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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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