Orientable Manifold with Boundary

  • Thread starter kakarukeys
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  • #26
mathwonk
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well you are asking the world's most difficult question, getting minus signs correct. anyway the case of a one dimensional manifold is a little special, because orientation is usually defined by choosing an ordered basis of the vector space.

now if the space has positive dimension then every ordered basis can be changed by replacing the first vector by minus itself to get the opposite orientation. the boundary of a one dimensional manifold however has dimesnino zero, and there is only one basis of that space, the empty basis.

so in that one case the definition of orientation is different, just the choice of +1 or -1.

now in the usual case we define orientation of a boundary by saying, take any basis for the boundary such that preceding it by an outward pointing normal vector gives an oriented basis of the manifold.

now if we precede the empty set by an outward pointing normal vector, we do get an oriented basis for the manifold [0,1] at the point 1 but not at the point 0.

so perhaps we should say that preceding by the number +1 means keeping the orientation and preceding by the number -1 means changing it.

then we would get that the correct orientation on the boundary of [0,1] is +1 at 1, and -1 at 0, since the outward normal is positively oriented at 1 but not at 0.

notice the arguments about determinants of matrices make no sense in this case, and assume that the dimension of the boundary is at least one.
 
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  • #27
mathwonk
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and it makes no sense to ask to prove FTC from stokes, since stokes equals the FTC in dimension one.

oh i see you are asking why your version of stokes is giving the wrong answer to FTC, well the only possibility is that your procedure for deciding the orientation is backwards, as indicated above.
 
  • #28
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I got it.
Thanks to both of you.
 

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