I have been wandering among inverses of matrices and tensors, determinants, Levi-Civita symbol and the Laplacian and found a wonderful formula for an inverse tensor (4). This leads to the differential equation connecting the determinant of a tensor, the determinant of its inverse, its inverse and itself at (5) and (6). The former can be used to prove the equivalence of Wikipedia's (7) Carroll's (8) formulas for the Laplacian which was where this all started.

I have presented the chain of reasoning as a set of questions and answers:

## Questions

Consider the following:

For a square matrix ##A## and its inverse ##A^{-1}##

\begin{align}

A^{-1}=\frac{1}{\det{\left(A\right)}}adj{\left(A\right)}&\phantom {10000}(1)\nonumber

\end{align}The adjugate is the transpose of the cofactor matrix. The cofactor matrix is the matrix whose ##i,j##th element is the ##\left(-1\right)^{i+j}## times the determinant of the matrix formed by removing row ##i## column ##j## from ##A##.

**a) Prove that equation (1) is true. Easy.**

**b) Rewrite it in component form. Very easy.**

Carroll's 2.66 for the determinant of a matrix ##A## with components ##A_{\ \ \ \ \ {\mu\prime}_1}^{\mu_1}##, was

\begin{align}{\widetilde{\epsilon}}_{{\mu\prime}_1{\mu\prime}_2\ldots{\mu\prime}_n}\det{\left(A\right)}={\widetilde{\epsilon}}_{\mu_1\mu_2\ldots\mu_n}A_{\ \ \ \ \ {\mu\prime}_1}^{\mu_1}A_{\ \ \ \ \ {\mu\prime}_2}^{\mu_2}\ldots A_{\ \ \ \ \ {\mu\prime}_n}^{\mu_n}&\phantom {10000}(2)\nonumber

\end{align}

Vanhees71's PF post uses the first simplification of (2) in four dimensions and then says, slightly rearranged, that components of ##g^{\mu_11}## are

\begin{align}

gg^{\mu_11}=\left(-1\right)^{\mu_1}\epsilon^{\mu_1\mu_2\mu_3\mu_4}g_{\mu_22}g_{\mu_33}g_{\mu_44}&\phantom {10000}(3)\nonumber

\end{align}I find a more general formula

\begin{align}

\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}&\phantom {10000}(4)\nonumber

\end{align}Vanhees71 used (3) to get

\begin{align}

\partial_\nu g=gg^{\rho\sigma}\partial_\nu g_{\rho\sigma}&\phantom {10000}(5)\nonumber

\end{align}**c) Are (3) and (4) (in 4 dimensions) the same? Easy.**

**d) Use (2) to prove that the correct one really is correct. Hard.**

**e) Use the correct one to prove (5). Middling.**

**f) Prove the generalization of (5):**

\begin{align}

\left|\bar{A}\right|\partial_\nu\left|A\right|={\bar{A}}^{\rho\sigma}\partial_\nu A_{\sigma\rho}&\phantom {10000}(6)\nonumber

\end{align}There is a Wikipedia formula in "arbitrary curvilinear coordinates in N dimensions" for the Laplacian:

\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}(7)\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}(8)\nonumber

\end{align}**g) Use (5) to prove that (7) and (8) are the same. Middling to hard.**

## Answers

c) No, d) The correct one is (4). The rest of the answers are in the first six pages of

The remaining six pages contain examples to test the answers and check progress.