The definition of torsion tensor is the following:

$$ \mathbf{T}(\mathbf{X},\mathbf{Y})=\nabla_{\mathbf{X}}\mathbf{Y}-\nabla_{\mathbf{Y}}\mathbf{X} -\left[\mathbf{X},\mathbf{Y}\right]. $$

In an holonomic base $\left[\mathbf{e}_a,\mathbf{e}_b\right]=0$ the coordinates are

$$ T{^a_{\ \ bc}}e_a=\nabla_b e_c-\nabla_c e_b=(\Gamma^a_{\ cb}-\Gamma^a_{\ bc})e_a $$

due to $ \nabla_b\mathbf{e}_c=\Gamma^a_{\ \ cb}\mathbf{e}_a$. This result is different from the general result

$$ T^a_{\ \ bc}= \Gamma^a_{bc}-\Gamma^a_{cb}. $$

Where am I wrong in this calculations?

  • $\begingroup$ This looks fine to me. $\endgroup$ Jul 10, 2017 at 19:39
  • $\begingroup$ The definition is actually $\nabla_b e_c = \Gamma^a_{bc}$ $\endgroup$
    – Slereah
    Jul 10, 2017 at 19:51
  • $\begingroup$ @Slereah according to Wikipedia it is not, if i'm not mistaken wheeler too uses my convention $\endgroup$ Jul 10, 2017 at 20:03
  • $\begingroup$ @Slereah hawking in the structure of space time uses my $\endgroup$ Jul 10, 2017 at 20:04
  • 1
    $\begingroup$ The answer is of course that there are varying sign conventions for torsion/curvature and varying conventions for the definition of $\Gamma^i{}_{jk}$. $\endgroup$
    – Ryan Unger
    Jul 11, 2017 at 0:24

2 Answers 2


It is just a matter of convention. MTW uses the convention that you have followed. Others, like Wald, use different one. Just a nuisance.

  • $\begingroup$ Ok, but if I want to use my convention $\nabla_b e_c = \Gamma^a_{cb}$ and to be self consistent with the definition of $\mathbf{T}$ with covariant derivative what I have to do? Because if I leave the definition as it is written in the question I have two different expression of the torsion tensor. May I have to put a minus in front of the definition of $\mathbf{T}$? (Indeed, I have to define the torsion tensor with the opposite sign convention $\mathbf{T}(\mathbf{X},\mathbf{Y})=-\left(\nabla_{\mathbf{X}}\mathbf{Y}-\nabla_{\mathbf{Y}}\mathbf{X} -\left[\mathbf{X},\mathbf{Y}\right]\right)$) $\endgroup$ Jul 11, 2017 at 8:46
  • $\begingroup$ Can be a correct solution? O it's wrong? $\endgroup$ Jul 11, 2017 at 8:47

The connection (in this case the Levi-Civita conn.) $\Gamma^\alpha{}_\beta$ is a one-form. We can express it as (in a coordinate (holonomic) basis) $$\Gamma^\alpha{}_\beta \equiv \Gamma_\mu{}^\alpha{}_\beta\, dx^\mu \tag 1$$ on another hand in the case of the spin-connection we have $$\omega^I{}_J \equiv \omega_\mu{}^I{}_J\, dx^\mu \tag 2$$ There are two types of the indices on the above equations. I would call the indices that appearLHS of the equations only: $\{\alpha,\beta\}$ for eqn(1) and $\{ I,J \}$ for eqn (2) the vector space indices or transformation indices (not the standard words), and I would call the index $\mu$ on both eqn. the form index.

Consider an equation write with only the vector space indices $$ \nabla {\bf e}^a = \Gamma^a{}_b {\bf e}_a \;.$$ There are two conventions to write in term with the form index

CONVENTION 1 $$ \nabla_c {\bf e}^a = \Gamma_c{}^a{}_b {\bf e}_a =: \Gamma^a_{\underline{c}b}{\bf e}_a\;,$$ where the underline indicate that it is a form index.

CONVENTION 2 $$ \nabla_c {\bf e}^a = \Gamma^a{}_{bc} {\bf e}_a =: \Gamma^a_{b \underline{c}}{\bf e}_a\;,$$ where the underline indicate that it is a form index. In the both case $c$ is a form index.

If you notice that which index is a form index you will never confuse.

Your second equation $$T{^a_{\ \ bc}}e_a=\nabla_b e_c-\nabla_c e_b=(\Gamma^a_{\ cb}-\Gamma^a_{\ bc})e_a \tag 3$$ is obviously use the CONVENTION 2 , we can write $$T{^a_{\ \ bc}}e_a=\nabla_b e_c-\nabla_c e_b=(\Gamma^a_{\ c\underline b}-\Gamma^a_{\ b\underline c})e_a \tag 4$$ but your third equation us the CONVENTION 1 we can write it as $$ T^a_{\ \ bc}= \Gamma^a_{\underline b c}-\Gamma^a_{\underline c b}. \tag 5 $$

$T^a_{\ \ bc}$ in (4) and (5) are the same.

  • $\begingroup$ You give the covariant basis. You should write $\nabla_c e_a $ instead. $\endgroup$
    – Dimiter P
    Aug 17, 2021 at 6:18

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