Commutation for constraints

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?

• Can you please explain your notation $(\Phi_m,\Phi)$ ? – user10001 Sep 15 '12 at 6:20
• $(\Phi_m,\Phi)$ are like the momenta.For Lagrangian L, $\Phi_m=\frac{\partial L}{\partial \dot q}$. Same goes for $\Phi$ – aries0152 Sep 15 '12 at 8:06
• Then i think notation $(\Phi_1,\Phi_2)$ would be more appropriate. $m$ may be confused for a variable subscript. – user10001 Sep 15 '12 at 8:24

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.