come to think of it, $ds$ seems not $2$ -form. Is it still vaild argument?
          $endgroup$
          – Lev Ban
          Dec 26 '18 at 19:48












          $begingroup$
          @LeB: What do you mean by $ds$? I thought you meant a two-form on $S^2$ (not the exterior derivative of a function, which is not a two-form).
          $endgroup$
          – Michael Albanese
          Dec 26 '18 at 21:29




          $begingroup$
          @LeB: What do you mean by $ds$? I thought you meant a two-form on $S^2$ (not the exterior derivative of a function, which is not a two-form).
          $endgroup$
          – Michael Albanese
          Dec 26 '18 at 21:29












          $begingroup$
          @MichaelAlbanese I am sorry! I got a bit confused. If it is not too much, could you let me know which kind of 2 form I can use of for this problem?
          $endgroup$
          – Lev Ban
          Dec 26 '18 at 23:38






          $begingroup$
          @MichaelAlbanese I am sorry! I got a bit confused. If it is not too much, could you let me know which kind of 2 form I can use of for this problem?
          $endgroup$
          – Lev Ban
          Dec 26 '18 at 23:38






          1




          1




          $begingroup$
          The two form $omega = x,dywedge dz - y,dxwedge dz + z,dxwedge dy$ on $mathbb{R}^3$ restricts to a closed form on $S^2$ which generates $H^2_{text{dR}}(S^2)$, and satisfies $I_2^*omega = -omega$.
          $endgroup$
          – Michael Albanese
          Dec 27 '18 at 1:19






          $begingroup$
          The two form $omega = x,dywedge dz - y,dxwedge dz + z,dxwedge dy$ on $mathbb{R}^3$ restricts to a closed form on $S^2$ which generates $H^2_{text{dR}}(S^2)$, and satisfies $I_2^*omega = -omega$.
          $endgroup$
          – Michael Albanese
          Dec 27 '18 at 1:19




















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          To store a contact into the json file from server.js file using a class in NodeJS

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          0 I have created three files -contactPresentation.js, contactService.js, contacts.json. I will run the contactService.js file. And it will navigate to contactPresentation.js file that will only choose an option to either add or display a file. If add(), then add the data into add of service file.And the data will be added to the contacts.json file . And if display(), then display the display() of service file. And the display will take the data from contacts.json file. However, i know my code is flawed. Please correct it . I am also getting an error - this.contacts.push(new ContactService(userName, contactNumber, emailId)); ^ TypeError: Cannot read property 'push' of undefined at ContactService.add (e:NodeJSProjectcontactManager2contactService.js:57:23) at new C
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          Redirect URL with Chrome Remote Debugging Android Devices

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          1 How can i redirect an URL with Chrome Remote Debugging Android Devices? like Requestly Chrome extension on desktop. google-chrome debugging url redirect testing share | improve this question edited Nov 24 '18 at 8:16 sachinjain024 14.7k 23 73 129 asked Nov 23 '18 at 16:09 Momen Momen 91 10
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          Dieringhausen

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          Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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