- #1
ianhoolihan
- 145
- 0
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?
Cheers
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?
Cheers
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