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Problem with solution Torsion tensor & Christoffel symbols with torsion

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

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}