Let $M$ be a (pseudo-)Riemannian manifold with metric $g_{ab}$. Let $\nabla_a$ be the Levi-Civita connection on $M$. It's well-known that the Laplace—Beltrami operator can be defined in this context as $$\nabla^2 = \nabla^a \nabla_a = g^{ab} \nabla_a \nabla_b$$ where $g^{ab}$ is the dual metric and repeated indices are summed. However, we also have the coordinate formula $$\nabla^2 \phi = \frac{1}{\sqrt{| \det g |}} \partial_a \left( \sqrt{| \det g |} g^{ab} \partial_b \phi \right)$$ which, as I understand, comes from using the formula for the Hodge dual.

Without invoking advanced machinery, what is the easiest way to directly prove the equivalence of the two definitions? I can see that if the partial derivatives of $g_{ab}$ vanish, then $$\nabla_{a} \left( g^{ab} \nabla_b \phi \right) = \partial_a \left( g^{ab} \partial_b \phi \right) = \frac{1}{\sqrt{| \det g |}} \partial_a \left( \sqrt{| \det g |} g^{ab} \partial_b \phi \right)$$ and in the general case it suffices to prove that $$\Gamma^{b}_{\phantom{b}ab} = \frac{1}{\sqrt{| \det g |}} \partial_a \left( \sqrt{| \det g |} \right)$$ but then it is seems to be necessary to compute the derivative of a determinant. Is there a trick which can be used to avoid this calculation?

  • $\begingroup$ I don't know of a trick to avoid this, but (from a linear algebra or introductory differential geometry class) you probably know how to compute the derivative of $det$ - right? $\endgroup$ – Gerben May 2 '11 at 18:17
  • $\begingroup$ @Gerben: Yes, I can finish the proof that way. But I think it's not unfair to say that the derivative of $\det$ is esoteric knowledge in the context this problem arose. (It came up in a undergraduate general relativity exam, and I suspect candidates are not expected to be analysts.) $\endgroup$ – Zhen Lin May 2 '11 at 18:59
  • $\begingroup$ I didn't see this comment before I wrote my answer. Still, I think it is quite reasonable to expect people to know how to differentiate the determinant; it is after all apparent from Cramer's rule. $\endgroup$ – Glen Wheeler May 3 '11 at 19:54

For general relativity students I think this might be a reasonable proof. One needs to know that your second expression for $\nabla^2\phi$ (the one using $\sqrt{|\det g|}$) is independent of the choice of coordinates (*). For any point you can then choose coordinate system such that $\partial_a g_{bc}=0$ and the two formulas (as you notice) give the same result.

(*) means (as you also notice - you just want us to restate it in a simpler language - and I'm not sure whether I succeed to do it) that if $u^a$ is a vector-valued density, i.e. if $u^a=v^a \sqrt{|\det g|}$ where $v^a$ is a vector field, then $\partial_a u^a$ is a density, i.e. $\partial_a u^a=f\sqrt{|\det g|}$ for some function $f$ (with $f$ independent of the choice of coordinates - which implies $f=\nabla_av^a$). If we really want to avoid differential forms then we can invoke Gauss theorem (in coordinates), notice that the flow of $u^a$ though a hypersurface is independent of the coordinates, and hence its divergence $\partial_a u^a$ is a well-defined (independent of coordinates) density.

edit: here is (really the same, but) a bit more sensible argument why $f$ (see above) is independent of coordinates. If $\psi$ is a function with compact support then $\int (\partial_a\psi)\, v^a \sqrt{|\det g|} dx^1\dots dx^n$ is independent of the choice of coordinates, and it is equal to (by per partes) $$-\int \psi \partial_a\Bigl( v^a \sqrt{|\det g|}\Bigr)\sqrt{|\det g|}^{-1}\,\sqrt{|\det g|} dx^1\dots dx^n$$ which shows (by choosing $\psi$ with smaller and smaller support and with $\int\psi\sqrt{|\det g|} dx^1\dots dx^n=1$) that $$f=\partial_a\Bigl( v^a \sqrt{|\det g|}\Bigr)\sqrt{|\det g|}^{-1}$$ is independent of coordinates.

  • $\begingroup$ Your first paragraph seems like the most plausible intended solution. I can't help but feel it's circular reasoning though — in some sense, this proof is how we know the second formula is independent of the choice of coordinates, isn't it? $\endgroup$ – Zhen Lin May 2 '11 at 19:57
  • $\begingroup$ @Zhen Lin: you are certainly right - I tried to mumble something in the second paragraph why the formula is independent of coordinates, and now I added (+-) a proof of the fact $\endgroup$ – user8268 May 2 '11 at 20:48

This does not need any fancy trickery or complicated machinery. The moral of the story is: do not be scared of differentiating determinants! Formally, the derivative of a determinant is the trace. This actually happens quite often in geometric analysis, because the measure on a Riemannian manifold is given by $d\mu = \sqrt{\det g} \mathcal{H}^n$, where $n$ is the dimension of the manifold and $\mathcal{H}^n$ is $n$-dimensional Hausdorff measure.

We have $$ \partial_k \det A = (\partial_k A_{ij})A^{ij} \det A, $$ so in particular $$ \partial_i \sqrt{\det g} = \frac{(\partial_ig_{pq})g^{pq}}{2} \sqrt{\det g}. $$ Thus $$\begin{align*} \frac{1}{\sqrt{\det g}}\partial_i\Big(g^{ij}\sqrt{\det g}\partial_j\phi\Big) &= (\partial_ig^{ij})\partial_j\phi + g^{ij}\frac{1}{\sqrt{\det g}}\Big(\partial_i\sqrt{\det g}\partial_j\phi\Big) + g^{ij}\partial_{ij}\phi\\ &= (\partial_ig^{ij})\partial_j\phi + g^{ij}\frac{1}{\sqrt{\det g}}\Big(\frac{(\partial_ig_{pq})g^{pq}}{2} \sqrt{\det g}\partial_j\phi\Big) + g^{ij}\partial_{ij}\phi\\ &= (\partial_ig^{ij})\partial_j\phi + \frac{1}{2}g^{ij}(\partial_ig_{pq})g^{pq}\partial_j\phi + g^{ij}\partial_{ij}\phi\\ &= g^{ij}\partial_{ij}\phi - g^{ij}\Gamma_{ij}^p(\partial_ig_{pq})g^{pq}\partial_j\phi, \end{align*}$$ using the standard expression for the coefficients of the Levi-civita connection (the Christoffel symbols) in terms of the metric.


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