text: ## Tuesday, 19 January 2021
  
  ### Inverses, determinants, Levi-Civita symbol and Laplacian
  
  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.
explanation: The document is a detailed exploration of mathematical concepts related to inverses of matrices and tensors, determinants, the Levi-Civita symbol, and the Laplacian. It presents a series of questions and answers, delving into proofs and derivations of various formulas. The questions range from proving basic properties to more complex derivations involving tensor calculus and differential equations.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: ## Tuesday, 19 January 2021
  
  ### Inverses, determinants, Levi-Civita symbol and Laplacian
  
  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.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a detailed exploration of mathematical concepts related to inverses of matrices and tensors, determinants, the Levi-Civita symbol, and the Laplacian. It presents a series of questions and answers, delving into proofs and derivations of various formulas. The questions range from proving basic properties to more complex derivations involving tensor calculus and differential equations.
  
  Label: multiple-question-and-answer