Suppose that we have a complex Lie algebra $\mathfrak g$. My (topology:)) book says (author gives this construction as an abstract example of cochain complex) that we can construct a cochain complex $C$: $C^n = \text{Hom}(\bigwedge^n\mathfrak g, \mathbb C)$ and $d: C^n \to C^{n + 1}, \omega \to d\omega$, where
$d\omega(x_1,$ $...$ $, x_{n +1}) = \sum\limits_{1 \leq i < j \leq n + 1} (-1)^{i + j + 1} \omega([x_i, x_j],$ $x_1,$ $\ldots$ , $\hat{x_i}, \ldots$ $\hat{x_j}$, $\ldots$, $x_n$)
(here the hat over a symbol means that it's omitted).
After that the author proofs that $d$ is a differential and introduces cohomologies of Lie algebra.
It's all nice and simple... But he doesn't say anything about $H^0$ (the definition that was given above isn't correct for $d_0$, of course).
So my question is: how to calculate $H^0(\mathfrak g)$?
The definition is correct, you just have to interpret $d_0$ as the zero map. Hence $H^0(\mathfrak g) = \Bbb C$. That looks tautological but keep in mind cohomology makes sense for any $\mathfrak g$-module $M$, and in this case we have $H^0(\mathfrak g, M) = M^{\mathfrak g} := \{m \in M : xm = 0, \forall x \in \mathfrak g\}$. This is because the differential is a bit more complicated in general, but it simplifies here because your module is the trivial module, that is $xm = 0$ for all $x \in \mathfrak g, m \in \Bbb C$. For more information, look at "Chevalley-Eilenberg complex".