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:

(Note that $$A_{jj}(t) > 0$$ since $$g$$ is also positive definite $$\Leftrightarrow$$ the eigenavalues of $$g$$ are strictly positiv)

\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}

Carl (talk) 11:51, 28 June 2015 (CEST)

There was some stuff I forgot at the end. I added the full proof to the solution. I used dots in instead of derivatives in most places, this is probably bad style but it looks less confusing this way.

Carl (talk) 15:50, 28 June 2015 (CEST)

you forgot to change the indicies in b) of solution (so you use j for the derivative and also for to sum over, which is confusing)

I've also got an alternative proof for $$tr(g^{-1}\dot{g})=tr(A^{-1}\dot{A})$$:

\begin{align} g^{-1}\dot{g}&=(T^{-1}AT)^{-1}\dot{(T^{-1}AT)} \\ &=T^{-1}A^{-1}T(\dot{T^{-1}}AT+T^{-1}\dot{A}T+T^{-1}A\dot{T}) \\ &=T^{-1}(-A^{-1}\dot{T}T^{-1}A+A^{-1}\dot{A}+\dot{T}T^{-1})T \end{align}

As $$\dot{TT^{-1}}=\dot{T}T^{-1}+T\dot{T^{-1}}=0$$.

Since the trace is linear and invariant under conjugacy, $$tr(g^{-1}\dot{g})=tr(T^{-1}(-A^{-1}\dot{T}T^{-1}A+A^{-1}\dot{A}+\dot{T}T^{-1})T)=tr(-A^{-1}\dot{T}T^{-1}A+A^{-1}\dot{A}+\dot{T}T^{-1})=-tr(-A^{-1}\dot{T}T^{-1}A)+tr(A^{-1}\dot{A})+tr(\dot{T}T^{-1})=tr(A^{-1}\dot{A})$$

--Brynerm (talk) 09:53, 1 July 2015 (CEST)

thanks for pointing that out with the j's. And yes it is a bit faster using the fact that the trace is invariant under conjugacy instead of always using $$tr(AB) = tr(BA)$$.

Carl (talk) 11:19, 1 July 2015 (CEST)

I think there is a little typo in Cravens solution. When he calculates $$L_{g}$$ in part b), I believe he meant to write $$\frac{1}{2\det{\boldsymbol{g}}}$$ instead of $$\frac{1}{2\det{\boldsymbol{f}}}$$

--Simon (talk) 10:57, 22 July 2015 (CEST)