Inequality of the laplacian involving the Ricci curvature












1












$begingroup$


I am reading Eschenburg and Heintze's proof of the Cheeger-Gromoll splitting theorem. Lemma 1 states:




Let $fin C^infty(M)$ with $||grad(f)||=1$. If c is an integral curve of the gradient, then it is a minimising geodesic and
$$begin{align}
-Ricc(c', c') & =(Delta fcirc c)'+||Hess_fcirc c||^2 \
& le (Delta fcirc c)' +frac{1}{n-1}(Delta fcirc c)^2
end{align}$$




The first part, about the minimising geodesics, has already been answered here



They begin the proof by choosing some $t_0in mathbb{R}$ and they take a neighbourhood $U$ of $c(t_0)$, where they choose and orthonormal frame ${ E_1,E_2,...,E_n}$ such that $E_n=gradf$ and $E_i$ are parallel along $gradf$.
Then the calculation give
$$begin{align}
Ricc(E_n,E_n) &=sum_{i=1}^{n} left< R(E_i,E_n)E_n,E_iright> \
&=sum_ileft(
-left< nabla_{E_n}nabla _{E_i}E_n,E_i right>
-left< nabla_{nabla_{E_i}E_n}E_n,E_iright> right) \
&=-E_nleft(sum_i left< nabla_{E_i}E_n,E_iright>right)
-sum_ileft< nabla_{E_i}E_n,nabla_{E_i}E_nright> \
&=-E_n(Delta f)-||Hess_f||^2
end{align}$$



Tis way we prooved th equality of the Lemma.
I understand intuitively the existence of such a frame ${ E_1,E_2,...,E_n}$ but I would like to see a rigorous proof. Mainly I don't understand how one gets from this to the inequality. All the authors say is that from the Schwarz inequality we have
$$begin{align}
Ricc(E_n,E_n) &
leq -E_n(Delta f)-sum_{i=1}^{n-1}left<Hess_f(E_i),E_iright>\
& leq -E_n(Delta f) -frac{1}{n-1}(Delta f)^2
end{align}$$
But I can't see how it is done






riemannian-geometry laplacian







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I am reading Eschenburg and Heintze's proof of the Cheeger-Gromoll splitting theorem. Lemma 1 states:




    Let $fin C^infty(M)$ with $||grad(f)||=1$. If c is an integral curve of the gradient, then it is a minimising geodesic and
    $$begin{align}
    -Ricc(c', c') & =(Delta fcirc c)'+||Hess_fcirc c||^2 \
    & le (Delta fcirc c)' +frac{1}{n-1}(Delta fcirc c)^2
    end{align}$$




    The first part, about the minimising geodesics, has already been answered here



    They begin the proof by choosing some $t_0in mathbb{R}$ and they take a neighbourhood $U$ of $c(t_0)$, where they choose and orthonormal frame ${ E_1,E_2,...,E_n}$ such that $E_n=gradf$ and $E_i$ are parallel along $gradf$.
    Then the calculation give
    $$begin{align}
    Ricc(E_n,E_n) &=sum_{i=1}^{n} left< R(E_i,E_n)E_n,E_iright> \
    &=sum_ileft(
    -left< nabla_{E_n}nabla _{E_i}E_n,E_i right>
    -left< nabla_{nabla_{E_i}E_n}E_n,E_iright> right) \
    &=-E_nleft(sum_i left< nabla_{E_i}E_n,E_iright>right)
    -sum_ileft< nabla_{E_i}E_n,nabla_{E_i}E_nright> \
    &=-E_n(Delta f)-||Hess_f||^2
    end{align}$$



    Tis way we prooved th equality of the Lemma.
    I understand intuitively the existence of such a frame ${ E_1,E_2,...,E_n}$ but I would like to see a rigorous proof. Mainly I don't understand how one gets from this to the inequality. All the authors say is that from the Schwarz inequality we have
    $$begin{align}
    Ricc(E_n,E_n) &
    leq -E_n(Delta f)-sum_{i=1}^{n-1}left<Hess_f(E_i),E_iright>\
    & leq -E_n(Delta f) -frac{1}{n-1}(Delta f)^2
    end{align}$$
    But I can't see how it is done






    riemannian-geometry laplacian







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am reading Eschenburg and Heintze's proof of the Cheeger-Gromoll splitting theorem. Lemma 1 states:




      Let $fin C^infty(M)$ with $||grad(f)||=1$. If c is an integral curve of the gradient, then it is a minimising geodesic and
      $$begin{align}
      -Ricc(c', c') & =(Delta fcirc c)'+||Hess_fcirc c||^2 \
      & le (Delta fcirc c)' +frac{1}{n-1}(Delta fcirc c)^2
      end{align}$$




      The first part, about the minimising geodesics, has already been answered here



      They begin the proof by choosing some $t_0in mathbb{R}$ and they take a neighbourhood $U$ of $c(t_0)$, where they choose and orthonormal frame ${ E_1,E_2,...,E_n}$ such that $E_n=gradf$ and $E_i$ are parallel along $gradf$.
      Then the calculation give
      $$begin{align}
      Ricc(E_n,E_n) &=sum_{i=1}^{n} left< R(E_i,E_n)E_n,E_iright> \
      &=sum_ileft(
      -left< nabla_{E_n}nabla _{E_i}E_n,E_i right>
      -left< nabla_{nabla_{E_i}E_n}E_n,E_iright> right) \
      &=-E_nleft(sum_i left< nabla_{E_i}E_n,E_iright>right)
      -sum_ileft< nabla_{E_i}E_n,nabla_{E_i}E_nright> \
      &=-E_n(Delta f)-||Hess_f||^2
      end{align}$$



      Tis way we prooved th equality of the Lemma.
      I understand intuitively the existence of such a frame ${ E_1,E_2,...,E_n}$ but I would like to see a rigorous proof. Mainly I don't understand how one gets from this to the inequality. All the authors say is that from the Schwarz inequality we have
      $$begin{align}
      Ricc(E_n,E_n) &
      leq -E_n(Delta f)-sum_{i=1}^{n-1}left<Hess_f(E_i),E_iright>\
      & leq -E_n(Delta f) -frac{1}{n-1}(Delta f)^2
      end{align}$$
      But I can't see how it is done






      riemannian-geometry laplacian







      share|cite|improve this question









      $endgroup$




      I am reading Eschenburg and Heintze's proof of the Cheeger-Gromoll splitting theorem. Lemma 1 states:




      Let $fin C^infty(M)$ with $||grad(f)||=1$. If c is an integral curve of the gradient, then it is a minimising geodesic and
      $$begin{align}
      -Ricc(c', c') & =(Delta fcirc c)'+||Hess_fcirc c||^2 \
      & le (Delta fcirc c)' +frac{1}{n-1}(Delta fcirc c)^2
      end{align}$$




      The first part, about the minimising geodesics, has already been answered here



      They begin the proof by choosing some $t_0in mathbb{R}$ and they take a neighbourhood $U$ of $c(t_0)$, where they choose and orthonormal frame ${ E_1,E_2,...,E_n}$ such that $E_n=gradf$ and $E_i$ are parallel along $gradf$.
      Then the calculation give
      $$begin{align}
      Ricc(E_n,E_n) &=sum_{i=1}^{n} left< R(E_i,E_n)E_n,E_iright> \
      &=sum_ileft(
      -left< nabla_{E_n}nabla _{E_i}E_n,E_i right>
      -left< nabla_{nabla_{E_i}E_n}E_n,E_iright> right) \
      &=-E_nleft(sum_i left< nabla_{E_i}E_n,E_iright>right)
      -sum_ileft< nabla_{E_i}E_n,nabla_{E_i}E_nright> \
      &=-E_n(Delta f)-||Hess_f||^2
      end{align}$$



      Tis way we prooved th equality of the Lemma.
      I understand intuitively the existence of such a frame ${ E_1,E_2,...,E_n}$ but I would like to see a rigorous proof. Mainly I don't understand how one gets from this to the inequality. All the authors say is that from the Schwarz inequality we have
      $$begin{align}
      Ricc(E_n,E_n) &
      leq -E_n(Delta f)-sum_{i=1}^{n-1}left<Hess_f(E_i),E_iright>\
      & leq -E_n(Delta f) -frac{1}{n-1}(Delta f)^2
      end{align}$$
      But I can't see how it is done






      riemannian-geometry laplacian




      riemannian-geometry laplacian






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 8 at 2:26









      George NtouliosGeorge Ntoulios

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          $begingroup$

          For the second question, note that $operatorname{Hess}f$ is a symmetric matrix and thus can be diagonalized by some orthonormal basis ${E_1, cdots, E_n}$. We can also choose $E_n = c' = nabla f$ since $c'$ is an eigenvector with eigenvalue $0$:
          $$ operatorname{Hess} f(c') = c' c' f - nabla_{nabla_{c'}c'} f = 0$$



          Thus we write



          $$ operatorname{Hess} f(E_i, E_j) = delta_{ij} lambda_i.$$



          with $lambda_n = 0$. In terms of $lambda_i$'s we have
          $$ Delta f = sum_{i=1}^{n-1} lambda_i, | operatorname{Hess} (f)|^2 = sum_{i=1}^{n-1} lambda_i^2.$$



          Thus we have (By Cauchy Schwarz inequality)



          begin{align}
          Delta f &= sum_{i=1}^{n-1} lambda_i\
          &= (lambda_1, cdots, lambda_{n-1}) cdot (1, cdots, 1) \
          &le |(lambda_1, cdots, lambda_{n-1})| | (1, cdots, 1)| \
          &= sqrt{lambda_1^2 + cdots + lambda_{n-1}^2} sqrt{n-1} \
          Rightarrow (Delta f)^2 &le (n-1) |operatorname{Hess} f|^2
          end{align}






          share|cite|improve this answer









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            $begingroup$

            For the second question, note that $operatorname{Hess}f$ is a symmetric matrix and thus can be diagonalized by some orthonormal basis ${E_1, cdots, E_n}$. We can also choose $E_n = c' = nabla f$ since $c'$ is an eigenvector with eigenvalue $0$:
            $$ operatorname{Hess} f(c') = c' c' f - nabla_{nabla_{c'}c'} f = 0$$



            Thus we write



            $$ operatorname{Hess} f(E_i, E_j) = delta_{ij} lambda_i.$$



            with $lambda_n = 0$. In terms of $lambda_i$'s we have
            $$ Delta f = sum_{i=1}^{n-1} lambda_i, | operatorname{Hess} (f)|^2 = sum_{i=1}^{n-1} lambda_i^2.$$



            Thus we have (By Cauchy Schwarz inequality)



            begin{align}
            Delta f &= sum_{i=1}^{n-1} lambda_i\
            &= (lambda_1, cdots, lambda_{n-1}) cdot (1, cdots, 1) \
            &le |(lambda_1, cdots, lambda_{n-1})| | (1, cdots, 1)| \
            &= sqrt{lambda_1^2 + cdots + lambda_{n-1}^2} sqrt{n-1} \
            Rightarrow (Delta f)^2 &le (n-1) |operatorname{Hess} f|^2
            end{align}






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              For the second question, note that $operatorname{Hess}f$ is a symmetric matrix and thus can be diagonalized by some orthonormal basis ${E_1, cdots, E_n}$. We can also choose $E_n = c' = nabla f$ since $c'$ is an eigenvector with eigenvalue $0$:
              $$ operatorname{Hess} f(c') = c' c' f - nabla_{nabla_{c'}c'} f = 0$$



              Thus we write



              $$ operatorname{Hess} f(E_i, E_j) = delta_{ij} lambda_i.$$



              with $lambda_n = 0$. In terms of $lambda_i$'s we have
              $$ Delta f = sum_{i=1}^{n-1} lambda_i, | operatorname{Hess} (f)|^2 = sum_{i=1}^{n-1} lambda_i^2.$$



              Thus we have (By Cauchy Schwarz inequality)



              begin{align}
              Delta f &= sum_{i=1}^{n-1} lambda_i\
              &= (lambda_1, cdots, lambda_{n-1}) cdot (1, cdots, 1) \
              &le |(lambda_1, cdots, lambda_{n-1})| | (1, cdots, 1)| \
              &= sqrt{lambda_1^2 + cdots + lambda_{n-1}^2} sqrt{n-1} \
              Rightarrow (Delta f)^2 &le (n-1) |operatorname{Hess} f|^2
              end{align}






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                For the second question, note that $operatorname{Hess}f$ is a symmetric matrix and thus can be diagonalized by some orthonormal basis ${E_1, cdots, E_n}$. We can also choose $E_n = c' = nabla f$ since $c'$ is an eigenvector with eigenvalue $0$:
                $$ operatorname{Hess} f(c') = c' c' f - nabla_{nabla_{c'}c'} f = 0$$



                Thus we write



                $$ operatorname{Hess} f(E_i, E_j) = delta_{ij} lambda_i.$$



                with $lambda_n = 0$. In terms of $lambda_i$'s we have
                $$ Delta f = sum_{i=1}^{n-1} lambda_i, | operatorname{Hess} (f)|^2 = sum_{i=1}^{n-1} lambda_i^2.$$



                Thus we have (By Cauchy Schwarz inequality)



                begin{align}
                Delta f &= sum_{i=1}^{n-1} lambda_i\
                &= (lambda_1, cdots, lambda_{n-1}) cdot (1, cdots, 1) \
                &le |(lambda_1, cdots, lambda_{n-1})| | (1, cdots, 1)| \
                &= sqrt{lambda_1^2 + cdots + lambda_{n-1}^2} sqrt{n-1} \
                Rightarrow (Delta f)^2 &le (n-1) |operatorname{Hess} f|^2
                end{align}






                share|cite|improve this answer









                $endgroup$



                For the second question, note that $operatorname{Hess}f$ is a symmetric matrix and thus can be diagonalized by some orthonormal basis ${E_1, cdots, E_n}$. We can also choose $E_n = c' = nabla f$ since $c'$ is an eigenvector with eigenvalue $0$:
                $$ operatorname{Hess} f(c') = c' c' f - nabla_{nabla_{c'}c'} f = 0$$



                Thus we write



                $$ operatorname{Hess} f(E_i, E_j) = delta_{ij} lambda_i.$$



                with $lambda_n = 0$. In terms of $lambda_i$'s we have
                $$ Delta f = sum_{i=1}^{n-1} lambda_i, | operatorname{Hess} (f)|^2 = sum_{i=1}^{n-1} lambda_i^2.$$



                Thus we have (By Cauchy Schwarz inequality)



                begin{align}
                Delta f &= sum_{i=1}^{n-1} lambda_i\
                &= (lambda_1, cdots, lambda_{n-1}) cdot (1, cdots, 1) \
                &le |(lambda_1, cdots, lambda_{n-1})| | (1, cdots, 1)| \
                &= sqrt{lambda_1^2 + cdots + lambda_{n-1}^2} sqrt{n-1} \
                Rightarrow (Delta f)^2 &le (n-1) |operatorname{Hess} f|^2
                end{align}







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 8 at 3:05









                Arctic CharArctic Char

                10318




                10318






























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