# Co-spinors and contra-spinors

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?

• The terminology is rather uncommon. What do they mean? – DanielC Oct 13 '19 at 14:42
• He described co-spinors as $$u^{a'} =X^{a}_{\ {b}} u^b$$ and contra-spinors as the complex conjugate but with subscripts – user243882 Oct 13 '19 at 15:43
• This is odd and confusing, since it's the opposite of tensors! – Cham Oct 13 '19 at 15:45

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) $$\rho$$.