Decomposition of the gravitino into helicity $\pm \frac{3}{2}$ and $\pm \frac{1}{2}$ components I'm reading this book on string theory. When they decompose two dimensional gravitino (formula 7.16)
$$
\chi_\alpha = \frac{1}{2}\rho^\beta \rho_\alpha \chi_\beta + \frac{1}{2}\rho_\alpha \rho^\gamma \chi_\gamma,   
$$
they interpret summands as helicity $\pm \frac{3}{2}$ and $\pm \frac{1}{2}$ components. Here $\rho^\alpha$ are generator of the $1+1$ Clifford algebra. I have the following questions about this decomposition


*

*How helicity operator is mathematically implemented in this case?

*Why the above claim is correct?

*Is it the same thing as decomposition of tensor product of vector and spinor representations of the Clifford algebra into two irreducible representations of "spin $\frac{3}{2}$ and $\frac{1}{2}$"?

 A: Here the object $\chi_\alpha$ has an explicit 2D vector index, as well as an implicit 2D spinor index. There for it is in the $\textbf{1}\otimes\frac{\textbf{1}}{\textbf{2}} =\frac{\textbf{1}}{\textbf{2}}\oplus \frac{\textbf{3}}{\textbf{2}} $ representation of the $SO(1,1)$ group.
Now the question is how do we isolate the $\frac{\textbf{1}}{\textbf{2}}$ representation in the sum, from the $\frac{\textbf{3}}{\textbf{2}}$ representation? we will succeed if we can form an object out of $\chi_\alpha$ which is in the $\frac{\textbf{1}}{\textbf{2}}$ representation of $SO(1,1)$ and remove that object from the original $\chi_\alpha$.
You can easily see that $\rho^\alpha \chi_\alpha \equiv \xi$ does the job elegantly, because the vector index is contracted and all is left is the spinor index. So $\rho^\alpha \chi_\alpha\equiv \xi$ extracts the spin half component of the gravitino, in much the same way as the trace of a tensor extracts the scalar component of a tensor.
Now we want to form an object $\psi_\alpha$ out of $\chi_\alpha$, which satisfies $\rho^\alpha\psi_\alpha =0$, because as we understand now such object would have the spin half component removed! you can see using simple gamma matrix identity that $\psi_\alpha \equiv \frac{1}{2} \rho^\beta \rho_\alpha \chi_\beta = \chi_\alpha - \frac{1}{2}\rho_\alpha \xi$, does the job and satisfies the condition. Which means it is the way to project the spin three halves component out of $\chi_\alpha$.
So schematically $\chi = \xi\oplus\psi$ is the consequence of $\textbf{1}\otimes\frac{\textbf{1}}{\textbf{2}} =\frac{\textbf{1}}{\textbf{2}}\oplus \frac{\textbf{3}}{\textbf{2}} $
Note: in two dimensions the helicity operator is $\propto \rho^0\rho^1$ just like the $\gamma^5$ in four dimensions
Note: this is the exactly same procedure to understand the Rarita-Schwinger condition on spin $3/2$ field in four dimensions
