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I'm computing the de Rham cohomology of the group $SU(2)$, with $n_g$ generators, making use of the base of left invariant 1-forms $\eta^i, i = \{1, ..., n_g\}$, in order to apply the following relation:

$$d\eta^i = -\frac{1}{2}c_{ijk}\eta^j\wedge\eta^k\tag1$$

where $c_{ijk}$ are the structure constants of the algebra $\mathfrak su(2)$. So with Eq. (1) you can see what p-forms are closed and exact and compute the de Rham left invariant cohomology, $H_L^p$, that is isomorphic to the de Rham cohomology, $H^p$, because $SU(2)$ is connected and compact.

But due to $\{\eta^i\}$ are 1-form,

1) How can I compute $H_L^0$? I know that $B^0_L = Im(d_{-1})$ is zero by definition, but is this zero a matrix or not? What about $Z^0_L = ker(d)$)?
2) If the set of generators is the equivalence of tangent vectors and $\{\eta^i\}$ the equivalence of $\{dx^i\}$, the cotangent vectors, what is $\{\eta^i\}$? My question arises from the fact that we can find explicit expressions for generators but I've never seen the same for the set $\{\eta^i\}$

Thanks in advance!

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