Co-spinors and contra-spinors

Question

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?

Answer 1

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 $$g~~\mapsto~~ (\rho(g)^{-1})^T~=~\epsilon \rho(g)\epsilon^{-1},\qquad g~\in~SL(2,\mathbb{C}),$$ 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) $\rho$.

Answer 2

As alluded to in commentary, there is a metric for spinors. (Edit: which is underlies the distinction between covariant and contravariant.) Watch, as I do a little magic and pull one out of the hat ... and a lot more, besides that!

In a Lorentzian space-time, one can always choose a frame of the form $\left(h^0, h^1, h^2, h^3\right)$, such that the components of the dual metric with respect to it are given by: $$g^{-1}\left(h^a, h^b\right) = h^{ab}, \hspace 1em \left(h^{ab}: a, b = 0, 1, 2, 3\right) = \text{diag}(+, -, -, -).$$ The line element of the metric $g$, with respect to this frame is: $$g = h^0 h^0 - h^1 h^1 - h^2 h^2 - h^3 h^3.$$ I am, here, denoting tensor products by juxtaposition (e.g. $h^0 h^0$ for $h^0 ⊗ h^0$).

For clarity, the elements of the frame are co-vectors, i.e. $h^a = h^a_μ dx^μ$. Thus, $g_{μν} = h^a_μ h^b_ν h_{ab}$, where $h_{ab} = h^{ab}$ (the frame components of the metric, as a matrix, is equal to its own inverse).

Now ... we'll define $$ L = \frac{h^0 + h^3}{\sqrt{2}}, \hspace 1em M = \frac{h^1 - i h^2}{\sqrt{2}}, \hspace 1em \bar{M} = \frac{h^1 + ih^2}{\sqrt{2}}, \hspace 1em N = \frac{h^0 - h^3}{\sqrt{2}}. $$ This is a skew-frame, not orthonormal, nor even orthogonal. Substituting, we find: $$ LN = \frac{h^0 h^0 - h^0 h^3 + h^3 h^0 - h^3 h^3}{2}, \\ M\bar{M} = \frac{h^1 h^1 + i h^1 h^2 - i h^2 h^1 + h^2 h^2}{2}, \\ \bar{M}M = \frac{h^1 h^1 - i h^1 h^2 + i h^2 h^1 + h^2 h^2}{2}, \\ NL = \frac{h^0 h^0 + h^0 h^3 - h^3 h^0 - h^3 h^3}{2}. $$ Thus: $$LN - M\bar{M} - \bar{M}M + NL = h^0 h^0 - h^1 h^1 - h^2 h^2 - h^3 h^3 = g.$$

Now ... we will adopt the following axiom, that the frame splits: $$L = ο\bar{ο}, \hspace 1em M = ο\bar{ι}, \hspace 1em \bar{M} = ι\bar{ο}, \hspace 1em N = ι\bar{ι}.$$ This involves a split of $𝟰: (L, M, \bar{M}, N)$ into $𝟮: (ο, ι)$ and $\bar{𝟮}: (\bar{ο}, \bar{ι})$. We will also postulate that the products of the $𝟮$ frame commute with those of the $\bar{𝟮}$ frame: $$ ο\bar{ο} = \bar{ο}ο, \hspace 1em ο\bar{ι} = \bar{ι}ο, \hspace 1em ι\bar{ο} = \bar{ο}ι, \hspace 1em ι\bar{ι} = \bar{ι}ι. $$

Then, it follows that $$\begin{align} LN - M\bar{M} - \bar{M}M + NL &= ο\bar{ο}ι\bar{ι} - ο\bar{ι}ι\bar{ο} - ι\bar{ο}ο\bar{ι} + ι\bar{ι}ο\bar{ο} \\ &= οι\bar{ο}\bar{ι} - οι\bar{ι}\bar{ο} - ιο\bar{ο}\bar{ι} + ιο\bar{ι}\bar{ο} \\ &= \left(οι - ιο\right)\left(\bar{ο}\bar{ι} - \bar{ι}\bar{ο}\right). \end{align}$$

Thus, $$g = ε\bar{ε},$$ where $$ε ≡ ο∧ι ≡ οι - ιο, \hspace 1em \bar{ε} ≡ \bar{ο}∧\bar{ι} ≡ \bar{ο}\bar{ι} - \bar{ι}\bar{ο}.$$ The metric $g$ over $𝟰$ splits into a metric $ε$ over $𝟮$ and a metric $\bar{ε}$ over $\bar{𝟮}$.

But, wait ... there's more!

$L$, $M$, $\bar{M}$ and $N$ are not just co-vectors, but one-forms. What about two-forms? Set $$ Z^+ ≡ οο, \hspace 1em Z^0 ≡ \frac{οι + ιο}{\sqrt{2}}, \hspace 1em Z^- ≡ ιι, \\ \bar{Z^+} ≡ \bar{ο}\bar{ο}, \hspace 1em \bar{Z^0} ≡ \frac{\bar{ο}\bar{ι} + \bar{ι}\bar{ο}}{\sqrt{2}}, \hspace 1em \bar{Z^-} ≡ \bar{ι}\bar{ι}. $$ Then $$ L∧M = LM - ML = ο\bar{ο}ο\bar{ι} - ο\bar{ι}ο\bar{ο} = οο\bar{ο}\bar{ι} - οο\bar{ι}\bar{ο} = οο\left(\bar{ο}\bar{ι} - \bar{ι}\bar{ο}\right) = Z^+ \bar{ε}. $$ Similarly, $$L∧\bar{M} = ε \bar{Z^+}.$$ For the other frame 2-forms, one gets: $$M∧N = ε \bar{Z^-}, \hspace 1em \bar{M}∧N = Z^- \bar{ε},$$ and $$ \frac{L∧N + M∧\bar{M}}{\sqrt{2}} = ε \bar{Z^0}, \hspace 1em \frac{L∧N + \bar{M}∧M}{\sqrt{2}} = Z^0 \bar{ε}. $$

Let's not forget the symmetric combinations - using the notation $a∨b = ab + ba$: $$ LL = Z^+ \bar{Z^+}, \hspace 1em \frac{L∨M}{\sqrt{2}} = Z^+ \bar{Z^0}, \hspace 1em MM = Z^+ \bar{Z^-}, \\ \frac{L∨\bar{M}}{\sqrt{2}} = Z^0 \bar{Z^+}, \hspace 1em \frac{L∨N + M∨\bar{M}}{2} = Z^0 \bar{Z^0}, \hspace 1em \frac{M∨N}{\sqrt{2}} = Z^0 \bar{Z^-}, \\ \bar{M}\bar{M} = Z^- \bar{Z^-}, \hspace 1em \frac{\bar{M}∨N}{\sqrt{2}} = Z^- \bar{Z^0}, \hspace 1em NN = Z^- \bar{Z^-}. $$ That realizes the following decompositions: $$ 𝟰∧𝟰 = \left(𝟯⊗\bar{𝟭}\right) ⊕ \left(𝟭⊗\bar{𝟯}\right), \hspace 1em 𝟰∨𝟰 = \left(𝟯⊗\bar{𝟯}\right) ⊕ \left(𝟭⊗\bar{𝟭}\right), \hspace 1em 𝟰⊗𝟰 = \left(𝟯 ⊕ 𝟭\right)⊗\left(\bar{𝟯} ⊕ \bar{𝟭}\right), $$ with the frame bases: $$ 𝟭:(ε), \hspace 1em 𝟯:\left(Z^+, Z^0, Z^-\right), \\ \bar{𝟭}:(\bar{ε}), \hspace 1em \bar{𝟯}:\left(\bar{Z^+}, \bar{Z^0}, \bar{Z^-}\right). $$

That's the full decomposition that the decomposition of the metric is part of.