Talk:Aufgaben:Problem 9

My idea is to transform \( g^{ij} \nabla_i(\partial_j f) \) into \(\Delta f \). From there we have to show \( L_g(f) = \Delta(f) \) as in notes_new at pg 82.

The first part should be:

$$ g^{ij} \nabla_i \partial_j + \underbrace{ \nabla_i g_{ij}}_\text{= 0, from a)} = g^{ij} \nabla_i \partial_j + \underbrace{( \nabla_i g_{ij}) \partial_j}_\text{= 0} = g^{ij} \nabla_i \partial_j + \underbrace{( \nabla_i g^{ij})\partial_j}_\text{= 0} = \nabla_i (g^{ij} \partial_j) = \nabla_i \partial^i = \Delta$$

f is not compactly supported so I don't think you can use this, although maybe they just forgot to write that. It does makes sense that we should use this identity, considering exercise (a) Carl (talk) 15:57, 20 June 2015 (CEST)

There is a simple proof for the derivative of the determinant:

We know that \(g\) is symmetric and therefore diagonalisable. \(\det g = \det (T^{-1}AT) = \det A\) where \(A\) is diagonal:

\begin{align} \frac{d}{dt} \det g(t) &= \frac{d}{dt}\det A(t) = \frac{d}{dt} \prod_{i=1}^n A_{ii}(t) = \sum_{j = 1}^n \prod_{i=1}^n \frac{A_{ii}(t)}{A_{jj}(t)} \dot{A}_{jj}(t) = \prod_{i=1}^n A_{ii}(t) \sum_{j = 1}^n \frac{\dot{A}_{jj}(t)}{A_{jj}(t)}\\ &= \det A(t) \sum_{j = 1}^n \frac{\dot{A}_{jj}(t)}{A_{jj}(t)} = \det A(t) \sum_{j = 1}^n (A(t)^{-1}\dot A(t))_{jj} = \det A(t) tr( A(t)^{-1} \dot{A}(t)) \end{align}