In my QFT lecture notes, it is written that the Lorentz group elements can be written as \begin{equation*} \Lambda = e^{i\vec{\theta}\cdot\vec{J} + i\vec{\eta}\cdot\vec{K}} \end{equation*}

where $\Big\{\vec{J}, \vec{K}\Big\}$ are the generators of the Lorentz algebra.

Ater this, they write that Weyl spinors transform under a representation of the Lorentz group, as \begin{equation*} \phi' = e^{i\frac{\vec{\sigma}}{2}\cdot\left(\vec{\theta} - i\vec{\eta}\right)}\phi \end{equation*}

Here, as $\Big\{\frac{\vec{\sigma}}{2}, -i\frac{\vec{\sigma}}{2}\Big\}$ indeed satisfies the Lorentz algebra commutation relations, it is indeed a representation of the Lorentz algebra $\Big\{\vec{J}, \vec{K}\Big\}$. However, not all representations of Lie algebras lead to a representation of the Lie group by exponentiation. So, for the case of the Weyl spinor, how can we show that the transformation rule is indeed a representation of the Lorentz group?

  • $\begingroup$ Elementary Weyl spinors transform either in the (1/2,0) or (0,1/2) projective representation of the restricted Lorentz group or in the true (1/2,0) or (0,1/2) representation of $\text{SL}(2,\mathbb C)$. $\endgroup$
    – DanielC
    Oct 1, 2021 at 9:36

1 Answer 1


TL;DR: To discuss non-projective group representations of spinors we need to go to the universal covering group.

In detail:

  1. First define a (left) Weyl spinor $\phi$ to transform in the defining group representation of $SL(2,\mathbb{C})$, which is the double cover of the restricted Lorentz group $SO^+(1,3;\mathbb{R})$.

  2. Only thereafter, we should identify the corresponding Lie algebra $sl(2,\mathbb{C})\cong so(1,3;\mathbb{R})$, the Lie algebra representation, and their 6 generators of boosts and rotations.


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