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The torsion tensor is defined as (Hawking p.34) \begin{equation} \mathbf{T}(\mathbf{X},\mathbf{Y}) = \nabla_{\mathbf{X}}\mathbf{Y} - \nabla_{\mathbf{Y}}\mathbf{X} - [\mathbf{X},\mathbf{Y}]. \end{equation} The connection is defined as (Hawking p.31) \begin{equation} \Gamma^a{}_{bc} = \langle\mathbf{E}^a,\nabla_{\mathbf{E}_b}\mathbf{E}_c\rangle, \end{equation} where $\{\mathbf{E}_a\}$ is any vector basis. So we have \begin{equation} T^c{}_{ab} = \langle \mathbf{E}^c,\mathbf{T}(\mathbf{E}^a,\mathbf{E}^b) \rangle = \Gamma^c{}_{ab}-\Gamma^c{}_{ba} - \langle \mathbf{E}^c,[\mathbf{E}^a,\mathbf{E}^b] \rangle \end{equation} In coordinate basis the Lie bracket (commutator) vanishes, but in general the commutator coefficients do not vanish, e.g. in the non-coordinate basis.

Let $\mathbf{T}=0$. Does it mean that the connection is symmetric only in the coordinate basis?

On the other hand, we can calculate The torsion tensor in holonomic coordinate as \begin{equation} T^\sigma{}_{\mu\nu} = \Gamma^{\sigma}{}_{\mu\nu} - \Gamma^{\sigma}{}_{\nu\mu}, \end{equation} or in an orthonormal frame (indicating with Latin indices) \begin{equation} T^c{}_{ab} = \Gamma^{c}{}_{ab} - \Gamma^{c}{}_{ba}-e^\mu_ae^\nu_b(e^c_{\mu,\nu}-e^c_{\nu,\mu}). \end{equation} Can we show they are equivalent under change of basis? The problem is usually we write tensor equation with abstract index notation, and I just found that they give different results with different a priori chosen basis. Does it mean that we have to express the tensor equation with non-coordinate basis, since it seems to be more general?

EDIT

This should be a fairy straightforward tensor manipulation (I should have put more effort).

Define $\partial_a \mathbf{e}_b = \Gamma^c{}_{ab}\mathbf{e}_c$ where the Latin indices denote orthonormal frame. Then \begin{equation} \partial_a \mathbf{e}_b = e^\mu_a \nabla_\mu (e^\nu_b e_\nu) = [e^\nu_a(\partial_\mu e^\sigma_b) + e^\mu_a e^\nu_b \Gamma^\sigma{}_{\mu\nu}]e^c_\sigma e_c, \end{equation} that is \begin{equation} \Gamma^c{}_{ab} = e^\mu_a e^c_\sigma \partial_\mu e^\sigma_b + e^\mu_a e^\nu_b e^c_\sigma \Gamma^\sigma{}_{\mu\nu}. \end{equation} Plug this into $e^\mu_a e^\nu_b e^c_\sigma T^\sigma{}_{\mu\nu}$ we can show it is equal to $T^c{}_{ab}$.

The question remains: to manipulate these tensor object we use abstract index notation. Given a tensor equation like $\mathbf{T}(\mathbf{X},\mathbf{Y}) = \nabla_{\mathbf{X}}\mathbf{Y} - \nabla_{\mathbf{Y}}\mathbf{X} - [\mathbf{X},\mathbf{Y}]$, how do we write it in abstract index notation? The notation of symmetric relies on the index placement of the component. For $\mathbf{T}=0$ it give symmetric $\Gamma^\sigma{}_{\mu\nu}$ but not to the $\Gamma^a{}_{bc}$. Are they all equal for whatever basis we choose to express, and differ only in physical expression (like what observer sees)? However, on page 24 of the book by Wald, the author states (relating to abstract index notation)

However, in some cases it will be convenient to use a particular type of basis, e.g., a coordinate basis adapted to the symmetries of a particular spacetime. If we do this, then the equations we write down for the tensor components may be valid only in this basis.

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  • $\begingroup$ Well, you just wrote down the proof of that, didn't you? I'm really not sure what the question is. $\endgroup$ – ACuriousMind Apr 29 '16 at 9:47
  • $\begingroup$ I think the question is I do not know if an equation is only valid in a specified basis or the equations are a true tensor equations independent of basis. $\endgroup$ – Henry Apr 29 '16 at 10:05
  • $\begingroup$ Your first equation is manifestly a true tensor equation. The second and third express the first in an arbitrary basis. Then the fourth and fifth depend on your specific choice of basis. What is the question? Is the confusion that most physics texts only work in coordinate bases and you have hence never seen the non-holonomic version? $\endgroup$ – ACuriousMind Apr 29 '16 at 10:10
  • $\begingroup$ So I should express the commutator as $\langle \mathbf{E}^c,[\mathbf{E}^a,\mathbf{E}^b] \rangle = \gamma^c{}_{ab}$ and leave it there, just to be independent of basis? You are right, I've never seen the non-holonomic version. What do we mean by "abstract index notation"? $\endgroup$ – Henry Apr 29 '16 at 10:20
  • $\begingroup$ Can you recommend any text that treats non-holonomic basis? For example to show every usual geometric object of GR in non-holonomic basis? $\endgroup$ – Henry Apr 29 '16 at 10:41
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Non-coordinate bases are always used when coupling to fermions. You could simply look at the maths literature, but as a physicist you'll want to look at textbooks on supergravity. Take a look at words such as "vielbein", "frame" and "spin connection". As to literature, there's Freedman/van Proeyen "Supergravity", chapter 7. Alternatively Ortin's "Gravity and Strings". Also note that torsion shows up in supergravity due to the presence of the Fermions. So that's also nicely discussed in those books.

I would have posted this as a comment, yet apparently my reputation score here is insufficient for doing so.

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  • $\begingroup$ Thanks for the reply. Why is it different to use coordinate or non-coordinate basis in describing a tensor object at each point on a manifold? $\endgroup$ – Henry Apr 30 '16 at 0:40
  • $\begingroup$ At the level of classical GR there is no fundamental difference. These are just two tools used to describe the same thing. You could say that non-coordinate basis approach is a little more general, but it is also a little more involved. If you are not going "deep" into GR, coordinate bases are simpler and completely suficient for you. On the other hand, in some specific cases use of right basis might lead to tremendous simplification of calculations. $\endgroup$ – Blazej Apr 30 '16 at 7:44
  • $\begingroup$ Just to add: If you have coordinates $x^m$ you would write the basis of tangent space as $\partial_m$, that of cotangent space as $dx^m$. If you have a non-coordinate basis in tangent space you would call this $E_a$. In terms of a coordinate basis, $E_a = E_a^m \partial_m$. The cotangent bundle then has the basis $e^a = e^a_m dx^m$ and as matrices the $e^a_m$ and $E_a^m$ are inverse. For a tensor $T_m^n$ you can write $T_m^n = e_m^a E_b^n T_a^b$ etc. So from this point there is no difference. Things change when you take derivatives. $[E_a, E_b] \neq 0$, $[\partial_m, \partial_n] = 0$. $\endgroup$ – jws Apr 30 '16 at 14:22

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