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Jul5-05, 07:26 AM
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#2
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Hurkyl is
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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.
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Jul5-05, 01:10 PM
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#3
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mathwonk is
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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.
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Jul5-05, 01:11 PM
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#4
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mathwonk is
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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.
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Jul6-05, 12:11 AM
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Last edited by kakarukeys; Jul6-05 at 12:15 AM..
#5
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kakarukeys is
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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.
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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  be the coordinates, suppose  is the boundary.
Let  be the positive orientation.
then since  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

implies
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.
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Jul6-05, 01:27 AM
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#6
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mathwonk is
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did my comments mean nothing at all?
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Jul6-05, 01:28 AM
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#7
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mathwonk is
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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.
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Jul6-05, 03:27 AM
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#8
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kakarukeys is
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Yes your comments explained the basic idea behind the proof.
I am still struggling with the steps of the proof.
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Jul6-05, 05:43 AM
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#9
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kakarukeys is
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ya sure. If you have notes, please send to me
kakarukeys@gmail.com
thank you.
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Jul6-05, 01:06 PM
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Last edited by mathwonk; Jul6-05 at 03:35 PM..
#10
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mathwonk is
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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"
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Jul6-05, 01:10 PM
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#11
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mathwonk is
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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.
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Jul6-05, 11:53 PM
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#12
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mathwonk is
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did i answer your question? (about the matrices) in the first half of post 10?
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Jul8-05, 05:59 AM
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#13
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kakarukeys is
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Originally Posted by mathwonk
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.
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Please don't if it is too laborious, I don't want to waste your precious time.
Originally Posted by mathwonk
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.
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In my case, I only have
There may not be zeroes in the last row and last column.
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Jul8-05, 10:17 AM
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#14
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mathwonk is
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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?
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Jul10-05, 02:10 AM
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Last edited by kakarukeys; Jul10-05 at 02:18 AM..
#15
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kakarukeys is
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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?
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Jul10-05, 02:20 AM
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#16
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kakarukeys is
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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.
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