# Aufgaben:Problem 9

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## Contents

Let $$U \subset \mathbb{R}^n$$ be a domain. Assume that $$g$$ is a Riemannian metric field on $$U$$.

a) The covariant derivative $$\nabla_k$$ acts on a $$(0, 2)$$-tensorfield $$T_{lm}$$ by $$\nabla_k T_{lm} = \partial_k T_{lm} − T_{im}\Gamma^i_{kl} − T_{li}\Gamma^i_{km},$$ where $$\Gamma^l_{ij}$$ are the Christoffel symbols defined as $$\Gamma^l_{ij}=\frac{1}{2}g^{lk}(\partial_i g_{jk} + \partial_j g_{ik} − \partial_k g_{ij} ).$$ Show that $$\nabla_k g_{ij} = 0, \forall i, j, k \in \{1, ..., n\}.$$

b) Recall the Laplace-Beltrami operator $$L_g =\frac{1}{\sqrt{\det(g)}}\sum_{i,j}{\partial_j (\sqrt{\det(g)} g^{ij}\partial_i)}$$. Show that $$L_g(f) = g^{ij}\nabla_i(\partial_j f)$$ where $$f$$ is a smooth function on $$U$$.

Note: the covariant derivative $$\nabla_k$$ acts on a covector field $$v_l(x)$$ by $$\nabla_k v_l(x):=\partial_k v_l − v_i\Gamma^i_{kl}$$.

c) Let $$\chi: U \to \mathbb{R}_{>0}$$ be a strictly positive smooth function on $$U$$, and let $$\tilde{g} = \chi^2 g$$.

Show that the Christoffel symbols $$\tilde{\Gamma}^l_{ij}$$ (tilde refers to metric $$\tilde{g}$$) are given by $$\tilde{\Gamma}^l_{ij}= \Gamma^l_{ij} + (\partial_i \log{ \chi})\delta^l_j + (\partial_j \log{\chi})\delta^l_i − g^{lk}(\partial_k \log{ \chi})g_{ij} .$$ Conclude that, if $$\bar{g} := \frac{4}{(1−|x|^2)^2}\mathbb{I}$$, then $$\bar{\Gamma}^l_{ij}=\frac{2}{1 − |x|^2}(x_i\delta_{jl} + x_j\delta_{il} − x_l\delta_{ij} )$$

## Solution

### a)

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

from the symmetry of $$g_{ij}$$ and from: $$g_{ki}g^{ij} = g^{ji}g_{ik} = \delta^j_k$$ 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)

Claim: $$\partial_j \det g = (\det g) tr \Big(g^{-1}\partial_j g \Big)$$

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

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

\begin{align} \partial_j \det g &= \partial_j \det A = \partial_j \prod_{i=1}^n A_{ii} = \sum_{k = 1}^n \prod_{i=1}^n \frac{A_{ii}}{A_{kk}} \partial_j {A}_{kk} = \prod_{i=1}^n A_{ii} \sum_{k = 1}^n \frac{\dot{A}_{kk}}{A_{kk}}\\ &= \det A \sum_{k = 1}^n \frac{\dot{A}_{kk}}{A_{kk}} = \det A \sum_{k = 1}^n (A^{-1}\dot A)_{kk} = \det A\ tr( A^{-1} \partial_j{A})\\ &= \det g\ tr(A^{-1} \partial_j{A}) \overset{!}{=} \det g\ tr(g^{-1} \partial_j{g}) \end{align}

proving the last part separately (You can easily verify that the product rule holds for matrices):

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

where I used that $$T\dot{T^{-1}} = -\dot{T} T^{-1}$$. And the fact that the trace is invariant under conjugacy.

$$\square$$

Now we can prove (b):

$$L_g(f) = \frac{1}{\sqrt{\det g}} \partial_j(\sqrt{\det g} g^{ij} \partial_i) f$$

using the product rule (and swappig the terms around):

$$= \frac{1}{\sqrt{\det g}} \Big( \sqrt{\det g} g^{ij} \partial_j \partial_i +\sqrt{\det g} \partial_j (g^{ij}) \partial_i +\partial_j\Big(\sqrt{\det g}\Big) g^{ij} \partial_i)\Big) f$$

using the Claim: $$\partial_j \det g = (\det g) tr \Big(g^{-1}\partial_j g \Big) = (\det g) (g^{kl}\partial_j g_{lk})$$

$$= g^{ij} \partial_j \partial_i f + (\partial_j g^{ij})\partial_i f + \frac{1}{2} (g^{kl}\partial_j g_{lk}) g^{ij} \partial_i f$$

with the product rule: $$(\partial_j g^{-1}) = - g^{-1}(\partial_j g) g^{-1}$$. In particular: $$\partial_j g^{ij} = -g^{ik} (\partial_j g_{kl}) g^{lj}$$ (Notice that calling the summation indecies $$k$$ and $$l$$ is allowed, as it just combines the sums, which is possible as all indices run from $$1$$ to $$n$$)

$$= g^{ij} \partial_i \partial_j f -g^{ik} (\partial_j g_{kl}) g^{lj}\partial_i f + \frac{1}{2} g^{kl}(\partial_j g_{lk}) g^{ij} \partial_i f$$

Now some indecie swapping: In the second part: $$l \leftrightarrow i$$ and in the third part $$l \leftrightarrow i$$ and $$k \leftrightarrow j$$ (We can do this by seperating the sums, swapping indecies and then putting them back together)

$$= g^{ij} \partial_i \partial_j f -g^{lk} (\partial_j g_{ki}) g^{ij}\partial_l f + \frac{1}{2} g^{ji}(\partial_k g_{ij}) g^{lk} \partial_l f$$

$$= g^{ij} \Big(\partial_i \partial_j f -g^{lk} (\partial_j g_{ki}) \partial_l f + \frac{1}{2} (\partial_k g_{ij}) g^{lk} \partial_l f \Big)$$

$$= g^{ij} \Big(\partial_i \partial_j f - \partial_l f \frac{1}{2} g^{lk} \big( (\partial_j g_{ki}) + (\partial_j g_{ki}) -(\partial_k g_{ij}) \big) \Big)$$

notice that if we pull all the sums apart again we can switch $$(\partial_j g_{ki})$$ to $$(\partial_i g_{kj})$$ as the only other $$i,j$$ term in that sum would be $$g^{ij}$$ which is symmetric. Then the second term is Cristoffel:

$$= g^{ij} \Big(\partial_i \partial_j f - \partial_l f \Gamma^l{}_{ij}\Big) = g^{ij} \nabla_i (\partial_j f)$$

## Problem 9 (Craven)

$$\partial_k g_{lm} := g_{klm}$$

### a

\begin{align} \nabla_k g_{lm} &= g_{klm} - g_{mi}\Gamma^{i}_{km} \\ &= g_{klm} - \frac{1}{2} \overbrace{g_{mi}g^{ip}}^{\delta^{p}_{m}}\left(g_{klp}+g_{lpk}-g_{pkl})\right) - \frac{1}{2} \overbrace{g_{li}g^{ip}}^{\delta^{p}_{i}}\left(g_{kmp} + g_{mpk} - g_{pkm}\right) \\ &= g_{klm} - \frac{1}{2}\left(g_{klm} + g_{lmk} - g_{mkl}\right)-\frac{1}{2}\left(g_{klm}+g_{mlk}-g_{lkm}\right) = 0 \end{align}

### b

Insert the definition of the covariant derivative, define $$\partial_j f := f_j$$:

$$g^{ij}\nabla_i f_j = g^{ij}\partial_i f_j - \frac{1}{2}g^{lk}g^{ij}\left(g_{ijl}+g_{jli} - g_{lij}\right)f_k$$

Since we sum over i, j by symmetry of the indices $$g^{ij}g_{ijl} = g^{ij}g_{jli}$$ thus this is equal to the expression

$$g^{ij}\partial_i f_i - g^{ij}g_{ijl}g^{lk}f_k + \frac{1}{2}g^{ij}g_{lij}g^{lk}f_k$$

We now calculate with the other expression given:

$$\sqrt{\det{g}}^{-1}\partial_l\left(\sqrt{\det{g}}g^{lk}f_k\right)=\frac{1}{2\det{g}}\left(\partial_l \det{g}\right)g^{lk}f_k + \partial_l g^{lk}f_k + g^{lk}\partial_l f_k$$

Now what is left is to calculate the partial derivative of the determinant. By the chain rule we get: $$\partial_l \det{g} = d\det_{g}{\partial_lg}$$ Now we have to determine the linear map $$d\det_g{X}$$ that takes an element of the tangential space $$X$$ and maps it to a scalar. For this we use a suitable curve. By the chain rule if $$\phi(0) = \psi(0), \phi'(0) = \psi'(0)$$ then $$\frac{d}{dt}_{t=0}\det{\phi(t)}$$. Thus we pick the curve $$ge^{g^{-1}Xt} = \phi(t)$$. Inserting, using matrix identities and taking the derivative gives us $$d\det_g{\partial_l g} = \det{g}~\text{tr}(g^{-1}\partial_lg) = \det{g}g^{ij}g_{lij}$$. Thus we get the expression: $$\frac{1}{2} g^{ij}g_{lij}g^{lk}f_k+\partial_lg^{lk}f_k + g^{lk}\partial_lf_k$$ Thus all that is left to show is that $$\partial_lg^{lk}f_k = -g^{ij}g_{ijl}g^{lk}f_k$$.

$$gg^{-1}$$ and from product rule we get $$(\partial_mg)g^{-1} + g(\partial_mg^{-1}) = 0 \Rightarrow -g^{-1}\partial_mgg^{-1}=\partial_mg^{-1}$$. Inserting indices gives us $$-g^{ij}g_{mjk}g^{kl} = \partial_mg^{il}$$. Inserting $$m=i$$ and summing over the index i gives us $$-g^{ij}g_{jkl}g^{kl}=\partial_ig^{il}$$ thus the identity is proven.

### c

\begin{align} \begin{split} \tilde \Gamma_{ij}^{l} &= \frac{1}{2}\chi^{-2}g^{lk}\left(\partial_i\chi^2g_{jk} + \chi^{2}g_{ijk} + \partial_j\chi^{2}g_{ki} + \chi^2g_{jki}-\partial_k\chi^2g_{ij}- \chi^2g_{kij}\right) \\ &=\overbrace{\frac{1}{2}g^{lk}\left(g_{ijk} + g_{jki}-g_{kij}\right)}^{\Gamma^{l}_{ij}} + \overbrace{\frac{1}{2}\chi^{-2}\partial_i\chi^2}^{\partial_i\ln{\chi}}\overbrace{g^{lk}g_{jk}}^{\delta^l_j}+\overbrace{\frac{1}{2}\chi^{-2}\partial_j\chi^2}^{\partial_j \ln{\chi}}\overbrace{g^{lk}g_{ki}}^{\delta^{l}_{i}} - \overbrace{-\frac{1}{2}\chi^{-2}\partial_k\chi^2}^{\partial_k \ln{\chi}}g^{lk}g_{ij} \\ &= \Gamma^{l}_{ij} + \partial_i\ln{\chi}\delta^l_j + \partial_j\ln{\chi}\delta^l_i - \partial_k \ln{\chi}g^{lk}g_{ij} \end{split} \end{align} $$g_{ij} = \delta_{ij}, \chi = \frac{2}{1-|x|^2}, g^{ij}=\delta^{ij}$$ just as in the example given. $$\Gamma^l_{ij} = 0$$, since the metric $$I$$ is constant. Now $$\partial_i \ln{\chi}=\partial_i(\ln{2}-\ln{(1-|x|^2)}) = \frac{2x_i}{1-|x|^2}$$. We conclude: $$\tilde \Gamma^l_{ij} = \frac{2}{1-|x|^2}\left(x_i\delta^l_j + x_j\delta^l_i - x_k\delta^{lk}\delta_{ij}\right) = \frac{2}{1-|x|^2}\left(x_i\delta_{jl} + x_j\delta_{li} - x_l\delta_{ij}\right)$$