What is a good formula for the Laplace operator?

[George Keeling]

TL;DR Summary

I have found a few different formulas for the Laplace operator. They don't seem to be the same. Or maybe they are and I just can't prove it.

I have found various formulations for the Laplacian and I want to check that they are all really the same. Two are from Wikipedia and the third is from Sean Carroll. They are:

A Wikipedia formula in $n$ dimensions:
\begin{align}
\nabla^2=\frac{1}{\sqrt{\left|g\right|}}\frac{\partial}{\partial x^i}\left(\sqrt{\left|g\right|}g^{ij}\frac{\partial}{\partial x^j}\right)&\phantom {10000}(2)\nonumber
\end{align}A Wikipedia formula in "in 3 general curvilinear coordinates $(x^1,x^2,x^3)$":
\begin{align}
\nabla^2=g^{\mu\nu}\left(\frac{\partial^2}{\partial x^\mu\partial x^\nu}-\Gamma_{\mu\nu}^\lambda\frac{\partial}{\partial x^\lambda}\right)&\phantom {10000}(3)\nonumber
\end{align}And Carroll's formula (from exercise 3.4) :
\begin{align}
\nabla^2=\nabla_\mu\nabla^\mu=g^{\mu\nu}\nabla_\mu\nabla_\nu&\phantom {10000}(4)\nonumber
\end{align}The Wikipedia also gives a formula for the Laplacian in spherical polar coordinates:
\begin{align}
\nabla^2f&=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial f}{\partial r}\right)+\frac{1}{r^2\sin{\theta}}\frac{\partial}{\partial\theta}\left(\sin{\theta}\frac{\partial f}{\partial\theta}\right)+\frac{1}{r^2\sin^2{\theta}}\frac{\partial^2f}{\partial\phi^2}&\phantom {10000}(5)\nonumber\\
&=\frac{1}{r}\frac{\partial^2}{\partial r^2}\left(rf\right)+\frac{1}{r^2\sin{\theta}}\frac{\partial}{\partial\theta}\left(\sin{\theta}\frac{\partial f}{\partial\theta}\right)+\frac{1}{r^2\sin^2{\theta}}\frac{\partial^2f}{\partial\phi^2}&\phantom {10000}(6)\nonumber
\end{align}where $\phi$ represents the azimuthal angle and $\theta$ the zenith angle or co-latitude. So the metric will be
\begin{align}
g_{\mu\nu}=\left(\begin{matrix}1&0&0\\0&r^2&0\\0&0&r^2\sin^2{\theta}\\\end{matrix}\right)&\phantom {10000}(7)\nonumber
\end{align}I assumed that the coordinates are ordered $r,\theta,\phi$ although Wikipedia does not say that.

I want to prove that
A) (2), (3) and (4) both give (5) or (6) the Laplacian in spherical polar coordinates.
B) (4) is equivalent to (3) the general 3-dimensional expression.
C) (4) is equivalent to (2) the general $n$-dimensional expression.
A was quite easy. B follows immediately from the formula for the covariant derivative. I could not prove C. I also tried it for a diagonal metric and failed.

Does anybody have a view on how to calculate a Laplacian? I am new to them. Maybe it is possible to prove C. Is there some secret about the determinant of the metric which I have forgotten or don't know?

 

[Orodruin]

Start by proving
$$
\frac{1}{\sqrt g}\partial_\mu \sqrt g = \frac 12 \partial_\mu\ln g = \Gamma_{\mu\nu}^\nu.
$$

 

[vanhees71]

I think the most elegant way to derive the Laplacian in terms of general generalized coordinates is to use the action principle. To that end we can start from the most simple coordinates possible, i.e., Cartesian ones, where $\Delta=\delta^{ab} \partial_a \partial_b$. Defining $\Delta \phi=j$ with an arbitrary source field $j$, the Laplace operator in Cartesian coordinates follows from the action with the Lagrangian
$$L=\frac{1}{2} (\partial_a \phi) (\partial_a \phi)+j \phi.$$
The action is
$$A=\int \mathrm{d}^n x L.$$
In arbitrary generalized coordinates the metric is given by
$$\mathrm{d} x^a \mathrm{d} x^b \delta_{ab} = \delta_{ab} \frac{\partial x^a}{\partial q^j} \frac{\partial x^b}{\partial q^k} \mathrm{d} q^j \mathrm{d} q^k =g_{jk} \mathrm{d} q^j \mathrm{d} q^k.$$
The Jacobian for the invariant volume element has to be evaluated with the Levi-Civita tensor. In Cartesian coordinates this tensor's components are given by $\epsilon_{a_1 \cdots a_n}$, which is antisymmetric under exchange of any index pair and $\epsilon_{12\ldots n}=1$. In arbitrary coordinates the components transform covariantly (as the lower indices indicate):
$$\epsilon_{k_1\ldots k_n}' = \prod_{j=1}^n \frac{\partial x^{a_j}}{\partial q^{k_j}} \epsilon_{a_1\cdots a_n}=J \epsilon_{k_1\ldots k_n}.$$
Now it's simple to see that the Jacobian determinant of the transformation is $J=\sqrt{g}$, where $g=\det(g_{\mu \nu})$.

Thus the action in general coordinates reads
$$A=\int \mathrm{d}^n q \sqrt{g} [g^{jk} (\partial_j \phi)(\partial_k \phi)+j \phi].$$
Variation with respect to $\phi$ and partial integration in the resulting expression for $\delta A$ (or equivalently using the Euler-Lagrange equations) leads to
$$-\partial_j (\sqrt{g} g^{jk} \partial_k \phi)+\sqrt{g} j=0 \; \Rightarrow \; j=\Delta \phi=\frac{1}{\sqrt{g}} \partial_j (\sqrt{g} g^{jk} \partial_k \phi).$$
This is indeed (2).

That (2) is the same as (3) and (4) follows from standard formula of vector analysis in Riemann spaces (see, e.g., Landau-Lifshitz vol. 2 for a straight-forward treatment of the only slightly more general case of 4D general-relativistic spacetime, which is a pseudo-Riemannian space).

 

[George Keeling]

Start by proving$\frac{1}{\sqrt g}\partial_\mu \sqrt g = \frac 12 \partial_\mu\ln g = \Gamma_{\mu\nu}^\nu$.

I could do the first part but not the second and if I had done the second I would still have to prove that $$
\frac{1}{2}g^{\mu\lambda}g^{\nu\rho}\partial_\rho g_{\mu\lambda}\partial_\nu f-g^{\mu\lambda}g^{\nu\rho}\partial_\lambda g_{\mu\rho}\partial_\nu f=\frac{1}{2}g^{\mu\nu}g^{\lambda\rho}\partial_\mu g_{\rho\lambda}\partial_\nu f+\partial_\mu g^{\mu\nu}\partial_\nu f
$$which I got to when I was trying with a diagonal metric. But now I know that the above must be true, which might be useful, and also the surprising$$
\frac{1}{2}\partial_\mu\ln{\left|g\right|}=\Gamma_{\mu\nu}^\nu
$$Thanks also to vanhees71. We're doing the principle of least action in the next section of the book I have. I obviously also need

standard formulas of vector analysis in Riemann spaces (see, e.g., Landau-Lifshitz vol. 2 for a straight-forward treatment of the only slightly more general case of 4D general-relativistic spacetime, which is a pseudo-Riemannian space).

perhaps I would find similar in 'Mathematical Methods for Physics and Engineering' by Mattias Blennow but, shockingly, it is not in the Berlin library system and I hesitate to fork out 90€ for the paperback. I want a full preview from the library!

 

[Orodruin]

I could do the first part but not the second

A useful relation is that ln(det(A)) = tr(ln(A)) for any matrix A.

Once you have done this, you should be able to show that
$$
\frac 1{\sqrt g} \partial_\mu (\sqrt g V^\mu) = \nabla_\mu V^\mu
$$
for any vector field $V$. Then use this relation for $V^\mu = g^{\mu\nu} \partial_\nu f$.

 

[George Keeling]

Well, I was able to prove that

ln(det(A)) = tr(ln(A))

so I learned a bit about logs of matrices (OMG!) and I was able to prove that $$\frac 1{\sqrt g} \partial_\mu (\sqrt g V^\mu) = \nabla_\mu V^\mu$$ but only for a diagonal metric for which I used the product symbol as in $$
\left(\prod_{\lambda} g^{\lambda\lambda}\right)\partial_\mu\prod_{\nu} g_{\nu\nu}
$$for the first time ever. And then using $V^\mu = g^{\mu\nu} \partial_\nu f$ showing that Wikipedia in n-dimensions and Carroll were the same was easy. But only for a diagonal metric . Fortunately I have never seen a non-diagonal metric.

 

[Orodruin]

Why only for a diagonal metric? What part are you missing to make it general?

Edit: The Minkowski metric is non-diagonal if you pick (for example) light-cone coordinates.

 

[George Keeling]

Here's what I did. I was not able to exploit the bit about determinants and traces and logs. I suspect the proper proof might have something to do with diagonalising matrices, but that didn't work out neatly.
Lemma 1
For a diagonal metric:
\begin{align}
g^{-1}\partial_\mu g&=\left(\prod_{\lambda} g^{\lambda\lambda}\right)\partial_\mu\prod_{\nu} g_{\nu\nu}&\phantom {10000}(1)\nonumber\\

&=\left(\prod_{\lambda} g^{\lambda\lambda}\right)\left(\sum_{\rho}{\partial_\mu g_{\rho\rho}\prod_{\nu}^{\nu\neq\rho}g_{\nu\nu}}\right)&\phantom {10000}(2)\nonumber\\

&=\sum_{\rho}{g^{\rho\rho}\partial_\mu g_{\rho\rho}}&\phantom {10000}(3)\nonumber
\end{align}and
\begin{align}
g^{\nu\lambda}\partial_\mu g_{\nu\lambda}&=\sum_{\lambda}{g^{\lambda\lambda}\partial_\mu g_{\lambda\lambda}}=g^{-1}\partial_\mu g&\phantom {10000}(4)\nonumber
\end{align}Lemma 2
\begin{align}
\frac{1}{\sqrt g}\partial_\mu\left(\sqrt g V^\mu\right)&=\partial_\mu V^\mu+V^\mu\frac{1}{\sqrt g}\partial_\mu\left(\sqrt g\right)=\partial_\mu V^\mu+V^\mu\frac{1}{2g}\partial_\mu g&\phantom {10000}(5)\nonumber\\
\nabla_\mu V^\mu&=\mathrm{\partial}_\mu V^\mu+\Gamma_{\mu\nu}^\mu V^\nu=\mathrm{\partial}_\mu V^\mu+\frac{1}{2}g^{\mu\lambda}\left(\partial_\mu g_{\lambda\nu}+\partial_\nu g_{\mu\lambda}-\partial_\lambda g_{\mu\nu}\right)V^\nu&\phantom {10000}(6)\nonumber\\

&=\mathrm{\partial}_\mu V^\mu+\frac{1}{2}g^{\mu\lambda}V^\nu\partial_\nu g_{\mu\lambda}=\mathrm{\partial}_\mu V^\mu+V^\mu\frac{1}{2}g^{\nu\lambda}\partial_\mu g_{\nu\lambda}&\phantom {10000}(7)\nonumber
\end{align}using (4)
\begin{align}
g^{-1}\partial_\mu g=g^{\nu\lambda}\partial_\mu g_{\nu\lambda}&\phantom {10000}(8)\nonumber
\end{align}So for a diagonal metric
\begin{align}
\frac{1}{\sqrt g}\partial_\mu\left(\sqrt g V^\mu\right)=\nabla_\mu V^\mu&\phantom {10000}(9)\nonumber
\end{align}Wikipedia N dimensions

The Wikipedia formula in n dimensions was the same as
\begin{align}
\nabla^2f&=\frac{1}{\sqrt{\left|g\right|}}\left(\partial_\mu\sqrt{\left|g\right|}g^{\mu\nu}\partial_\nu f\right)&\phantom {10000}(10)\nonumber
\end{align}Using (9) with $V^\mu=g^{\mu\nu}\partial_\nu f$ as suggested
\begin{align}
\nabla^2f&=\nabla_\mu g^{\mu\nu}\partial_\nu f=g^{\mu\nu}\nabla_\mu\partial_\nu f=\nabla^\nu\nabla_\nu f&\phantom {10000}(11)\nonumber
\end{align}

 

[Orodruin]

Lemma 1
For a diagonal metric:

So, this is not using the hint that ln(det(A)) = tr(ln(A)).

However, you can also use your result that it holds for a diagonal metric to show that the relation must hold for any metric. Try letting $G = P G_d P^{-1}$ be a matrix containing the elements of the non-diagonal metric, where $G_d$ is a diagonal matrix. What is det(G)? What is $\operatorname{tr}(G^{-1} \partial_\mu G)$? (Note that, generally, P will depend on the coordinates as well as $G_d$.)

 

[vanhees71]

[WRONG in view of #20 !]

Method 1: Using the minors of matrices to calculate the inverse with help of determinants:

You can use the fact that $g^{\mu \nu}$ is the inverse matrix of $g_{\mu \nu}$ and (also using that the matrix is symmetric)
$$g=\epsilon^{\mu_1\ldots \mu_4} g_{\mu_1 1} g_{\mu_2 2} g_{\mu_3 3} g_{\mu_4 4},$$
where I use the convention that $\epsilon^{\mu_1\ldots \mu_4}$ is the Levi-Civita symbol (not tensor components in GR!) with $\epsilon^{0123}=+1$. Then
$$\partial_{\nu} g = \epsilon^{\mu_1 \ldots \mu_4} (\partial_{\nu} g_{\mu_1 1} g_{\mu_2 2} g_{\mu_3 3} g_{\mu_4 4} + g_{\mu_1 1} \partial_{\nu} g_{\mu_2 2} g_{\mu_3 3} g_{\mu_4 4} +g_{\mu_1 1} g_{\mu_2 2} \partial_{\nu} g_{\mu_3 3} g_{\mu_4 4} + g_{\mu_1 1} g_{\mu_2 2} g_{\mu_3 3} \partial_{\nu} g_{\mu_4 4}).$$
Now the contraction of the Levi-Civita symbol with the metric components without the partial derivative are components of $g g^{\mu \nu}$, e.g.,
$$\epsilon^{\mu_1 \ldots \mu_4} g_{\mu_2 2} g_{\mu_3 3} g_{\mu_4 4}=(-1)^{\mu_1} g g^{\mu_1 1}.$$
Taking all 4 terms of this kind together you get
$$\partial_{\nu} g = g g^{\rho \sigma} \partial_{\nu} g_{\rho \sigma}.$$

 

[George Keeling]

I remember abandoning this thread in a frustrated hissy fit. Almost exactly one year later I have found a decent proof of the Voss-Weyl formula:$$
\nabla_\mu F^\mu=\frac{1}{\sqrt{\left|g\right|}}\frac{\partial}{\partial x^\mu}\left(\sqrt{\left|g\right|}F^\mu\right)
$$replacing $F^\mu$ by $\nabla^\mu f=\partial^\mu f$ shows immediately that (2) and (4) in the original post are the same. Job done!
Along the way the proof also shows that

$$\partial_{\nu} g = g g^{\rho \sigma} \partial_{\nu} g_{\rho \sigma}$$

without using the horrid Levi-Civita symbol. This quite amazing formula is true for any rank 2 tensor I believe and can be written$$
Q^{-1}\partial_kQ=Q^{ij}\partial_kQ_{ij}
$$where $Q^{-1}$ is the determinant of the inverse of $Q_{ij}$ which has determinant $Q$. It is also well known that $Q^{-1}=1/Q$ so the -1 sort of has a double meaning.
Edit: That formula isn't right see below.

 

[George Keeling]

I know I shouldn't argue with my nice teacher but I think that the $\left(-1\right)^{\mu_1}$ should be removed from

e.g.,
$\epsilon^{\mu_1 \ldots \mu_4} g_{\mu_2 2} g_{\mu_3 3} g_{\mu_4 4}=(-1)^{\mu_1} g g^{\mu_1 1}.$

the other factors of terms on the RHS of $\partial_\nu g=\ldots$ would be $$
\epsilon^{\mu_1\mu_2\mu_3\mu_4}g_{\mu_11}g_{\mu_33}g_{\mu_44}=\left(-1\right)^{\mu_2}gg^{\mu_22}
$$$$
\epsilon^{\mu_1\mu_2\mu_3\mu_4}g_{\mu_11}g_{\mu_22}g_{\mu_44}=\left(-1\right)^{\mu_3}gg^{\mu_32}
$$$$
\epsilon^{\mu_1\mu_2\mu_3\mu_4}g_{\mu_11}g_{\mu_22}g_{\mu_33}=\left(-1\right)^{\mu_4}gg^{\mu_44}
$$Adding them up we get$$
\partial_\nu g=\left(-1\right)^{\mu_1}gg^{\mu_11}\partial_\nu g_{\mu_11}+\left(-1\right)^{\mu_2}gg^{\mu_22}\partial_\nu g_{\mu_22}+\left(-1\right)^{\mu_3}gg^{\mu_32}\partial_\nu g_{\mu_33}+\left(-1\right)^{\mu_4}gg^{\mu_44}\partial_\nu g_{\mu_44}
$$The $\mu_i$'s in that are all dummy variables so set them all to $\rho$ and compact the sum over the second indices to $\sigma$ and we have$$
\partial_\nu g=\left(-1\right)^\rho gg^{\rho\sigma}\partial_\nu g_{\rho\sigma}
$$now write out the summation over $\rho$ $$
\partial_\nu g=-gg^{1\sigma}\partial_\nu g_{1\sigma}+gg^{2\sigma}\partial_\nu g_{3\sigma}-gg^{3\sigma}\partial_\nu g_{3\sigma}+gg^{4\sigma}\partial_\nu g_{4\sigma}
$$which is not quite right.

After much labour I showed that, for any tensor $A$ with inverse ${\bar{A}}$,$$
\left|A\right|{\bar{A}}^{\alpha\mu_\alpha}=\epsilon^{\mu_1\mu_2\mu_3\ldots\mu_n}\prod_{\beta\neq\alpha} A_{\mu_\beta\beta}
$$which in four dimensions, for a symmetric tensor like the metric whose inberse its itself with indices raised, is the same as the 'e.g.' without the pesky $\left(-1\right)^{\mu_1}$ and produces a nice summation and the desired result.

The formula I wrote also has an error it. It should be $$
Q^{-1}\partial_kQ={\bar{Q}}^{ij}\partial_kQ_{ji}
$$The second $ij$ indices were the wrong way round. ${\bar{Q}}^{ij}$ is the inverse of $Q^{ij}$.

 

[vanhees71]

I don't see, where there is a mistake in my formula. Also note that $g^{\mu \nu}=g^{\nu \mu}$ is the inverse matrix to $g_{\mu \nu}=g_{\nu \mu}$: $g_{\mu \nu} g^{\nu \rho}=\delta_{\mu}^{\rho}$.

 

[George Keeling]

When I used your formula in post #12 it gives $$
\partial_\nu g=-gg^{1\sigma}\partial_\nu g_{1\sigma}+gg^{2\sigma}\partial_\nu g_{3\sigma}-gg^{3\sigma}\partial_\nu g_{3\sigma}+gg^{4\sigma}\partial_\nu g_{4\sigma}
$$They should all be + signs to make the sum work. The guess is that the $\left(-1\right)^{\mu_1}$ is unnecessary and I showed it 'after much labour'... I have a twelve page document on that labour and other things. I could cut the other stuff out and put it up here...

I understand the things about the metric.

 

[vanhees71]

Is it maybe, that I used the convention that the indices run from 0 to 3 and you the one where they run from $1$ to $4$? Then the overall sign for the first term should of course be $(-1)^{\mu_1+1}$ and you get the correct sign. It's always a bit of a pain to get the signs right with the expansion of determinants in terms of rows or columns :-(.

 

[George Keeling]

I thought of that but I don't think it helps. One gets alternating signs in both cases. Its either -+-+ or +-+- and what we need is ++++.

 

[vanhees71]

Ok, let's see. We have
$$g=\epsilon^{\mu_1 \mu_2 \mu_3 \mu_4} g_{\mu_10} g_{\mu_21} g_{\mu_3 2} g_{\mu_4 3}.$$
Further $g^{\mu \nu}$ is the inverse matrix of $g_{\mu \nu}$ and thus
$$g g^{\mu0}=(-1)^{\mu} \epsilon^{\mu \mu_1 \mu_2 \mu_3} g_{\mu_1 1} g_{\mu_2 2} g_{\mu_3 3}$$
then
$$g g^{\mu 1}=(-1)^{\mu+1} \epsilon^{\mu \mu_1 \mu_2 \mu_3} g_{\mu_1 0} g_{\mu_2 2} g_{\mu_3 3}$$
and so on. [EDIT: formula corrected in response to #18]

Now we have
$$\partial_{\alpha} g = \epsilon^{\mu_1 \mu_2 \mu_3 \mu_4} (\partial_{\alpha} g_{\mu_10} g_{\mu_21} g_{\mu_3 2} g_{\mu_4 3} +g_{\mu_10} \partial_{\alpha} g_{\mu_21} g_{\mu_3 2} g_{\mu_4 3} + \cdots. \qquad (*)$$
Using the above relation of the minors of the matrix with the inverse we find
$$\partial_{\alpha} g=g (g^{\mu_1 0} \partial_{\alpha} g_{\mu_1 0} + g^{\mu_2 1} \partial_{\alpha} g_{\mu_2 1}+\cdots$$
So there are no signs. You have to note that there's an additional sign from changing the order in the summation indices in the Levi-Civita symbol. E.g., for the 2nd term in (*) you have
$$\partial_{\alpha} g_{\mu_2 1} \epsilon^{\mu_1 \mu_2 \mu_3 \mu_4} g_{\mu_1 0} g_{\mu_3 2} g_{\mu_4 3} = -\partial_{\alpha} g_{\mu_2 1} \epsilon^{\mu_2 \mu_1 \mu_3 \mu4} g_{\mu_1 0} g_{\mu_3 2} g_{\mu_4 3} = - \partial_{\alpha} g_{\mu_2 1} (-g g^{\mu_2 1})=+g g^{\mu_2 1} \partial_{\alpha} g_{\mu_2 1}.$$
I hope this clarifies my derivation. I should have put in these index-sign battles in my original posting. Sorry!

 

[George Keeling]

Is there a typo in $$
g g^{\mu 1}=(-1)^{\mu+1} \epsilon^{\mu \mu_1 \mu_2 \mu_3} g_{\mu1 0} g_{\mu_2 2} g_{\mu_3 3}
$$? I will return tomorrow. Thanks.

 

[vanhees71]

[WRONG in view of #20 !]

Sigh. Yes, I forgot an underscore in the LaTeX:
$$g g^{\mu 1}=(-1)^{\mu+1} \epsilon^{\mu \mu_1 \mu_2 \mu_3} g_{\mu_1 0} g_{\mu_2 2} g_{\mu_3 3}$$
I've corrected it also in the posting above.

 

[George Keeling]

So we have these two formulas

Further $g^{\mu\nu}$ is the inverse matrix of $g_{\mu\nu}$ and thus$$
g g^{\mu0}=(-1)^{\mu} \epsilon^{\mu \mu_1 \mu_2 \mu_3} g_{\mu_1 1} g_{\mu_2 2} g_{\mu_3 3}
$$then$$
g g^{\mu 1}=(-1)^{\mu+1} \epsilon^{\mu \mu_1 \mu_2 \mu_3} g_{\mu_1 0} g_{\mu_2 2} g_{\mu_3 3}
$$and so on.

Say we had a 3-dimensional metric to make things a bit simpler the two formulas would presumably be$$
gg^{\mu0}=\left(-1\right)^\mu\epsilon^{\mu\mu_1\mu_2}g_{\mu_11}g_{\mu_22}
$$$$
gg^{\mu1}=\left(-1\right)^{\mu+1}\epsilon^{\mu\mu_1\mu_2}g_{\mu_10}g_{\mu_22}
$$Try that on a metric$$
g_{\mu\nu}=\left[\begin{matrix}a&b&c\\b&d&e\\c&e&f\\\end{matrix}\right]
$$Using the famous inverse matrix formula $gg^{\mu\nu}=c^{\nu\mu}$ where $c^{\nu\mu}$ are the cofactors. That gives me $$
gg^{11}=c^{11}=af-c^2
$$using the $gg^{\mu1}$ above$$
gg^{11}=\left(-1\right)^2\epsilon^{1\mu_1\mu_2}g_{\mu_10}g_{\mu_22}=\epsilon^{102}g_{00}g_{22}+\epsilon^{120}g_{20}g_{02}=-af+c^2
$$They disagree by a sign.

I think that is correct. If you agree read on.
-------------------------------------------------------------------
Taking the two formulas again the second is the same as $$
gg^{\mu1}=\left(-1\right)^{\mu+1}\epsilon^{\mu\mu_0\mu_2\mu_3}g_{\mu_00}g_{\mu_22}g_{\mu_33}
$$Note that the indexed $\mu$ is indexed 0,1,2,3

Can we write the two more generally as$$
gg^{\mu\nu}=\left(-1\right)^{\mu+\nu}\epsilon^{\mu\prod_{\lambda\neq\nu}\mu_\lambda}\prod_{\lambda\neq\nu} g_{\mu_\lambda\lambda} \ \ \ \ \ ?
$$The $\epsilon^{\mu\prod_{\lambda\neq\nu}\mu_\lambda}$ is not a multiplication! It just means a list which might look like a multiplication. It will soon go away.

Note that there is no $\mu_\nu$ in that so we can set $\mu=\mu_\nu$ and it becomes$$
gg^{\mu_\nu\nu}=\left(-1\right)^{\mu_\nu}\left(-1\right)^\nu\epsilon^{\mu_\nu\prod_{\lambda\neq\nu}\mu_\lambda}\prod_{\lambda\neq\nu} g_{\mu_\lambda\lambda}
$$we can now push the $\mu_\nu$ back into the list $\epsilon^{\mu_\nu\prod_{\lambda\neq\nu}\mu_\lambda}$ which changes the sign by $\left(-1\right)^\nu$ so that is$$
gg^{\mu_\nu\nu}=\left(-1\right)^{\mu_\nu}\epsilon^{\mu_0\mu_1\mu_2\mu_3}\prod_{\lambda\neq\nu} g_{\mu_\lambda\lambda}
$$I believe that the $\left(-1\right)^{\mu_\nu}$ should be removed from that. As I attempt to show in the attached document. If you agree with that then proceed to post 12!

 

[vanhees71]

I think you are right! Sorry for the confusion.

Let's check. I take the right-hand side of your equation and contract it with $g_{\mu_\nu \rho}$:
$$g_{\mu_{\nu} \rho} \epsilon^{\mu_0 \mu_1 \mu_2 \mu_3} \prod_{\lambda \neq \nu} g_{\mu_{\lambda} \lambda} = \delta^{\nu}_{\rho} g.$$

For a general matrix $A$ it's of course (now written with all indices as lower indices for simplicity)
$$\mathrm{det} A \bar{A}_{k \mu_{k}}=\epsilon_{\mu_1 \ldots \mu_n} \prod_{\lambda \neq k} A_{\mu_{\lambda} \lambda}.$$
Here, I also use the notation $\bar{A}_{jk}$ as the matrix elements of $\hat{A}^{-1}$.

 

[George Keeling]

Sorry for the confusion.

Not at all! Your pushed me along and were extremely tolerant and patient. Thank you very much.

 

[vanhees71]

I have to thank you! I learned finally, how to clearly and efficiently write the inversion formula for matrices using determinants with the Levi-Civita symbol.