Do scalar integrals correspond to integration of differential forms?
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
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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
add a comment |
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
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
multivariable-calculus differential-geometry differential-topology smooth-manifolds differential-forms
asked Jun 14 '17 at 8:38
edmedm
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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.
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1 Answer
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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.
add a comment |
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.
add a comment |
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.
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.
answered Jun 15 '17 at 0:00
Jack LeeJack Lee
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