Evaluate integral (Chern article)

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My question is evaluate some integral of the article "A Simple Intrinsic Proof of the Gauss-Bonnet Formula for Closed Riemannian Manifolds" write by Chern.



Let's go:



If $(M^n,g)$ is a closed even dimension Riemann manifold with $nabla$ Levi-Civita connection, we can write locally $nabla_X V = theta^i(X)e_i$, where $V = v^ie_i$, $theta^i = dv^i(X) + v^jomega_j^i$ and $omega_i^j$ connection forms. In the same way for Riemann curvature, we have $Omega_i^j$ curvature forms, satisfying $domega_i^j = omega_i^k wedge omega_k^j + Omega_i^j$.



So, pulling-back $theta_i$ and $Omega_i^j$ by $rho: SM rightarrow M$, where $SM$ is the unit-sphere bundle, we define two kind of intrinsic forms in $SM$, namely



$$Phi_k = sum_sigma in S_n sgn(sigma)v_sigma(1)theta_sigma(2) wedge cdots wedge theta_sigma(n-2k)wedgeOmega_sigma(n-2k+1)^sigma(n-2k+1)wedge cdots wedge Omega_sigma(n-1)^sigma(n)$$
and



$$ Psi_k = sum_sigma in S_n sgn(sigma) Omega_sigma(1)^sigma(2)wedgetheta_sigma(3)wedge cdots wedge theta_sigma(n-2k)wedgeOmega_sigma(n-2k+1)^sigma(n-2k+1)wedge cdots wedge Omega_sigma(n-1)^sigma(n)$$



$k = 0, cdots fracn2 -1 $. It's not too hard to show the following recurrent relation: $$ dPhi_k = Psi_k-1 + fracn - 2k - 12(k+1)Psi_k$$



Where $Psi_-1 equiv 0$.



Define the form, in $M$, $Omega = displaystyle (-1)^fracn2-1frac1(2pi)^fracn2Pf(Omega_i^j)$ (called Euler form), definition of Pfaffian polynomial here , obviously $rho^*Omega = displaystyle (-1)^fracn2-1frac12^npi^fracn2left(fracn2right)!Psi_fracn2-1$, write $Psi_fracn2-1$ in terms of $dPhi_k's$ we obtain $dPi = Omega$ in $SM$, with $Pi = displaystyle frac1pi^fracn2sum_t=0^fracn2-1(-1)^t frac11 cdot 3 cdots (n - 2t - 1)t!2^fracn2+tPhi_t$.



With some tricks and Stokes' theorem we show $$displaystyle int_M Omega = sum_i=1^sind_x_s(nabla_gf)int_SM_x_sPi|_SM_x_s $$



for $x_1, cdots, x_s$ singularities of $nabla_g f$, $f$ a Morse's function. I'd like $int_SM_x_s Pi|_SM_x_s = 1$ to use the Hopf index theorem.



In the paper, Chern claims $$ int_SM_x_s Pi|_SM_x_s = frac11cdot3 cdots (n-1)(2pi)^fracn2 int_SM_x_sPhi_0|_SM_x_s$$



and of course it's equal 1, essentially, I don't know why $ displaystyle int_SM_x_s Phi_k = 0$, for $k geq 1$.



Thanks!







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    up vote
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    My question is evaluate some integral of the article "A Simple Intrinsic Proof of the Gauss-Bonnet Formula for Closed Riemannian Manifolds" write by Chern.



    Let's go:



    If $(M^n,g)$ is a closed even dimension Riemann manifold with $nabla$ Levi-Civita connection, we can write locally $nabla_X V = theta^i(X)e_i$, where $V = v^ie_i$, $theta^i = dv^i(X) + v^jomega_j^i$ and $omega_i^j$ connection forms. In the same way for Riemann curvature, we have $Omega_i^j$ curvature forms, satisfying $domega_i^j = omega_i^k wedge omega_k^j + Omega_i^j$.



    So, pulling-back $theta_i$ and $Omega_i^j$ by $rho: SM rightarrow M$, where $SM$ is the unit-sphere bundle, we define two kind of intrinsic forms in $SM$, namely



    $$Phi_k = sum_sigma in S_n sgn(sigma)v_sigma(1)theta_sigma(2) wedge cdots wedge theta_sigma(n-2k)wedgeOmega_sigma(n-2k+1)^sigma(n-2k+1)wedge cdots wedge Omega_sigma(n-1)^sigma(n)$$
    and



    $$ Psi_k = sum_sigma in S_n sgn(sigma) Omega_sigma(1)^sigma(2)wedgetheta_sigma(3)wedge cdots wedge theta_sigma(n-2k)wedgeOmega_sigma(n-2k+1)^sigma(n-2k+1)wedge cdots wedge Omega_sigma(n-1)^sigma(n)$$



    $k = 0, cdots fracn2 -1 $. It's not too hard to show the following recurrent relation: $$ dPhi_k = Psi_k-1 + fracn - 2k - 12(k+1)Psi_k$$



    Where $Psi_-1 equiv 0$.



    Define the form, in $M$, $Omega = displaystyle (-1)^fracn2-1frac1(2pi)^fracn2Pf(Omega_i^j)$ (called Euler form), definition of Pfaffian polynomial here , obviously $rho^*Omega = displaystyle (-1)^fracn2-1frac12^npi^fracn2left(fracn2right)!Psi_fracn2-1$, write $Psi_fracn2-1$ in terms of $dPhi_k's$ we obtain $dPi = Omega$ in $SM$, with $Pi = displaystyle frac1pi^fracn2sum_t=0^fracn2-1(-1)^t frac11 cdot 3 cdots (n - 2t - 1)t!2^fracn2+tPhi_t$.



    With some tricks and Stokes' theorem we show $$displaystyle int_M Omega = sum_i=1^sind_x_s(nabla_gf)int_SM_x_sPi|_SM_x_s $$



    for $x_1, cdots, x_s$ singularities of $nabla_g f$, $f$ a Morse's function. I'd like $int_SM_x_s Pi|_SM_x_s = 1$ to use the Hopf index theorem.



    In the paper, Chern claims $$ int_SM_x_s Pi|_SM_x_s = frac11cdot3 cdots (n-1)(2pi)^fracn2 int_SM_x_sPhi_0|_SM_x_s$$



    and of course it's equal 1, essentially, I don't know why $ displaystyle int_SM_x_s Phi_k = 0$, for $k geq 1$.



    Thanks!







    share|cite|improve this question
























      up vote
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      My question is evaluate some integral of the article "A Simple Intrinsic Proof of the Gauss-Bonnet Formula for Closed Riemannian Manifolds" write by Chern.



      Let's go:



      If $(M^n,g)$ is a closed even dimension Riemann manifold with $nabla$ Levi-Civita connection, we can write locally $nabla_X V = theta^i(X)e_i$, where $V = v^ie_i$, $theta^i = dv^i(X) + v^jomega_j^i$ and $omega_i^j$ connection forms. In the same way for Riemann curvature, we have $Omega_i^j$ curvature forms, satisfying $domega_i^j = omega_i^k wedge omega_k^j + Omega_i^j$.



      So, pulling-back $theta_i$ and $Omega_i^j$ by $rho: SM rightarrow M$, where $SM$ is the unit-sphere bundle, we define two kind of intrinsic forms in $SM$, namely



      $$Phi_k = sum_sigma in S_n sgn(sigma)v_sigma(1)theta_sigma(2) wedge cdots wedge theta_sigma(n-2k)wedgeOmega_sigma(n-2k+1)^sigma(n-2k+1)wedge cdots wedge Omega_sigma(n-1)^sigma(n)$$
      and



      $$ Psi_k = sum_sigma in S_n sgn(sigma) Omega_sigma(1)^sigma(2)wedgetheta_sigma(3)wedge cdots wedge theta_sigma(n-2k)wedgeOmega_sigma(n-2k+1)^sigma(n-2k+1)wedge cdots wedge Omega_sigma(n-1)^sigma(n)$$



      $k = 0, cdots fracn2 -1 $. It's not too hard to show the following recurrent relation: $$ dPhi_k = Psi_k-1 + fracn - 2k - 12(k+1)Psi_k$$



      Where $Psi_-1 equiv 0$.



      Define the form, in $M$, $Omega = displaystyle (-1)^fracn2-1frac1(2pi)^fracn2Pf(Omega_i^j)$ (called Euler form), definition of Pfaffian polynomial here , obviously $rho^*Omega = displaystyle (-1)^fracn2-1frac12^npi^fracn2left(fracn2right)!Psi_fracn2-1$, write $Psi_fracn2-1$ in terms of $dPhi_k's$ we obtain $dPi = Omega$ in $SM$, with $Pi = displaystyle frac1pi^fracn2sum_t=0^fracn2-1(-1)^t frac11 cdot 3 cdots (n - 2t - 1)t!2^fracn2+tPhi_t$.



      With some tricks and Stokes' theorem we show $$displaystyle int_M Omega = sum_i=1^sind_x_s(nabla_gf)int_SM_x_sPi|_SM_x_s $$



      for $x_1, cdots, x_s$ singularities of $nabla_g f$, $f$ a Morse's function. I'd like $int_SM_x_s Pi|_SM_x_s = 1$ to use the Hopf index theorem.



      In the paper, Chern claims $$ int_SM_x_s Pi|_SM_x_s = frac11cdot3 cdots (n-1)(2pi)^fracn2 int_SM_x_sPhi_0|_SM_x_s$$



      and of course it's equal 1, essentially, I don't know why $ displaystyle int_SM_x_s Phi_k = 0$, for $k geq 1$.



      Thanks!







      share|cite|improve this question














      My question is evaluate some integral of the article "A Simple Intrinsic Proof of the Gauss-Bonnet Formula for Closed Riemannian Manifolds" write by Chern.



      Let's go:



      If $(M^n,g)$ is a closed even dimension Riemann manifold with $nabla$ Levi-Civita connection, we can write locally $nabla_X V = theta^i(X)e_i$, where $V = v^ie_i$, $theta^i = dv^i(X) + v^jomega_j^i$ and $omega_i^j$ connection forms. In the same way for Riemann curvature, we have $Omega_i^j$ curvature forms, satisfying $domega_i^j = omega_i^k wedge omega_k^j + Omega_i^j$.



      So, pulling-back $theta_i$ and $Omega_i^j$ by $rho: SM rightarrow M$, where $SM$ is the unit-sphere bundle, we define two kind of intrinsic forms in $SM$, namely



      $$Phi_k = sum_sigma in S_n sgn(sigma)v_sigma(1)theta_sigma(2) wedge cdots wedge theta_sigma(n-2k)wedgeOmega_sigma(n-2k+1)^sigma(n-2k+1)wedge cdots wedge Omega_sigma(n-1)^sigma(n)$$
      and



      $$ Psi_k = sum_sigma in S_n sgn(sigma) Omega_sigma(1)^sigma(2)wedgetheta_sigma(3)wedge cdots wedge theta_sigma(n-2k)wedgeOmega_sigma(n-2k+1)^sigma(n-2k+1)wedge cdots wedge Omega_sigma(n-1)^sigma(n)$$



      $k = 0, cdots fracn2 -1 $. It's not too hard to show the following recurrent relation: $$ dPhi_k = Psi_k-1 + fracn - 2k - 12(k+1)Psi_k$$



      Where $Psi_-1 equiv 0$.



      Define the form, in $M$, $Omega = displaystyle (-1)^fracn2-1frac1(2pi)^fracn2Pf(Omega_i^j)$ (called Euler form), definition of Pfaffian polynomial here , obviously $rho^*Omega = displaystyle (-1)^fracn2-1frac12^npi^fracn2left(fracn2right)!Psi_fracn2-1$, write $Psi_fracn2-1$ in terms of $dPhi_k's$ we obtain $dPi = Omega$ in $SM$, with $Pi = displaystyle frac1pi^fracn2sum_t=0^fracn2-1(-1)^t frac11 cdot 3 cdots (n - 2t - 1)t!2^fracn2+tPhi_t$.



      With some tricks and Stokes' theorem we show $$displaystyle int_M Omega = sum_i=1^sind_x_s(nabla_gf)int_SM_x_sPi|_SM_x_s $$



      for $x_1, cdots, x_s$ singularities of $nabla_g f$, $f$ a Morse's function. I'd like $int_SM_x_s Pi|_SM_x_s = 1$ to use the Hopf index theorem.



      In the paper, Chern claims $$ int_SM_x_s Pi|_SM_x_s = frac11cdot3 cdots (n-1)(2pi)^fracn2 int_SM_x_sPhi_0|_SM_x_s$$



      and of course it's equal 1, essentially, I don't know why $ displaystyle int_SM_x_s Phi_k = 0$, for $k geq 1$.



      Thanks!









      share|cite|improve this question













      share|cite|improve this question




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      edited Aug 30 at 23:51

























      asked Aug 27 at 12:32









      Carlos Alcantara

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          It's not so scary, after all :) You're integrating the forms $Phi_k$ over the unit sphere bundle at a fixed point $x_0$ of $M$ (your notation is different from his, since for him $M$ is the unit sphere bundle of the manifold $R$). For $kge 1$, the form $Phi_k$ will involve at least one curvature form $Omega_i^j$. The curvature forms are horizontal for the fibration $SMto M$, and you're integrating over a fiber. So those integrals all vanish.






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            It's not so scary, after all :) You're integrating the forms $Phi_k$ over the unit sphere bundle at a fixed point $x_0$ of $M$ (your notation is different from his, since for him $M$ is the unit sphere bundle of the manifold $R$). For $kge 1$, the form $Phi_k$ will involve at least one curvature form $Omega_i^j$. The curvature forms are horizontal for the fibration $SMto M$, and you're integrating over a fiber. So those integrals all vanish.






            share|cite|improve this answer
























              up vote
              2
              down vote



              accepted










              It's not so scary, after all :) You're integrating the forms $Phi_k$ over the unit sphere bundle at a fixed point $x_0$ of $M$ (your notation is different from his, since for him $M$ is the unit sphere bundle of the manifold $R$). For $kge 1$, the form $Phi_k$ will involve at least one curvature form $Omega_i^j$. The curvature forms are horizontal for the fibration $SMto M$, and you're integrating over a fiber. So those integrals all vanish.






              share|cite|improve this answer






















                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                It's not so scary, after all :) You're integrating the forms $Phi_k$ over the unit sphere bundle at a fixed point $x_0$ of $M$ (your notation is different from his, since for him $M$ is the unit sphere bundle of the manifold $R$). For $kge 1$, the form $Phi_k$ will involve at least one curvature form $Omega_i^j$. The curvature forms are horizontal for the fibration $SMto M$, and you're integrating over a fiber. So those integrals all vanish.






                share|cite|improve this answer












                It's not so scary, after all :) You're integrating the forms $Phi_k$ over the unit sphere bundle at a fixed point $x_0$ of $M$ (your notation is different from his, since for him $M$ is the unit sphere bundle of the manifold $R$). For $kge 1$, the form $Phi_k$ will involve at least one curvature form $Omega_i^j$. The curvature forms are horizontal for the fibration $SMto M$, and you're integrating over a fiber. So those integrals all vanish.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 27 at 16:30









                Ted Shifrin

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