Why are the gamma-matrices invariant?

[samalkhaiat]

We already had this discussion, and I asked you a question that you never answered. I will ask you again. How $\gamma^{\mu}$ transforms under general coordinate transformations in curved spacetime? Until you answer that question, you will not convince me that you understand the things sufficiently well.

To be honest, I never answered because I came to the conclusion that you were talking about a subject you don’t fully understand. I also recall telling you that $\gamma^{\mu}(x)$, with $\mu$ being a world-index, is a short-hand notation, i.e., a definition which stands for $e^{\mu}{}_{a}(x) \gamma^{a}$. The Diffeomorphism group as well as the local Lorentz group transform the veirbeins $e^{\mu}{}_{a}(x)$ and leave $\gamma^{a}$ invariant.

 

[samalkhaiat]

To understand this, group theory is not enough. Mathematically, one also needs some stuff from differential geometry, such as fibre bundles. Or in a physical language, here one deals with two spaces. One is the spacetime itself (which sees the vector index $\mu$ ), and another is the tangential internal space (which sees the spinor indices $\alpha, \beta$ ). So if you are doing a Lorentz transformation in spacetime, it does not affect the indices in the internal space.

You are doing it again. Please speak about the stuff you fully understand and leave the rest for the experts. It is all about Lie groups:
A spin structure over a Lorentzian 4D spacetime (flat or curved) is a principal $SL(2, \mathbb{C})$ bundle, say $\mathcal{E}$, together with a bundle map $\pi$ from $\mathcal{E}$ onto the principal $SO(1,3)$ bundle of oriented orthonormal frames (the veibeins). For this to work, the map $\pi$ has to be compatible with both $SL(2,\mathbb{C})$ and $SO(1,3)$ actions, i.e., $$\pi (mA) = \pi(m) \Lambda (A) ,$$ where $m$ is any point in the bundle space of $\mathcal{E}$, $A$ is any group element in $SL(2,\mathbb{C})$, and $mA$ is the spin-frame into which $m$ is sent by the action of $A$. Moreover, $\pi(m) \Lambda(A)$ is the orthonormal frame into which $\pi(m)$ is taken under the action of $\Lambda(A) \in SO(1,3)$. A Dirac bi-spinor is then the cross section of the vector bundle associated with the reducible (Dirac) representation of $SO_{+}^{\uparrow}(1,3)$ on $\mathbb{C}^{4}$:
$$\mathcal{S}(\mathbb{C}^{4}) \equiv (S_{2} , \epsilon_{\alpha\beta}) \oplus (\bar{S}_{2} , \epsilon_{\dot{\alpha}\dot{\beta}}) .$$
In fact, the whole Bi-spinor calculus in a Lorentzian 4D spacetime, $M^{4}$, (curved or flat) rests on the isomorphism between the tangent space at a point, $T_{p}(M^{4})$, and the tensor product of the two fundamental carrier spaces of the spin group $SL(2,\mathbb{C})$, the 2-fold covering of $SO(1,3)$: $$T_{p}(M^{4}) \cong (S_{2} , \epsilon_{\alpha\beta}) \otimes (\bar{S}_{2} , \epsilon_{\dot{\alpha}\dot{\beta}}) .$$

Anyway, before claiming again that I am wrong, make sure that you first learned something about fibre bundles,

Yeah, I know “a little bit” of fibre bundle, spin-manifold, cohomology classes, etc.

spinors in curved spacetime, etc from the literature.

Of this I know enough to write a textbook on super-gravity.

Are you familiar with Yang-Mills theories in flat spacetime?

Yes, I learned about Yang-Mills when I was a little boy.

(Please say you are, otherwise we have a problem.)

I can certainly give you a lot of problems in Yang-Mills.

The point is that transformations in the internal space are independent on transformations in spacetime.

This is because the compact gauge group $G$ and the Lorentz group $SO(1,3)$ are two independent, commuting Lie groups. Or, in the language of the stuff you wanted me to learn about, the bundle $M^{4} \times G$ (and its associated connection) admits a global cross-section, i.e., it is a trivial bundle on Minkowski spacetime $M^{4}$. So, the isometry group of $M^{4}$, $ISO(1,3)$ acts trivially on the gauge group $G$.

the internal SO(1,3) transformation is independent from the spacetime SO(1,3) transformation.

Okay, why don't you prove that statement for us. This is how you may be able to do it: construct all the representation of this “internal” $SO(1,3)$ (whatever that means) and show that there exists no 1-to-1 correspondence with the known representations of $SO(1,3)$.
I will be very interested in such proof.

 

[Demystifier]

I also recall telling you that $\gamma^{\mu}(x)$, with $\mu$ being a world-index, is a short-hand notation, i.e., a definition which stands for $e^{\mu}{}_{a}(x) \gamma^{a}$. The Diffeomorphism group as well as the local Lorentz group transform the veirbeins $e^{\mu}{}_{a}(x)$ and leave $\gamma^{a}$ invariant.

I agree with that. But then $\gamma^{\mu}(x)$ does transform as a vector, right? OK, I am fine with your claim that $\gamma^{\mu}(x)$ is "just a notation" for something. But in my paper I use the notation $\Gamma^{\mu}$, which is also a notation for something (see Eq. (59)) which transforms as a vector. If you are allowed to use the notation above, why am I not allowed to use my notation?

You are suggesting that I am not an expert in this stuff. Fine, perhaps I am not. But then let me quote somebody else, say Steven Weinberg. (Is he a good expert enough?) In his book "Gravitation and Cosmology", Sec. 12.5 The Tetrad Formalism, he says the following:
"... Dirac field of the electron is a coordinate scalar and a Lorentz spinor ..." (page 367, second paragraph, my italics).
Are you saying that Weinberg is wrong? Fine, if Weinberg is wrong, then I am wrong too. Or are you saying that it is just a notation? Fine, but if Weinberg can use a notation, why can't I?

 

[Demystifier]

Okay, why don't you prove that statement for us. This is how you may be able to do it: construct all the representation of this “internal” $SO(1,3)$ (whatever that means) and show that there exists no 1-to-1 correspondence with the known representations of $SO(1,3)$.
I will be very interested in such proof.

As in my post above, let me again refer to Weinberg, the same book, the same section, the same quote, only with different italics:
"... Dirac field of the electron is a coordinate scalar and a Lorentz spinor ..."
How can something transform as both scalar and spinor at the same time? Only if the two transformations are independent. Yet, Lorentz group is a subgroup of the group of all coordinate transformations and the two groups don't commute. How would you explain that? Are you saying that Weinberg is wrong? If he is wrong then so am I.

 

[PatrickUrania]

No, no no. $S(\Lambda)$ cannot be unitary. It is a finite-dimensional (matrix) representation of the Lorentz group $SO(1,3)$ which is a non-compact Lie group. The fundamental theorem say: a non-compact group has no finite-dimensional unitary representations. Notice that, I always attached the phrase “infinite-dimensional unitary representation” when I dealt with $U(\Lambda , a)$.

OK. But the reasoning actually works for any invertable matrix U = S^-1, with the same result.

 

[Demystifier]

show that there exists no 1-to-1 correspondence with the known representations of SO(1,3).
I will be very interested in such proof.

Here is one funny example. Assume that we live on a 3-brane in some higher dimensional universe. Assume also that there is another 3-brane which does not interact with our 3-brane. On each 3-brane there is SO(1,3) symmetry. There is, of course, a mathematical 1-to-1 correspondence between the representations of these two groups. Yet, SO(1,3) transformation on our 3-brane is physically independent on SO(1,3) transformation on the other 3-brane.

A more mathematical way to express this is to say that the full symmetry is not SO(1,3) but a product $SO(1,3)\times SO(1,3)$. This is a mathematical way to express the fact that the first SO(1,3) transformation is independent from the second one, without invoking branes. In the same way, Dirac spinor can be viewed as an object that lives not in a representation of SO(1,3) but in a representation of $SO(1,3)\times SO(1,3)$. In the first SO(1,3) it is a scalar and in the second SO(1,3) it is a spinor. Of course, the first SO(1,3) refers to the symmetry of base manifold (spacetime) and the second SO(1,3) refers to the symmetry of the fibre. Is anything wrong with that description?

 

[haushofer]

The above gave me a wild idea. Consider a quantity (operator, matrix, or something like that) $I(x)$ satisfying
$$I^2(x)=g(x)$$
where $g(x)$ is the determinant of the metric tensor with signature (-+++). In Minkowski spacetime this reduces to
$$I^2(x)=-1$$
so $I$ can be represented by the imaginary unit $I=i$. But in curved spacetime $I(x)\neq i$.

What if in quantum theory we replace $i$ with $I(x)$? Could it have something to do with merging quantum theory with gravity?

As I said, at this level this is only a wild speculation. But has anybody thought about something like that?

I guess this would only make sense in an odd number of spatial dimensions (and one time direction). E.g., in 4+1 (flat) dimensions $I =\pm 1$, and I'm pretty sure you need complex numbers in QM :P

 

[haushofer]

By the way, this is how I look upon this stuff (working with supergravity) : in GR one has general coordinate transformations on the spacetime manifold, and local Lorentz transformations in the tangent space. Spin 1/2 fields are scalars under gct's (they don't carry a curved index, only spinor indices) but transform as spinors under local Lorentz transformations. I guess this is the view Demystifier also holds.

So gamma matrices with a curved index are affected by gct's due to the vielbein involved.

 

[Demystifier]

I guess this would only make sense in an odd number of spatial dimensions (and one time direction). E.g., in 4+1 (flat) dimensions $I =\pm 1$, and I'm pretty sure you need complex numbers in QM :P

In the (-+++...) signature it works in any number of spatial dimensions (and one time dimension).

 

[haushofer]

Yes, how silly of me, of course.

 

[ShayanJ]

I guess this would only make sense in an odd number of spatial dimensions (and one time direction). E.g., in 4+1 (flat) dimensions $I =\pm 1$, and I'm pretty sure you need complex numbers in QM :P

What about using split-complex numbers? (In case anyone is wondering different conventions should be equivalent in such senses!)

 

[PatrickUrania]

By the way, this is how I look upon this stuff (working with supergravity) : in GR one has general coordinate transformations on the spacetime manifold, and local Lorentz transformations in the tangent space. Spin 1/2 fields are scalars under gct's (they don't carry a curved index, only spinor indices) but transform as spinors under local Lorentz transformations. I guess this is the view Demystifier also holds.

So gamma matrices with a curved index are affected by gct's due to the vielbein involved.

Why involve a vielbein at all? To me this looks like giving some special status to the linear term in a Taylor series. As is shown in the article by A. Weldon mentioned above (see here: cds.cern.ch/record/466101/files/0009086.pdf), you can do just as well without them. If spacetime is flat this simplifies to what has been said above by Demystifier and me. If spacetime is curved, things are more complicated because of non-zero connections, but the same idea still works. As Einstein said: make things as simple as possible - but not simpler.

 

[Demystifier]

By the way, this is how I look upon this stuff (working with supergravity) : in GR one has general coordinate transformations on the spacetime manifold, and local Lorentz transformations in the tangent space. Spin 1/2 fields are scalars under gct's (they don't carry a curved index, only spinor indices) but transform as spinors under local Lorentz transformations. I guess this is the view Demystifier also holds.

So gamma matrices with a curved index are affected by gct's due to the vielbein involved.

Perhaps @samalkhaiat is trying to say that only the invariant gamma matrices are the "true" gamma matrices, while the vector gamma matrices are merely "a notation". If this is what he is trying to say, then the source of our disagreement is not in mathematics (and certainly not in physics) but in philosophy of mathematics.
https://en.wikipedia.org/wiki/Philosophy_of_mathematics
Namely, the belief that there is such a thing as "true" gamma matrices looks like a kind of mathematical Platonism to me. By contrast, I would describe myself as a mathematical formalist, viewing gamma matrices as objects that I can define in any way I want, provided that they serve as a useful tool for doing physics (with not contradicting experiments, of course). Platonists perhaps are better in understanding mathematics at a deeper level, but formalists can be more flexible in using mathematics for practical purposes. If pure mathematicians are mostly Platonists, then theoretical physicists are mostly formalists. Mathematical physicists (which @samalkhaiat seems to be) may be somewhere in between.

 

[Demystifier]

Of this I know enough to write a textbook on super-gravity.

As you are an expert in super-gravity, are you also interested in string theory? If so, you might also want say something about my paper
http://arxiv.org/abs/hep-th/0702060
In particular, there I present a different argument why Klein-Gordon current is a natural object for Dirac fields. In QFT one has separate currents for different types of particles, but in string theory all these particles are just different states of the same unifying object - the superstring. I find that there is a natural unifying current for the superstring itself, which turns out to be a generalized Klein-Gordon current. So, in the particle limit of the superstring current, the natural current for Dirac particles turns out to be the Klein-Gordon current, rather than the Dirac current.

 

[haushofer]

What about using split-complex numbers? (In case anyone is wondering different conventions should be equivalent in such senses!)

Didn't know these numbers, so I have to think about that. :)

 

[haushofer]

Why involve a vielbein at all? To me this looks like giving some special status to the linear term in a Taylor series. As is shown in the article by A. Weldon mentioned above (see here: cds.cern.ch/record/466101/files/0009086.pdf), you can do just as well without them. If spacetime is flat this simplifies to what has been said above by Demystifier and me. If spacetime is curved, things are more complicated because of non-zero connections, but the same idea still works. As Einstein said: make things as simple as possible - but not simpler.

Interesting paper, i'll look at it. :)

I'm not sure I understand his comparison with the Coulomb gauge. I like to consider GR as a gauge theory of the Poincaré algebra (and similarly N=1 SUGRA and Newton-Cartan theory as gauge theories); the Vielbeins are then the gauge fields of the local translations. I'm not sure how that relates to the paper by Weldon, but then I have to take a closer look at it. But this is offtopic.

 

[Spinnor]

In curved spacetime, it's not only that $\gamma^{\mu}$ is a vector, but is also a function $\gamma^{\mu}(x)$ depending on the spacetime point $x$. So no, the γ-matrices do not have to be just complex numbers.

In fact, the γ-matrices satisfy the algebra
$$\{\gamma^{\mu}(x), \gamma^{\nu}(x) \} =2g^{\mu\nu}(x)$$
where $g^{\mu\nu}(x)$ is the spacetime metric. In Minkowski spacetime this can be reduced to
$$\{\gamma^{\mu}, \gamma^{\nu} \} =2\eta^{\mu\nu}$$
Hence γ-matrices can be chosen to be just numbers only when the metric tensor is just numbers.

If we were to use the Schwarzschild metric are the gamma matrices hard to solve for?

Thanks.

 

[haushofer]

If we were to use the Schwarzschild metric are the gamma matrices hard to solve for?

Thanks.

No. You just need to write the Schwarzschild metric in terms of the vielbeins, which is easy enough (the solutions are then given modulo a local Lorentz transformation, of course).

 

[PatrickUrania]

As you are an expert in super-gravity, are you also interested in string theory? If so, you might also want say something about my paper
http://arxiv.org/abs/hep-th/0702060

I have been reading your paper and I noticed that for example in eq. 16 you use the γ-matrices in two distinct ways. On one hand you have a four-vector γ^μ that transforms nicely. On the other hand you have γ⁰ which ruins the tensor character of the equation. However, you can in all those places replace the frame dependent γ⁰ by its value in the rest frame which I call ε. By construction this is a Lorentz-scalar and the adjoint spinor can be defined as $$\Psi^\dagger \epsilon$$
This spinor is a Lorentz-scalar in the same way as the original spinor is. It is also a solution to the adjoint Dirac-equation (hope that's the right term but the idea should be clear). It is easy to prove that this continues to work for any representation of the γ-matrices.
This way you now have a complex vectorspace equipped with a scalar product. All equations are now tensor equations in the spinor space as well.

 

[vanhees71]

Sure, that's the true meaning of $\gamma^0$, when it is used to define the pseudo-unitary "bispinor product". That the kovariant spinor product is "pseudo-unitary" rather than a usual scalar product in a complex vector space, is due to the fact that the proper orthochronous Lorentz group is not compact and thus has no non-trivial finite-dimensional unitary representations.

 

[samalkhaiat]

he says the following:
"... Dirac field of the electron is a coordinate scalar and a Lorentz spinor ..." (page 367, second paragraph, my italics).
Are you saying that Weinberg is wrong?

He is absolutely correct, but you are wrong. Weinberg did not, does not and will never say that “Dirac spinor is a Lorentz scalar”, not even in his dreams. You said that not Weinberg.

Or are you saying that it is just a notation?

What notation is that? Weinberg is telling you how you should deal with Dirac spinor in pseudo-Riemannian spacetime.

 

[samalkhaiat]

As in my post above, let me again refer to Weinberg, the same book, the same section, the same quote, only with different italics:
"... Dirac field of the electron is a coordinate scalar and a Lorentz spinor ..."
How can something transform as both scalar and spinor at the same time?

Yes, at the same time and place, i.e., at the same point.

Only if the two transformations are independent.

Of course they are independent. One sees the spin index, the other does not: $SO(1,3)$ acts on $\psi_{\alpha} \in \mathcal{S}(\mathbb{C}^{4}) \cong S_{2} \oplus \bar{S}_{2}$ because, the pair $\left( \mathcal{S} , \pi (SL(2, \mathbb{C})) \right)$ is a representation of its simply connected covering group $\mbox{Spin} (1,3) \cong SL(2, \mathbb{C})$. But, the group of general coordinate transformations, $GL(4, \mathbb{R})$, does not act (or acts trivially) on $\psi_{\alpha}$ because, the simply connected covering group of $GL(4, \mathbb{R})$ is not a spin group, i.e., it does not have representation on $\mathcal{S}(\mathbb{C}^{4})$.

Yet, Lorentz group is a subgroup of the group of all coordinate transformations. How would you explain that?

So what, it is a subgroup? Dirac spinor belongs to the representation space $\mathcal{S}(\mathbb{C}^{4})$ which is clearly different from the representation spaces of $GL(4 , \mathbb{R})$.

Are you saying that Weinberg is wrong?

Again, he is correct you are not.

 

[samalkhaiat]

Here is one funny example. Assume that we live on a 3-brane in some higher dimensional universe. Assume also that there is another 3-brane which does not interact with our 3-brane. On each 3-brane there is SO(1,3) symmetry. There is, of course, a mathematical 1-to-1 correspondence between the representations of these two groups. Yet, SO(1,3) transformation on our 3-brane is physically independent on SO(1,3) transformation on the other 3-brane.

A more mathematical way to express this is to say that the full symmetry is not SO(1,3) but a product $SO(1,3)\times SO(1,3)$. This is a mathematical way to express the fact that the first SO(1,3) transformation is independent from the second one, without invoking branes. In the same way, Dirac spinor can be viewed as an object that lives not in a representation of SO(1,3) but in a representation of $SO(1,3)\times SO(1,3)$. In the first SO(1,3) it is a scalar and in the second SO(1,3) it is a spinor. Of course, the first SO(1,3) refers to the symmetry of base manifold (spacetime) and the second SO(1,3) refers to the symmetry of the fibre. Is anything wrong with that description?

Yes very funny, very funny indeed. You certainly did not understand what I said about spin structure. And you do not seem to know anything about fibre bundles.
Descriptively, a fibre bundle $E$ is a twisted product of two spaces $M \times F$ together with an action on the fibre $F$ by the structure group $G(TM)$ of the tangent bundle of $M$.
If the fibre $F$ is a vector space, then $E$ is a vector bundle.
If $G(TM)$ is not simply connected, and $\hat{G}$ (the simply connected universal covering group of $G$) has a representation on $F$, then the homomorphism $\pi : \hat{G} \to G(TM)$ induces a non-trivial action of $G(TM)$ on the index space of elements of $F$.
This is exactly what is happening to the Dirac field: $M$ is the Minkowski spacetime, $G(TM) = SO(1,3)$, $\hat{G} = \mbox{Spin}(1,3) \cong SL(2, \mathbb{C})$, $SO^{\uparrow}(1,3) \cong SL(2, \mathbb{C}) / Z_{2}$, $F = \mathbb{C}^{4}$, and $\mathcal{S}(\mathbb{C}^{4}) \cong S_{2} \oplus \bar{S}_{2}$ is the (Dirac) representation of $\mbox{Spin}(1,3)$ on $\mathbb{C}^{4}$. So, the homomorphism $\pi : SL(2 , \mathbb{C}) \to SO(1,3)$ induces a non-trivial action of the Lorentz group $SO(1,3)$ on the index of Dirac spinor $\psi_{\alpha} \in \mathcal{S}(\mathbb{C}^{4})$.

 

[samalkhaiat]

OK. But the reasoning actually works for any invertable matrix U = S^-1, with the same result.

No, Similarity transformation preserves both the invariance of $\gamma^{\mu}$ and the covariance of the Dirac equation under $SO(1,3)$.

 

[samalkhaiat]

Why involve a vielbein at all? To me this looks like giving some special status to the linear term in a Taylor series.

People worked hard and proved theorems which allow us to throw away the rest of “Taylor series” whatever that means. In this context, the famous and decisive theorem is that of Geroch (1968):
A non-compact spacetime $M^{4}$ has a spin structure if and only if there exist 4 continuous vector fields on $M^{4}$ which constitute a Minkowski tetrad, $e^{a}{}_{\mu}(x)$, in the tangent space at each point of $M^{4}$. [*]
The point is this: without spin structure, the very concept of a spinor field does not exist [See Penrose & Rindler, Vol 1, “Spinor Calculus and Relativistic Fields”].

Your questions in the first post were answered mathematically in my first post. So, instead of hanging on to a dead fish (which does not buy you anything and nobody buy it of you), spend some of your time learning about Lie groups and their representation theory. Not just it is a beautiful subject, in fact, if undergraduate students can’t do without calculus, theoretical physicists can not do without group theory.
[*] By definition, a 4-dimensional space-time is a pair $( M , g )$ consisting of a connected, 4-dimentional, Hausdroff $C^{\infty}$ manifold $M$, together with a Lorentz metric $g$ on $M$: It can be shown that a manifold admits Lorentzian metric if and only if there exists a globally defined (timelike) vector field $X : M \to T (M)$, non-vanishing at each point of the manifold, i.e. $M$ must be time-orientable [Any non-compact manifold admits a Lorentzian metric. For a compact orientable manifold, existence of a Lorentzian metric is equivalent to the fact that the manifold has zero Euler characteristic]. Thus, two Lorentzian metrics on $M$ are considered equivalent if they are related by the diffeomorphism group of $M$. Thus, our spacetime is modeled mathematically by equivalence classes of pairs $( M , g )$. When a space-time metric has been introduced, one can define (in addition to the diffeomorphism group) an action on $M$ of the so-called local Lorentz group as follow. Let $\{ e_{a} (p) = e_{a}{}^{\mu} ( x ) \partial_{\mu}\}$, $a = 0 , 1 , 2 , 3$ be a set of 4 smooth vector fields on $M$, forming an orthonormal frame at each point $p \in M$, $$\langle e_{a} | e_{b} \rangle_{p} \equiv e_{a}{}^{\mu} (x) e_{b}{}^{\nu} (x) g_{\mu \nu} (x) = \eta_{a b} ,$$ where $\eta$ is the Minkowski metric. The set $\{ e_{a} (p) \}$ is called “veibein”, frame field or Minkowski tetrad in the tangent space at each point of $M$. The components $e^{a}{}_{\mu}$ of the inverse veibein are related to the metric tensor $g_{\mu \nu}$ by $g_{\mu \nu} ( x ) = e^{a}{}_{\mu} ( x ) e^{b}{}_{\nu} ( x ) \eta_{a b} .$ Since $e^{a}{}_{\mu}$ has $n^{2} = 16$ independent components while $g_{\mu \nu}$ has $n ( n + 1 ) / 2 = 10$, the veibein contains $n ( n – 1 ) / 2 = 6$ extra degrees of freedom. They are nothing but the freedom to carry out (local) Lorentz transformations in the tangent spaces.

 

[dextercioby]

Sam, you definitely know geometry. Perhaps you can help me with the following question whose explicit answer I haven't seen anywhere: Let's take Minkowski flat spacetime M₄ and its de Rham differential complex built on the cotangent bundle by the exterior derivative d. How exactly do you put an electromagnetic field (one form field $A_{\mu}$ judged in terms of the de Rham complex) in/on spacetime by means of the so-called U(1) gauge bundle, or more precisely why is the space-time differentiation (flat Levi-Civita connection) "equal" or "similar" to the connection in the gauge bundle?

 

[samalkhaiat]

Sam, you definitely know geometry. Perhaps you can help me with the following question whose explicit answer I haven't seen anywhere: Let's take Minkowski flat spacetime M₄ and its de Rham differential complex built on the cotangent bundle by the exterior derivative d. How exactly do you put an electromagnetic field (one form field $A_{\mu}$ judged in terms of the de Rham complex) in/on spacetime by means of the so-called U(1) gauge bundle, or more precisely why is the space-time differentiation (flat Levi-Civita connection) "equal" or "similar" to the connection in the gauge bundle?

And you definitely know how to put me in trouble. Well, it is not really difficult, but to put a connection on principal bundle requires many definitions and then few theorems take your structure closer to the Maurer-Cartan form and Cartan structure equation.
Good account with applications to Dirac and ‘t Hooft-Polykov monopoles, and instantons can be found in
[1] M. Gockeler & T. Schucker : “Differential Geometry, gauge theories, and gravity”, Cambridge University Press. 1990. Chapter 9 & 10.
And my favourite
[2] J. A. de Azcarraga & J. M. Izquierdo: “Lie groups, Lie algebras: cohomology and some application in physics”, Cambridge University Press. 1995. Chapter 2.

 

[Demystifier]

He is absolutely correct, but you are wrong. Weinberg did not, does not and will never say that “Dirac spinor is a Lorentz scalar”, not even in his dreams.

Statement 1 (Weinberg): Dirac field is a coordinate scalar.
Statement 2 (me): A coordinate transformation may be a Lorentz transformation, in which case Dirac field is a coordinate Lorentz scalar.

Please explain how can it be that Statement 1 is right and Statement 2 wrong? Are you saying that a coordinate transformation cannot be a Lorentz transformation?

 

[vanhees71]

A bispinor is a bispinor.

I'm referring only to the usual flat Minkowski spacetime, because I'm not very familiar with spinors in GR. As far as I remember there you have to introduce vierbeins (tetrads), and there can be spinors only in spacetimes where you have tetrades.

In Minkowski space a bispinor (Dirac spinor) behaves under a Lorentz transformation as
$$\psi'(x')=S(\Lambda) \psi(\Lambda^{-1} x'),$$
where
$$S=\exp \left (-\frac{\mathrm{i}}{4} \omega_{\mu \nu} \sigma^{\mu \nu} \right ) \quad \text{with} \quad \sigma^{\mu \nu} = \frac{\mathrm{i}}{2} [\gamma^{\mu},\gamma^{\nu}].$$
The $\gamma^{\mu}$ matrices are "Minkowski vectors" in the sense that
$$S^{-1}(\Lambda) \gamma^{\mu} S(\Lambda)={\Lambda^{\mu}}_{\nu} \gamma^{\nu}.$$

 

[PatrickUrania]

The $\gamma^{\mu}$ matrices are "Minkowski vectors" in the sense that
$$S^{-1}(\Lambda) \gamma^{\mu} S(\Lambda)={\Lambda^{\mu}}_{\nu} \gamma^{\nu}.$$

Indeed. The left part of your last equation again satisfies the Clifford algebra, hence it can be used as a new set of γ-matrices. Using that set requires a transformation of the wavefunction that recovers the original wavefunction, only this time there was no frame change. Also, by your last equation that new set is just the Lorentz-transformed original set. That is what has been argued here.

 

[samalkhaiat]

The $\gamma^{\mu}$ matrices are "Minkowski vectors" in the sense that
$$S^{-1}(\Lambda) \gamma^{\mu} S(\Lambda)={\Lambda^{\mu}}_{\nu} \gamma^{\nu}.$$

Even though you used a quotation mark “Minkowski vector”, the statement is still misleading:
1) As we all know, objects carrying space-time indices need not be space-time tensors. For example the Levi-Civita connection $\Gamma^{\mu}_{\nu\rho}$ is not a type-(1,2) tensor, and certainly the Dirac $\gamma^{\mu}$ is not a vector.
2) The above equation is not a Lorentz (group) transformation $(\mbox{LT})$ equation: Notice that on the left-hand-side you have an (matrix) action on the spin indices
$$S^{-1}_{mp}\gamma^{\mu}_{pq}S_{qn},$$ while on the right-hand-side the action is on the vector index only
$$\Lambda^{\mu}{}_{\nu}\gamma^{\nu}_{mn}.$$
So, the correct transformation of $\gamma^{\mu}_{mn}$ under the matrix spin group of Lorentz must be
$$\mbox{LT}(\gamma^{\mu}_{mn}) = \Lambda^{\mu}{}_{\nu} \ S^{-1}_{mp} \ \gamma^{\nu}_{pq} \ S_{qn} = \gamma^{\mu}_{mn} .$$
This is exactly what I proved in my first post in this thread: Under Lorentz transformations, the Dirac gamma’s are invariant numerical matrices.

 

[Demystifier]

under the matrix spin group of Lorentz ...
the Dirac gamma’s are invariant numerical matrices.

I think that nobody here doubts that. But it looks as if you fail to realize that there is a thing called the group of Lorentz coordinate transformations, which is not the same thing as matrix spin group of Lorentz. Even though the group is the same, the corresponding transformations are not. (Your mathematics is very sophisticated, in fact much more sophisticated than mine, so I'm sure you know that, in abstract algebra, the concept of abstract group is one thing, while realization of group as a group of transformations of some concrete objects is another. By choosing different objects on which a transformation will act, one obtains different realizations of the same group.)

The Lorentz coordinate transformation is just a special case of a general coordinate transformation (the diffeomorphism group), so what is true for general coordinate transformations must also be true for Lorentz coordinate transformations. So if the Dirac gamma transforms as a vector under general coordinate transformations (and Weinberg says it does), then the same Dirac gamma transforms as a vector under Lorentz coordinate transformations (which is what I repeat over and over again).

And this is not in a conflict with your correct claim that Dirac gamma is invariant under matrix spin group of Lorentz. We are both right, and the conflict is only apparent because
(i) we talk about different realizations of the same Lorentz group, and
(ii) we use a somewhat different language (admittedly, yours being more sophisticated than mine, creating an illusion that your statements sound "more correct" than mine).

 

[vanhees71]

Indeed. The left part of your last equation again satisfies the Clifford algebra, hence it can be used as a new set of γ-matrices. Using that set requires a transformation of the wavefunction that recovers the original wavefunction, only this time there was no frame change. Also, by your last equation that new set is just the Lorentz-transformed original set. That is what has been argued here.

There are no wave functions in relativistic QT, only field operators!

 

[vanhees71]

Even though you used a quotation mark “Minkowski vector”, the statement is still misleading:
1) As we all know, objects carrying space-time indices need not be space-time tensors. For example the Levi-Civita connection $\Gamma^{\mu}_{\nu\rho}$ is not a type-(1,2) tensor, and certainly the Dirac $\gamma^{\mu}$ is not a vector.
2) The above equation is not a Lorentz (group) transformation $(\mbox{LT})$ equation: Notice that on the left-hand-side you have an (matrix) action on the spin indices
$$S^{-1}_{mp}\gamma^{\mu}_{pq}S_{qn},$$ while on the right-hand-side the action is on the vector index only
$$\Lambda^{\mu}{}_{\nu}\gamma^{\nu}_{mn}.$$
So, the correct transformation of $\gamma^{\mu}_{mn}$ under the matrix spin group of Lorentz must be
$$\mbox{LT}(\gamma^{\mu}_{mn}) = \Lambda^{\mu}{}_{\nu} \ S^{-1}_{mp} \ \gamma^{\nu}_{pq} \ S_{qn} = \gamma^{\mu}_{mn} .$$
This is exactly what I proved in my first post in this thread: Under Lorentz transformations, the Dirac gamma’s are invariant numerical matrices.

Well, I used the usual physicist's slang, according to which a "Lorentz transformation" of a spinor or tensor field is defined by the representation these fields live on. For the Dirac-spinor field this means
$$[\hat{U}(\Lambda) \hat{\psi}(x) \hat{U}^{\dagger}(\Lambda)]^a={S^a}_{b}(\Lambda) \hat{\psi}^b(\Lambda^{-1} x).$$
Of course the Dirac matrices act in spinor space (indices $a$ and $b$ in the formula).

 

[PatrickUrania]

There are no wave functions in relativistic QT, only field operators!

In QFT you are absolutely right. In Dirac theory it's still about a wave function.