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Quotient set of equivalence class in de Rham cohomology

  1. Jun 5, 2012 #1
    Hi all,

    So the equivalence class [itex]X/\sim[/itex] is the set of all equivalences classes [itex][x][/itex]. I was wondering if there was a way of writing it in terms of the usual quotient operation:
    [tex]G/N=\{gN\ |\ g\in G\}?[/tex]

    From what I've read, it would be something like [itex]X/\sim = X/[e][/itex]. But, since I'm looking at the de Rham cohomology group [itex]H^k = \{ closed\ k\ forms\}/\sim[/itex] so
    [tex]H^k = \{ closed\ k\ forms\}/[0] = \{ \omega [0]\ |\ \omega\ is\ a\ closed\ form\}[/tex]
    doesn't work, as the the operation [itex]\omega [0][/itex] doesn't seem to make sense.

    It's also defined [itex]H^k = Z^k/B^k[/itex] if you're familiar with that notation.

    Any thoughts?

    Last edited: Jun 5, 2012
  2. jcsd
  3. Jun 7, 2012 #2
    Ideas anyone?
  4. Jun 13, 2012 #3
    Quotients of vector spaces can be written E/F = {x+F | x∈E} as the group law involved in such quotients is the law of vector addition.
    Any problem ?
  5. Jun 13, 2012 #4
    Yes, still a problem. First, [itex]B^k = [0] [/itex], which I should have mentioned before. Second, by your definition
    [tex] H^k = Z^k/B^k = \{z + B^k\ |\ z\in Z^k\} [/tex]

    But this appears to me to include no information about the equivalence relation [itex]\sim[/itex], which states that [itex]u[/itex] and [itex]v[/itex] are equivalent if there exists some [itex]w[/itex] such that [itex]u - v = dw[/itex].

    Ok, but since each equivalence class can be defined by its specific [itex]dw[/itex] (and [itex]d[/itex] is a unique (one-to-one???) map) then it is defined by its [itex]w[/itex]. Hence it makes more sense that
    [tex] H^k = Z^k/B^k = \{z - B^k\ |\ z\in Z^k\}. [/tex]

    If [itex]z - b = dw[/itex], then [itex]z= d(w+a)[/itex] since [itex]b=da \in B^k[/itex], which also implies that [itex]z\in B^k[/itex]. But what if there is no [itex]w[/itex] such that [itex]z-b = dw[/itex]? How does this give the rest of the equivalence classes?
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