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Level 4
Level 4 requires the knowledge of vector calculus, (multidimensional) differential and integral calculus. Suitable for advanced students.

# Problem with solutionTorsion tensor & Christoffel symbols with torsion

Let's consider the case where the covariant derivative operator $$\nabla$$ does not fulfill the condition for torsion-freeness.

1. Show that there exists a so called torsion tensor $$T^{c}_{\;ab}$$, such that for all smooth functions $$f$$ the following condition is fulfilled:$$\nabla_a \, \nabla_b \, f ~-~ \nabla_b \, \nabla_a \, f ~=~ - T^{c}_{\;ab} \, \nabla_c \, f$$
2. Given a metric $$g^{ab}$$, we can calculate Christoffel Symbols without torsion in the following way:$$\Gamma^{c}_{\;ab} ~=~ \frac{1}{2} \, g^{cs} \, \left( \partial_a \, g_{bs} ~+~ \partial_b \, g_{as} - \partial_s \, g_{ab} \right)$$Derive the Christoffel Symbols WITH torsion.
Solution tips

Use properties of the covariant derivative operator $$\nabla$$, like linearity, the Leibniz product rule and commutativity with contraction. Also relabel the indices to get different equations and combine the equations.

## Solution

Solution for (a)

Two covariant derivatives $$\nabla$$ and $$\widetilde{\nabla}$$ differ by a tensor field $$C^{c}_{\;ab}$$ if they're applied to a dual vector $$\omega_b$$:1$$\nabla_a \, \omega_b ~=~ \widetilde{\nabla}_a \, \omega_b ~-~ C^{c}_{\;ab} \, \omega_c$$

This relation was derived without using the torsion freeness property!

The equation 1 must be also fulfilled if we relabel the indices $$a \rightarrow b$$ and $$b \rightarrow a$$:2$$\nabla_b \, \omega_a ~=~ \widetilde{\nabla}_b \, \omega_a ~-~ C^{c}_{\;ba} \, \omega_c$$

Subtract eq. 2 from 1:3\begin{align} \nabla_a \, \omega_b ~-~ \nabla_b \, \omega_a &~=~ \widetilde{\nabla}_a \, \omega_b ~-~ \widetilde{\nabla}_b \, \omega_a \\\\ &~-~ C^{c}_{\;ab} \, \omega_c ~-~ C^{c}_{\;ba} \, \omega_c \end{align}

Write dual vectors as $$\omega_b = \nabla_b \, f$$, $$\omega_a = \nabla_a \, f$$ and $$\omega_c = \nabla_c \, f$$ and then factor out $$f$$:4\begin{align} \left( \nabla_a \, \nabla_b ~-~ \nabla_b \, \nabla_a \right) \, f &~=~ \left( \widetilde{\nabla}_a \, \nabla_b ~-~ \widetilde{\nabla}_b \, \nabla_a \right) \, f \\\\ &~-~ \left( C^{c}_{\;ab} ~-~ C^{c}_{\;ba} \right) \, \nabla_c \, f \end{align}

We assumed that $$\nabla$$ is not torsion-free, so the left hand side is not zero. But we can choose $$\widetilde{\nabla}$$ as torsion-free, so that the term with $$\widetilde{\nabla}$$ vanishes:5$$\left( \nabla_a \, \nabla_b ~-~ \nabla_b \, \nabla_a \right) \, f ~=~ - \left( C^{c}_{\;ab} ~-~ C^{c}_{\;ba} \right) \, \nabla_c \, f$$

Define:6$$T^{c}_{\;ab} ~:=~ C^{c}_{\;ab} ~-~ C^{c}_{\;ba}$$and insert into 5:

7$$\left( \nabla_a \, \nabla_b ~-~ \nabla_b \, \nabla_a \right) \, f ~=~ - T^{c}_{\;ab} \, \nabla_c \, f$$
Solution for (b)

We use the definition of the covariant derivative operator $$\nabla$$ aplied to a tensor (in this case metric $$g_{ab}$$) and metric compatibility property $$\nabla_c \, g_{ab} = 0$$:8$$0 ~=~ \nabla_c \, g_{ab} ~=~ \partial_c \, g_{ab} ~-~ \Gamma^{d}_{\;ca} \, g_{db} ~-~ \Gamma^{d}_{\;cb} \, g_{ad}$$

Rearrange for $$\partial_c \, g_{ab}$$:9\begin{align} \partial_c \, g_{ab} &~=~ \Gamma^{d}_{\;ca} \, g_{db} ~+~ \Gamma^{d}_{\;cb} \, g_{ad} \\\\ &~=~ \Gamma_{bca} ~+~ \Gamma_{acb} \end{align}

This equation must also be fulfilled if we relable the indices $$c \rightarrow a$$ and $$a \rightarrow c$$:10$$\partial_a \, g_{cb} ~=~ \Gamma_{bac} ~+~ \Gamma_{cab}$$

And to get a third equation we permute the indices $$a \rightarrow c$$, $$c \rightarrow b$$ and $$b \rightarrow a$$:11$$\partial_b \, g_{ca} ~=~ \Gamma_{abc} ~+~ \Gamma_{cba}$$

Now, add equations 1 and 2 and then subtract 3:12\begin{align} \partial_c \, g_{ab} ~+~ \partial_a \, g_{cb} ~-~ \partial_b \, g_{ca} &~=~ \Gamma_{bca} ~+~ \Gamma_{acb} ~+~ \Gamma_{bac} \\\\ &~+~ \Gamma_{cab} ~-~ \Gamma_{abc} ~-~ \Gamma_{cba} \end{align}

Substitute the definition of the torsion tensor from (a) into eq. 12:13\begin{align} \partial_c \, g_{ab} ~+~ \partial_a \, g_{cb} ~-~ \partial_b \, g_{ca} &~=~ \Gamma_{bca} ~+~ \Gamma_{bac} \\\\ &~+~ T^{a}_{\;cb} ~+~ T^{c}_{\;ab} \end{align}

We can rewrite 13 using:13.1$$\Gamma_{bca} ~+~ \Gamma_{bac} ~=~ T^{b}_{\;ca} ~+~ 2 \, \Gamma_{bac}$$

Then weg get:14\begin{align} \partial_c \, g_{ab} ~+~ \partial_a \, g_{cb} ~-~ \partial_b \, g_{ca} &~=~T^{b}_{\;ca} ~+~ 2 \, \Gamma_{bac} \\\\ &~+~ T^{a}_{\;cb} ~+~ T^{c}_{\;ab} \end{align}

Rearrange for $$\Gamma_{bac}$$:15\begin{align} \Gamma_{bac} &~=~ \frac{1}{2} ( \partial_a \, g_{cb} ~+~ \partial_c \, g_{ab} ~-~ \partial_b \, g_{ca} \\\\ &~-~ T^{b}_{\;ca} ~-~ T^{d}_{\;cb} ~-~ T^{c}_{\;ab} ) \end{align}

Raise index $$b$$:16\begin{align} \Gamma^{b}_{\;ac} &~=~ \frac{1}{2} g^{bs} \, ( \partial_a \, g_{cs} ~+~ \partial_c \, g_{as} ~-~ \partial_s \, g_{ca} \\\\ &~-~ T^{s}_{\;ca} ~-~ T^{d}_{\;cs} ~-~ T^{c}_{\;as} ) \end{align}

Relabel $$b \rightarrow c$$ and $$c \rightarrow b$$ to get:

Christoffel symbols with torsion17\begin{align} \Gamma^{c}_{\;ab} &~=~ \frac{1}{2} \, g^{cs} \, ( \partial_a \, g_{bs} ~+~ \partial_b \, g_{as} - \partial_s \, g_{ab} \\ &~~~~~~~~~~~~~~~~ ~-~ T^{s}_{\;ba} ~-~ T^{d}_{\;bs} ~-~ T^{b}_{\;as}) \end{align}