Aufgaben:Problem 9

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Solution

a)

$$\nabla_k g_{ij} = \partial_k g_{ij} - g_{nj} \Gamma^n_{kj} - g_{in}\Gamma^n_{kj} =$$ $$ = \partial_k g_{ij} - g_{nj}(\frac{1}{2} g^{np}(\partial_k g_{ip} + \partial_i g_{kp} - \partial_p g_{ki})) - g_{in}(\frac{1}{2} g^{nq}(\partial_k g_{jq} + \partial_j g_{kq} - \partial_q g_{kj})) =$$

from the symmetry of \( g_{ij} \) and from: \( g_{ki}g^{ij} = g^{ji}g_{ik} = \delta^i_k \) we obtain:

$$ = \partial_k g_{ij} - \frac{1}{2} \delta^p_j(\partial_k g_{ip} + \partial_i g_{kp} - \partial_p g_{ki}) - \frac{1}{2} \delta^q_i(\partial_k g_{jq} + \partial_j g_{kq} - \partial_q g_{kj}) = $$

$$ = \partial_k g_{ij} - \frac{1}{2} (\partial_k g_{ij} + \partial_i g_{kj} - \partial_j g_{ki}) - \frac{1}{2} (\partial_k g_{ji} + \partial_j g_{ki} - \partial_i g_{kj}) = $$

$$ = \partial_k g_{ij} - \frac{1}{2} (\partial_k g_{ij} + \partial_k g_{ji}) - \frac{1}{2} (\partial_j g_{ki} - \partial_i g_{kj} + \partial_i g_{kj} - \partial_j g_{ki} ) = 0 $$

again we used the symmetry of \( g_{ij} \) in the first bracket.


b)

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

The first part should be:

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


b)

\begin{align} \begin{split} L_g(f) &= \frac{1}{\sqrt{\det g}} \sum_{i,j} \partial_j(\sqrt{\det g} g^{ij} \partial_i) f \\ &= \frac{1}{\sqrt{\det g}} \sum_{i,j} \partial_j\Big(\sqrt{\det g}\Big) g^{ij} \partial_i) f + \sqrt{\det g} \partial_j (g^{ij}) \partial_i f + \sqrt{\det g}) g^{ij} \partial_j \partial_i f \\ &= \sum_{i,j} \partial_j g^{ij} \partial_i f + g^{ij} \partial_j \partial_i f + \frac{1\det g }{2 \det g} tr \Big(g^{lm}\partial_j g_{mk} \Big) g^{ij} \partial_i f \\ &= \sum_{i,j} g^{ij} \partial_i \partial_j f + \partial_j g^{ij} \partial_i f + \frac{1}{2}g^{ij} g^{km}\partial_j g_{mk} \partial_i f \\ &= \sum_{i,j} g^{ij} \partial_i \partial_j f + \partial_i g^{ij} \partial_j f + \frac{1}{2}g^{ij} g^{km}\partial_i g_{mk} \partial_j f \\ &= \sum_{i,j} g^{ij} \partial_i \partial_j f - g^{lk} \partial_l g_{ki} g^{ij} \partial_j f + \frac{1}{2}g^{ij} g^{km}\partial_i g_{mk} \partial_j f \\ &= g^{ij} \partial_i \partial_j f - g^{lk} \partial_l g_{ki} g^{ij} \partial_j f + \frac{1}{2}g^{ij} g^{km}\partial_i g_{mk} \partial_j f \end{split} \end{align} Now we change indicies this is possible because sum over all indicies $$j \leftrightarrow l$$, $$i \leftrightarrow k$$.

\begin{align} \begin{split} &= g^{ij} \partial_i \partial_j f - g^{ji} \partial_j g_{ik} g^{kl} \partial_l f + \frac{1}{2}g^{kl} g^{ij}\partial_k g_{ij} \partial_l f \\ &= g^{ij}\Big(\partial_i f_j - g^{lk} \partial_j g_{ik} f_l + \frac{1}{2}g^{kl} \partial_k g_{ij} f_l \Big)\\ &=g^{ij}\Big(\partial_i f_j -\frac{1}{2} f_l g^{lk}(\partial_i g_{jk} + \partial_j g_{ik} - \partial_k g_{ij})\Big) \\ &=g^{ij}\nabla_i\partial_j f \end{split} \end{align}