Metric on $M/G$ which makes $pi: Mto M/G$ a Riemannian submersion












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$begingroup$



Let $G$ be a Lie group acting isometrically, freely and properly on a Riemannian manifold $(M,g)$ and $pi:Mto N:=M/G$ the natural projection. Show that there is a unique metric $h$ on $N$ which makes the projection $pi:(M,g)to (N,h)$ a Riemannian submersion.




For $xin N$ and $pinpi^{-1}(x)$, we take the vertical/horizontal spaces $V_p:=ker dpi_p$ and $H_p:=V_p^perp$, so that $T_pM=V_poplus H_p$. Since $dpi_p$ is surjective, then $dpi_p|_{H_p}:H_pto T_{pi(p)}N$ is bijective. For $v_1,v_2in T_xN$, we define:
$$h_x(v_1,v_2)=g_p(dpi|_{H_p}^{-1}v_1,dpi|_{H_p}^{-1}v_2)$$
(in particular, this satisfies $|v|_g=|dpi(v)|_h$)



I was able to prove $h_x$ doesn't depend on the choice of $p$ using the fact that $G$ acts isometrically. The uniqueness comes easy from the fact that another $h'$ would be such that $|v|_{h'}=|v|_h$ for all $vin T_xN$, which is enough to conclude $h'=h$.



I'm having trouble to prove $h$ is smooth. I wish to argue that if $X,Y$ are a fields in $N$, we may find horizontal fields $X_H, Y_H$ in $M$ such that $dpi(X_H)=X,dpi(Y_H)=Y$, so $h(X,Y)=g(X_H,Y_H)$. I only need to show that such fields $X_H,Y_H$ exist (I believe they do), but I don't know how.



Any suggestions?






lie-groups riemannian-geometry smooth-manifolds







share|cite|improve this question











$endgroup$

















    0












    $begingroup$



    Let $G$ be a Lie group acting isometrically, freely and properly on a Riemannian manifold $(M,g)$ and $pi:Mto N:=M/G$ the natural projection. Show that there is a unique metric $h$ on $N$ which makes the projection $pi:(M,g)to (N,h)$ a Riemannian submersion.




    For $xin N$ and $pinpi^{-1}(x)$, we take the vertical/horizontal spaces $V_p:=ker dpi_p$ and $H_p:=V_p^perp$, so that $T_pM=V_poplus H_p$. Since $dpi_p$ is surjective, then $dpi_p|_{H_p}:H_pto T_{pi(p)}N$ is bijective. For $v_1,v_2in T_xN$, we define:
    $$h_x(v_1,v_2)=g_p(dpi|_{H_p}^{-1}v_1,dpi|_{H_p}^{-1}v_2)$$
    (in particular, this satisfies $|v|_g=|dpi(v)|_h$)



    I was able to prove $h_x$ doesn't depend on the choice of $p$ using the fact that $G$ acts isometrically. The uniqueness comes easy from the fact that another $h'$ would be such that $|v|_{h'}=|v|_h$ for all $vin T_xN$, which is enough to conclude $h'=h$.



    I'm having trouble to prove $h$ is smooth. I wish to argue that if $X,Y$ are a fields in $N$, we may find horizontal fields $X_H, Y_H$ in $M$ such that $dpi(X_H)=X,dpi(Y_H)=Y$, so $h(X,Y)=g(X_H,Y_H)$. I only need to show that such fields $X_H,Y_H$ exist (I believe they do), but I don't know how.



    Any suggestions?






    lie-groups riemannian-geometry smooth-manifolds







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      Let $G$ be a Lie group acting isometrically, freely and properly on a Riemannian manifold $(M,g)$ and $pi:Mto N:=M/G$ the natural projection. Show that there is a unique metric $h$ on $N$ which makes the projection $pi:(M,g)to (N,h)$ a Riemannian submersion.




      For $xin N$ and $pinpi^{-1}(x)$, we take the vertical/horizontal spaces $V_p:=ker dpi_p$ and $H_p:=V_p^perp$, so that $T_pM=V_poplus H_p$. Since $dpi_p$ is surjective, then $dpi_p|_{H_p}:H_pto T_{pi(p)}N$ is bijective. For $v_1,v_2in T_xN$, we define:
      $$h_x(v_1,v_2)=g_p(dpi|_{H_p}^{-1}v_1,dpi|_{H_p}^{-1}v_2)$$
      (in particular, this satisfies $|v|_g=|dpi(v)|_h$)



      I was able to prove $h_x$ doesn't depend on the choice of $p$ using the fact that $G$ acts isometrically. The uniqueness comes easy from the fact that another $h'$ would be such that $|v|_{h'}=|v|_h$ for all $vin T_xN$, which is enough to conclude $h'=h$.



      I'm having trouble to prove $h$ is smooth. I wish to argue that if $X,Y$ are a fields in $N$, we may find horizontal fields $X_H, Y_H$ in $M$ such that $dpi(X_H)=X,dpi(Y_H)=Y$, so $h(X,Y)=g(X_H,Y_H)$. I only need to show that such fields $X_H,Y_H$ exist (I believe they do), but I don't know how.



      Any suggestions?






      lie-groups riemannian-geometry smooth-manifolds







      share|cite|improve this question











      $endgroup$





      Let $G$ be a Lie group acting isometrically, freely and properly on a Riemannian manifold $(M,g)$ and $pi:Mto N:=M/G$ the natural projection. Show that there is a unique metric $h$ on $N$ which makes the projection $pi:(M,g)to (N,h)$ a Riemannian submersion.




      For $xin N$ and $pinpi^{-1}(x)$, we take the vertical/horizontal spaces $V_p:=ker dpi_p$ and $H_p:=V_p^perp$, so that $T_pM=V_poplus H_p$. Since $dpi_p$ is surjective, then $dpi_p|_{H_p}:H_pto T_{pi(p)}N$ is bijective. For $v_1,v_2in T_xN$, we define:
      $$h_x(v_1,v_2)=g_p(dpi|_{H_p}^{-1}v_1,dpi|_{H_p}^{-1}v_2)$$
      (in particular, this satisfies $|v|_g=|dpi(v)|_h$)



      I was able to prove $h_x$ doesn't depend on the choice of $p$ using the fact that $G$ acts isometrically. The uniqueness comes easy from the fact that another $h'$ would be such that $|v|_{h'}=|v|_h$ for all $vin T_xN$, which is enough to conclude $h'=h$.



      I'm having trouble to prove $h$ is smooth. I wish to argue that if $X,Y$ are a fields in $N$, we may find horizontal fields $X_H, Y_H$ in $M$ such that $dpi(X_H)=X,dpi(Y_H)=Y$, so $h(X,Y)=g(X_H,Y_H)$. I only need to show that such fields $X_H,Y_H$ exist (I believe they do), but I don't know how.



      Any suggestions?






      lie-groups riemannian-geometry smooth-manifolds




      lie-groups riemannian-geometry smooth-manifolds






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 4 '18 at 22:48







      rmdmc89

















      asked Dec 4 '18 at 21:49









      rmdmc89rmdmc89

      2,0831922




      2,0831922






















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          $begingroup$

          Let $yin N$, and $xin M$ such that $p(x)=y$. Let $U$ be an open subset containing $x$ whose adherence $V$ is compact. Since the action is proper, $Orb_xcap V$ has a finite number of connected components. We can shrink $U$ such that $U$ contains a unique connected component of $Orb_xcap V$, $C$, let $T$ be a submanifold transverse to $C$, the restriction $p$ of $pi$ to $T$ is a diffeomorphism onto its image, for every $zin T$, we can define $Y_H(z)=dp^{-1}_{pi(z)}(X_H(pi(z))$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Tselmo, thanks for your answer! I've just had another idea, I'd like to know what you think: since $pi$ is a submersion, we can find charts $(U,phi)$ at $p$ and $(V,psi)$ at $x$ such that $$psicirc picirc phi^{-1}:(x_1,...,x_m)mapsto (x_1,...,x_n)$$ Therefore, $H_q=text{span}left(left.frac{partial}{partial phi_1}right|_q,...,left.frac{partial}{partial phi_m}right|_qright)$ for all $qin U$. So if $X=sum_{i=1}^nf_ifrac{partial}{partial psi_i}$, we define $X_H:=sum_{i=1}^m(f_icircpi)frac{partial}{partial phi_i}$. Thoes this look correct?
            $endgroup$
            – rmdmc89
            Dec 5 '18 at 13:07











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          $begingroup$

          Let $yin N$, and $xin M$ such that $p(x)=y$. Let $U$ be an open subset containing $x$ whose adherence $V$ is compact. Since the action is proper, $Orb_xcap V$ has a finite number of connected components. We can shrink $U$ such that $U$ contains a unique connected component of $Orb_xcap V$, $C$, let $T$ be a submanifold transverse to $C$, the restriction $p$ of $pi$ to $T$ is a diffeomorphism onto its image, for every $zin T$, we can define $Y_H(z)=dp^{-1}_{pi(z)}(X_H(pi(z))$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Tselmo, thanks for your answer! I've just had another idea, I'd like to know what you think: since $pi$ is a submersion, we can find charts $(U,phi)$ at $p$ and $(V,psi)$ at $x$ such that $$psicirc picirc phi^{-1}:(x_1,...,x_m)mapsto (x_1,...,x_n)$$ Therefore, $H_q=text{span}left(left.frac{partial}{partial phi_1}right|_q,...,left.frac{partial}{partial phi_m}right|_qright)$ for all $qin U$. So if $X=sum_{i=1}^nf_ifrac{partial}{partial psi_i}$, we define $X_H:=sum_{i=1}^m(f_icircpi)frac{partial}{partial phi_i}$. Thoes this look correct?
            $endgroup$
            – rmdmc89
            Dec 5 '18 at 13:07
















          1












          $begingroup$

          Let $yin N$, and $xin M$ such that $p(x)=y$. Let $U$ be an open subset containing $x$ whose adherence $V$ is compact. Since the action is proper, $Orb_xcap V$ has a finite number of connected components. We can shrink $U$ such that $U$ contains a unique connected component of $Orb_xcap V$, $C$, let $T$ be a submanifold transverse to $C$, the restriction $p$ of $pi$ to $T$ is a diffeomorphism onto its image, for every $zin T$, we can define $Y_H(z)=dp^{-1}_{pi(z)}(X_H(pi(z))$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Tselmo, thanks for your answer! I've just had another idea, I'd like to know what you think: since $pi$ is a submersion, we can find charts $(U,phi)$ at $p$ and $(V,psi)$ at $x$ such that $$psicirc picirc phi^{-1}:(x_1,...,x_m)mapsto (x_1,...,x_n)$$ Therefore, $H_q=text{span}left(left.frac{partial}{partial phi_1}right|_q,...,left.frac{partial}{partial phi_m}right|_qright)$ for all $qin U$. So if $X=sum_{i=1}^nf_ifrac{partial}{partial psi_i}$, we define $X_H:=sum_{i=1}^m(f_icircpi)frac{partial}{partial phi_i}$. Thoes this look correct?
            $endgroup$
            – rmdmc89
            Dec 5 '18 at 13:07














          1












          1








          1





          $begingroup$

          Let $yin N$, and $xin M$ such that $p(x)=y$. Let $U$ be an open subset containing $x$ whose adherence $V$ is compact. Since the action is proper, $Orb_xcap V$ has a finite number of connected components. We can shrink $U$ such that $U$ contains a unique connected component of $Orb_xcap V$, $C$, let $T$ be a submanifold transverse to $C$, the restriction $p$ of $pi$ to $T$ is a diffeomorphism onto its image, for every $zin T$, we can define $Y_H(z)=dp^{-1}_{pi(z)}(X_H(pi(z))$.






          share|cite|improve this answer









          $endgroup$



          Let $yin N$, and $xin M$ such that $p(x)=y$. Let $U$ be an open subset containing $x$ whose adherence $V$ is compact. Since the action is proper, $Orb_xcap V$ has a finite number of connected components. We can shrink $U$ such that $U$ contains a unique connected component of $Orb_xcap V$, $C$, let $T$ be a submanifold transverse to $C$, the restriction $p$ of $pi$ to $T$ is a diffeomorphism onto its image, for every $zin T$, we can define $Y_H(z)=dp^{-1}_{pi(z)}(X_H(pi(z))$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 4 '18 at 22:50









          Tsemo AristideTsemo Aristide

          56.6k11444




          56.6k11444












          • $begingroup$
            Tselmo, thanks for your answer! I've just had another idea, I'd like to know what you think: since $pi$ is a submersion, we can find charts $(U,phi)$ at $p$ and $(V,psi)$ at $x$ such that $$psicirc picirc phi^{-1}:(x_1,...,x_m)mapsto (x_1,...,x_n)$$ Therefore, $H_q=text{span}left(left.frac{partial}{partial phi_1}right|_q,...,left.frac{partial}{partial phi_m}right|_qright)$ for all $qin U$. So if $X=sum_{i=1}^nf_ifrac{partial}{partial psi_i}$, we define $X_H:=sum_{i=1}^m(f_icircpi)frac{partial}{partial phi_i}$. Thoes this look correct?
            $endgroup$
            – rmdmc89
            Dec 5 '18 at 13:07


















          • $begingroup$
            Tselmo, thanks for your answer! I've just had another idea, I'd like to know what you think: since $pi$ is a submersion, we can find charts $(U,phi)$ at $p$ and $(V,psi)$ at $x$ such that $$psicirc picirc phi^{-1}:(x_1,...,x_m)mapsto (x_1,...,x_n)$$ Therefore, $H_q=text{span}left(left.frac{partial}{partial phi_1}right|_q,...,left.frac{partial}{partial phi_m}right|_qright)$ for all $qin U$. So if $X=sum_{i=1}^nf_ifrac{partial}{partial psi_i}$, we define $X_H:=sum_{i=1}^m(f_icircpi)frac{partial}{partial phi_i}$. Thoes this look correct?
            $endgroup$
            – rmdmc89
            Dec 5 '18 at 13:07
















          $begingroup$
          Tselmo, thanks for your answer! I've just had another idea, I'd like to know what you think: since $pi$ is a submersion, we can find charts $(U,phi)$ at $p$ and $(V,psi)$ at $x$ such that $$psicirc picirc phi^{-1}:(x_1,...,x_m)mapsto (x_1,...,x_n)$$ Therefore, $H_q=text{span}left(left.frac{partial}{partial phi_1}right|_q,...,left.frac{partial}{partial phi_m}right|_qright)$ for all $qin U$. So if $X=sum_{i=1}^nf_ifrac{partial}{partial psi_i}$, we define $X_H:=sum_{i=1}^m(f_icircpi)frac{partial}{partial phi_i}$. Thoes this look correct?
          $endgroup$
          – rmdmc89
          Dec 5 '18 at 13:07




          $begingroup$
          Tselmo, thanks for your answer! I've just had another idea, I'd like to know what you think: since $pi$ is a submersion, we can find charts $(U,phi)$ at $p$ and $(V,psi)$ at $x$ such that $$psicirc picirc phi^{-1}:(x_1,...,x_m)mapsto (x_1,...,x_n)$$ Therefore, $H_q=text{span}left(left.frac{partial}{partial phi_1}right|_q,...,left.frac{partial}{partial phi_m}right|_qright)$ for all $qin U$. So if $X=sum_{i=1}^nf_ifrac{partial}{partial psi_i}$, we define $X_H:=sum_{i=1}^m(f_icircpi)frac{partial}{partial phi_i}$. Thoes this look correct?
          $endgroup$
          – rmdmc89
          Dec 5 '18 at 13:07


















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