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

      637




      637






















          1 Answer
          1






          active

          oldest

          votes


















          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









          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065740%2finequality-of-the-laplacian-involving-the-ricci-curvature%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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












              $begingroup$

              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






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3065740%2finequality-of-the-laplacian-involving-the-ricci-curvature%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    An IMO inspired problem

                    Image
                    14 4 This problem from IMO 1988 is said to be one of the most elegant ones in functional equations. Problem : The function $f$ is defined on the set of all positive integers as follows: begin{align} f(1) = 1, f(3) &= 3, f(2n) = f(n), \ f(4n+1) &= 2f(2n+1) - f(n), \ f(4n+3) &= 3 f(2n+1) - 2f(n) end{align} Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$. The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$. My question is the following : How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a
                    Read more

                    Management

                    Image
                    Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
                    Read more

                    Has there ever been an instance of an active nuclear power plant within or near a war zone?

                    Image
                    36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
                    Read more