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There is a saying in Nakahara's Geometry, Topology and Physics P371 about principal bundles and associated vector bundles:

In general relativity, the right action corresponds to the local Lorentz transformation while the left action corresponds to the general coordinate transformation.

Because the structure group right acts on Principal bundles and left acts on associated vector bundles.

But I don't think that the local Lorentz transformation is general coordinate transformation. Since for local Lorentz transformation, the structure group is $O^{+}_{\uparrow}(1,3)$ while for general coordinate transformation, the structure group is $GL(4,\mathbb{R})$.

So is the book wrong? Or I didn't understand correctly.

Qmechanic
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  • I improved your title, but please check whether it's still accurate since I'm not sure I know this area well enough to write a good title. – David Z Feb 23 '16 at 04:11
  • Yes. Local Lorentz transformation is not the same thing as general coordinate transformations. What statement in the quote you have made you think that it should be so? – Prahar Feb 24 '16 at 00:27
  • Related: http://physics.stackexchange.com/q/155315/2451 – Qmechanic Aug 10 '16 at 21:52

3 Answers3

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You did not understand correctly, although Nakahara's statement is slightly wrong. I looked into the book, because your quote is insufficient to determine what you or Nakahara are talking about.

Nowhere is Nakahara talking about a local Lorentz transformation being a general coordinate transformation. In the context of your quote, the group in question is still the full $\mathrm{GL}(4,\mathbb{R})$ of the frame bundle. Additionally, he is not saying that every local $\mathrm{GL}(4)$ transformation is a coordinate transformation.

What he is saying is that the left action of the transition function of the frame bundle is the action of the Jacobian of a coordinate transformation, and that the action of local $\mathrm{GL}(4)$-valued functions on the frame bundle is the action of a local Lorentz transformation. The latter statement is incorrect, of course - not every matrix in $\mathrm{GL}(4)$ is an element of the Lorentz group.

ACuriousMind
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By a "frame" in GR we tend to mean a tetrad or vielbein $e^\mu_i$, $\mu,i\in\{0,1,2,3\}$. See here for the precise definition. Now, we can think of this object $e$ as a matrix. It is common to refer to the index $\mu$ as a "coordinate" index and $i$ as a "Lorentz" index. Now, as Nakahara talks about in Chap. 7, a tensor transforms with a term like $\partial x^\mu/\partial\tilde x^\nu$, which is a $\mathrm{GL}(4,\mathbb{R})$ matrix. The $i$ index transforms via Lorentz transformations, as explained in the linked post. So if we imagine the matrix $e$ being written as $(e)_{\mu i}$, it is clear that we must multiply by the appropriate $\mathrm{GL}(4,\mathbb{R})$ matrix on the left to get coordinate transformations. We multiply on the right by an $\mathrm{O}^+_\uparrow (1,3)$ matrix to get Lorentz transformations.

Community
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Ryan Unger
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Here I answer on the question from the title.

There is some feature which is related to the spinor representation of the proper Lorentz group. In fact, there is homomorphism $SL(2, C) \to SO(3,1)$, whose core contains two elements - unity and minus unity. The second one corresponds to representation $$ T(N)= -T(-N), \quad N\in SL(2,C) $$ and gives rise to applications of spinors in physics. For this representation we must introduce Lorentz group representation for which $$ T(\Lambda)=\pm T(N), \quad \Lambda (N)\in SO(3,1) $$ We can't throw out one of signs, since this breaks the continuity and the group property $T(\Lambda_{2})T(\Lambda_{1}) = T(\Lambda_{2}\Lambda_{1})$. So if we talk about the "usual" representations of the proper Lorentz group, we don't deal with spinor representations. If you mean the "usual" (unique) representations of the Lorentz group, then the statement that they are $GL(4,R)$ elements is correct. If you, however, talk about projective representations, which are realized in QFT and include "two-sign" representations, then the statement is incorrect: the group $GL(4, R)$ of general coordinate transformations doesn't contain representations similar to the spinor representation of the Lorentz group.

Name YYY
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  • The statement "there is a homomorphism $\mathrm{SO}(3,1)\to\mathrm{SL}(2,\mathbb{C})$" is false. Said homomorphism supposedly arises from a homomorphism of the algebras, and so only integrates to the connected component $\mathrm{SO}^+(3,1)$. The correct statement is that $\mathrm{SL}(2,\mathbb{C})\times\mathrm{SL}(2,\mathbb{C})$ is the double cover of the complexified Lorentz group $\mathrm{SO}(1,3)_\mathbb{C}$, where the latter is obtained as the simply connected Lie integration of $\mathfrak{so}(1,3)\otimes\mathbb{C}$. – ACuriousMind Feb 23 '16 at 17:04
  • Also, you seem to have answered a different question - this one is about whether or not local Lorentz transformations are general coordinate transformations, and has nothing to do with spinor or vector representations. – ACuriousMind Feb 23 '16 at 17:06
  • @ACuriousMind : I've meaned that there is a homomorphism $SO^{\uparrow}(3,1) \to SL(2,C)$, since before this I've written about proper group (I've forgotten $\uparrow$ symbol). If I understand correctly, you write that this statement is wrong, don't you? – Name YYY Feb 23 '16 at 17:22
  • @ACuriousMind : "... this one is about whether or not local Lorentz transformations are general coordinate transformations, and has nothing to do with spinor or vector representations...", - in my opinion, these questions are directly related, since "two-sign" representations of the proper Lorentz group are "unique" representations of its spinor group. – Name YYY Feb 23 '16 at 17:24
  • Ah, with $\uparrow$ (if that means proper orthochronous, i.e. the connected component), it is correct. I don't see any relation between general coordinate transformations and the spinor representations at all, though. This is completely classical GR, there are no spinors, and it is just a matter of fact that the Jacobian of general coordinate transformations need not be a Lorentz group element, hence the authors claim about the correspondence of general transformations and local Lorentz transformations is just wrong. – ACuriousMind Feb 23 '16 at 17:31
  • However, you have it the wrong way around for your homomorphism still, cf. point 4 in this answer by Qmechanic. The homomorphism is from $\mathrm{SL}(2,\mathbb{C})$ to the proper orthochronous Lorentz transformations, not the way you write (and you can't invert it precisely because the kernel is non-trivial). – ACuriousMind Feb 23 '16 at 17:35
  • @ACuriousMind : yes, here I'm wrong. – Name YYY Feb 23 '16 at 17:43
  • "...This is completely classical GR, there are no spinors, and it is just a matter of fact that the Jacobian of general coordinate transformations need not be a Lorentz group element..." But it seems that OP asked is the proper Lorentz group the subgroup of $GL(4,R)$. – Name YYY Feb 23 '16 at 18:05