# What are tetrads and the spin connection

I have a question about the nomenclature of the "spin connection".

To me "spin" implies quantum mechanics.

However, as I understand it, the spin connection is just a part of the covariant derivative of a vector that is written in tetrad (i.e., local frame) components. So isn't the "spin connection" a perfectly valid general relativistic concept that has nothing to do with quantum mechanics?

When I go to Wikipiedia and type in "spin connection", I get:

"In differential geometry and mathematical physics, a spin connection is a connection on a spinor bundle."

which seems to imply quantum mechanics (spinor).

Also, what use are tetrads in general relativity? Why can't general relativity be done using a basis induced by the coordinate chart of the spacetime manifold (i.e., the partial derivatives), instead of choosing a new "moving frame" at each point in spacetime?

## Answers and Replies

haushofer
Science Advisor
Hi RedX,

spin connections and tetrads usually show up when you want to do supergravity. Tensors only transform under general coordinate transformations. However, fermions are described by spinors, which don't transform under general coordinate transformations! If you want to describe spinors in spacetime, you have to come with something else. The solution is to introduce tetrads. These tetrads transform under local Lorentz transformations, and in this way you can describe at every point in spacetime how your spinor transforms under local lorentz transformations in the tangent space.

The spin connection pops up when you want to describe parallel transport also for spinors. For normal tensors it shouldn't make a difference if you parallel transport with the gamma connection or with the spin connection, and this is basically the tetrad postulate. From this you can also easily convert the usual Riemann tensor of Gamma into the curvature of the spin connection by using tetrads.

The tetrad and spin connection also pop up when you consider gravity as a gauge theory of the Poincare algebra. The gauge field associated to the translations are then the vielbeins, and the gauge field associated to the local Lorentz transformations are the spin connection. By imposing constraints you can arrive at ordinary general relativity again, but this can be a little bit subtle (after all, GR works with general coordinate transformations, NOT with local translations, so you have to do some rewriting and constraint-imposing).

Normal GR can ofcourse also be written down with tetrads if you want to, but as long as you're not considering fermions I don't know if this is really clarifying. Because at every point you can transform the metric into a flat one, you can define the tetrads as

$$g_{\mu\nu} = e_{\mu}^a e_{\nu}^b \eta_{ab}$$

The ab indices are called "flat", and only these indices transform under local Lorentz transformations. These tetrads can then be seen as the "coordinate transformations which make your metric flat". However, this also means in general that you're going from a coordinate basis to a non-coordinate basis. The metric has, in D dimensions D^2 components, but because of its symmetry it has $\frac{1}{2}D(D+1)$ independent components. The difference between these two is $\frac{1}{2}D(D-1)$, which is the dimension of SO(D-1,1), the Lie group of local Lorentz transformations.

I hope this helps, otherwise I'll be happy to try to answer more questions :)

I don't believe "spin" is a QM concept. However quantization rules apply in certain situations limiting spin possibilities.

Due to precession effects that relativistic situations force on an electron orbiting an atom that were discovered by L. Thomas, it was found (by teams working in a QM context) that such an electron must perform a rotation when subject to a Lorentz transformation.

haushofer
Science Advisor
I think it's called a spin-connection more or less because it allows you to couple "non-trivial spin matter", by which is probably meant "spin 1/2,3/2 etc", to gravity.

spin connections and tetrads usually show up when you want to do supergravity. Tensors only transform under general coordinate transformations. However, fermions are described by spinors, which don't transform under general coordinate transformations! If you want to describe spinors in spacetime, you have to come with something else. The solution is to introduce tetrads. These tetrads transform under local Lorentz transformations, and in this way you can describe at every point in spacetime how your spinor transforms under local lorentz transformations in the tangent space.
There seems to be a bit of asymmetry between fermions and bosons here. Bosons and fermions both transform under the Lorentz group SO(4). When adding general relativity with GL(4), bosons in their vector representations can transform under this group rather easily. But fermions in their spinor representations can't transform under GL(4). So you have to do this local thing with tetrads for fermions. Is this implying that local SO(4) is equivalent to GL(4)? But how can this be, since GL(4) is also local (you can define a linear transformation at each point). It seems to me that GL(4) is a lot bigger group than SO(4).

Assuming a local SO(4) is equivalent to local GL(4), then it would seem more symmetrical to have both fermions and bosons transform under local SO(4) rather than GL(4). So for a vector field V, have the covariant derivative be with the spin connection $$DV=\partial V+ \omega V$$ rather than the christoffel connection
$$DV=\partial V+\Gamma V$$, where the former V is written in lorentz indices and the latter V is written in GL(4) indices.
The spin connection pops up when you want to describe parallel transport also for spinors. For normal tensors it shouldn't make a difference if you parallel transport with the gamma connection or with the spin connection, and this is basically the tetrad postulate.
Is the tetrad postulate a mathematical postulate that says around each point of a Riemannian manifold, it is locally flat, so you can define a local flat frame at each point?

Or is it a physical postulate that says at every point in spacetime, an observer can have an inertial frame (the tetrad frame)? Isn't that the physical interpretation of tetrads, placing an observer at each point of spacetime?
The tetrad and spin connection also pop up when you consider gravity as a gauge theory of the Poincare algebra. The gauge field associated to the translations are then the vielbeins, and the gauge field associated to the local Lorentz transformations are the spin connection. By imposing constraints you can arrive at ordinary general relativity again, but this can be a little bit subtle (after all, GR works with general coordinate transformations, NOT with local translations, so you have to do some rewriting and constraint-imposing).
Do the gauge fields for Poincare transformations correspond to particles? I think there should only be one particle, the graviton, which corresponds to the metric field. Is the vielbein field or the spin connection field independent of the metric field? There might be relationships between the vielbein field, the spin connection field, and the metric field, but which one is the graviton?
However, this also means in general that you're going from a coordinate basis to a non-coordinate basis. The metric has, in D dimensions D^2 components, but because of its symmetry it has $\frac{1}{2}D(D+1)$ independent components. The difference between these two is $\frac{1}{2}D(D-1)$, which is the dimension of SO(D-1,1), the Lie group of local Lorentz transformations.
I am sort of confused about what you are saying here. A general linear transformation has D^2 components. An SO(D) transformation has $\frac{1}{2}D(D-1)$ components. How does this imply the metric has $\frac{1}{2}D(D+1)$ components? If you subtract the two, you do get that answer, but are you talking about translations here (is Poincare GL(4)+translations or is it SO(4)+translations?)?

haushofer
Science Advisor
There seems to be a bit of asymmetry between fermions and bosons here. Bosons and fermions both transform under the Lorentz group SO(4). When adding general relativity with GL(4), bosons in their vector representations can transform under this group rather easily. But fermions in their spinor representations can't transform under GL(4). So you have to do this local thing with tetrads for fermions. Is this implying that local SO(4) is equivalent to GL(4)? But how can this be, since GL(4) is also local (you can define a linear transformation at each point). It seems to me that GL(4) is a lot bigger group than SO(4).
I haven't really thought about this, but maybe this helps. Let's take a scalar field. Under global infinitesimal Poincare transformations it transforms as

$$\delta\phi(x) = \Bigl( a^{\mu}\partial_{\mu} - \frac{1}{2}\lambda^{\mu\nu}L_{\mu\nu}\Bigr) \phi(x)$$
The $a$ parametrizes the translations, and the $\lambda$ parametrizes Lorentz transformations, so we can also write $\delta(a,\lambda)$. The $\partial_{\mu}$ and $L_{\mu\nu}$ are the usual generators of the Poincare algebra in the coordinate representation. This transformation has the following parts:

*a translation in spacetime
*a Lorentz rotation in spacetime: the "orbital part"
*a spin-part

The first two parts are extrinsic (they are transformations in spacetime), and the third part is intrinsic. Now, in going from local Poincare transformations to general coordinate transformations, we can put the first two parts into a new parameter $\xi(x)$ and write

$$\delta_{gct}\phi(x) = \xi^{\mu}\partial_{\mu}\phi(x)$$

So, we've put the $a^{\mu}(x)$ and the orbital part of $\lambda^{\mu\nu}(x)$ into $\xi(x)$.

This transformation identity doesn't only hold for scalars, but also for spinors (because they're defined in the tangent space) and every "frame" tensor. A "frame" tensor is a tensor with "flat indices", for instance

$$V^a(x) = e^a_{\mu}V^{\mu}$$

But we then still have the spin-part of the local Lorentz transformations, and these transform under the appropriate representation. For instance,

$$\delta(\lambda) V^a = \lambda^a_{\ b}V^b$$

where this lambda is in the (1/2,1/2) representation of SO(3,1).

Does this clarify things? If you want to know more about these kind of things, maybe the lecture notes of Van Proeyen help. Personally, I'm not too fond of these notes, but explains these things for instance in his "A menu of supergravities".

Is the tetrad postulate a mathematical postulate that says around each point of a Riemannian manifold, it is locally flat, so you can define a local flat frame at each point?
No, I don't think so; this already follows from the very definition of how your tetrads are defined :)

Or is it a physical postulate that says at every point in spacetime, an observer can have an inertial frame (the tetrad frame)? Isn't that the physical interpretation of tetrads, placing an observer at each point of spacetime?
The tetrad postulate is a postulate which links the normal Gamma connection and the spin connection to each-other. Personally, I like the way Carroll (his GR-notes) or Samtleben (introduction to supergravity) treat this. Also, Samtleben's notes are really the most easy introduction you can find on these kinds of subjects, so if you haven't looked at them: try them! :)

Carroll for instance shows that the postulate follows from the demand that it shouldn't matter if one describes things via tetrads or in a coordinate basis. From this he derives that the tetrad is "covariantly constant";this is the tetrad postulate.

Do the gauge fields for Poincare transformations correspond to particles? I think there should only be one particle, the graviton, which corresponds to the metric field. Is the vielbein field or the spin connection field independent of the metric field?
Good point. If you gauge the Poincare algebra, you get TWO gauge fields ofcourse. The spin-connection shouldn't be a propagating degree of freedom! So you want to make the spin connection dependent of the vielbein, in the same way the Levi-Civita connection depends on the metric. You can do this by putting the curvature of the translations to zero:

$$R_{\mu\nu}^a(P) = 0$$

The number of equations of this constraint is exactly the same as the number of components of the spin connection. And surprisingly, this is also the constraint you need to eliminate the P-transformations in favour of the general coordinate transformations! This makes this constraint a very natural one.

There might be relationships between the vielbein field, the spin connection field, and the metric field, but which one is the graviton?
So eventually the thing is that with all the constraints and the equations of motion you cut down the number of degrees of freedom to the amount you know of normal GR, and the only propagating degree of freedom is the vielbein. Hence normal gravity arises.

I am sort of confused about what you are saying here. A general linear transformation has D^2 components. An SO(D) transformation has $\frac{1}{2}D(D-1)$ components. How does this imply the metric has $\frac{1}{2}D(D+1)$ components? If you subtract the two, you do get that answer, but are you talking about translations here (is Poincare GL(4)+translations or is it SO(4)+translations?)?
I'm sorry, I wasn't very clear here.

So, we can choose to describe normal GR with a metric or with a vielbein. The metric has $\frac{1}{2}D(D+1)$ components. But the vielbein has $D^2$ components. That looks like a redundancy, and it is. However, this is not surprising: with the vielbein you have the freedom of local Lorentz transformations (they carry a local Lorentz index), and so the physical degrees of freedom at this stage is

$$D^2 - \frac{1}{2}D(D-1) = \frac{1}{2}D(D+1)$$

which is what you expect :)

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I haven't really thought about this, but maybe this helps.

Thanks a lot for your responses. They have been very helpful.
*a translation in spacetime
*a Lorentz rotation in spacetime: the "orbital part"
*a spin-part

The first two parts are extrinsic (they are transformations in spacetime), and the third part is intrinsic.

I have a question about the terminology here. Usually the term extrinsic refers to the embedding space of a manifold, and intrinsic refers to the manifold itself. But in this context, you are using extrinsic to mean the spacetime manifold, and intrinsic to mean the tangent space? Is this standard terminology for physicists as opposed to mathematicians?

I think I get what you're saying. But in Kaku's QFT book he defines the Poincare transformation to include mixing of vector or spinor components in addition to acting on the spacetime coordinates of the fields:

$$U(\theta, a)\phi^A(x)U^{-1}(\theta,a)=\mathcal D^{AB}(-\theta)\phi_B(x')=\mathcal D^{AB}(-\theta)e^{(a^{\mu}\partial_{\mu} - \frac{1}{2}\lambda^{\mu\nu}L_{\mu\nu})}\phi_B(x)$$

So you called the local Poincare transformation $$\delta_{gct}$$, and called the other part the local Lorentz transformation $$\delta(\lambda)$$.

Kaku doesn't make such a distinction and combines them both.

But I like your picture better that the rotation of the spin or vector indices occurs in the tangent space and is therefore a local Lorentz transformation.

Also, I notice that you don't have an 'i' in front of your orbital angular momentum operator. Sometimes I see that, and I was just wondering about the conventions. For example Kaku writes:

$$e^{i\mathcal L}$$

and then defines L with an 'i' in it, say $$i (x\partial_y-y\partial_x)$$, whereas you write:

$$e^{\mathcal L}$$

and define L without an 'i' in it, say $$(x\partial_y-y\partial_x)$$.
So, we can choose to describe normal GR with a metric or with a vielbein. The metric has LaTeX Code: \\frac{1}{2}D(D+1) components. But the vielbein has LaTeX Code: D^2 components. That looks like a redundancy, and it is. However, this is not surprising: with the vielbein you have the freedom of local Lorentz transformations (they carry a local Lorentz index), and so the physical degrees of freedom at this stage is

LaTeX Code: <BR>D^2 - \\frac{1}{2}D(D-1) = \\frac{1}{2}D(D+1) <BR>

which is what you expect :)
I understand it now. That's neat. I'm used to describing the graviton 'h' as the peturbative expansion of the metric: $$g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$$. So since I choose to use the metric instead of vielbeins, how would I treat the Dirac Lagrangian, which is written with a vielbein? Can the Dirac Lagrangian be written without a vielbein? For example, could you solve for the vielbein in terms of the metric tensor, and substitute this expression into a Dirac Lagrangian?

edit:

actually, the section where Kaku's book talks about Poincare transformation he is talking about flat spacetime. for local poincare transformations, I think Kaku would agree that indices transform in the tangent space (local Lorentz), and spacetime coordinates transform in GL(4) (local Poincare).

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haushofer
Science Advisor
Thanks a lot for your responses. They have been very helpful.

I have a question about the terminology here. Usually the term extrinsic refers to the embedding space of a manifold, and intrinsic refers to the manifold itself. But in this context, you are using extrinsic to mean the spacetime manifold, and intrinsic to mean the tangent space? Is this standard terminology for physicists as opposed to mathematicians?
I don't really bother about the terminology to be honest :P In the context of gauge theories you can read "extrinsic" symmetries as "spacetime symmetries", and "intrinsic" as gauge transformations in internal spaces. Or, formulated in terms of fibre bundles: "extrinsic" refers to the base manifold, and "intrinsic" to the fibers. I don't know how standard this terminology is, but for me it's clear. :) Anyway, it's not like the mathematician's terminology, so if you like that more you can forget about mine.

Also, I notice that you don't have an 'i' in front of your orbital angular momentum operator. Sometimes I see that, and I was just wondering about the conventions. For example Kaku writes:

$$e^{i\mathcal L}$$ (*)

and then defines L with an 'i' in it, say $$i (x\partial_y-y\partial_x)$$, whereas you write:

$$e^{\mathcal L}$$

and define L without an 'i' in it, say $$(x\partial_y-y\partial_x)$$.

Ah, yes, I wasn't really paying attention to these kind of conventions. As far as I know physicists like to define your (*) like Kaku, while mathematicians like it without the "i". Because we're talking about unitary groups, the physicists notation means that the generators are Hermitian, while the mathematician's generator would be anti Hermitian.

I understand it now. That's neat. I'm used to describing the graviton 'h' as the peturbative expansion of the metric: $$g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$$. So since I choose to use the metric instead of vielbeins, how would I treat the Dirac Lagrangian, which is written with a vielbein? Can the Dirac Lagrangian be written without a vielbein? For example, could you solve for the vielbein in terms of the metric tensor, and substitute this expression into a Dirac Lagrangian?

I don't think so. If you want to write down a Dirac action in curved spacetime, you just need one vielbein. In flat spacetime the gamma matrices are $\gamma^a$ which satisfy the Clifford algebra $\{ \gamma^a, \gamma^b \}=2\eta^{ab}$, and if you define the "curved" gamma matrices in the intuitive way, $\gamma^{\mu} = e^{\mu}_a \gamma^a$, then you get the algebra $\{ \gamma^{\mu},\gamma^{\nu} \} = 2 g^{\mu\nu}$. This gamma matrix couples to the derivative and the spin connection term in the action. If you want to see this explicitly, you could refer to Nakahara, 7.10.3 :)

actually, the section where Kaku's book talks about Poincare transformation he is talking about flat spacetime. for local poincare transformations, I think Kaku would agree that indices transform in the tangent space (local Lorentz), and spacetime coordinates transform in GL(4) (local Poincare).
I agree.

I was reading the Carroll notes you mentioned (great notes by the way), and he mentions adding to each point of the manifold an internal vector space. At each point there were already tangent and cotangent spaces, but now he is adding more vector spaces. He calls the collection of all vector spaces at a point (whether they are the added internal spaces, or the original tangent and cotangent spaces) fibers.

What I'm not understanding is what he means by "structure group" in the sentence:

"Besides the base manifold (for us, spacetime) and the fibers, the other important ingredient in the definition of a fiber bundle is the "structure group," a Lie group which acts on the fibers to describe how they are sewn together on overlapping coordinate patches. Without going into details, the structure group for the tangent bundle in a four-dimensional spacetime is generally GL (4), the group of real invertible 4 × 4 matrices; if we have a Lorentzian metric, this may be reduced to the Lorentz group SO(3, 1). Now imagine that we introduce an internal three-dimensional vector space, and sew the fibers together with ordinary rotations; the structure group of this new bundle is then SO(3)."

What is meant by describing how fibers are sewn together on overlapping coordinate patches? The requirement for a differentiable manifold is that the overlapping coordinate patches are smooth, and how the patches are sewn together are given by the smooth transition functions. So I know how the coordinate patches are sewn together, but I never heard about fibers (the vector spaces) being sewn together.

Is this saying that you can define a transition function which instead of taking points in one coordinate patch to points in another, it takes vectors defined in one coordinate patch to another? Points can overlap on different coordinate patches, but vector spaces?

haushofer
Science Advisor
Now I look at it it also confuses me I must say. What Carroll calls here the structure group I would call (and mathematicians also) the transition function. The structure group is the Lie group G which acts on the fibre F. I'm not sure how much you know about fibre bundles, but it wouldn't make a lot of sense to write down here the whole formalism; a lot of other people have done that already in a much clearer way :). But the idea is the following:

A fibre bundle is a topological space which looks locally like a direct product of two topological spaces, just like a manifold locally looks like R^n. The fibre bundle consists of the following:

*A differentiable manifold E which acts as the total space
*A differentiable manifold M which acts as the base manifold
*A differentiable manifold F which acts as the fibre

With these manifolds we have to introduce some more structure which makes our description more precise. In GR, we take as base manifold our spacetime, the fibres are the tangent spaces (this fibre bundle is also called the tangent bundle) and the structure group would be SO(3,1). The local trivialization tells us how we can go locally from the direct product to the fibrebundle, just as the inverse of a coordinate function on a manifold tells us how to go from R^n to the manifold.

If we now have an overlap on our base manifold (in the case of GT: our spacetime), then there are two elements f_i and f_j in the fibre F associated to one point u in the fibrebundle E. The transition function t_{ij} tells us then how these two elements of F are related:

$$f_i = t_{ij}f_j$$

This transition function will depend on the point p on the base manifold M associated to the point u in the fibre. These two points p and u are related by the projection map $\pi: E \rightarrow M$.

Nakahara's explanation in chapter 9 is quite formal and not exceptionally clear; maybe you could first try the following link,

duffell.org/chapter_4.pdf

where your question is answered from equation (4.13) onwards. Basically, what this transition function does for you is to relate the different local trivializations on an overlap on the base manifold M.

I suggest to take a look at the pdf, and maybe Nakahara's chapter 9.1-9.2.2, that should make a lot more things clear.

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The duffell.org notes were really good. You sure know how to find good notes!
With these manifolds we have to introduce some more structure which makes our description more precise. In GR, we take as base manifold our spacetime, the fibres are the tangent spaces (this fibre bundle is also called the tangent bundle) and the structure group would be SO(3,1).

Why SO(3,1)? In the duffell.org notes, the structure group for a vector bundle is GL(4).
I guess you can have any group for your structure group, so long as the group maps the fiber onto itself. So for example, the fiber bundle (0,1) over the circle is either the cylinder or the mobius strip, depending on whether you choose the group of one element (identity), or the group of two elements (i.e., parity, which preserves the fiber but is responsible for the 180 twist). So you are free to choose any subgroups of the group structure since the entire group maps the fiber onto itself including the subgroup, but your fiber bundle will look different depending on which group you choose. So is there a reason for choosing SO(3,1) over GL(4)?

The notes make the comment that you don't have to use all elements of a group for your structure group. So for example for the cylinder you can use any group you want as the structure group, but only use the identity element of that group. So I guess you can have your structure group as GL(4) but only use the subgroup SO(3,1).

The local trivialization tells us how we can go locally from the direct product to the fibrebundle, just as the inverse of a coordinate function on a manifold tells us how to go from R^n to the manifold.

I think the local trivialization should go from the fiberbundle to the direct product, not the other way around. I checked Wikipedia, and Wikipedia has equation 4.3 of the duffel.org notes going the other way around. When I read the notes, it made more sense to me when it goes the other away around.
If we now have an overlap on our base manifold (in the case of GT: our spacetime), then there are two elements f_i and f_j in the fibre F associated to one point u in the fibrebundle E. The transition function t_{ij} tells us then how these two elements of F are related:

$$f_i = t_{ij}f_j$$

This transition function will depend on the point p on the base manifold M associated to the point u in the fibre. These two points p and u are related by the projection map $\pi: E \rightarrow M$.
I think in trying to understand the notes I've lost sight of the bigger picture. Wasn't the whole purpose of introducing local trivializations to decompose the total manifold E into the product space of the base and fiber manifold (M, F). So given a point q of E, you want it to correspond to a certain (p,f) of the product space, but you don't know what f is as f can have some symmetry and be rotated relative to p. So you need to know where q is on the local patch (p,f) you have chosen to describe your manifold at that point. So to me, when you collapse q with the projection map $\pi: E \rightarrow M$, you lose knowledge of f, defeating the purpose.

Or is the point of collapsing q to p to allow you to select a coordinate patch Ui on M that will tell you the orientation of the fiber F, and then you would feed the Ui into $$\phi_i:\pi^{-1}(U_i) \rightarrow U_i \times F$$ which gives you only a picture of the local product space of MxF of E, and then you still have the original q, so then you can figure out how q decomposes to (p,f) in this local region?

haushofer
Science Advisor
Hey, sorry for the late reply :)

The duffell.org notes were really good. You sure know how to find good notes!

Thanks. For me it's always a pleasure to find notes which can make difficult things comprehensible. Also when reading articles I sometimes get the feeling that people are overcomplexifying things.

Why SO(3,1)?
Because at every point in spacetime you have the Lorentz group at your disposal in the tangent space, not GL(4). Via coordinate transformations we can go from one point to the other, but at every point we have the tangent space attached, and in these tangent spaces we can define different representations of the Lorentz group SO(3,1). And this enables us to define fermions. This wouldn't make a lot of sense using GL(4), right?

So I guess you can have your structure group as GL(4) but only use the subgroup SO(3,1).
I guess you could, but why not only take SO(3,1) then?

I think the local trivialization should go from the fiberbundle to the direct product, not the other way around. I checked Wikipedia, and Wikipedia has equation 4.3 of the duffel.org notes going the other way around. When I read the notes, it made more sense to me when it goes the other away around.

You mean because in the case of manifolds your coordinate functions also go from the manifold M to R^n (so from the manifold to its "local realization")?

I think in trying to understand the notes I've lost sight of the bigger picture. Wasn't the whole purpose of introducing local trivializations to decompose the total manifold E into the product space of the base and fiber manifold (M, F).
I must say it has been a while since I studied fibre bundles and I basically never use them, so I have to reread this kind of stuff to give a decent answer. But I understand local trivializations as the mappings which describe the transformations on the fiber induced by coordinate transformations on the base manifold. I will take another look at it and if I have possible useful insights I will post them here :)

atyy
Science Advisor
duffell.org/chapter_4.pdf

The duffell.org notes were really good.

Thanks! Those are nice notes!

I must say it has been a while since I studied fibre bundles and I basically never use them, so I have to reread this kind of stuff to give a decent answer. But I understand local trivializations as the mappings which describe the transformations on the fiber induced by coordinate transformations on the base manifold. I will take another look at it and if I have possible useful insights I will post them here :)

You've been very helpful and I don't wish to bother you anymore, especially on something you never use :).

I'm taking a "math methods in physics course", and I have an end-of-semester presentation in a few days, and I was going to choose to present two topics: relativistic quantum mechanics on an arbitrary manifold, and the geometry of gauge theories (fiber bundles). But I found out I have too much material on relativistic quantum mechanics, so I'm not going to present fiber bundles, which is anyways too mathematical for me!

As for the original question, here's the mathematical answer that you were saying (ignore the (0) at the bottom: I don't know why it does that):

$$\begin{eqnarray*} D_\mu V^a=\partial_\mu V^a+\omega^{a}_{\mu b}V^b=\partial_\mu V^a+\omega^{ab}_{\mu}V_b \\ \nonumber =\partial_\mu V^a+\frac{\omega^{ls}_{\mu}}{2}[\delta^{a}_{l}\delta^{b}_{s}-\delta^{b}_{l}\delta^{a}_{s}]V_b =\partial_\mu V^a+\frac{\omega^{ls}_{\mu}}{2}[\delta^{a}_{l}g_{bs}-\delta^{a}_{s}g_{bl}]V^b \\ \nonumber =\partial_\mu V^a-\frac{i}{2}\omega^{ls}_{\mu}(M_{ls})^{a}_{\mbox{ }b}V^b \end{eqnarray*}$$

So the spin connection $$\omega$$ is acting on a vector V (written in flat-space indices), and due to antisymmetry of the spin connection, it can be rewritten in terms of the Lorentz generators M for the vector representation of the Lorentz group. Replace M by the representation for spinors (the product of gamma matrices), and you have a connection for spinors.

Again, thank you for all the time you spent helping me.

edit: so for example, the Dirac equation in curved space follows from this covariant derivative:

$$D_\mu = \partial_\mu - \frac{i}{2} \eta_{ac} \omega^c_{b\mu} \frac{\sigma^{ab}}{2}$$

and the generators are:

$$\frac{\sigma^{ab}}{2}=\frac{i}{4} \left[\gamma^{a},\gamma^{b}\right]$$

which agrees with Wikpiedia:

http://en.wikipedia.org/wiki/Dirac_equation#Curved_spacetime_Dirac_equation

So that 1/2 before the generator thing is really weird, but that's what's derived.

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haushofer
Science Advisor
In Samtleben's SUGRA notes on page 8 you see that he defines the Lorentz covariant derivative as

$$D_{\mu} = \partial_{\mu} + \frac{1}{2}\omega_{\mu}^{ab}M_{ab}$$

with M the generators of the lie algebra of SO(3,1). It's nice to see it explicitly (it differs a factor -i, but I guess that comes from the convention of the Lie algebra).