Computing Fubini-Study metric from the formal definition












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Definition: the Fubini-Study metric $g_{FB}$ on $mathbb{CP}^n$ is the only metric which makes the projection $pi:(mathbb{S}^{2n+1},g)to(mathbb{CP}^n,g_{FB})$ a Riemannian submersion (where $g$ is the standard metric)




I'm trying to deduce the coefficients of $g_{FB}$ in charts, i.e.:
begin{align*}
g_{FB}left(frac{partial}{partial x_alpha},frac{partial}{partial x_beta}right)=g_{FB}left(frac{partial}{partial y_alpha},frac{partial}{partial y_beta}right)&=frac{1}{(1+|z|^2)^2}((1+|z|^2)delta_{alphabeta}-(x_alpha x_beta+y_alpha y_beta))\
g_{FB}left(frac{partial}{partial x_alpha},frac{partial}{partial y_beta}right)&=frac{-1}{(1+|z|^2)^2}(x_alpha y_beta-y_alpha x_beta)
end{align*}



Since $x_alpha x_beta+y_alpha y_beta=text{Re}(z_alphaoverline{z}_beta)$ and $-(x_alpha y_beta-y_alpha x_beta)=text{Im}(z_alphaoverline{z}_beta)$, I thought maybe it would be a good idea to use complex variables, but I don't know how to do that from the condition that $pi$ is a Riemannian submersion.



Any suggestions?






riemannian-geometry complex-geometry projective-space







share|cite|improve this question











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    1












    $begingroup$



    Definition: the Fubini-Study metric $g_{FB}$ on $mathbb{CP}^n$ is the only metric which makes the projection $pi:(mathbb{S}^{2n+1},g)to(mathbb{CP}^n,g_{FB})$ a Riemannian submersion (where $g$ is the standard metric)




    I'm trying to deduce the coefficients of $g_{FB}$ in charts, i.e.:
    begin{align*}
    g_{FB}left(frac{partial}{partial x_alpha},frac{partial}{partial x_beta}right)=g_{FB}left(frac{partial}{partial y_alpha},frac{partial}{partial y_beta}right)&=frac{1}{(1+|z|^2)^2}((1+|z|^2)delta_{alphabeta}-(x_alpha x_beta+y_alpha y_beta))\
    g_{FB}left(frac{partial}{partial x_alpha},frac{partial}{partial y_beta}right)&=frac{-1}{(1+|z|^2)^2}(x_alpha y_beta-y_alpha x_beta)
    end{align*}



    Since $x_alpha x_beta+y_alpha y_beta=text{Re}(z_alphaoverline{z}_beta)$ and $-(x_alpha y_beta-y_alpha x_beta)=text{Im}(z_alphaoverline{z}_beta)$, I thought maybe it would be a good idea to use complex variables, but I don't know how to do that from the condition that $pi$ is a Riemannian submersion.



    Any suggestions?






    riemannian-geometry complex-geometry projective-space







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$



      Definition: the Fubini-Study metric $g_{FB}$ on $mathbb{CP}^n$ is the only metric which makes the projection $pi:(mathbb{S}^{2n+1},g)to(mathbb{CP}^n,g_{FB})$ a Riemannian submersion (where $g$ is the standard metric)




      I'm trying to deduce the coefficients of $g_{FB}$ in charts, i.e.:
      begin{align*}
      g_{FB}left(frac{partial}{partial x_alpha},frac{partial}{partial x_beta}right)=g_{FB}left(frac{partial}{partial y_alpha},frac{partial}{partial y_beta}right)&=frac{1}{(1+|z|^2)^2}((1+|z|^2)delta_{alphabeta}-(x_alpha x_beta+y_alpha y_beta))\
      g_{FB}left(frac{partial}{partial x_alpha},frac{partial}{partial y_beta}right)&=frac{-1}{(1+|z|^2)^2}(x_alpha y_beta-y_alpha x_beta)
      end{align*}



      Since $x_alpha x_beta+y_alpha y_beta=text{Re}(z_alphaoverline{z}_beta)$ and $-(x_alpha y_beta-y_alpha x_beta)=text{Im}(z_alphaoverline{z}_beta)$, I thought maybe it would be a good idea to use complex variables, but I don't know how to do that from the condition that $pi$ is a Riemannian submersion.



      Any suggestions?






      riemannian-geometry complex-geometry projective-space







      share|cite|improve this question











      $endgroup$





      Definition: the Fubini-Study metric $g_{FB}$ on $mathbb{CP}^n$ is the only metric which makes the projection $pi:(mathbb{S}^{2n+1},g)to(mathbb{CP}^n,g_{FB})$ a Riemannian submersion (where $g$ is the standard metric)




      I'm trying to deduce the coefficients of $g_{FB}$ in charts, i.e.:
      begin{align*}
      g_{FB}left(frac{partial}{partial x_alpha},frac{partial}{partial x_beta}right)=g_{FB}left(frac{partial}{partial y_alpha},frac{partial}{partial y_beta}right)&=frac{1}{(1+|z|^2)^2}((1+|z|^2)delta_{alphabeta}-(x_alpha x_beta+y_alpha y_beta))\
      g_{FB}left(frac{partial}{partial x_alpha},frac{partial}{partial y_beta}right)&=frac{-1}{(1+|z|^2)^2}(x_alpha y_beta-y_alpha x_beta)
      end{align*}



      Since $x_alpha x_beta+y_alpha y_beta=text{Re}(z_alphaoverline{z}_beta)$ and $-(x_alpha y_beta-y_alpha x_beta)=text{Im}(z_alphaoverline{z}_beta)$, I thought maybe it would be a good idea to use complex variables, but I don't know how to do that from the condition that $pi$ is a Riemannian submersion.



      Any suggestions?






      riemannian-geometry complex-geometry projective-space




      riemannian-geometry complex-geometry projective-space






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 13 '18 at 17:44







      rmdmc89

















      asked Dec 12 '18 at 20:50









      rmdmc89rmdmc89

      2,1221922




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          It's indeed very natural to use the hermitian geometry structure here. Let $e_0;e_1,dots e_n$ be a unitary frame at the point $[Z]=[e_0]inBbb P^n$. Then
          $$de_0 = omega_{0bar 0}e_0 + sum omega_{0bar j}e_j$$
          and the hermitian metric on $Bbb P^n$ is given by $sum |omega_{0bar j}|^2 = sum omega_{0bar j}overline{omega_{0bar j}}$. In one of your charts, say $Z_0ne 0$, we take coordinates by setting $Z=(1,z)$ and $e_0 = Z/|Z|$.



          Now, note that $omega_{0bar 0} = i,dtheta$ where $e^{itheta}$ gives the fiber of your Riemannian submersion. Moreover, denoting the hermitian inner product by $(cdot,cdot)$,
          $$sum |omega_{0bar j}|^2 = (de_0,de_0) - |omega_{0bar 0}|^2 = (de_0,de_0)-|(de_0,e_0)|^2.$$
          Substituting $e_0 = dfrac{(1,z)}{|(1,z)|}$ and differentiating appropriately, you'll get your desired formula for the metric as a $2$-tensor in terms of $dz_j$ and $dbar z_j$.






          share|cite|improve this answer











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            $begingroup$

            It's indeed very natural to use the hermitian geometry structure here. Let $e_0;e_1,dots e_n$ be a unitary frame at the point $[Z]=[e_0]inBbb P^n$. Then
            $$de_0 = omega_{0bar 0}e_0 + sum omega_{0bar j}e_j$$
            and the hermitian metric on $Bbb P^n$ is given by $sum |omega_{0bar j}|^2 = sum omega_{0bar j}overline{omega_{0bar j}}$. In one of your charts, say $Z_0ne 0$, we take coordinates by setting $Z=(1,z)$ and $e_0 = Z/|Z|$.



            Now, note that $omega_{0bar 0} = i,dtheta$ where $e^{itheta}$ gives the fiber of your Riemannian submersion. Moreover, denoting the hermitian inner product by $(cdot,cdot)$,
            $$sum |omega_{0bar j}|^2 = (de_0,de_0) - |omega_{0bar 0}|^2 = (de_0,de_0)-|(de_0,e_0)|^2.$$
            Substituting $e_0 = dfrac{(1,z)}{|(1,z)|}$ and differentiating appropriately, you'll get your desired formula for the metric as a $2$-tensor in terms of $dz_j$ and $dbar z_j$.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              It's indeed very natural to use the hermitian geometry structure here. Let $e_0;e_1,dots e_n$ be a unitary frame at the point $[Z]=[e_0]inBbb P^n$. Then
              $$de_0 = omega_{0bar 0}e_0 + sum omega_{0bar j}e_j$$
              and the hermitian metric on $Bbb P^n$ is given by $sum |omega_{0bar j}|^2 = sum omega_{0bar j}overline{omega_{0bar j}}$. In one of your charts, say $Z_0ne 0$, we take coordinates by setting $Z=(1,z)$ and $e_0 = Z/|Z|$.



              Now, note that $omega_{0bar 0} = i,dtheta$ where $e^{itheta}$ gives the fiber of your Riemannian submersion. Moreover, denoting the hermitian inner product by $(cdot,cdot)$,
              $$sum |omega_{0bar j}|^2 = (de_0,de_0) - |omega_{0bar 0}|^2 = (de_0,de_0)-|(de_0,e_0)|^2.$$
              Substituting $e_0 = dfrac{(1,z)}{|(1,z)|}$ and differentiating appropriately, you'll get your desired formula for the metric as a $2$-tensor in terms of $dz_j$ and $dbar z_j$.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                It's indeed very natural to use the hermitian geometry structure here. Let $e_0;e_1,dots e_n$ be a unitary frame at the point $[Z]=[e_0]inBbb P^n$. Then
                $$de_0 = omega_{0bar 0}e_0 + sum omega_{0bar j}e_j$$
                and the hermitian metric on $Bbb P^n$ is given by $sum |omega_{0bar j}|^2 = sum omega_{0bar j}overline{omega_{0bar j}}$. In one of your charts, say $Z_0ne 0$, we take coordinates by setting $Z=(1,z)$ and $e_0 = Z/|Z|$.



                Now, note that $omega_{0bar 0} = i,dtheta$ where $e^{itheta}$ gives the fiber of your Riemannian submersion. Moreover, denoting the hermitian inner product by $(cdot,cdot)$,
                $$sum |omega_{0bar j}|^2 = (de_0,de_0) - |omega_{0bar 0}|^2 = (de_0,de_0)-|(de_0,e_0)|^2.$$
                Substituting $e_0 = dfrac{(1,z)}{|(1,z)|}$ and differentiating appropriately, you'll get your desired formula for the metric as a $2$-tensor in terms of $dz_j$ and $dbar z_j$.






                share|cite|improve this answer











                $endgroup$



                It's indeed very natural to use the hermitian geometry structure here. Let $e_0;e_1,dots e_n$ be a unitary frame at the point $[Z]=[e_0]inBbb P^n$. Then
                $$de_0 = omega_{0bar 0}e_0 + sum omega_{0bar j}e_j$$
                and the hermitian metric on $Bbb P^n$ is given by $sum |omega_{0bar j}|^2 = sum omega_{0bar j}overline{omega_{0bar j}}$. In one of your charts, say $Z_0ne 0$, we take coordinates by setting $Z=(1,z)$ and $e_0 = Z/|Z|$.



                Now, note that $omega_{0bar 0} = i,dtheta$ where $e^{itheta}$ gives the fiber of your Riemannian submersion. Moreover, denoting the hermitian inner product by $(cdot,cdot)$,
                $$sum |omega_{0bar j}|^2 = (de_0,de_0) - |omega_{0bar 0}|^2 = (de_0,de_0)-|(de_0,e_0)|^2.$$
                Substituting $e_0 = dfrac{(1,z)}{|(1,z)|}$ and differentiating appropriately, you'll get your desired formula for the metric as a $2$-tensor in terms of $dz_j$ and $dbar z_j$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 13 '18 at 23:31

























                answered Dec 13 '18 at 23:22









                Ted ShifrinTed Shifrin

                63.7k44591




                63.7k44591






























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