may be used to raise and lower indices. Because of this latter fact, there exist only two independent fundamental 2-dimensional representations of SL(2,$\mathbb C$), in direct correspondence with the two chiral Weyl spinors considered previously.

Another way of arguing the same conclusion is as follows. Given a matrix $M\in$ SL(2,$\mathbb C$), each of the matrices $M$, $M^{-1}$, $M^*$ and $(M^*)^{-1}$ defines [*a priori*]{} another 2-dimensional representation of the same group. As pointed out above, $M$ and $M^*$ are necessarily not unitarily equivalent. However, $M$ and $M^{-1}$ on the one hand, and $M^*$ and $(M^*)^{-1}$ on the other hand, are each unitarily equivalent in pairs, using the invariant tensors $\epsilon^{\alpha\beta}$ and $\epsilon_{\alpha\beta}$ because of the property ${\rm det}\,M=1$ for these $2\times 2$ matrices.

In conclusion, first we have the right-handed Weyl spinor representation $(1/2,0)$, $\psi_\alpha$ or $\psi^\alpha$, such that $$\psi^\alpha=\epsilon^{\alpha\beta}\,\psi_\beta\ \ ,\ \ 
\psi_\alpha=\epsilon_{\alpha\beta}\,\psi^\beta\ \ ,\ \ 
\epsilon^{12}=+1\ \ ,\ \ 
\epsilon_{12}=-1\ ,$$ and transforming under SL(2,$\mathbb C$) according to $${\psi'}_\alpha={M_\alpha}^\beta\,\psi_\beta\ \ ,\ \ 
{\psi'}^\alpha=\psi^\beta\,{\left(M^{-1}\right)_\beta}^\alpha\ .$$ Likewise, the left-handed Weyl spinor representation $(0,1/2)$, $\bar{\psi}_{\dot{\alpha}}$ or $\bar{\psi}^{\dot{\alpha}}$, is such that $$\overline{\psi}^{\dot{\alpha}}=\epsilon^{\dot{\alpha}\dot{\beta}}\,
\overline{\psi}_{\dot{\beta}}\ \ ,\ \ 
\overline{\psi}_{\dot{\alpha}}=\epsilon_{\dot{\alpha}\dot{\beta}}\,
\overline{\psi}^{\dot{\beta}}\ \ ,\ \ 
\epsilon^{\dot{1}\dot{2}}=+1\ \ ,\ \ 
\epsilon_{\dot{1}\dot{2}}=-1\ ,$$ each of these spinors transforming according to $${\overline{\psi}'}_{\dot{\alpha}}={M^*_{\dot{\alpha}}}^{\dot{\beta}}\,
\overline{\psi}_{\dot{\beta}}\ \ ,\ \ 
{\overline{\psi}'}^{\dot{\alpha}}=\overline{\psi}^{\dot{\beta}}\,
{\left((M^*)^{-1}\right)_{\dot{\beta}}}^{\dot{\alpha}} \ .$$ Here, in order to disti