Why \(S^+\) and \(S^-\) Are Conjugates of Each Other

This note explains why the two irreducible chiral spinor representations of \(SO(2n)\), denoted \(S^+\) and \(S^-\), are either conjugate to each other or conjugate to themselves, depending on whether \(n\) is odd or even.

1. What it means for two representations to be complex conjugates

A representation \(R\) is complex conjugate to a representation \(R'\) if there exists a matrix \(C\) such that for every group element \(g\),

\[ C^{-1} R^*(g) C = R'(g). \]

So the problem is to determine what happens to the spinor representation after complex conjugation.

2. Spinor transformation law

For the two chiral spinor representations, the representation matrices are

\[ R_\pm = e^{\frac{i}{4}\omega_{ij}\sigma_{ij}} P_\pm, \]

where the chirality projectors are

\[ P_\pm = \frac{1}{2}(1 \pm \gamma_F). \]

The representations \(S^+\) and \(S^-\) are precisely the subspaces selected by \(P_+\) and \(P_-\).

3. Complex conjugating the representation

Taking the complex conjugate gives

\[ R_\pm^* = e^{-\frac{i}{4}\omega_{ij}\sigma_{ij}^*} P_\pm^*. \]

To compare this with the original representation, we use a conjugation matrix \(C\).

4. Two key identities

The needed identities are:

\[ C^{-1} \sigma_{ij}^* C = -\sigma_{ij} \] and \[ C^{-1} \gamma_F C = (-1)^n \gamma_F. \]

The first identity tells us how the generators behave under conjugation. The second identity tells us what happens to chirality.

5. What happens to the projector

Since

\[ P_\pm = \frac{1}{2}(1 \pm \gamma_F), \]

we obtain

\[ C^{-1} P_\pm^* C = \frac{1}{2}\bigl(1 \pm (-1)^n \gamma_F\bigr). \]

Therefore:

\[ C^{-1} P_\pm^* C = \begin{cases} P_\pm, & n \text{ even}, \\ P_\mp, & n \text{ odd}. \end{cases} \]

6. Putting everything together

Apply \(C^{-1}(\cdot)C\) to the complex conjugate representation:

\[ C^{-1} R_\pm^* C = C^{-1} \left(e^{-\frac{i}{4}\omega_{ij}\sigma_{ij}^*} P_\pm^*\right) C. \]

Using \(C^{-1}\sigma_{ij}^*C=-\sigma_{ij}\), the exponential becomes

\[ C^{-1} e^{-\frac{i}{4}\omega_{ij}\sigma_{ij}^*} C = e^{\frac{i}{4}\omega_{ij}\sigma_{ij}}. \]

So we get

\[ C^{-1} R_\pm^* C = e^{\frac{i}{4}\omega_{ij}\sigma_{ij}} \left(C^{-1} P_\pm^* C\right). \]

Now substitute the result for the projector.

7. Even and odd \(n\)

If \(n\) is even, then \(C^{-1}P_\pm^*C=P_\pm\), so

\[ C^{-1} R_\pm^* C = e^{\frac{i}{4}\omega_{ij}\sigma_{ij}} P_\pm = R_\pm. \]

Thus each chiral representation is equivalent to its own complex conjugate:

\[ (S^\pm)^* \cong S^\pm \qquad (n \text{ even}). \]

If \(n\) is odd, then \(C^{-1}P_\pm^*C=P_\mp\), so

\[ C^{-1} R_\pm^* C = e^{\frac{i}{4}\omega_{ij}\sigma_{ij}} P_\mp = R_\mp. \]

Therefore the complex conjugate of one chiral spinor is the other:

\[ (S^+)^* \cong S^-, \qquad (S^-)^* \cong S^+ \qquad (n \text{ odd}). \]

8. The main idea

The crucial point is that complex conjugation acts on the chirality operator through

\[ C^{-1}\gamma_F C = (-1)^n \gamma_F. \]

So:

When \(n\) is even, chirality is preserved under conjugation, so \(S^+\) and \(S^-\) are each self-conjugate.

When \(n\) is odd, chirality is flipped under conjugation, so \(S^+\) and \(S^-\) are conjugates of each other.

9. Final summary

\[ \boxed{ \begin{aligned} (S^\pm)^* &\cong S^\pm, && n \text{ even}, \\ (S^+)^* &\cong S^-, \quad (S^-)^* \cong S^+, && n \text{ odd}. \end{aligned}} \]

So the statement that \(S^+\) and \(S^-\) are conjugates of each other is true in the odd-\(n\) case. In the even-\(n\) case, each one is instead conjugate to itself.

This is additional material to complement page 415 of Group Theory in a Nutshell by Anthony Zee.

Created with assistance from ChatGPT.