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

      1,0821317




      1,0821317






















          1 Answer
          1






          active

          oldest

          votes


















          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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          1 Answer
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          active

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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


















          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

          63.9k1599309




          63.9k1599309








          • 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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          0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19
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          “minimization” problem in Euclidean space related to orthonormal basis

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          3 $begingroup$ I have stumbled upon an exercice for second year undegraduate student majoring in economics which I find quite demanding. I have an idea for the solution, but it seems awfully complicated, and I am wondering if there was a simpler way to solve it. This goes as follows: Let $(E, langle, rangle)$ be a euclidean space of dimension $n geq 2$ . 1) For any $x in E$ , show that $||x|| = sqrt{n}$ if and only if there exists $(e_1, dots, e_n)$ an orthonormal basis of $E$ such that $x =e_1+ ldots + e_n$ . 2) For any $(x,y) in E$ , show that $||x|| = sqrt{n}$ , $||y|| = sqrt{frac{n(n+1)(2n+1)}{6}}$ and $langle x,y rangle = frac{n(n+1)}{2}$ if and only if there eixsts $(e_1, ldots, e_n)$ an orthonormal basis of $E$ such that $x = e_1 + ldots + e_n$ and $y = e_1 + 2e_2 + ldots + ne_n$ The first q
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