Undergrad Any surface bounded by the same curve in Stokes' theorem

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In Stokes' theorem, the closed line integral of a vector field equals the surface integral of its curl over any surface bounded by the same curve, highlighting the flexibility of curves in enclosing multiple surfaces. In contrast, Gauss' theorem relates the surface integral of a vector field to the volume integral of its divergence over a unique volume, emphasizing the uniqueness of volumes enclosed by surfaces in three-dimensional space. This asymmetry arises because, in three-dimensional space, a two-dimensional surface uniquely bounds a volume, a characteristic feature of R^3. The general Stokes formula remains symmetric across different dimensions, indicating that the properties of manifolds are not dependent on their embedding. The discussion also touches on the possibility of a two-dimensional surface enclosing multiple three-dimensional volumes, likening it to a one-dimensional curve enclosing multiple two-dimensional surfaces.
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In Stokes' theorem, the closed line integral of f=the surface integral of curl f on ANY surface bounded by the same curve. But in Gauss' theorem, the surface integral of f on a surface=the volume integral of div f on a unique volume bounded by the surface. A surface can only enclose 1 volume whereas a curve can enclose many surfaces. So why is the asymmetry?
 
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it is because in ##\mathbb{R}^3## if a 2D surface bounds some volume then this volume is unique. It is just a feature of ##\mathbb{R}^3##. The general Stokes formula ##\int_{\partial M}\omega=\int_Md\omega## is symmetric. And the manifold ##M## is not obliged to be embedded anywhere
 
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wrobel said:
it is because in ##\mathbb{R}^3## if a 2D surface bounds some volume then this volume is unique. It is just a feature of ##\mathbb{R}^3##. The general Stokes formula ##\int_{\partial M}\omega=\int_Md\omega## is symmetric. And the manifold ##M## is not obliged to be embedded anywhere
I'm curious on this property of ##\mathbb R ^3##. Do you mean that its homology is trivial?
 
wrobel said:
it is because in ##\mathbb{R}^3## if a 2D surface bounds some volume then this volume is unique. It is just a feature of ##\mathbb{R}^3##. The general Stokes formula ##\int_{\partial M}\omega=\int_Md\omega## is symmetric. And the manifold ##M## is not obliged to be embedded anywhere
Could you please explain why this asymmetry (1D-2D and 2D-3D) is in the language of calculus or on a similar level?
 
feynman1 said:
Could you please explain why this asymmetry (1D-2D and 2D-3D) is in the language of calculus or on a similar level?
If you live in 2d world there only one 2d-volume is contained inside a loop of a closed curve. But if you exit in 3d then you see that there are a lot of two dimensional films pulled up on this curve. The same if you come out from ##\mathbb{R}^3## to ##\mathbb{R}^4##. The general Stokes formula does not care in which ##\mathbb{R}^m## you live and which metric you use to consider vectors instead of differential forms. Have a patience it will be explained in more advanced courses later
 
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wrobel said:
If you live in 2d world there only one 2d-volume is contained inside a loop of a closed curve. But if you exit in 3d then you see that there are a lot of two dimensional films pulled up on this curve. The same if you come out from ##\mathbb{R}^3## to ##\mathbb{R}^4##. The general Stokes formula does not care in which ##\mathbb{R}^m## you live and which metric you use to consider vectors instead of differential forms. Have a patience it will be explained in more advanced courses later
I can think of only 1 possibility where a 2d surface can enclose multiple 3d volumes: the 3d volume inside or that outside (cavity). Does that count as an analogue of 1d curve enclosing multiple 2d surfaces?
 
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