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As i was reading my teacher's notes on $SU(2)$ and $SO(3)$, i have had this question. Why do co-spinors transform differently under a rotation than contra-spinors?

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    $\begingroup$ The terminology is rather uncommon. What do they mean? $\endgroup$ – DanielC Oct 13 '19 at 14:42
  • $\begingroup$ He described co-spinors as $$u^{a'} =X^{a}_{\ {b}} u^b$$ and contra-spinors as the complex conjugate but with subscripts $\endgroup$ – user243882 Oct 13 '19 at 15:43
  • $\begingroup$ This is odd and confusing, since it's the opposite of tensors! $\endgroup$ – Cham Oct 13 '19 at 15:45
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The 3D rotation group $SO(3)$ has double-cover $SU(2)$, which is a subgroup of $SL(2,\mathbb{C})$.

TL;DR: A spinor index of a 2-component spinor is raised and lowered with the 2D Levi-Civita symbol, which can informally be viewed as a "symplectic metric" for the symplectic group $Sp(2,\mathbb{C})\cong SL(2,\mathbb{C})$.

In more detail, in case of the Lie group $SL(2,\mathbb{C})$,

  • the fundamental/defining representation $\rho={\rm id}:SL(2,\mathbb{C})\to SL(2,\mathbb{C})$ is the left-handed Weyl spinor representation;

  • the dual/contragredient/transposed representation is $$(\rho(g)^{-1})^T=\epsilon \rho(g)\epsilon^{-1},\qquad g~\in~SL(2,\mathbb{C}),$$ and hence an equivalent representation to $\rho$;

  • the complex conjugate representation $\bar{\rho}$ is the right-handed Weyl spinor representation. If we restrict $\bar{\rho}$ to $SU(2)$, it is equivalent to (the restriction of) $\rho$.

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