# Quotient set of equivalence class in de Rham cohomology

1. Jun 5, 2012

### 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

Last edited: Jun 5, 2012
2. Jun 7, 2012

### ianhoolihan

Ideas anyone?

3. Jun 13, 2012

### 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 ?

4. Jun 13, 2012

### ianhoolihan

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?