For an arbitrary Lie group, is it always possible to chose a left-invariant Riemannian metric such that the Laplace-Beltrami operator $\Delta$ is given by 


$$\Delta f = \delta^{i j} X\_i X\_j f$$


for some orthonormal frame $\{X\_i\}$ of Lie vector fields?


Essentially, can we chose a left-invariant metric and orthonormal Lie frame such that the Christoffel symbols with respect to this frame satisfy $\delta^{ij} \Gamma\_{i j}^{~~~k}=0$ for all indices $k$?


In terms of structure constants, we want a frame of Lie vector fields such that the structure constants $\alpha\_{ij}^{~~~k}$ for this frame satisfy $\alpha\_{ij}^{~~~j}=0$ (summation implied) for all indices $i$. Then we can define a left-invariant metric that makes this frame orthonormal via pullbacks as usual.


I know that this is possible on $\mathbb{R}^n$ and the Heisenberg group, but is it possible on any Lie group? Is some additional hypothesis on the Lie group (such as unimodularity) required for this to hold?



Laplace-Beltrami operator on a Lie group
As José Figueroa-O'Farrill pointed out in his comment above, the condition $\alpha\_{i j}^{~~~j}$ is equivalent to unimodularity. So we can write the Laplace-Beltrami operator in this form iff the Lie group is unimodular.


