4
$\begingroup$

Given that the affine connection can be written as: $$\Gamma ^a{}_{bc}= \bigg\{ {a \atop bc} \bigg\} - \frac{1}{2}(T^a{}_{bc}+T_c{}^a{}_b-T_{bc}{}^a) $$

Where $\big\{ {a \atop bc} \big\}$ denotes the metric connection, given by: $$\bigg\{ {a \atop bc} \bigg\}=\frac{1}{2}g^{ad}(\partial_bg_{dc}+\partial_cg_{bd}-\partial_dg_{bc})$$

And the $T^\alpha{}_{\beta \gamma}$ are the torsion tensors, defined as $T^\alpha{}_{\beta \gamma}=\Gamma^\alpha{}_{\beta \gamma}-\Gamma^\alpha{}_{\gamma \beta}$.

Now, defining an index symmetrisation operation as $\Gamma^a{}_{(bc)}\equiv\frac{1}{2}(\Gamma^a{}_{bc}+\Gamma^a{}_{cb})$, I need to prove that: $$\Gamma ^a{}_{(bc)}= \bigg\{ {a \atop bc} \bigg\} + T_{(bc)}{}^a$$

I've tried proving it by just plugging in the symmetrisation operation the expressions for both the affine connections, and it's clear that the metric connection terms are by definition equal (assuming symmetry in the metric functions), so they just sum together. Instead, the part involving the torsion tensors really puzzle me, cause it seems by definition that they are antisymmetric, so they should just cancel out. I'm guessing it has something to do with the raised and lowered indexes that only allows the one with the rightmost index raised to not cancel... But I'm completely stuck. Any help will be much appreciated!

$\endgroup$
6
  • 2
    $\begingroup$ Yes your hunch is right: the symmetry properties only apply for the specific index combination, so its just a case of manipulating the indices to match the desired form. $\endgroup$
    – Eletie
    Mar 10 at 10:24
  • $\begingroup$ And... Should I do that multiplying by the covariant or contravariant components of the metric tensor? If so, would't they still cancel out since the symmetries in the metric tensor won't induce any extra sign changes? $\endgroup$ Mar 10 at 10:29
  • 1
    $\begingroup$ Just to check, do you have a source for your first equation? Usually one writes $\Gamma = \{ \} + K$, with K being the contorsion tensor defined here en.wikipedia.org/wiki/Contorsion_tensor this slightly differs from your definitions. $\endgroup$
    – Eletie
    Mar 10 at 10:40
  • $\begingroup$ I got it from Hobson's General Realativity: An introduction for physicists. It's equation 3.11, and this is exercise 3.8, which also requires to prove that equation yourself (something I thought I had done right) $\endgroup$ Mar 10 at 10:45
  • 1
    $\begingroup$ I'll write a full answer as it's just some pretty messy indices manipulations. $\endgroup$
    – Eletie
    Mar 10 at 11:20
1
$\begingroup$

So here's one way of doing it, probably a little clumsy but trying to be as explicit as possible. Also note that I use both symmetry and antisymmetry brackets at the same time, so it's just about being careful with both. Stating with $$ \Gamma^{a}{}_{bc} = \big\{{}^a_{bc}\big\} - \frac{1}{2} \big( T^{a}{}_{bc} + T_{c}{}^{a}{}_b - T_{bc}{}^{a} \big) \ , $$ we can write this in terms of the connection (using antisymmetry brackets) as $$ \Gamma^{a}{}_{bc} = \big\{{}^a_{bc}\big\} - \frac{1}{2} \big(2\Gamma^{a}{}_{[bc]} +2 g_{cd} g^{ae} \Gamma^{d}{}_{[eb]} - 2 g_{bd}g^{ae} \Gamma^{d}{}_{[ce]} \big) \ . $$ Then symmetrising one finds $$ \begin{align} \Gamma^{a}{}_{(bc)} &= \big\{{}^a_{bc}\big\} - \frac{1}{2} \Big(0+2g_{(c|d} g^{ae}\Gamma^{d}{}_{[e|b)]}+2g_{(b|d} g^{ae}\Gamma^{d}{}_{[e|c)]} \Big) \\ &= \big\{{}^a_{bc}\big\} - \frac{1}{2} g^{ae} \big( g_{(c|d}\Gamma^{d}{}_{e|b)}-g_{(c|d}\Gamma^{d}{}_{b)e}+g_{(b|d}\Gamma^{d}{}_{e|c)}+g_{(b|d}\Gamma^{d}{}_{c)e}\big) \\ &= \big\{{}^a_{bc}\big\} - \frac{1}{2} g^{ae} \big(2g_{(c|d}\Gamma^{d}{}_{e|b)} -2g_{(c|d}\Gamma^{d}{}_{b)e}\big) \\ &= \big\{{}^a_{bc}\big\} + g^{ae} g_{(c|d}\Gamma^{d} {}_{b)e} - g^{ae}g_{(c|d}\Gamma^{d}{}_{e|b)} \\ &= \big\{{}^a_{bc}\big\} +g^{ae} g_{d(c} T^{d}{}_{b)e} \\ &= \big\{{}^a_{bc}\big\} + T_{(bc)}{}^{a} \ \ , \end{align} $$ which gives the desired outcome. The first term on the first line vanishes from $[(ab)]=0$, and the Christoffel symbol of course stays the same. The rest of the lines are just manipulating indices whilst trying to stay clear and explicit. Hope this helps.

$\endgroup$
2
  • $\begingroup$ Yes! This got me through it as soon as I got how that "non consecutive indexes symetrization" worked. Thank you very much! $\endgroup$ Mar 11 at 7:38
  • 1
    $\begingroup$ @JorgeCasajus glad that helped! No worries $\endgroup$
    – Eletie
    Mar 11 at 9:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.