1

Let $(g_{ij})$ be the fundamental tensor of the Finsler metric $F$, that is $g_{ij}(x,y) := \frac{1}{2}\frac{\partial^{2}F^{2}}{\partial y^{i}\partial y^{j}}$.

$$D\Big(\log \big(\sqrt{\det (g_{ij})}\big) \Big)=\frac{1}{2}g^{ij}Dg_{ij}.$$

My Question: Why $D(\det(g_{ij})) = D(g_{ij})$?

GNUSupporter 8964民主女神 地下教會
  • 17,998
user
  • 1,370

1 Answers1

2

Use this, for example, to show that $[d_A(\log\det A)](H)=\operatorname{trace}(A^{-1}H)$ which hopefully agrees with your RHS if you put $g=A$, $Dg=H$, and use the expression for the inverse of metric tensor and the trace of product.

Community
  • 1
rych
  • 4,445