Is there a Kunneth Formula for Lie algebra cohomology?
Specifically, I want the second cohomology of the Lie algebra $\mathbb{R}\oplus \mathfrak{sl}(2,\mathbb{C})$ to vanish. I would like to say
$H^2\big(\mathbb{R} \oplus \mathfrak{sl}(2,\mathbb{C}), \mathbb{R}\big) \approx \big[H^2(\mathbb{R}, \mathbb{R}) \otimes H^0\big(\mathfrak{sl}(2,\mathbb{C}), \mathbb{R} \big)\big] \oplus \big[H^1(\mathbb{R}, \mathbb{R}) \otimes H^1\big(\mathfrak{sl}(2,\mathbb{C}), \mathbb{R} \big)\big]\oplus \big[H^0(\mathbb{R}, \mathbb{R}) \otimes H^2\big(\mathfrak{sl}(2,\mathbb{C}), \mathbb{R} \big)\big]$.
Then I should be able to conclude that this is zero because each direct sum term is zero, right?
There is indeed a Künneth formula for cohomology of Lie algebras. If $\mathfrak g$ and $\mathfrak h$ are two Lie algebras over a field $k$, there is an isomorphism $$H^\bullet(\mathfrak g\times\mathfrak h,k)\cong H^\bullet(\mathfrak g,k)\otimes H^\bullet(\mathfrak h,k)$$ of graded vector spaces (indeed, of graded algebras) Here all coefficient modules have trivial action.
How to prove this? Well, this is Theorem 3.1 in Chapter XI of the book on homological algebra by Cartan and Eilenberg: the isomorphism above is a special case of the fact that their product $\vee$ is an isomorphism in this situation. It does take a certain amount of decoding what they write to see this, of course. The observation that makes this work is that the trivial $(\mathfrak g\times\mathfrak h)$-module is isomorphic to the tensor product of the trivial $\mathfrak g$-module and the trivial $\mathfrak h$-module.