SL(2,C) to Lorentz in Carmeli's Theory of Spinors

[samalkhaiat]

A manifest covariant formalism is provided by Dirac spinors rather than Weyl spinors!

What is it non covariant in the following equations?
$$( \sigma^{ \mu } )_{ A \dot{ B } } \partial_{ \mu } \chi^{ A } = m \bar{ \eta }_{ \dot{ B } } ,$$
$$( \sigma^{ \mu } )_{ A \dot{ B } } \partial_{ \mu } \eta^{ A } = m \bar{ \chi }_{ \dot{ B } } .$$
Irreducible representations of the Lorentz group know absolutely nothing about Dirac’s bispinors. In fact, Dirac bispinor is a Parity invariant, 4-Dimensional Reducible Representation of $SO( 1 , 3 ) + \mbox{ Parity }$, i.e.,
$$\psi = \left( \begin{array}{c} \chi_{ A } \\ \bar{ \eta }^{ \dot{ A } } \end{array} \right) \sim ( \frac{ 1 }{ 2 } , 0 ) \oplus ( 0 , \frac{ 1 }{ 2 } ) .$$
Here is a question to the readers. What prevents us from representing Dirac bispinor by the $( 1/2 , 0 ) \oplus ( 1/2 , 0 )$ or $( 0 , 1/2 ) \oplus ( 0 , 1/2 )$ Reducible Representations of $SO( 1 , 3 ) + \mbox{ Parity }$?

Sam

 

[samalkhaiat]

... Tr in the formula is not invariant under SL(2,C) transformations...

Can you prove this (ridiculous) claim of yours?

The Tr in the formula, when written in full details, means
$$\Lambda^{ \alpha }{}_{ \beta } ( g ) = \frac{ 1 }{ 2 } \ g^{ A }{}_{ C } \ \bar{ g }^{ \dot{ B } }{}_{ \dot{ D } } \ ( \sigma^{ \alpha } )^{ C \dot{ D } } \ ( \sigma_{ \beta } )_{ A \dot{ B } } .$$
When all the indices of $SL( 2 , C )$ are contracted, we say that the object in question is (scalar) invariant under $SL( 2 , C )$ transformations. So, I insist that you show us how is it that the trace is not invariant?

I also asked you ( in post # 54) to prove few more (false) claims you made in this thread, but you have not replied yet.

Sam

 

[arkajad]

Can you prove this (ridiculous) claim of yours?

Of course I can. Take as an example

$$A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$$

Then $\det(A)=1$ therefore $A\in SL(2,C).$

We have

$$A A^\dagger =\begin{pmatrix}2&1\\1&1\end{pmatrix}$$

and $\mbox{Tr}( A A^\dagger )=3$

With $Q=I$ (the identity matrix) we have

$$\mbox{Tr}(AQA^\dagger)=3\neq 2=\mbox{Tr}(Q)$$

This means Tr is not SL(2,C)-invariant. It is SU(2) ( and even U(2) )-invariant.

 

[vanhees71]

I think there is so much confusion in this thread that I cannot help further. Just some final remarksfrom my side.

(i) The Dirac equation is indeed a reducible representation of $\mathrm{SO}(1,3)^{\uparrow}$ ($\mathrm{D}(1/2,0) \otimes \mathrm{D}(0,1/2)$ but an irreducible representation of $\mathrm{O}(1,3)^{\uparrow}$, i.e., it contains a non-trivial reprsentation of space reflections.

The correct decomposition of the Dirac equation into coupled Weyl-spinor, leading to the chiral representation of the Dirac matrices reads
$$\mathrm{i} \sigma^{\mu} \partial_{\mu} \eta=m \chi,$$
$$\mathrm{i} \overline{\sigma}^{\mu} \partial_{\mu} \chi=m \eta.$$
Here $\chi$ and $\eta$ are two two-component spinors, called the righthanded and lefthanded components of the four-component Dirac spinor and
$$\sigma^{\mu}=(\mathbb{1}_2,+\vec{\sigma}), \quad \overline{\sigma}^{\mu}=(\mathbb{1}_2,-\vec{\sigma}).$$
Note again that also here the index on the $\overline{\sigma}$ matrices is a contravariant (upper) index!

(ii) In the Weyl-spinor representations $\mathrm{D}(1/2,0)$ and $\mathrm{D}(0,1/2)$ the skew-symmetric bilinear forms $\xi_A \xi^A=\eta_{AB} \xi^A \xi^B$ and $\chi_{\dot{A}} \chi^{\dot{A}}=\eta_{\dot{A} \dot{B}} \chi^{\dot{A}} \chi^{\dot{B}} =$ are invariant under the corresponding $\mathrm{SL}(2,\mathbb{C})$ transformations, where the spinors with the dotted indices transform with the conjugate complex matrix:
$$\xi'=A \xi, \quad \chi'=A^{*} \chi.$$
Not the trace of the $Q=x^{\mu} \sigma_{\mu}$ is conserved, because the transformation reads
$$Q'=g Q g^{\dagger}$$
and $g$ is in general not unitary. The trace is only invariant under the subgroup $\mathrm{SU}(2)$, which provides the usual spin-1/2 representation of rotations for Weyl spinors. Also note that
$$\mathrm{tr} Q=2 x^0,$$
which of course changes under general Lorentz transformations and is invariant only under rotations, and thus the representation $g$ of a general Lorentz transformation changes the trace but unitary ones don't. The unimodular transformations $g$, however, keep $\mathrm{det} Q=x_{\mu} x^{\mu}$ invariant as it must be.

(iii) A very good book on this topic is

Sexl, Urbandtke, Relativity, Groups, Particles, Springer (2001)

 

[arkajad]

Not the trace of the $Q=x^{\mu} \sigma_{\mu}$ is conserved

I guess you wanted to say:

The trace of the $Q=x^{\mu} \sigma_{\mu}$ is not conserved

 

[arkajad]

There is an additional remark that can be useful for the readers of this thread. Namely, if the homomorphism from SL(2,C) to the restricted Lorentz group is defined as in the first Carmaeli's book "Group Theory and General Relativity" (correctly):
$${\Lambda(g)^\alpha}_{\phantom{\alpha}\beta}=\mbox{Tr }(\sigma^\alpha g\sigma^\beta g^\dagger)$$
then
$$\mbox{Tr }\Lambda(g)=|\mbox{Tr }g|^2.$$
This is Eq. (3.85) in "Theory of Spinors" and not numbered equation on p. 36 in Carmeli's "Group theory and General Relativity".

 

[arkajad]

(i) The Dirac equation is indeed a reducible representation of $\mathrm{SO}(1,3)^{\uparrow}$ ($\mathrm{D}(1/2,0) \otimes \mathrm{D}(0,1/2)$ but an irreducible representation of $\mathrm{O}(1,3)^{\uparrow}$, i.e., it contains a non-trivial reprsentation of space reflections.

Perhaps it may be useful here, using simple matrix algebra, to understand why massless two-component spinor equation is SL(2,C) invariant.

So, suppose $\Psi$ is a spinor field transforming according to the contragradient representation of SL(2,C):
$$\Psi\mapsto \tilde{\Psi}=g^{*-1}\Psi.$$
while the coordinates transform in a standard way:
$$\tilde{x}^\mu={\Lambda^\mu}_{\phantom{\mu}\nu}x^\nu$$
so that
$$\tilde{\partial}_\nu={\Lambda^\rho}_{\phantom{\rho}\nu}\partial_\rho$$
Then
$$\sigma^\nu\tilde{\partial}_\nu\tilde{\Psi}=g\tilde{\partial}_\nu g^{-1}\sigma^\nu g^{-1*}\tilde{\partial}_\nu\Psi$$
$$=g{{\Lambda^{-1}}^\nu}_{\phantom{\nu}\mu}\sigma^\mu\tilde{\partial}_\nu\Psi$$
$$=g\sigma^\mu\partial_\mu\Psi$$
Therefore, owing to the fact that $g$ is invertible, $\sigma^\mu\partial_\mu\Psi=0$ if and only if $\sigma^\nu\tilde{\partial}_\nu\tilde{\Psi}=0$

 

[samalkhaiat]

Just some final remarksfrom my side.
(i) The Dirac equation is indeed a reducible representation of $\mathrm{SO}(1,3)^{\uparrow}$ ($\mathrm{D}(1/2,0) \otimes \mathrm{D}(0,1/2)$

$( 1/2 , 0 ) \otimes ( 0 , 1/2 )$ is the 4-vector irreducible representation. May be you meant to write $\oplus$.

but an irreducible representation of $\mathrm{O}(1,3)^{\uparrow}$, i.e., it contains a non-trivial reprsentation of space reflections.

No, I meant exactly what I said: Dirac “spinor” is parity-invariant, Reducible representation of $SO( 1 , 3 ) + P$. By this I meant the piece $\Lambda_{ P } SO^{ \uparrow } ( 1 , 3 )$, sometimes written as $\Lambda_{ P } L_{ + }^{ \uparrow }$, which is the $L_{ - }^{ \uparrow }$ piece of the pseudo-orthognal group $O( 1 , 3 )$.
Dirac bispinor does not exist in the Lorentz group $SO^{ \uparrow } ( 1 , 3 )$. Only when we allow for parity, we can construct the parity-invariant direct sum $( 1/2 , 0 ) \oplus ( 0 , 1/2 )$ with the generator written in block diagonal, i.e., Reducible: Parity action on the generators does not form an in-equivalent class of representations. This is the reason for the question, I asked at the end of post # 61, and nobody cared to answer.

The correct decomposition of the Dirac equation into coupled Weyl-spinor, leading to the chiral representation of the Dirac matrices reads

In the representation theory, we do not decompose the Dirac equation. On the contrary, we construct the Dirac equation from the equations satisfied by the two in-equivalent fundamental representations of $SO^{ \uparrow }( 1 , 3 )$, and demanding Parity-Invariance.

Not the trace of the $Q=x^{\mu} \sigma_{\mu}$ is conserved, because the transformation reads
$$Q'=g Q g^{\dagger}$$
and $g$ is in general not unitary

I don’t know the meaning of this statement! As a matrix, $Q = x_{ \mu } \sigma^{ \mu }$, has nothing to do with group $SL( 2 , C )$. Therefore, it is meaningless to ask about the invariance of $\mbox{Tr} Q$ with respect to $SL( 2 , C )$ transformations. In fact ( and I said this before), the $SL( 2 , C )$ trace of the mixed spinor-tensor $Q_{ A \dot{ B } } = x_{ \mu } ( \sigma^{ \mu } )_{ A \dot{ B } }$ is not even defined in the group.

Sam

 

[samalkhaiat]

Of course I can. Take as an example

$$A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$$

Then $\det(A)=1$ therefore $A\in SL(2,C).$

We have

$$A A^\dagger =\begin{pmatrix}2&1\\1&1\end{pmatrix}$$

and $\mbox{Tr}( A A^\dagger )=3$

With $Q=I$ (the identity matrix) we have

$$\mbox{Tr}(AQA^\dagger)=3\neq 2=\mbox{Tr}(Q)$$

This means Tr is not SL(2,C)-invariant. It is SU(2) ( and even U(2) )-invariant.

Thank you very much, you made me laugh. Is this your understanding of the invariance of the trace operation? I ground my 12 years old kid for a week, if she reproduces your argument. Who said $\mbox{Tr} ( A Q A^{ \dagger } )$ must equal to $\mbox{Tr} Q$? And why should they be equal?
Look, I don’t know how many different ways I have explained this for you, here is another way. Consider the Matrix $Q = x_{ \mu } \sigma^{ \mu }$, square it by ordinary matrix multiplication and take (1/2) of its trace, you find
$$\frac{ 1 }{ 2 } \mbox{Tr} ( Q^{ 2 } ) = \frac{ 1 }{ 2 } Q_{ a c } Q_{ c a } = x_{ 0 }^{ 2 } + x_{ 1 }^{ 2 } + x_{ 2 }^{ 2 } + x_{ 3 }^{ 2 } .$$
This is the invariant metric of the Orthogonal group $SO( 4 )$. So, why didn’t we get the Invariant Lorentz metric? The reason is this: As a matrix, $Q$ has nothing to do with the group $SL( 2 , C )$ and because we calculated the $SO( 4 )$-invariant trace, we got Euclidean metric.
To avoid this embarrassing situation, people often calculate $\det | Q |$ in order to generate the Lorentz metric with correct signature. Even in this case, we get the correct signature because of the presence of the $SL( 2 , C )$ metric $\epsilon$ in the definition of $\det | Q |$. To see the beauty of $SL( 2 , C )$ tensor calculas, let us repeat the calculation by treating $Q$ as the mixed spinor-tensor $Q_{ A \dot{ B } } = x_{ \mu } ( \sigma^{ \mu } )_{ A \dot{ B } }$ and evaluate the $SL( 2 , C )$-invariant trace
$$\mbox{Tr} ( Q^{ 2 } ) = Q^{ A \dot{ B } } Q_{ A \dot{ B } } = x_{ \mu } x_{ \nu } ( \sigma^{ \mu } )^{ A \dot{ B } } ( \sigma^{ \nu } )_{ A \dot{ B } } = 2 \left( x_{0}^{2} - x_{1}^{2} - x_{2}^{2} - x_{3}^{2}\right) .$$
Here the Lorentz invariant metric with correct signature arises naturally because the invariant trace requires the presence of $SL( 2 , C )$ spinor metric $\epsilon$:
$$( \sigma^{ \mu } )^{ A \dot{ B } } ( \sigma^{ \nu } )_{ A \dot{ B } } = \epsilon^{ A C } \epsilon^{ \dot{ B } \dot{ D } } ( \sigma^{ \mu } )_{ C \dot{ D } } ( \sigma^{ \nu } )_{ A \dot{ B } } .$$
Now, look the two calculations and recall my warning: Do not confuse Matrix Multiplication with Tensor Contraction. Get It?

Sam

 

[arkajad]

As a matrix, $Q$ has nothing to do with the group $SL( 2 , C )$

Sam

It does, as is indicated, for instance, in this part Carmeli's monograph (Eq. (3.80)):

----------------------------------------------------------------------------------
$\det(Q')=\det(Q)$ so "determinant is SL(2,C) invariant".
But $\mbox{Tr }(Q')\neq \mbox{Tr }(Q)$, so "trace is not SL(2,C) invariant" (what is also evident from the fact that $\mbox{Tr }(Q)=2x^0$).

 

[vanhees71]

I think you should forget these books by Carmeli and use another one. As already said, a great book on the representation theory of the Lorentz and Poincare groups and their Lie algebras is

Sexl, Urbandtke, Relativity, Groups, Particles, Springer

My own attempt to explain these issues, you find in my QFT manuscript (Appendix B):

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

There I avoid the introduction of matrices and work with the SL(2,C) tensor formalism with undotted and dotted indices, which is much safer to avoid all kinds of mistakes, as indicated by samalkhaiat in his last posting.

 

[arkajad]

There I avoid the introduction of matrices

There is nothing wrong with working with matrices - provided you know what you are doing. And working with matrices is not that difficult. In fact, it is straightforward and elementary. There are no reasons to complicate things that can be explained and calculated in an elementary way. Easier, quicker, less prone to errors.
After all this is all about SL(2,C) and SL(2,C) is a group of matrices.

And if for some reasons you have aversion to the books by Carmeli, here is p. 305 from "Differential Geometry and Lie groups for Physicists" by Marian Fecko, Cambridge 2006.

Here is some little complication as Fecko first introduces sigma with tilde, and then gets rid of this tilde by using eta! But it is all the same.

 

[arkajad]

Essentially the same reasoning about the matrix group SL(2,C) is also in the online book by Jean Gallier "http://www.seas.upenn.edu/~jean/diffgeom.pdf", p. 227 in July 10, 2013 edition. Gallier book is also valuable because on p. 102 he warns against serious errors concerning SL(2,C) in the monograph "Matrix groups. An introduction to Lie Groups" by Andrew Baker, Springer 2002

 

[vanhees71]

I'd be utmost suspicious against books that write $\Lambda^{\mu}_{\nu}$ instead of ${\Lambda^{\mu}}_{\nu}$, but I don't know the book by Fecko. So I don't want to say anything against it only from this one quote.

I have nothing against matrices. You only have to define them properly. Using the SL(2,C)-tensor index formalism, there is, however no such trouble reflected in this thread with a lot of misunderstandings and trivial mistakes. If you write
$${\sigma^{\mu}}_{A \dot{B}}$$
it's clear from the very beginning, what transforms how. Then you only have to take into account that index raising and lowering in the SL(2,C) business is done with the skew-symmetric bilinear form
$$\epsilon_{AB}=\epsilon_{\dot{A} \dot{B}}=\begin{pmatrix}<br /> 0 & 1 \\ -1 & 0<br /> \end{pmatrix}, \quad \epsilon_{A \dot{B}}=0.$$
Then all the discussions in this thread most probably would have been avoided. That's all I'm saying.

Of course, many calculations can be abbreviated by using the matrix-vector notation, but one must make sure to define properly, which matrix denotes which SL(2,C)-spinor or SL(2,C)-transformation matrix elements!

 

[arkajad]

Of course, many calculations can be abbreviated by using the matrix-vector notation, but one must make sure to define properly, which matrix denotes which SL(2,C)-spinor or SL(2,C)-transformation matrix elements!

But the discussion is not about spinors. The discussion is about matrix groups: $SL(2,C)$ and $SO^+(3,1)$. Spinors is a different subject. It comes later after you first understand simple matrix groups. To understand matrix groups you have to operate on matrices. How to multiply them, conjugate them, take determinants, traces, exponentials etc. This is all elementary algebra and group theory. Spinors come later, when we learn differential geometry, frame bundles, spin bundles etc. Of course one can talk about tensors and spinors even without differential geometry, in a flat space, or as a part of the theory of groups representations (or even "Clifford modules"), but, as I mentioned above, first you need to understand groups that are at work. And these groups are simple matrix groups that can be studied with matrix algebra methods. Nothing more is needed (well, also some topology, but that can be also made simple).

 

[arkajad]

For those who have aversion to matrices, to ease their sufferings, here is a piece from "Matrix groups: An Introduction to Lie Group Theory" by Andrew Baker, Springer 2002

He has $S\mapsto ASA^*$ instead of $Q\mapsto gQg^\dagger$. Notation is different, but the content is always the same: it is all about properties of the matrix group SL(2,C).

 

[vanhees71]

The matrices live on a vector space, which in the case of SU(2) or SL(2,C) is called spinor space. Also the various finite-dimensional (irreducible) representations define spinors spaces. In the case of SU(2) or SL(2,C) one can show that all finite-dimensional representations can be found by reduction of representations induced by the fundamental representations on tensor products.

 

[arkajad]

The matrices live on a vector space

The matrices are just matrices - tables of numbers, with definite rules of multiplication etc.
To calculate determinant of a matrix you do not have to use vector space (though you can, it becomes more complicated). To calculate trace of a matrix you do not have to use vector space, though you can, it becomes more complicated.

Matrices can be interpreted as operating on vector space, but such an interpretation is not always needed. It is of course good to know, but you do not have always use all that you know. You use that what is needed. Killing a mosquito with a gun is of course possible, but is it always wise?

 

[arkajad]

Some comments of more general nature:

I think much of the confusion comes from the fact that some physicists fail to distinguish between two different mathematical objects. For instance "Lorentz group". How we define it? There are two different ways of defining it. One way: it is the group of 4x4 real matrices with the property that ... etc.
Second way: Suppose we have a four-dimensional real vector space V endowed with a scalar product of signature (+---). The Lorentz group is the group of isometries of this vector space.
The point is that these two definitions define objects in two different categories. Then comes a theorem:
Given an orthonormal basis in V there is an isomorphism from one mathematical object to another. Lorentz groups of two definitions becomes isomorphic. But one has to remember that if we chose a different orthonormal basis in V, the identification of the two objects will change! So, they are isomorphic, but there are many different isomorphisms and no one is better than other. The same, with appropriate changes, can be applied to SL(2,C).

I am talking here about the first kind. Others want to talk about the second. Once we understand the difference between these two categories, there will be no place for confusion.

 

[samalkhaiat]

The matrices are just matrices - tables of numbers, with definite rules of multiplication etc.
To calculate determinant of a matrix you do not have to use vector space (though you can, it becomes more complicated). To calculate trace of a matrix you do not have to use vector space, though you can, it becomes more complicated.

Matrices can be interpreted as operating on vector space, but such an interpretation is not always needed. It is of course good to know, but you do not have always use all that you know. You use that what is needed. Killing a mosquito with a gun is of course possible, but is it always wise?

You shouldn't read Carmeli's texts then. Find a book that suit your "understanding". Believe me, there are a plenty of them.

 

[arkajad]

You shouldn't read Carmeli's texts then. Find a book that suit your "understanding". Believe me, there are a plenty of them.

Indeed, there are good books. Carmeli's book are god. Also Frankel "The Geometry of Physics", Cambridge 1997. Also Naimark "Linear Representations of the Lorentz Group", Pergamon Press, 1964. Penrose and Rindler "Spinors and Space-Time" is not so good, as there are no clear definitions and theorems - just one long talk. It confuses those physicists who do not have deep enough mathematical background, so that they may not know how to make vague things mathematically precise. I think Penrose and Rindler are partly responsible for not understanding the difference between groups of matrices and groups of transformations.

Also Huggett and Tod, "An Introduction to Twistors Theory", Cambridge 1994 (2nd ed) is not bad:

Though it is not free of errors.