## Wednesday, 6 January 2021

### The Voss-Weyl formula

 Hermann Weyl
The Voss-Weyl formula is for the contraction of the covariant derivative of a vector field:$$\nabla_\mu F^\mu=\frac{1}{\sqrt{\left|g\right|}}\frac{\partial}{\partial x^\mu}\left(\sqrt{\left|g\right|}F^\mu\right)$$It immediately solved an outstanding problem from a year ago that I had with the Laplacian. While proving it I find three other interesting formulas:
Jacobi's formula is$$\partial_kQ={\widetilde{Q}}^{ji}\partial_kQ_{ij}$$where $Q_{ij}$ is a tensor, $Q$ is its determinant and what might be new is ${\widetilde{Q}}^{ij}$ which is the cofactor of $Q_{ij}$ and thus as a matrix, ${\widetilde{Q}}^{ji}$ is the transpose of the cofactors of $Q_{ij}$. The cofactor of a component $Q_{ij}$ is the determinant of the matrix made by removing row $i$, column $j$ then multiplying by $\left(-1\right)^{i+j}$. The important thing is that the transpose of the cofactors is called the adjugate, so ${\widetilde{Q}}^{ji}$ is the adjugate and the inverse of a matrix is the adjugate divided by its determinant. We write that:$$Q^{ij}=\frac{1}{Q}{\widetilde{Q}}^{ji}$$Finally on the way to proving the Voss-Weyl formula we find the marvellously symmetrical $$Q^{-1}\partial_kQ=Q^{ij}\partial_kQ_{ij}$$where $Q^{-1}$ is the determinant of the inverse of a rank 2 tensor $Q_{ij}$ which has determinant $Q$. It is also well known that $Q^{-1}=\frac{1}{Q}$.

I must thank David A. Clarke of Saint Mary’s University, Halifax NS, Canada for his guidance.