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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?

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Unless I'm mistaken, your condition $\alpha_{ij}{}^j=0$ is precisely unimodularity, assuming that this notation means you sum over $j$ for each $i$. – José Figueroa-O'Farrill Aug 10 '14 at 0:23
Wow you are right! I didn't think of that for some reason. Thanks so much for pointing that out. – ಠ_ಠ Aug 10 '14 at 1:04
up vote 1 down vote accepted

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.

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