How to prove Weyl spinors transform as a representation of Lorentz group?

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?

• 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 $\text{SL}(2,\mathbb C)$. Oct 1, 2021 at 9:36

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.