An indecomposable Lie group with Riemannian bi-invariant metric is always simple and hence Einstein. Indefinite bi-invariant metrics are not necessarily Einstein, not even on simple Lie groups. We study the question of whether a semi-Riemannian bi-invariant metric is conformal to an Einstein metric. We obtain results for all three cases in the structure theorem by Medina and Revoy for indecomposable metric Lie algebras: the case of simple Lie algebras, and the cases of double extensions of metric Lie algebras by ℝ or a simple Lie algebra. Simple Lie algebras are conformally Einstein precisely when they are Einstein, or when equal to \( {\mathfrak{sl}}_2\mathrm{\mathbb{C}} \) and conformally flat. Double extensions of metric Lie algebras by simple Lie algebras of rank greater than one are never conformally Einstein, and neither are double extensions of Lorentzian oscillator algebras, whereas the oscillator algebras themselves are conformally Einstein. Our results give a complete answer to the question of which metric Lie algebras in Lorentzian signature and in signature (2, n − 2) are conformally Einstein.

具有黎曼双不变度量的不可分解的李群总是很简单，因此也很容易成为爱因斯坦。不确定的双不变度量不一定是爱因斯坦，甚至在简单的李群上也是如此。我们研究半黎曼双不变度量是否符合爱因斯坦度量的问题。我们在Medina和Revoy的结构定理中针对不可分解的度量Lie代数获得了所有这三种情况的结果：简单Lie代数的情况，以及度量Li或简单Lie代数的双重Lie代数的情况。简单的Lie代数恰好在它们是爱因斯坦时或在等于\（{\ mathfrak {sl}} _ 2 \ mathrm {\ mathbb {C}} \）时都是保形的爱因斯坦。并保形平坦。度量Lie代数被秩大于1的简单Lie代数的双重扩展永远不是保形的爱因斯坦，也不是Lorentzian振荡代数的双重扩展，而振荡代数本身就是保形的爱因斯坦。我们的结果给出了一个完全答案，即洛伦兹签名和签名（2，n -2）中的哪个度量李代数是保形爱因斯坦的。