This thread shows how easily one gets into troubles if one confuses TENSOR CONTRACTION, i.e., A_{ a r } B^{ c r }, with the corresponding MATRIX MULTIPLICATION, i.e., A_{ a r } B^{ r c }. The two expressions become identical, if and only if the followings are true:
1) the first and the second index both belong to identical or equivalent representations of the group in the question, and
2) B^{ c r } = B^{ r c }.
We will see that none of these conditions is satisfied in SL( 2 , C ). But, before doing the tensor notations of SL( 2 , C ), I would like to derive the relation between the Lorentz transformation,
x^{ \mu } \rightarrow \Lambda^{ \mu }{}_{ \nu } \ x^{ \nu } , \ \ \Lambda \in SO( 1 , 3 ) , \ \ (1)
and the corresponding SL( 2 , C ) transformation of the Hermitian matrix x_{ \mu } \sigma^{ \mu } = \eta_{ \mu \tau } \ x^{ \mu } \ \sigma^{ \tau }:
<br />
\eta_{ \mu \tau } \ x^{ \mu } \ \sigma^{ \tau } \rightarrow \eta_{ \nu \tau } \ x^{ \nu } \ g \ \sigma^{ \tau } \ g^{ \dagger } , \ \ ( g , g^{ \dagger } ) \in SL( 2 , C ) \ (2)<br />
I choose to define
<br />
\eta^{ \tau \mu } \ \sigma_{ \mu } = \sigma^{ \tau } \equiv ( I , \vec{ \sigma } ) ,<br />
I also define
<br />
\eta^{ \tau \mu } \ \bar{ \sigma }_{ \mu } = \bar{ \sigma }^{ \tau } \equiv ( I , - \vec{ \sigma } ) .<br />
Notice that \sigma_{ \mu } and \bar{ \sigma }^{ \nu } are the same matrices. However, writing \sigma_{ \mu } = \bar{ \sigma }^{ \mu } would mess up your covariant expressions (some thing that you would not want to do). So, to stay out of troubles we will treat \sigma^{ \mu } and \bar{ \sigma }^{ \nu } as two independent objects.
Now, by direct matrix multiplication, we can easily verify that
\mbox{ Tr } ( \sigma^{ \tau } \bar{ \sigma }^{ \rho } ) = 2 \ \eta^{ \tau \rho } . \ \ (3)
Contracting both side of Eq(1) with \eta_{ \mu \tau } \sigma^{ \tau } gives
\eta_{ \mu \tau } \ x^{ \mu }\ \sigma^{ \tau } \rightarrow \Lambda^{ \mu }{}_{ \nu } \ \eta_{ \mu \tau } \ x^{ \nu } \ \sigma^{ \tau } . \ \ \ (4)
Comparing the RHS of Eq(4) with that of Eq(2) leads to
\Lambda^{ \mu }{}_{ \nu } \ \eta_{ \mu \tau } \ \sigma^{ \tau } = \eta_{ \nu \tau } \ g \ \sigma^{ \tau } \ g^{ \dagger } .
Multiplying both sides by \bar{ \sigma }^{ \rho } and taking the trace, we find (using Eq(3)) that
\Lambda^{ \mu }{}_{ \nu } \ \eta_{ \mu \tau } ( 2 \ \eta^{ \tau \rho }<br />
) = \mbox{ Tr } (g \ \eta_{ \nu \tau } \ \sigma^{ \tau } \ g^{ \dagger } \ \bar{ \sigma }^{ \rho } ) ,
or
2 \Lambda^{ \mu }{}_{ \nu } \ \delta^{ \rho }_{ \mu } = \mbox{ Tr } ( g \ \sigma_{ \nu } \ g^{ \dagger } \ \bar{ \sigma }^{ \rho } ) .
Thus
\Lambda^{ \rho }{}_{ \nu } ( g ) = \frac{ 1 }{ 2 } \mbox{ Tr } ( \bar{ \sigma }^{ \rho } \ g \ \sigma_{ \nu } \ g^{ \dagger } ).
***********
Tensor Contraction & Trace in SL( 2 , C)
The matrices \sigma^{ \mu } behave like mixed spinor-tensors indexed by A = 1,2 and \dot{B} = \dot{1}, \dot{2} which label the fundamental and anti-fundamental (conjugate) representations respectively,
( \sigma^{ \mu } )_{ A \dot{B} } = ( I , \vec{ \sigma } )_{ A \dot{B} } .
Complex conjugation turns undotted indices into dotted ones and vice versa. For example, the complex conjugate of the spinor-tensor \Psi_{ A B \dot{C}} is given by
\overline{ \Psi_{ A B \dot{C} } } = \bar{ \Psi }_{ \dot{A} \dot{B} C } .
Therefore, the hermiticity of \sigma^{ \mu } means, in the SL( 2 , C ) index notations, that
\overline{ ( \sigma^{ \mu } )_{ A \dot{B} } } \equiv ( \bar{ \sigma }^{ \mu } )_{ \dot{A} B } = ( \sigma^{ \mu } )_{ B \dot{A} } . \ \ (5)
Raising and lowering of the indices is done here by the invariant (metric) spinors
\epsilon_{ A B } = \epsilon^{ A B } = \epsilon_{ \dot{A} \dot{B} } = <br />
\epsilon^{ \dot{A} \dot{B} } , \ \ (6a)
which are the 2-dimensional Levi-Civita symbol, i.e. ,
<br />
\epsilon_{ A B } = - \epsilon_{ B A } = \left( \begin{array} {rr} 0 & 1 \\ -1 & 0 \end{array} \right) . \ \ (6b)<br />
They satisfy the relations
\epsilon_{ A B } \epsilon^{ C B } = \epsilon^{ C }{}_{ A } = - \epsilon_{ A }{}^{ C } = \delta^{ C }_{ A } . \ \ (6c)
The invariance of the metric spinor \epsilon_{ A B } under SL( 2 , C), which is a necessary condition for the consistency of the relations (6), follows from the definition of a determinant
\det | g | \epsilon_{ A B } = g_{ A }{}^{ C } \ g_{ B }{}^{ D } \ \epsilon_{ C D } ,
and the fact that, for all g \in SL( 2 , C ), \det | g | = 1.
The invariant trace of tensors is, therefore, obtained by contracting all indices with the appropriate spinor metric. For example, the trace of the tensor T^{ A B } is given by
\mbox{Tr}_{ sl(2,C) } ( T ) = \epsilon_{ A B } T^{ A B } . \ \ \ (7)
It is clear from this that a traceless matrix, if interpreted as tensor, needs not be traceless tensor. For instance, as an anti-symmetric matrix, the Levi-Civita symbol is traceless, i.e. ,
\mbox{Tr}_{ M } ( \epsilon ) = 0 ,
while the tensor \epsilon^{ A B }, as seen from Eq(7) and Eq(6c), has a non-vanishing trace,
\mbox{Tr}_{ sl(2,C) } ( \epsilon ) = \epsilon^{ A }{}_{ A } = - \epsilon_{ A }{}^{ A } = 2 .
Another example is the Pauli matrices \sigma^{ i } themselves. While \mbox{Tr}_{ M } ( \sigma^{ i } ) = 0, \mbox{Tr}_{ sl(2,C) } ( \sigma^{ i } ) does not exist (not even defined). However, the tensor ( \sigma^{ \mu } \sigma^{ \nu }<br />
)_{ A B } has a trace given by
<br />
\mbox{Tr}_{ sl(2,C) } ( \sigma^{ \mu } \sigma^{ \nu } ) = ( \sigma^{ \mu } )_{ A \dot{ B } } ( \sigma^{ \nu } )^{ A \dot{ B } } = 2 \eta^{ \mu \nu } . \ \ \ (8)<br />
This follows directly from the Dirac algebra (upon contracting the indices A and C):
<br />
\{ \sigma^{ \mu } , \sigma^{ \nu } \}^{ C }_{ A } \equiv ( \sigma^{ \mu} )_{ A \dot{ B } } ( \sigma^{ \nu } )^{ C \dot{ B } } + ( \sigma^{ \nu } )_{ A \dot{ B } } ( \sigma^{ \mu } )^{ C \dot{ B } } = 2 \ \delta^{ C }_{ A } \eta^{ \mu \nu } .<br />
The hermiticity condition Eq(5) allows us to rewrite the trace in Eq(8) as that of matrix multiplication
<br />
\mbox{Tr}_{ sl(2,C) } ( \sigma^{ \mu } \sigma^{ \nu } ) = ( \bar{ \sigma }^{ \mu } )_{ \dot{ B } A } ( \sigma^{ \nu } )^{ A \dot{ B } } = \mbox{Tr}_{ M } ( \bar{ \sigma }^{ \mu } \sigma^{ \nu } ) .<br />
Let us now show that the complex conjugation defined in Eq(5) is consistent with our (index free) definition of \bar{ \sigma }^{ \mu } = ( I , - \vec{ \sigma } ). Consider
( \sigma^{ \mu } )^{ A \dot{ B } } = \epsilon^{ A C } \ \epsilon^{ \dot{ B } \dot{ D } } \ ( \sigma^{ \mu } )_{ C \dot{ D } } ,
now take the complex conjugate of both sides,
<br />
\overline{ ( \sigma^{ \mu } )^{ A \dot{ B } } } = \overline{ \epsilon^{ A C } } \ \overline{ \epsilon^{ \dot{ B } \dot{ D } } } \ \overline{ ( \sigma^{ \mu } )_{ C \dot{ D } } } .<br />
Using the rule of complex conjugation and Eq(5), we find
<br />
( \bar{ \sigma }^{ \mu } )^{ \dot{ A } B } = - \epsilon^{ B D } \ ( \sigma^{ \mu } )_{ D \dot{ C } } \ \epsilon^{ \dot{ C } \dot{ A } } = - ( \epsilon \ \sigma^{ \mu } \ \epsilon )^{ B \dot{ A } } .<br />
Therefore
\bar{ \sigma }^{ \mu } = - ( \epsilon \ \sigma^{ \mu } \ \epsilon )^{ t } .
Since \epsilon^{ t } = \epsilon^{ -1 } = - \epsilon, then
\bar{ \sigma }^{ \mu } = - \epsilon \ ( \sigma^{ \mu } )^{ t } \ \epsilon = ( I , - \vec{ \sigma } ) .
I finish this post by writing the completeness relations with respect to the trace
\eta_{ \mu \nu } \ ( \sigma^{ \mu } )_{ A \dot{ B } } \ ( \sigma^{ \nu } )^{ C \dot{ D } } = 2 \ \delta^{ C }_{ A } \ \delta^{ \dot{ D } }_{ \dot{ B } } ,
which follow from
\eta_{ \mu \nu } \ ( \sigma^{ \mu } )_{ A \dot{ B } } \ ( \sigma^{ \nu } )_{ C \dot{ D } } = 2 \ \epsilon_{ A C } \ \epsilon_{ \dot{ B } \dot{ D } } .
Sam
But for what it's worth, I'll summarize it all over again.
You've repeatedly been given full explanations, all the way back to dextercioby's #2. So I can't imagine what good it would do to repeat it again, since you are apparently only interested in publicizing Carmeli's "mistake" rather than understanding it.
the point, once again, is that Tr(σμσν) is not a covariant quantity unless you do one of two things: a) replace one of the σ's, by defining a barred quantity σ−μ which reverses the sign of one component, and writing Tr(σ−μσν), or b) redefine the Tr operation itself to implicitly include this replacement. There is nothing especially strange about redefining Tr. Doing this is similar to what we do by redefining the dot product operation u·v to include the minus sign when dealing with 4-vectors.