Do scalar integrals correspond to integration of differential forms?












2














I have read through part of Spivak's Calculus on Manifolds up to the chapter Integration on Chains. Now I know that given a smooth vector field on $Bbb R^3$, doing line integral over a parametrised smooth curve, and doing surface integral over a parametrised smooth surface, correspond to integrating a $1$-form over a $1$-chain, and integrating a $2$-form over a $2$-chain, respectively.



But what about if we are given a smooth scalar field and then we do scalar line integral or scalar surface integral? Do they correspond to integrating some differential forms? If not, can we generalise scalar integral in some other ways?










share|cite|improve this question



























    2














    I have read through part of Spivak's Calculus on Manifolds up to the chapter Integration on Chains. Now I know that given a smooth vector field on $Bbb R^3$, doing line integral over a parametrised smooth curve, and doing surface integral over a parametrised smooth surface, correspond to integrating a $1$-form over a $1$-chain, and integrating a $2$-form over a $2$-chain, respectively.



    But what about if we are given a smooth scalar field and then we do scalar line integral or scalar surface integral? Do they correspond to integrating some differential forms? If not, can we generalise scalar integral in some other ways?










    share|cite|improve this question

























      2












      2








      2


      3





      I have read through part of Spivak's Calculus on Manifolds up to the chapter Integration on Chains. Now I know that given a smooth vector field on $Bbb R^3$, doing line integral over a parametrised smooth curve, and doing surface integral over a parametrised smooth surface, correspond to integrating a $1$-form over a $1$-chain, and integrating a $2$-form over a $2$-chain, respectively.



      But what about if we are given a smooth scalar field and then we do scalar line integral or scalar surface integral? Do they correspond to integrating some differential forms? If not, can we generalise scalar integral in some other ways?










      share|cite|improve this question













      I have read through part of Spivak's Calculus on Manifolds up to the chapter Integration on Chains. Now I know that given a smooth vector field on $Bbb R^3$, doing line integral over a parametrised smooth curve, and doing surface integral over a parametrised smooth surface, correspond to integrating a $1$-form over a $1$-chain, and integrating a $2$-form over a $2$-chain, respectively.



      But what about if we are given a smooth scalar field and then we do scalar line integral or scalar surface integral? Do they correspond to integrating some differential forms? If not, can we generalise scalar integral in some other ways?







      multivariable-calculus differential-geometry differential-topology smooth-manifolds differential-forms






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jun 14 '17 at 8:38









      edmedm

      3,6131425




      3,6131425






















          1 Answer
          1






          active

          oldest

          votes


















          6














          The reason differential forms are appropriate objects to integrate over manifolds is that their integrals can be defined without assuming any additional structure on the manifold (except an orientation), and they're invariant under diffeomorphisms.



          However, if you add a Riemannian metric $g$ to your oriented $n$-manifold $M$, then you can define the integral of any continuous function. Here's how that works. The Riemannian metric and the orientation uniquely determine a canonical $n$-form on the manifold, which is called the Riemannian volume form, and frequently denoted by $dV_g$. One way to characterize it is that it is the unique $n$-form that satisfies $dV_g(E_1,dots, E_n) = 1$ whenver $(E_1,dots,E_n)$ is an oriented orthonormal basis for any tangent space. (The notation $dV_g$ is just chosen to echo the standard notation for multiple integrals over subsets of $mathbb R^n$; it's not meant to imply that $dV_g$ is an exact form, which is rarely the case.)



          Having defined the Riemannian volume form, we can now define the integral of a scalar function $f$ over $M$ to be the integral of the $n$-form $f, dV_g$:
          $$
          text{integral of $f$ over $M$} = int_M f, dV_g.
          $$
          This makes sense on any oriented Riemannian manifold provided $f$ is compactly supported
          (or decays fast enough for the integral to converge). (You can also extend this to nonorientable manifolds using densities, but that's not particularly relevant here.)



          The examples you mentioned -- line integrals and surface integrals of scalar functions -- are special cases of this construction. Any smooth curve or smooth surface in $mathbb R^3$ inherits a Riemannian metric by restricting the Euclidean dot product to vectors tangent to the curve or surface. The Riemannian volume form for this metric is typically denoted by $ds$ in the case of a curve and $dA$ in the case of a surface, and if you unwind the calculus definitions of $int_c f,ds$ or $int_S f, dA$, you'll find that it's exactly what I just described.






          share|cite|improve this answer





















            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%2f2322049%2fdo-scalar-integrals-correspond-to-integration-of-differential-forms%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









            6














            The reason differential forms are appropriate objects to integrate over manifolds is that their integrals can be defined without assuming any additional structure on the manifold (except an orientation), and they're invariant under diffeomorphisms.



            However, if you add a Riemannian metric $g$ to your oriented $n$-manifold $M$, then you can define the integral of any continuous function. Here's how that works. The Riemannian metric and the orientation uniquely determine a canonical $n$-form on the manifold, which is called the Riemannian volume form, and frequently denoted by $dV_g$. One way to characterize it is that it is the unique $n$-form that satisfies $dV_g(E_1,dots, E_n) = 1$ whenver $(E_1,dots,E_n)$ is an oriented orthonormal basis for any tangent space. (The notation $dV_g$ is just chosen to echo the standard notation for multiple integrals over subsets of $mathbb R^n$; it's not meant to imply that $dV_g$ is an exact form, which is rarely the case.)



            Having defined the Riemannian volume form, we can now define the integral of a scalar function $f$ over $M$ to be the integral of the $n$-form $f, dV_g$:
            $$
            text{integral of $f$ over $M$} = int_M f, dV_g.
            $$
            This makes sense on any oriented Riemannian manifold provided $f$ is compactly supported
            (or decays fast enough for the integral to converge). (You can also extend this to nonorientable manifolds using densities, but that's not particularly relevant here.)



            The examples you mentioned -- line integrals and surface integrals of scalar functions -- are special cases of this construction. Any smooth curve or smooth surface in $mathbb R^3$ inherits a Riemannian metric by restricting the Euclidean dot product to vectors tangent to the curve or surface. The Riemannian volume form for this metric is typically denoted by $ds$ in the case of a curve and $dA$ in the case of a surface, and if you unwind the calculus definitions of $int_c f,ds$ or $int_S f, dA$, you'll find that it's exactly what I just described.






            share|cite|improve this answer


























              6














              The reason differential forms are appropriate objects to integrate over manifolds is that their integrals can be defined without assuming any additional structure on the manifold (except an orientation), and they're invariant under diffeomorphisms.



              However, if you add a Riemannian metric $g$ to your oriented $n$-manifold $M$, then you can define the integral of any continuous function. Here's how that works. The Riemannian metric and the orientation uniquely determine a canonical $n$-form on the manifold, which is called the Riemannian volume form, and frequently denoted by $dV_g$. One way to characterize it is that it is the unique $n$-form that satisfies $dV_g(E_1,dots, E_n) = 1$ whenver $(E_1,dots,E_n)$ is an oriented orthonormal basis for any tangent space. (The notation $dV_g$ is just chosen to echo the standard notation for multiple integrals over subsets of $mathbb R^n$; it's not meant to imply that $dV_g$ is an exact form, which is rarely the case.)



              Having defined the Riemannian volume form, we can now define the integral of a scalar function $f$ over $M$ to be the integral of the $n$-form $f, dV_g$:
              $$
              text{integral of $f$ over $M$} = int_M f, dV_g.
              $$
              This makes sense on any oriented Riemannian manifold provided $f$ is compactly supported
              (or decays fast enough for the integral to converge). (You can also extend this to nonorientable manifolds using densities, but that's not particularly relevant here.)



              The examples you mentioned -- line integrals and surface integrals of scalar functions -- are special cases of this construction. Any smooth curve or smooth surface in $mathbb R^3$ inherits a Riemannian metric by restricting the Euclidean dot product to vectors tangent to the curve or surface. The Riemannian volume form for this metric is typically denoted by $ds$ in the case of a curve and $dA$ in the case of a surface, and if you unwind the calculus definitions of $int_c f,ds$ or $int_S f, dA$, you'll find that it's exactly what I just described.






              share|cite|improve this answer
























                6












                6








                6






                The reason differential forms are appropriate objects to integrate over manifolds is that their integrals can be defined without assuming any additional structure on the manifold (except an orientation), and they're invariant under diffeomorphisms.



                However, if you add a Riemannian metric $g$ to your oriented $n$-manifold $M$, then you can define the integral of any continuous function. Here's how that works. The Riemannian metric and the orientation uniquely determine a canonical $n$-form on the manifold, which is called the Riemannian volume form, and frequently denoted by $dV_g$. One way to characterize it is that it is the unique $n$-form that satisfies $dV_g(E_1,dots, E_n) = 1$ whenver $(E_1,dots,E_n)$ is an oriented orthonormal basis for any tangent space. (The notation $dV_g$ is just chosen to echo the standard notation for multiple integrals over subsets of $mathbb R^n$; it's not meant to imply that $dV_g$ is an exact form, which is rarely the case.)



                Having defined the Riemannian volume form, we can now define the integral of a scalar function $f$ over $M$ to be the integral of the $n$-form $f, dV_g$:
                $$
                text{integral of $f$ over $M$} = int_M f, dV_g.
                $$
                This makes sense on any oriented Riemannian manifold provided $f$ is compactly supported
                (or decays fast enough for the integral to converge). (You can also extend this to nonorientable manifolds using densities, but that's not particularly relevant here.)



                The examples you mentioned -- line integrals and surface integrals of scalar functions -- are special cases of this construction. Any smooth curve or smooth surface in $mathbb R^3$ inherits a Riemannian metric by restricting the Euclidean dot product to vectors tangent to the curve or surface. The Riemannian volume form for this metric is typically denoted by $ds$ in the case of a curve and $dA$ in the case of a surface, and if you unwind the calculus definitions of $int_c f,ds$ or $int_S f, dA$, you'll find that it's exactly what I just described.






                share|cite|improve this answer












                The reason differential forms are appropriate objects to integrate over manifolds is that their integrals can be defined without assuming any additional structure on the manifold (except an orientation), and they're invariant under diffeomorphisms.



                However, if you add a Riemannian metric $g$ to your oriented $n$-manifold $M$, then you can define the integral of any continuous function. Here's how that works. The Riemannian metric and the orientation uniquely determine a canonical $n$-form on the manifold, which is called the Riemannian volume form, and frequently denoted by $dV_g$. One way to characterize it is that it is the unique $n$-form that satisfies $dV_g(E_1,dots, E_n) = 1$ whenver $(E_1,dots,E_n)$ is an oriented orthonormal basis for any tangent space. (The notation $dV_g$ is just chosen to echo the standard notation for multiple integrals over subsets of $mathbb R^n$; it's not meant to imply that $dV_g$ is an exact form, which is rarely the case.)



                Having defined the Riemannian volume form, we can now define the integral of a scalar function $f$ over $M$ to be the integral of the $n$-form $f, dV_g$:
                $$
                text{integral of $f$ over $M$} = int_M f, dV_g.
                $$
                This makes sense on any oriented Riemannian manifold provided $f$ is compactly supported
                (or decays fast enough for the integral to converge). (You can also extend this to nonorientable manifolds using densities, but that's not particularly relevant here.)



                The examples you mentioned -- line integrals and surface integrals of scalar functions -- are special cases of this construction. Any smooth curve or smooth surface in $mathbb R^3$ inherits a Riemannian metric by restricting the Euclidean dot product to vectors tangent to the curve or surface. The Riemannian volume form for this metric is typically denoted by $ds$ in the case of a curve and $dA$ in the case of a surface, and if you unwind the calculus definitions of $int_c f,ds$ or $int_S f, dA$, you'll find that it's exactly what I just described.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jun 15 '17 at 0:00









                Jack LeeJack Lee

                27.1k54565




                27.1k54565






























                    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.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • 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.


                    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%2f2322049%2fdo-scalar-integrals-correspond-to-integration-of-differential-forms%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

                    Plaza Victoria

                    Puebla de Zaragoza

                    Musa