Left and right Weyl representations are inequivalent representations We introduce the two-component spinors in the following representations:
$$
\psi_\alpha \rightarrow\psi'_\alpha=\mathcal{M}_\alpha^\beta\psi_\beta$$
$$
\bar\psi_\dot{\alpha}\rightarrow\bar\psi_\dot{\alpha}'=\mathcal{M^*}_\dot{\alpha}^\dot{\beta}\bar{\psi}_\dot{\beta}
$$
Where the matrix $\mathcal{M}\in SL(2,\mathbb{C})$ 
These two representations are not equivalent because there is no matrix for which $\mathcal{M^*}=\mathcal{CMC^{-1}}$. I don't quite see how that is true. Since the entries in $\mathcal{M^*}$ are just the complex conjugates of $\mathcal{M}$, isn't it trivial to find a matrix $\mathcal{C}$ that satisfies that condition?
In other words, how can we prove that there is no such matrix $\mathcal{C}$?
 A: This is the claim that left-handed Weyl spinors and right-handed Weyl spinors are not the same.


*

*An abstract way to see that there cannot be such a matrix is to first show that the two representations correspond to $(1/2,0)$ and $(0,1/2)$ in the usual parlance of expressing representations of the complexification $\mathfrak{sl}(2,\mathbb{C})_\mathbb{C}\cong \mathfrak{sl}(2,\mathbb{C})\oplus\mathfrak{sl}(2,\mathbb{C})$ by representations of $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$, classified by pairs of half-integers $(j_1,j_2)$. Since the $j_i$ are values of Casimirs but an equivalence of representations must preserve Casimirs, the two representations cannot be equivalent.

*(This is cribbed from a deleted answer by Chiral Anomaly here) A concrete way to see this as follows:
$$ M^\ast = CMC^{-1}$$
implies that $\mathrm{tr}(M^\ast) = \mathrm{tr}(M)^\ast = \mathrm{tr}(M)$ by cyclicity of the trace, and so $\mathrm{tr}(M)\in\mathbb{R}$. But $M=\mathrm{diag}(z,z^{-1})\in\mathrm{SL}(2,\mathbb{C})$ for arbitrary $z\in\mathbb{C}$, and $z + z^{-1}$ is clearly not real for such arbitrary $z$.
