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A very short question: Does the spin connection that we encounter in General Relativity $$\omega_{\mu,ab}$$ have a geometric meaning? I am supposing it does because it comes from mathematical terms that take geometric parts on a manifold but I am not sure how to visualize that?

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The spin connection has a geometric meaning - it is a connection associated to a particular non-coordinate frame, which diagonalizes the metric. Here's how:

Let $M$ be our spacetime. The tangent bundle $TM$ may be thought of as the associated bundle to an $\mathrm{SO}(n)$-principal bundle, where the $\mathrm{SO}(n)$ matrices represent ordered orthonormal bases at every point (every column vector is orthonormal to every other in such a matrix, which is the way in which it represents a basis).

The spin connection is now just a $\mathfrak{so}(n)$-valued connection 1-form $\omega$ on $TM$, which may locally be expanded as $$ \omega = \omega_\mu \mathrm{d}x^\mu = {{\omega_\mu}^a}_b {T^b}_a\mathrm{d}x^\mu$$ and the ${{\omega_\mu}^a}_b$ the the connection coefficients physicists usually deal with, and the ${T^b}_a$ are a basis for the $\mathfrak{so}(n)$ matrices, usually the simple antisymmetric matrices with two non-zero entries own would always write down.

Usually, we think of tangent vectors as being expanded as $v = v^\mu \partial_\mu$, so the natural basis at every point is given by the coordinates, which may be arbitrarily ugly. in particular, the metric is $g_{\mu\nu}$. We now want to (locally) change frames such that the metric becomes the standard diagonal metric $\eta_{\mu\nu}$ 1 because that one is evidently easier to work with. Such a (local) change of frames is given by a linear invertible map $$ e : TM \to TM$$ which is given in components by ${e^a}_\mu$ with $b^a = e^a{}_\mu v^\mu$ for $v$ the components in the coordinate basis and $b$ the components in the diagonal basis. $e$ is called the vielbein. Since $TM$ carries the natural Levi-Civita connection given by the Christoffel symbols $\Gamma$, we get a connection on the bundle by $$ \omega = e \Gamma e^{-1} + e \mathrm{d}e^{-1}$$ or, in components, $$ {{\omega_\mu}^a}_b = {e^a}_\lambda {\Gamma^\lambda}_{\mu\nu}{e^\nu}_b + {e^a}_\mu \partial_\lambda {e^\mu}_b$$ which is how one obtains the spin connection. We may think of the spin connection as describing the Levi-Civita connection in a "moving frame" whose motion is given by the vielbein such that the metric takes the simple form we are used to from Euclidean/Minkowski space.


1$\mathrm{SO}(n)$ is the Riemannian, $\mathrm{SO}(1,n-1)$ the Lorentzian case, but there's not much of a difference in the description we have here.

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  • $\begingroup$ "Usually, we think of tangent vectors as being expanded as $v=v^{\mu}\partial_{\mu}$, so the natural basis at every point is given by the "coordinates", which may be arbitrarily "ugly". in particular, the metric is gμν." I do not understand why ugly? Sorry I could not follow with that. "We now want to (locally) change frames such that the metric becomes the standard diagonal metric $\eta_{\mu\nu}$ because that one is evidently easier to work with." Why easier to work with? I also did not understad the argument here, if you may elaborate on it. Thank you for your answer. $\endgroup$ – PhilosophicalPhysics Aug 13 '15 at 0:36
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    $\begingroup$ @PhilosophicalPhysics: "ugly" because, well, coordinates need not be chosen to be easy to work with. As for the metric, is it not evident that it is easier to work with a diagonal matrix with unit entries that with a full matrix like $g_{\mu\nu}$? Both of these are just heuristic motivations though, the "real" reason we need the spin connection is that spin is phrased in terms of the Lorentz group $\mathrm{SO}(1,3)$, so to describe spinors we need a connection that can act on those, and this is the spin connection (hence the name). $\endgroup$ – ACuriousMind Aug 13 '15 at 1:01
  • $\begingroup$ Hello, I'm reviving this thread after I read Robin's answer here:physics.stackexchange.com/questions/108523/…. He said " A spin 1/2 field cannot be described by anything built from 4-vector fields. " and you said "the "real" reason we need the spin connection is that spin is phrased in terms of the Lorentz group". Does this cause a contradicition, well because LT act on 4 vectors? I would appreciate your clarification because Robin's answer got be confused. $\endgroup$ – PhilosophicalPhysics Jan 27 '16 at 19:01
  • $\begingroup$ @PhilosophicalPhysics: When I said that spin is phrased in terms of the Lorentz group, I meant that spinors arise from the projective representations of the the Lorentz group. When RobinEkman says that spinors cannot be described in terms of four-vectors, he is saying that they are not a linear representation of the Lorentz group. As for why projective representations are relevant in quantum theories, I refer you to this Q&A of mine. $\endgroup$ – ACuriousMind Jan 28 '16 at 0:24
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    $\begingroup$ @Craig Two ways to see it: 1. When you think of $\Gamma$ as a $\mathfrak{gl}(n)$ gauge field, then $e$ is a gauge transformation and that equation is just how a gauge field transforms under gauge transformations. 2. If $e$ was the Jacobian of a coordinate transformation, then that would be how $\Gamma$ would transform. The only difference here is that there's no coordinate transformation underlying $e$. $\endgroup$ – ACuriousMind Dec 9 '19 at 17:54

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