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This question, metric determinant and its partial and covariant derivative, seems to indicate $$\nabla_a \sqrt{g}=0.$$ Why is this the case? I've always learned that $$\nabla_a f= \partial_a f,$$ hence surely $$\nabla_a \sqrt{g}= \partial_a\sqrt{g} \neq 0. $$

Where's the hole in my logic?

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    $\begingroup$ Notice that in their own definition $\nabla_{\mu}f\neq \partial_{\mu}f$. $\endgroup$ Apr 16, 2016 at 12:50
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    $\begingroup$ Why would we have the term "covariant derivative" if it was just the same as the normal partial derivative? $\endgroup$
    – Danu
    Apr 16, 2016 at 13:09
  • $\begingroup$ @Danu. I was under the impression that this is always true, see p39 here: damtp.cam.ac.uk/user/hsr1000/lecturenotes_2012.pdf . How is its action normally defined on a function rather than a tensor? $\endgroup$ Apr 16, 2016 at 21:25
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    $\begingroup$ @user138901 Ah, that comment makes it clear what you're confused about: $\sqrt{\lvert g\rvert}$ is not a scalar (as pointed out in the answer by Qmechanic)! I initially misinterpreted your question. $\endgroup$
    – Danu
    Apr 16, 2016 at 21:36
  • $\begingroup$ Both you and the older poster of the question you link to are right to ask about whether the fact that this is not quite a tensor, but rather a tensor density, makes a difference. But as it happens, it does not invalidate this equation, it is still zero. $\endgroup$ Dec 7, 2020 at 6:45

4 Answers 4

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  1. With $g:=\det(g_{\mu\nu})$ note that $\sqrt{|g|}$ transforms as a density $$\begin{align}\rho^{(x)}\quad\longrightarrow&\quad \rho^{(y)}~=~\frac{\rho^{(x)}}{|J|}, \cr J~:=~&\det\frac{\partial y^{\nu}}{\partial x^{\mu}}, \end{align}$$ rather than a scalar $$s^{(x)}\quad\longrightarrow\quad s^{(y)}~=~s^{(x)},$$ under general coordinate transformations $$x^{\mu}\quad\longrightarrow\quad y^{\nu}~=~f^{\nu}(x).$$ In particular, the covariant derivative of $\sqrt{|g|}$ does not necessarily coincide with the partial derivative of $\sqrt{|g|}$.

  2. It is consistent to define that a connection $\nabla$ acts on a density $\rho>0$ as $$ \nabla_{\mu}\ln\rho~:=~\partial_{\mu}\ln\rho-\Gamma^{\nu}_{\mu\nu},$$ because it then transforms covariantly $$ \frac{\partial y^{\nu}}{\partial x^{\mu}}\nabla^{(y)}_{\nu}\ln\rho^{(y)}~=~\nabla^{(x)}_{\mu}\ln\rho^{(x)}.$$ A connection $\nabla$ and a density $\rho$ are by definition compatible if $$\nabla_{\mu}\ln\rho~=~0\quad\Leftrightarrow\quad\partial_{\mu}\ln\rho~=~\Gamma^{\nu}_{\mu\nu} .$$ Using linearity and Leibniz rule, we conclude that $$\nabla_{\mu}f(\ln\rho) ~=~0$$ for a sufficiently nice (e.g. real analytic) function $f$.

  3. A connection $\nabla$ is by definition compatible with a metric $$\mathbb{\bf g}~=~g_{\mu\nu}\mathrm{d}x^{\mu}\odot \mathrm{d}x^{\nu}$$ if $$\nabla_{\lambda}\mathbb{\bf g} ~=~0 \quad\Leftrightarrow\quad (\nabla_{\lambda}\mathbb{\bf g})_{\mu\nu}~=~0.$$ E.g. the Levi-Civita connection is compatible with the metric $\mathbb{\bf g}$. Metricity implies that$^1$ $$\begin{align} 2\Gamma^{\kappa}_{\lambda\kappa} ~=~&g^{\mu\nu}\partial_{\lambda}g_{\nu\mu}\cr ~=~&g^{\mu\nu}\eta_{\nu\sigma}\partial_{\lambda}\eta^{\sigma\tau}g_{\tau\mu}\cr ~=~&\partial_{\lambda}{\rm tr}{\rm Ln}(\eta^{\sigma\tau}g_{\tau\mu})\cr ~=~&\partial_{\lambda}\ln\det(\eta^{\sigma\tau}g_{\tau\mu})\cr ~=~&\partial_{\lambda}\ln|g|,\end{align}$$ which in turn shows that the density $\rho=\sqrt{|g|}$ is compatible with a metric connection $\nabla$. This proves OP's title question. Note that it is unnecessary to assume that the connection $\nabla$ is torsionfree. $\Box$

  4. Below is an alternative approach that doesn't use the transformation rule for densities: A metric connection satisfies $$\nabla_{\lambda}(\mathbb{\bf g}^{\otimes n}) ~=~0\quad\Leftrightarrow\quad (\nabla_{\lambda}(\mathbb{\bf g}^{\otimes n}))_{\mu_1\nu_1\ldots\mu_n\nu_n}~=~0.$$ Using linearity and Leibniz rule, the covariant derivative $\nabla_{\lambda}$ then annihilates any sufficiently nice (e.g. real analytic) function $f(g_{00},g_{01}, \ldots)$ of the metric. In particular, the square root of the determinant $\sqrt{|g|}$, so that $$\nabla_{\lambda}\sqrt{|g|}~=~0.$$ $\Box$

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$^1$ The above metric $\eta_{\mu\nu}$ is merely a trick to deal with the absolute value $|g|$ of $g$. The metric $\eta_{\mu\nu}$ is assumed to

  • (i) have the same signature as $g_{\mu\nu}$,

  • (ii) have $\det(\eta_{\mu\nu})=\pm 1$, and

  • (iii) have constant components in the pertinent local coordinate neighborhood.

There is an analogous trick to deal with the absolute value $|J|$ of the Jacobian determinant $J$.

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  • $\begingroup$ Qmechanic I do not really understand your heuristic explanation. Following your logic it would seem that any simple product of metric components should be covariantly constant. But it's not $\endgroup$
    – magma
    Apr 28, 2016 at 7:42
  • $\begingroup$ @magma : Which simple counterexample do you have in mind? $\endgroup$
    – Qmechanic
    Apr 28, 2016 at 9:04
  • $\begingroup$ We probably misunderstand each other, but -from what you write - it would seem that any polynomial function of metric components , such as $f=g_{00}^2$, is covariantly constant which is not the case. I do not think at the moment that there is a simple heuristic/intuitive justification for the constancy of the determinant. You just have to take the definition and make the calculations $\endgroup$
    – magma
    Apr 28, 2016 at 9:43
  • $\begingroup$ In the example $f=g_{00}g_{00}$, one could apply the covariant derivative $\nabla_{\lambda}$ to the $(0,4)$ tensor $T=(g_{\mu\nu}\mathrm{d}x^{\mu} \odot \mathrm{d}x^{\nu})\odot (g_{\rho\sigma}\mathrm{d}x^{\rho} \odot \mathrm{d}x^{\sigma})$, and compare the $0000$-component before and after. $\endgroup$
    – Qmechanic
    Apr 28, 2016 at 10:48
  • $\begingroup$ Any question that is formulated covariantly can be answered by choosing geodetic coordinates and replacing all covariant derivatives by ordinary partial derivatives of the coefficients. But some questions which you can write down are not covariantly formulated. In particular, asking about $g_{00}^2$ is not asking a covariantly formulated question. This is something Einstein understood quite well. $\endgroup$ Dec 7, 2020 at 6:36
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OK. Let us take ordinary derivative of determinant of some covariant 2-tensor $A_{\mu\nu}$. Let call it $A$. But it is more convenient to allow us to think about $A_{\mu\nu}$ like a matrix with covariant indices. So $$\det{A_{\mu\nu}} = A$$ Next, let's do the following calculations: $$\delta\ln{\det{A_{\mu\nu}}} = \ln{\det{(A_{\mu\nu}}+\delta A_{\mu\nu})}-\ln{\det{A_{\mu\nu}}} = \ln{\det({A^{\mu\sigma}(A_{\sigma\nu}+\delta A_{\sigma\nu}))}},$$ where $\delta$ is like differential and $A^{\mu\sigma}$ denotes contravriant 2-tensor with the following property: $A^{\mu\sigma}A_{\sigma\nu} = \delta^{\mu}_{\,\nu}$, in other words, "inverse" tensor.

Let's continue $$\ln{\det({A^{\mu\sigma}(A_{\sigma\nu}+\delta A_{\sigma\nu}))}} = \ln{\det{(I+A^{\mu\sigma}\delta A_{\sigma\nu})}} = \ln{(1 + \mathrm{Tr}\,{A^{\mu\sigma}\delta A_{\sigma\nu}})} = \mathrm{Tr}\,{A^{\mu\sigma}\delta A_{\sigma\nu}}.$$ But $$\mathrm{Tr}\,{A^{\mu\sigma}\delta A_{\sigma\nu}} = A^{\mu\sigma}\delta A_{\sigma\mu}$$

Divided by $dx^{\lambda}$ it gives $$\partial_{\lambda}\ln{\det{A_{\mu\nu}}} = A^{\mu\sigma}\partial_{\lambda} A_{\sigma\mu}.$$

Therefore, $$\frac{\partial_{\lambda}A}{A} = A^{\mu\sigma}\partial_{\lambda} A_{\sigma\mu}$$

Or $$\partial_{\lambda}g = g g^{\mu\sigma}\partial_{\lambda} g_{\sigma\mu}$$ The following step is pretty fun. Let's replace all ordinary partials by absolute (covariant). So we have $$\nabla_{\lambda}g = g g^{\mu\sigma}\nabla_{\lambda} g_{\sigma\mu}.$$ But $$\nabla_{\lambda} g_{\sigma\mu} = 0.$$ QED. The last is not hard exercise. Indeed, in the geodesic coordinates it is always true because in these coordinates $\nabla_{\nu} = \partial_{\nu}$. But if some tensor is equal to zero in one reference frame then it is zero in every reference frame.

But I am not sure about the same trick with arbitrary matrix (although it may turn out the same). It would be better to use the following. Since $$\det{g^{\mu\nu}A_{\nu\sigma}}$$ is a scalar, we can use ordinary derivative for this. But on the other hand, we could use covariant derivative for it. For scalar it is the same. So $$\nabla_{\nu}(\det{g^{\mu\nu}A_{\mu\nu}}) = g^{-1}\nabla_{\nu}A + A\nabla_{\nu}g^{-1} = g^{-1}\partial_\nu A + A\partial_\nu g^{-1}$$ Let us continue calculations $$\nabla_{\nu}A = \partial_{\nu} A - A\frac{\partial_\nu g}{g}$$ Where we used $\nabla_\nu g = 0$. Partial derivatives we can find from the previous equations.

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  • $\begingroup$ By the way, I am not sure about the same trick with arbitrary matrix. See the following. $\endgroup$
    – LRDPRDX
    Oct 21, 2016 at 12:37
  • $\begingroup$ @7919, I have chosen $\nabla_{\nu}$ for the covariant derivative. $\endgroup$
    – LRDPRDX
    Feb 21, 2018 at 5:18
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Here's a heuristic calculation: Let $\{E_i\}$ be an orthonormal frame ($g(E_i,E_j)=\epsilon_i\delta_{ij}, \epsilon_i=\pm 1$). Then $\mu$ is the canonical volume form $\sqrt{g}\,\mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n$ iff $\mu(E_1,\dotsc, E_n)=1$. Then $$(\nabla_X\mu)(E_1,\dotsc,E_n)=\nabla_X(\mu(E_1,\dotsc,E_n))-\sum \mu(E_1,\dotsc,\nabla_X E_i,\dotsc,E_n)=-\sum \epsilon_ig(E_i,\nabla_X E_i)\mu(E_1,\dotsc,E_i,\dotsc,E_n)=-\sum \epsilon_ig(E_i,\nabla_X E_i)=-\frac{1}{2}\sum \epsilon_i\nabla_X g(E_i,E_i)=0$$ for all vector fields $X$. Then using the derivation property of the connection, and $\nabla_X(\mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n)=0$ for all $X$, one has $$\nabla_X\mu=(\nabla_X\sqrt{g})\,\mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n=0$$ whence $\nabla_X\sqrt{g}=0$ for all $X$ and in some chart.

To be absolutely pedantic, one should adapt the defintion of the connection in terms of parallel transport to tensor densities. This is done in e.g. Straumann, General Relativity (2013). For a scalar density $\rho$ one finds in local coordinates $\nabla_i\rho=(\partial_i-\Gamma^l{}_{il})\rho$. From the standard expression for $\Gamma^l{}_{li}$ it is easy to verify that $\nabla_i\sqrt{g}=0$.

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  • $\begingroup$ My diff. geometry is very rusty, but I was wondering whether it would suffice to show that $\nabla_i\sqrt{g}=0$ in normal coordinates (where it is trivial)? Wouldn't the general case automatically follow? $\endgroup$ Apr 19, 2016 at 18:28
  • $\begingroup$ @AccidentalFourierTransform That's what I do in the "to be absolutely pedantic" section. (Not in normal coordinates, but in normal coordinates one has $\Gamma\equiv 0$ so it is indeed trivial.) The issue is that a density always refers to a coordinate system, that's what makes it a density. So I'm not sure if doing the calculation in one coordinate system is sufficient -- this object $\nabla_i\sqrt{g}$ is not a tensor. $\endgroup$
    – Ryan Unger
    Apr 19, 2016 at 19:40
  • $\begingroup$ IIRC, local coordinates$\neq$normal coordinates. What I mean is, in normal coordinates at $p$ we have $g_{ij}(p)=\delta_{ij}$ and $\Gamma^i_{jk}(p)=0$. Therefore, $g=1$ and $\nabla_i g=\partial_i g=0$. As $\nabla_i g$ is a (pseudo)tensor, if its zero in these coordinates, it must be zero in any coordinate system. [The terminology is probably very wrong because I studied these issues a couple of years ago]. I would phrase my reasoning as "we found a system of reference where $\nabla_i g=0$, and therefore, by covariance, it must be zero in any system of reference".Does this make sense to you? $\endgroup$ Apr 19, 2016 at 19:47
  • $\begingroup$ @AccidentalFourierTransform Normal coordinates are a special kind of local coordinates. (All coordinates are local coordinates.) The issue is that a priori, it is not clear what $\nabla_ig$ even means. $\endgroup$
    – Ryan Unger
    Apr 19, 2016 at 20:53
  • $\begingroup$ @AccidentalFourierTransform The axioms of the connection allow us only to take derivatives of functions and vector fields. We can extend this to one-forms (and simple tensors) by using the Leibnitz rule and to arbitrary tensors by factoring into simple tensors. It turns out one has to define an extra structure to make sense of $\nabla_i\rho$, where $\rho$ is a tensor density. $\endgroup$
    – Ryan Unger
    Apr 19, 2016 at 21:23
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Actually, the very notation $\nabla_\alpha \sqrt{g}$ is wrong, because $g = \det g_{\mu\nu}$ is coordinate dependent, and thus does not form a scalar field.

To make it more concrete, consider a 2-dimensional Euclidean space with polar coordinates $(r, \theta)$. $g = \mathrm{diag} (1, r^2)$. So $\sqrt{g} = r$. If $\nabla_r \sqrt{g}$ were meaningful, we would have $\nabla_r \sqrt{g} = \partial_r r = 1$, contradicting the claim in your original question.

I can only guess that you asked this question when you were trying to calculate derivatives of tensor fields such as $\epsilon_{\mu\nu\rho\sigma}$, in which $\sqrt{g}$ is used to turn a tensor density into a proper tensor. My answer to another question might be helpful.

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