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Differential forms question

  1. Nov 28, 2004 #1
    I'm just learning about differential forms and I've noticed something in my homework assignment. We have to evaluate
    zdx + xdy + ydz, over directed line segments in R-three by the method of pullback. Let a, b, and c be vectors in R-three. I noticed that Integral from a to c does NOT equal integral from a to b + integral from b to c, as it does with normal integrals. I think that this would make sense, since the meaning of zdx +xdy +ydz, which you are integrating, depends on the line segment in question. Is it true that this rule does not apply with these kinds of integrals, or have I simply made a mistake somewhere in my calculations? Thanks.
  2. jcsd
  3. Nov 29, 2004 #2


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    you are right! there is however a special class of integrals, i.e. differential forms, that DO give the same integral over any path joining the same two points. these are called "exact" differentials, and are precisely those of form df for some function in that region, i.e. a gradient.

    another related concept is of a "closed" differential, one such that its curl is zero. these are in fact the same as the exact differentials in any "simply connected" region.

    thus to measure how far a region is from being simply connected, one can ask how many closed differentials fail tro be exact.

    for example, if we remove n points from the plane, there will be exctly an n dimesnional vector space of closed forms in that region, if we consider all exact forms to be zero.

    \this measuring device isa big tool in topology called rerham cohomology.

    so you have just noticed one of the most imporatnt question in the subject!

    work done by gravity for example is exact so does not depend on the path taken by the object.
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