I am trying to vary the laplace-Beltrami operator with respect to the metric. Using the following two rules \begin{align} \frac{\delta g^{\alpha \beta}}{\delta g^{\mu \nu}} &=\frac{1}{2} \left[\delta^\alpha_\mu \delta^\beta_\nu + \delta^\alpha_\nu \delta^\beta_\mu \right]\\ \frac{\delta g_{\alpha \beta}}{\delta g^{\mu \nu}} &=- \frac{1}{2} \left[g_{\mu \alpha} g_{\nu \beta} + g_{\mu \beta} g_{\nu \alpha} \right], \end{align} I have come across the following conundrum when taking the metric variation of the Laplace-Beltrami operator $\square_x = g^{\mu \nu} \partial^x_\mu \partial^x_\nu$

\begin{align} \frac{\delta \square}{\delta g^{\mu \nu}} &= \frac{\delta}{\delta g^{\mu \nu}} \left[g_{\alpha \beta} \partial^\alpha \partial^\beta \right]=-\frac{1}{2} \left[g_{\mu \alpha} g_{\nu \beta} + g_{\mu \beta} g_{\nu \alpha} \right] \partial^\alpha \partial^\beta \\ &=-\partial_\mu \partial_\nu\\ &=\frac{\delta}{\delta g^{\mu \nu}} \left[g^{\alpha \beta} \partial_\alpha \partial_\beta \right] = + \frac{1}{2} \left[\delta^\alpha_\mu \delta^\beta_\nu + \delta^\alpha_\nu \delta^\beta_\mu \right] \partial_\alpha \partial_\beta\\ &=+ \partial_\mu \partial_\nu. \end{align} Where did I make a mistake ? I assume it has something to do with the gradient being a covariant object, but I am yet to spot the missing minus sign - where does it come into play ? Thanks in advance for your help.

Edit: How I derived the formulas for the metric variation: Starting from $g_{\mu \alpha} g^{\alpha \nu} = \delta^\nu_\mu$, I find that \begin{align} \delta g_{\mu \alpha} g^{\alpha \nu} + g_{\mu \alpha} \delta g^{\alpha \nu} &=0 \\ \Rightarrow \quad g_{\lambda \nu} g^{\alpha \nu} \delta g_{\mu \alpha} = \delta g_{\mu \lambda} &= -g_{\lambda \nu} g_{\mu \alpha} \delta g^{\alpha \nu} \\ \Leftrightarrow \quad \frac{\delta g_{\mu \lambda}}{\delta g^{\alpha \nu}} &= - g_{\lambda\nu} g_{\mu \alpha}, \end{align} which I subsequently symmetrized in the Lorentz indices. Then I used that $g^{\mu \nu} g_{\mu \nu}=d$ to derive that \begin{align} \delta(g^{\mu \nu} g_{\mu \nu}) &=0 = g^{\mu \nu } \delta g_{\mu \nu} + \delta g^{\mu \nu} g_{\mu \nu} \\ \Rightarrow \quad g^{\mu \nu} \delta g_{\mu \nu} &=- \delta g^{\mu \nu} g_{\mu \nu}, \end{align} such that the other rule follows analogously with a minus sign and raised indices. Is this correct ?

  • $\begingroup$ Where did you find your first two relations? $\endgroup$ Sep 27, 2021 at 11:49
  • $\begingroup$ I edited my post to answer your question. Does that make sense ? $\endgroup$ Sep 27, 2021 at 13:26

2 Answers 2


It is not true that $g_{ab} \partial^a \partial^b = g^{ab} \partial_a \partial_b$, because the left derivative in the left hand side acts on a metric too. The correct definition is $\square = g^{ab}\partial_a \partial_b$, from which we get

$$\begin{align} g_{ab} \partial^a \partial^b &= g_{ab} g^{ac} \partial_c (g^{bd} \partial_d) \\ &= g_{ab} g^{ac} g^{bd} \partial_c \partial_d + g_{ab} g^{ac} (\partial_c g^{bd}) \partial_d \\ &= g^{cd} \partial_c \partial_d + (\partial_b g^{bd}) \partial_d \end{align}$$

  • $\begingroup$ That makes sense to me, but why is $\square = g^{ab} \partial_a \partial_b$ the correct definition ? $\endgroup$ Sep 27, 2021 at 13:58
  • $\begingroup$ Also, $\partial_b g^{bd}$ vanishes in metric theories of gravity if I am not mistaken. So, where does the minus sign then come in ? $\endgroup$ Sep 27, 2021 at 14:12
  • $\begingroup$ @GenericPhysicsStudent: Are you using $\partial$ for the covariant derivative rather than the more standard $\nabla$? The covariant derivative of the (inverse) metric vanishes by definition, but it's definitely not the case that the coordinate derivatives of the (inverse) metric vanish. That doesn't even happen in flat space (e.g., the metric for Euclidean space in polar coordinates depends on $r$.) $\endgroup$ Sep 27, 2021 at 14:14
  • $\begingroup$ I am calculating the metric variation of the laplace-beltrami operator to get the energy-momentum tensor in flat space, i.e. I vary w.r.t some general metric and evaluate it on flat Minkowski space. Since I am considering the EM tensor of a scalar field, I always assumed to simply put $\square\phi = g^{ab} \nabla_a \nabla_b= g^{ab} \partial_a \partial_b \phi$. Since I have to first take the metric variation, you are probably right that I should also consider the contributions from the covariant derivative (i.e. christoffels). $\endgroup$ Sep 27, 2021 at 14:21
  • $\begingroup$ I see, so what I am gathering from your response is that: a) I should always be using $\square=g^{ab} \nabla_a \nabla_b$ and b) should be aware of non-trivial contributions from Christoffels despite considering a scalar field. $\endgroup$ Sep 27, 2021 at 14:23

The “raised-index derivative” $\partial^\mu$ depends implicitly on the inverse metric, since it stands for $$ g^{\mu \nu} \frac{\partial}{\partial x^\nu}. $$ The reason for this is that mathematically, the coordinate derivative with a lowered index can be defined without reference to a metric, and so it is natural to take it to be constant when doing a variation like this. By contrast, the definition of the “raised-index” coordinate derivative relies on the metric and so we can't “hold it fixed” while varying with respect to the metric.

Finally, note that there's a subtlety in your derivation that you haven't taken into account. If the background metric isn't flat, then the Laplace-Beltrami operator isn't $g^{\mu \nu} \partial_\mu \partial_\nu$ but $g^{\mu \nu} \nabla_\mu \nabla_\nu$. If this operates on some scalar field $f$ then we have $$ g^{\mu \nu} \nabla_\mu \nabla_\nu f = g^{\mu \nu} \left( \partial_\mu \partial_\nu f - \Gamma^{\rho} {}_{\mu \nu} \partial_\rho f \right) $$ for some scalar field $f$. Since the Christoffels $\Gamma^{\rho} {}_{\mu \nu}$ depend on the metric, they must be varied as well. The resulting terms will involve coordinate derivatives of $g^{\mu \nu}$ which must then be integrated by parts to find $\delta (\Box f)/\delta g^{\mu \nu}$. If the Laplace-Beltrami operator is applied to vector or tensor fields, additional terms involving the Christoffels will be present that must be varied as well.

  • $\begingroup$ Following up from @Javier post, the additional term is proportional to $\nabla_a g^{ab}$. Should this not vanish for a metric theory of gravity in general, and then in flat space as well ? $\endgroup$ Sep 27, 2021 at 14:43
  • $\begingroup$ @GenericPhysicsStudent: It's true that if you apply the logic in Javier's answer using the covariant derivative instead, you don't get the extra term. But that's a separate question from how to take the variation. Since the covariant derivative operator $\nabla_a$ is defined so that $\nabla_a g^{bc} = 0$, any variation of the metric leads to a change in the covariant derivative instead (similar to how you can't vary the metric without also varying the inverse metric, as you showed.) $\endgroup$ Sep 27, 2021 at 15:07
  • $\begingroup$ Thank you for your precise comments, they have cleared up a lot my confusions. To summarize then, would you agree with the statement that the answer to my original question is that my variational formulae (the first two) are correct, but the assumption $g^{ab} \partial_a \partial_b = g_{ab} \partial^a \partial^b$ is the culprit which lead to the contradiction $A=-A$ ? $\endgroup$ Sep 27, 2021 at 15:24
  • $\begingroup$ That's not quite right: those two operators are equal. Your error was in assuming that $\partial^a$ remains constant when you vary the metric. $\endgroup$ Sep 27, 2021 at 15:28
  • $\begingroup$ I see, thank you very much for your help ! $\endgroup$ Sep 28, 2021 at 6:38

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