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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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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.

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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.

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