Quotient set of equivalence class in de Rham cohomology

[ianhoolihan]

Hi all,

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

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

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

Any thoughts?

Cheers

 

[ianhoolihan]

Ideas anyone?

 

[spoirier]

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 ?

 

[ianhoolihan]

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 ?

Yes, still a problem. First, B^k = [0], which I should have mentioned before. Second, by your definition
H^k = Z^k/B^k = \{z + B^k\ |\ z\in Z^k\}

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

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

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