0

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.

mma
  • 2,165

1 Answers1

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 $n>\dim (L)$.

Dietrich Burde
  • 140,055
  • Thanks. And what about point 1? – mma Jan 17 '23 at 09:22
  • Point 1 follows from $H^2(\Bbb R,\Bbb R)=0$, so there is only the trivial equivalence class of central extensions of $\Bbb R$ by $\Bbb R$. – Dietrich Burde Jan 17 '23 at 17:08
  • Complete Lie algebras are new for me and I couldn't find anything about them in a textbook or Wikipedia. However I found a definition here as "A Lie algebra L is called a complete Lie algebra if its centre C(L) is zero and its derivations are all inner", while ChatGpt told that A Lie algebra is said to be "complete" when it is isomorphic to its own complexification." Could you show a resource that clarifies this? – mma Jan 18 '23 at 05:09
  • And could you please add a reference for the statement $H^n(L,M)=0$ for $n>\dim (L)$? – mma Jan 18 '23 at 05:42
  • 1
    The standard complex for Lie algebra cohomology has the space of cochains $C^k(L,M)={\rm Hom}_K(\Lambda^k(L),M)$, where the exterior algebra $\Lambda^k(L)$ has the dimension $\binom{n}{k}$, for $n=\dim (L)$. This is zero for $k>n$. This can be found in any reference listed in wikipeda. For complete Lie algebras you have the references given in the paper you have linked. The word "complete" also depends on the exact context, e.g., for complete metric spaces. – Dietrich Burde Jan 18 '23 at 08:54