# Cohomology and left invariant 1-forms

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 $$\{\eta^i\}$$