Proof of $H^2(\mathfrak g_1,\mathfrak g_1)=0$

Question

I want to prove $H^2(\mathfrak g_1,\mathfrak g_1)=0$ where $\mathfrak g_1=\mathbb R$, the $1$-dimensional (abelian) Lie algebra.

I think I need to show two things:

1. All central extensions $\mathbb R \overset{\iota}{\hookrightarrow} \mathbb R^2 \overset{\pi}{\twoheadrightarrow} \mathbb R$ are equivalent to each other.
2. There isn't any $\mathbb R \overset{\iota}{\hookrightarrow} \mathfrak{aff}(1,\mathbb R) \overset{\pi}{\twoheadrightarrow}\mathbb R$ central extension.

Originally, I thought that there isn't $\mathbb R \overset{\iota}{\hookrightarrow} \mathfrak{aff}(1,\mathbb R) \overset{\pi}{\twoheadrightarrow}\mathbb R$ extension at all, but it isn't true. I see now that the extension I found isn't central, but I have to prove that none of the $\mathbb R \overset{\iota}{\hookrightarrow} \mathfrak{aff}(1,\mathbb R) \overset{\pi}{\twoheadrightarrow}\mathbb R$ extensions are central. But I have no idea how to prove this and point 1.

Answer 1

Note that the center of the affine Lie algebra $L={\rm aff}(n,\Bbb R)$ is zero, since this algebra is complete, i.e., we have $$ 0=H^0(L,L)=Z(L),\; H^1(L,L)={\rm Out}(L)=0. $$ So there cannot be an injective map from $\Bbb R$ to the center of $L$. It is well known that complete Lie algebras $L$ with solvable radical ${\rm rad}(L)$ being abelian satisfy $$ H^n(L,L)=0, \; \forall \; n\ge 0. $$ For a reference see here. In particular, $H^2(L,L)=0$, so that $L$ is "rigid".

However, you have asked about the second adjoint cohomology for a $1$-dimensional Lie algebra, i.e., $\mathfrak{g}_1$ is not ${\rm aff}(1,\Bbb R)$, but $\Bbb R$. There we have $H^2(\Bbb R,\Bbb R)=0$, because in general $H^n(L,M)=0$ for $n>\dim (L)$.