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Orientable Manifold with Boundary

  1. Jul 5, 2005 #1
    How does the orientation on M induce an orientation on the boundary of M?

    I follow the book Lectures on Differential Geometry by Chern, do not understand the proof.

    The proof is
    the Jacobian Matrix of the transformation between coordinates of two charts has positive determinant (oriented charts), so the smaller Jacobian Matrix with one row and one column deleted (corresponding to the only one coordinate axis that runs away from the boundary) has positive determinant.

    Please do me a favour by explaining the proof clearly, or give me another easier proof.
     
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  3. Jul 5, 2005 #2

    Hurkyl

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    Hrm.

    I would have said "Because it does in Euclidean space", but I don't know how easy it is to prove you can patch together orientations of the pieces of the boundary when it spans multiple charts.
     
  4. Jul 5, 2005 #3

    mathwonk

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    how does counter clockwise orientation on the upper half plne induce an orientation on the real line?

    by noticing that at a point of the real line, one can distinguish between an arrow pointing into the upper half plane and one pointing out of it.

    so a vector v along the x axis is oriented positively if, when it is supplemented by an arrow w pointing into the upper half plane, we get an oriented (counterclockwise) basis <v,w> for the (tangent space to the) upper half plane.
     
  5. Jul 5, 2005 #4

    mathwonk

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    oh, that is the euliden space acse. but an oriented manifold is by definition one that admits an orientation preserving atlas, so those charts do preserve the euclidean orientation.
     
  6. Jul 5, 2005 #5
    Suppose we have charts that cover regions not containing the boundary and charts that cover regions containing the boundary. They are oriented (coordinate transformations among them have positive-determinant Jacobian matrix).

    for a region D, let [tex]{u^1, u^2, ...., u^m}[/tex] be the coordinates, suppose [tex]u^m = 0[/tex] is the boundary.

    Let [tex]du^1 \wedge du^2 \wedge ... \wedge du^m[/tex] be the positive orientation.

    then since [tex]du^1 \wedge du^2 \wedge ... \wedge du^{m-1}[/tex] is non-zero and gives an orientation on the boundary contained in region D. Minus sign gives another orientation.

    if we can transform to other charts on boundary, preserving the orientation, we can extend the orientation to the entire boundary.

    And thus we need to show

    [tex]\frac{\partial (v^1, ..., v^m)}{\partial (u^1, ..., u^m)} > 0[/tex]
    implies
    [tex]\frac{\partial (v^1, ..., v^{m-1})}{\partial (u^1, ..., u^{m-1})} > 0[/tex]

    I don't understand the proof of the above in the book, anyone has any idea?
    I can type the proof out, if you want.
     
    Last edited: Jul 5, 2005
  7. Jul 6, 2005 #6

    mathwonk

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    did my comments mean nothing at all?
     
  8. Jul 6, 2005 #7

    mathwonk

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    another suggestion: just ebcause chern is a great mathematiciazn does not mean he writes learnable boks. read guillemin and pollack instead. or my notes. i will send them to you if you give me an address.
     
  9. Jul 6, 2005 #8
    Yes your comments explained the basic idea behind the proof.
    I am still struggling with the steps of the proof.
     
  10. Jul 6, 2005 #9
  11. Jul 6, 2005 #10

    mathwonk

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    the point is that if you have an n-1 dimensional subspace spanned by the first n-1 vectors in a basis, and then the last vector points in a giuven direction, if you choose another such absis, the change of basis matrix will be in 2 blocks.

    there will be an n-1 by n-1 block in the upper left corner say, and a positive number in the lower right corner, and all zeroes otherwise in the last row and last column.

    hence the determinant of the whole matrix being positive forces also the determinant of the upper left block to be positive also.

    but i warn you, chern's writings are the worlds most terse and merciless. they are also non conceptual and highly compoutational. he was a great great mathematician but i do not recommend his works to anyone as the sole source for learning any subject at all.

    here is a review by another outstanding mathematician:
    ""This excellent and polished book may not be suitable for the very beginning student, but it is highly recommended for all mathematicians, from the advanced undergraduate student to the experienced professor. For many mathematicians it will be a work of reference for their research. It will be welcome by physicists."

    Prof. F Hirzebruch
    Max-Planck Institute, Bonn"
     
    Last edited: Jul 6, 2005
  12. Jul 6, 2005 #11

    mathwonk

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    i will try to pdf my notes and make them sendable. but they are just a rewrite of the treatment in guillemin and pollack which may be in your library.
     
  13. Jul 6, 2005 #12

    mathwonk

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    did i answer your question? (about the matrices) in the first half of post 10?
     
  14. Jul 8, 2005 #13
    Please don't if it is too laborious, I don't want to waste your precious time.

    In my case, I only have
    [tex]\frac{\partial v^m}{\partial u^m} > 0[/tex]
    [tex]\frac{\partial (v^1, ..., v^m)}{\partial (u^1, ..., u^m)} > 0[/tex]

    There may not be zeroes in the last row and last column.
     
  15. Jul 8, 2005 #14

    mathwonk

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    well if you think about the definition of a matrix of a linear transformation, the last row is hte image of the last vector, expanded in terms of the basis ofm the target space.

    now we have chsoen both bases to end with a vector pointing into the manifold, so i guess al i can say is that the last entry of the last comloumn is positive, since the component of an inner pointing vector is a positive multiple of another inner pointing vector.

    but if you look at the previous columns, you are expanding the images of the basis vectors for the boundaryu space, in terms of anmother absis for that same subspace. so there is no need to use the last inner pointing vector in that expansion. so every columnh except the last has a zero in the last entry.

    thus the last row is all zeroes except for the alst right bottom corner. then the same argument applies. i.e. the determinant is positive iff the determinant of the upper left block is positive. how about that?
     
  16. Jul 10, 2005 #15
    your argument is sound unless the following is true.

    a cotagent space at a point on the boundary is a direct sum of the cotagent space of that point when the boundary is considered an independent manifold and its complement.

    Every first m - 1 basis vectors given by any chart span the first subspace, and the last vector pointing into the manifold always spans the other subspace.

    I think it's true by definition of the chart. or is it true?
     
    Last edited: Jul 10, 2005
  17. Jul 10, 2005 #16
    only then can it be that

    when we transform from u basis to v basis,
    the first m-1 vectors are shuffled.
    and the last vector is only multiplied by a constant.
     
  18. Jul 10, 2005 #17
    Ok, forget about the cotangent space direct sum.
    Here's the argument

    We know given two charts, at a point on boundary
    the first m - 1 cotangent vectors obtained [tex]\{du^1, .... du^{m-1}\}, \{dv^1, .... dv^{m-1}\}[/tex]
    span the cotangent space of the boundary at the point, by definition of the chart.

    so entries of last column of the m by m Jacobian Matrix we considered are all zeros except the last one. (Because [tex]\{dv^1, .... dv^{m-1}\}[/tex] can be expressed solely as a linear combination of [tex]\{du^1, .... du^{m-1}\}[/tex]).
    And so the conclusion follows.

    P/S the last vector [tex]dv^m[/tex] need not be multiple of [tex]du^m[/tex], it can contains some linear combination of [tex]\{du^1, .... du^{m-1}\}[/tex].
     
    Last edited: Jul 10, 2005
  19. Jul 10, 2005 #18

    mathwonk

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    it depends on your definition of the complementary space. some people take the normal space to be the quotient of the full tangent space by the tangent space to the boundary. when you do that lkinear combinations of baoiundary tangent vectors are set to zero, so the normal vectors are scalar multiples of each other in that quotient space. that is the point of view in guillemin pollack, and hence also in the pdf notes i sent you.

    but you seem to understand it now.
     
  20. Jul 14, 2005 #19
    I thought I understand, but when I tried a simple example, I couldn't get what I want.

    Consider the Manifold, closed interval [0, 1]

    Using two charts:
    [0, 1) -> [0, 1) [tex]\phi_1(x) = x[/tex]

    (0, 1] -> [0, 1) [tex]\phi_2(x) = -x + 1[/tex]

    The first chart is positively oriented, the second chart is negatively oriented.

    So the orientation of the boundary point is +1 and -1 respectively.

    since [tex]+1 \wedge dx = dx =[/tex] gives the positive orientation on the first chart.

    since [tex]-1 \wedge dx = -dx [/tex] gives the "positive orientation" on the second chart.

    then applying Stokes' theorem we have,

    [tex]\int^1_0 df = \int_{0, 1} f(x) = f(0) - f(1)[/tex]

    since the 2nd chart is negatively oriented.
     
  21. Jul 14, 2005 #20

    The above example above doesn't count since you've chosen an atlas that does not induce an orientation on the interior manifold. Orientability implies there exists an atlas with coordinate charts whose Jacobians have pos. determinant, but it doesn't mean that any atlas of the manifold will have that property -- as you can see above. In order to use Stoke's theorem you need to settle on a particular orientation (you do after all have a choice of two) and use that.
     
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