Commutation for constraints

Question

Suppose from the Hamiltonian I got the Primary constraints $$(\Phi_m,\Phi)$$ And $\dot \Phi_m$ , $\dot \Phi$ leads to secondary constraints $$(\gamma_m,\gamma)$$ respectively. Now if the commutation of $\Phi_m$ with $\gamma_m$ is non zero (i.e they both are second class) does it terminate the commutation chain there?

Answer 1

OP wrote(v1):

Now if the (Poisson) commutation of $\Phi_m$ with $\gamma_m$ is non zero (i.e they both are second class) does it terminate the (Poisson) commutation chain there?

No, not necessarily. The rest of the $4\times 4$ matrix $\{\chi^i,\chi^j\}$ of the $4$ constraints still contains unspecified entries.

Finally, don't forget to check if there are tertiary constraints, quaternary constraints, etc.

Answer 2

To follow the process of deducation, \dot{\kappa}={\kappa,H_T}, to find whether there are some other new constraints. If there are no any new constraints, you can be taken as you are at the end.