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1

The definition (pay attention to not confuse generic tensor with tensor components) of the antisymmetric gamma tensor is: $$\gamma^{\mu_1 \mu_2 \dots \mu_r}=\gamma^{[\mu_1 \mu_2 \dots \mu_r]}$$ For the highest rank you have $r=D$, so you have to use all the possible indices. For example, in components, you have the identity: $$\gamma^{1 2 3}=\frac{1}{3!} ...


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Maybe I am missing something, but in your eq. (2) there are two equal terms on the right-hand side, and the factor is 1/2 to take that into account. In your eq.(1) there are also two equal terms on the right-hand side, and the factor is 1/2 to take that into account. There is no summation over all permutations in your formulas.


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Lorentz spinors appear as irreducible representations of the group SL(2,C). Elements of the group are 2x2 matrices with complex entries and unity determinant. A Lorentz spinor is a two component vector $\psi^{A},\chi^{A}\in V_{2}$ with $A=1,2$. The Levi-Civita tensor $\epsilon_{AB}$ is an invariant tensor under SL(2,C). This means that if we have an irrep ...


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I've clarified my doubt. The expression $ (\bar{\psi} \chi )^{T }=\chi^{T }\bar{\psi} ^{T } $ is simply the definition of trasposition (the same argument for the adjoint), independently on the nature of the objects in the parenthesis. What is subtle is that, the following is true: $ \bar{\psi} \chi =-\chi^{T }\bar{\psi} ^{T } =-(\bar{\psi} \chi )^{T }$ ...


2

Soliton is correct, there is a factor of two coming in when the indices are raised or lowered with the Levi-Civita tensors and then a contraction is made. Here is how it appears. Consider a four-vector as a Lorentz spinor, \begin{equation} x^{\mu}\rightarrow X^{\dot{A}B}= \left[ \begin{array}{cc} x^{0}+x^{3} & x^{1}-ix^{2} \\ ...


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One usually starts from the CCR for the creation/annihilation operators and derives from there the commutation rules for the fields. However, one can start from either (see for example here about this). Suppose we want then to start from the equal-time anticommutation rules for a Dirac field $\psi_\alpha(x)$: $$ \tag{1} \{ \psi_\alpha(\textbf{x}), ...


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I) Representation theory of the Lorentz$^1$ group is a fairly broad subject covered in many textbooks, see e.g. Ref. 1 for further information. II) The irreducible representation $$\tag{1} (j_L,j_R)~=~j_L\otimes_{\mathbb{C}} j_R, \qquad j_L, j_R~\in~ \frac{1}{2}\mathbb{N}_0,$$ is a tensor product of $V=V_L\otimes_{\mathbb{C}} V_R$ of two complex vector ...


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The matrices $\sigma^{\mu}$, being hermitian, satisfy:$(\sigma^{\mu}_{\alpha\dot{\beta}})^{*}$=$( \sigma^{\mu *})_{\dot{\alpha}\beta}$=$(\sigma^{\mu\dagger})_{\beta\dot{\alpha}}$=$(\sigma^{\mu})_{\beta\dot{\alpha}}$. So $(\xi\sigma^{\mu}\bar\psi)^{*}$=$(\xi^{\alpha}\sigma^{\mu}$$_{\alpha\dot{\beta}}$$\bar{\psi}$$^{\dot{\beta}})$$^{*}$=$\dots$, the rest ...


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I think you can consider the anti-commutator, then use the linearity of the trace, as follows: $$ \{\overline{\psi},\chi\} = 0$$ $$ \{\overline{\psi},\chi\}^T = 0$$ $$ (\overline{\psi}\chi)^T + (\chi\overline{\psi})^T = 0$$ $$ (\overline{\psi}\chi)^T + \overline{\psi}^T\chi^T = 0$$ $$ (\overline{\psi}\chi)^T - \chi^T\overline{\psi}^T = 0$$ $$ ...


2

Advanced Classical Field Theory (2009) by Giachetta, Mangiarotti, Sardanashvily remarks on p. 248: A non-compact world manifold admits a Dirac spinor structure if and only if it is parallelizable. For a compact world manifold $X$, its Euler characteristic and the second Stiefel-Whitney class $w_2$ must be zero, and its first Pontryagin number ...



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