The gauge transformation of the metric tensor is supposed to be $$\delta_\xi g_{\mu\nu}=\xi^\rho\partial_\rho g_{\mu\nu}+\partial_\mu\xi^\rho g_{\rho\nu}+\partial_\nu\xi^\rho g_{\mu\rho}$$ with $\delta_\xi x^\mu=-\xi^\mu(\tau)$, where $\xi^\mu$ is infinitesimal. What is the gauge transformation of the inverse metric $g^{\mu\nu}$?

My attempt:

We know $0=\delta(g^{-1}g)=\delta(g^{-1})g+g^{-1}\delta g\,$ and therefore $\delta(g^{-1})=-g^{-1}\delta (g) g^{-1}$

With $(g_{\mu\nu})^{-1}=g^{\mu\nu}$ I get: $$\begin{align} \delta_\xi g^{\mu\nu}&=-g^{\mu\nu}(\xi^\rho\partial_\rho g_{\mu\nu}+\partial_\mu\xi^\rho g_{\rho\nu}+\partial_\nu\xi^\rho g_{\mu\rho})g^{\mu\nu}\\ &=-g^{\mu\nu}(\xi^\rho(\partial_\rho g_{\mu\nu}) g^{\mu\nu}+\partial_\mu\xi^\rho g_{\rho\nu}g^{\mu\nu}+\partial_\nu\xi^\rho g_{\mu\rho}g^{\mu\nu})\\ &=-g^{\mu\nu}(\xi^\rho(\partial_\rho g_{\mu\nu}) g^{\mu\nu} + \partial_\mu\xi^\rho\delta_\rho^\mu +\partial_\nu\xi^\rho\delta_\rho^\nu)\\ &=-g^{\mu\nu}\xi^\rho(\partial_\rho g_{\mu\nu}) g^{\mu\nu}-g^{\nu}_\rho\partial_\mu\xi^\rho-g^{\mu}_\rho\partial_\nu\xi^\rho \end{align}$$

Is this the right way? The first term looks erroneous. And I am not quite sure if it is legal what I did with the deltas.


2 Answers 2


The way you manipulate indices doesn't make sense, for instance when you write $$g^{\mu \nu} (\partial_\rho g_{\mu \nu}) g^{\mu \nu}.$$ The same index should never appear more than two times. To fix this error: if you multiply matrices, you should add new dummy indices. For instance if $A,B,C$ are three matrices, then $$(ABC)_{ij} = A_{ik} B_{kl} C_{lj}.$$ So if you write $$(g^{-1} \delta(g) g^{-1})^{\mu \nu}$$ you really mean $$g^{\mu \alpha} \delta(g)_{\alpha \beta} g^{\beta \nu}.$$ Do you see that the flow of indices works now?

  • $\begingroup$ Yes, I see it now. I get now: $\delta g_{\alpha\beta} = g^{\mu\alpha}(\xi^\rho\partial_\rho g_{\alpha\beta}+\partial_\alpha\xi^\rho g_{\rho\beta}+\partial_\beta\xi^\rho g_{\alpha\rho})g^{\beta\nu}$. Is there more to simplify? $\endgroup$ Jun 9, 2019 at 7:34
  • $\begingroup$ @clearseplex You can further simplify by using the product rule $\partial_\rho(g^{\mu\alpha} g_{\alpha\beta}) = 0 = g^{\mu\alpha} \partial_\rho g_{\alpha\beta} + g_{\alpha\beta} \partial_\rho g^{\mu\alpha}$ (that's for the first term). In the other terms, you can also simplify by using the relation $g_{\rho\beta} g^{\beta\nu} = \delta_\rho^\nu$. $\endgroup$
    – Pxx
    Jun 9, 2019 at 12:25
  • $\begingroup$ Thanks. Final result is $\delta g^{\mu\nu}=-\xi^\rho\partial_\rho g^{\mu \alpha} \delta_\alpha^\nu + 2g^{\mu\alpha}\partial_\alpha\xi^\rho\delta_\alpha^\nu$. And in the first comment it should be $\delta g^{\mu\nu}$ not $\delta g_{\alpha\beta}$ . $\endgroup$ Jun 10, 2019 at 8:47
  • $\begingroup$ @clearseplex Mm I get something a bit different, starting from the expression that you gave above in the comments: $\delta g^{\mu\nu} = - \xi^\rho \partial_\rho g^{\mu\nu} + \partial_\rho \xi^\nu g^{\mu\rho} + \partial_\rho \xi^\mu g^{\rho\nu}$. Note that in the expression that you just gave, you can contract $g^{\mu\alpha} \delta_\alpha^\nu = g^{\mu\nu}$, since the $\delta_\alpha^\beta$ just means that $\alpha$ can be replaced by $\beta$ and vice versa. $\endgroup$
    – Pxx
    Jun 12, 2019 at 11:48
  • $\begingroup$ @clearseplex And what happened to the minus sign that you got from $\delta (g^{-1}) = - g^{-1} \delta g\ g^{-1}$? $\endgroup$
    – Pxx
    Jun 12, 2019 at 12:21

I would like to suggest a similar but slightly simpler calculation of the infinitesimal transformation of the inverse metric. I think it is also educative about how to deal with calculations involving indices. Starting as in the OP with:

$$\delta_\xi \left( g^{\mu\nu} g_{\mu\nu} \right) \overset{!}{=} 0 = \delta_\xi (g^{\mu\nu}) g_{\mu\nu} + g^{\mu\nu} \delta_\xi (g_{\mu\nu}) \tag{1}$$

We get:

$$\delta_\xi (g^{\mu\nu}) g_{\mu\nu} = - g^{\mu\nu} \left( \xi^\rho \partial_\rho g_{\mu\nu} + \partial_\mu \xi^\rho g_{\rho\nu} + \partial_\nu \xi^\rho g_{\mu\rho}\right)\tag{2}$$

where I used the infinitesimal transformation $\delta_\xi g_{\mu\nu}$ as given in the OP. Now the indices are fully contracted on both sides (as it should be), which means that we are free to rename them as long as we stay consistent. On the left hand side, we have a $g_{\mu\nu}$ that we would like to remove, so we are going to use our freedom for renaming the indices to try to factor out a $g_{\mu\nu}$ on the rhs.

Looking at the rhs of $(2)$, we see that the first term is problematic, since the $g_{\mu\nu}$ is "locked" inside the derivative. But we can use the product rule to write:

$$-g^{\mu\nu} \partial_\rho g_{\mu\nu} = - \partial_\rho \left( g^{\mu\nu} g_{\mu\nu} \right) + g_{\mu\nu} \partial_\rho g^{\mu\nu} \tag{3}$$

But $\partial_\rho \left( g^{\mu\nu} g_{\mu\nu} \right) = \partial_\rho \alpha^\mu_\mu = 0$! Using $(3)$, renaming the indices in the other terms and recalling that $g^{\alpha\beta} g_{\beta\gamma} = \delta^\alpha_\gamma$, we finally get:

$$\partial_\xi (g^{\mu\nu} ) g_{\mu\nu} = \xi^\rho \partial_\rho g^{\mu\nu} g_{\mu\nu} - \partial_\rho \xi^\mu g^{\rho\nu} g_{\mu\nu} - \partial_{\rho} \xi^{\nu} g^{\mu\rho} g_{\mu\nu} \tag{4}$$

We can now factorize $g_{\mu\nu}$ out on the rhs and read the result:

$$\partial_\xi (g^{\mu\nu} ) = \xi^\rho \partial_\rho g^{\mu\nu} - \partial_\rho \xi^\mu g^{\rho\nu} - \partial_{\rho} \xi^{\nu} g^{\mu\rho} \tag{5}$$


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