Does the inverse of the Dirac conjecture hold?

Question

In the theory of constrained Hamiltonian systems, one differentiates between primary and secondary constraints, where the former are constraints derived directly from the Hamiltonian in question and the latter are only realized 'on-shell', i.e. once the equations of motion are satisfied.

Further one can differentiate between first-class and second-class constraints, where first-class constraints have a vanishing Poisson bracket with all other constraints, and second-class constraints don't.

It can now be shown, that first-class primary constraints, generate gauge transformation.

The Dirac conjecture states that one could remove the requirement to have a first-class primary constraint and therefore that all first-class constraints (no matter if primary or secondary) generate gauge transformations.

The conjecture is shown to not hold in some very specific examples but is used in the literature nontheless.

My question is whether at least the inverse is true: can every gauge theory be formulated as a Hamiltonian problem with a first-class constraint?

The case of formulating a gauge theory as a hamiltonian with primary first-class constraint probably just means adding gauge fixing terms to the theory. In that case the question might be rather, if a closed-form gauge fixing term can be found for any gauge group. And if not, can gauge theories for which this fails be formulated as first-class secondary constraints?

Answer 1

Continuous gauge symmetries are generated by first-class constraints.

To show this, we only need to show it for the extended action $$S_E:=\int dt \, \dot q^n p_n - H +u^m \phi_m\,,$$ where $m$ runs over the first-class constraints. It is enough to show the statement for the extended action because the gauge symmetries of an action $S$ are the same as those of the total action $S_T$, whose gauge symmetries are the residual gauge symmetries of $S_E$.

Write the transformation as $$\delta q^n = Q^n\,,\qquad \delta p_n = P_n\,, \qquad \delta u^m = U^m\,.$$ Plugging in to the extended Lagrangian, $$\delta L_E = \dot q^n P_n-Q^n \dot p_n + \frac{\partial (H+u^m\phi_m)}{\partial q^n}Q^n+\frac{\partial (H+u^m\phi_m)}{\partial p_n}P_n+U^m \phi_m\,.$$ This is a total derivative $$= \frac{d}{dt}F$$ only if $$Q_n = \frac{\partial F}{\partial p_n}\,,\qquad P_n = \frac{\partial F}{\partial q^n}$$ $$U^m\phi_m=\{F,H\}+u^m \{F,\phi_m\}+\frac{D}{Dt}F$$ where $$\frac{D}{Dt}=\partial_t + u^m \frac{\partial}{\partial u^m} + \dot u^m \frac{\partial}{\partial \dot u^m}+\dots\,.$$

So we solve for $F$ and get $$F = f^m \phi_m + \bar{F}$$ where $f^m \phi_m$ is the inhomogenuous part of the solution and $\bar{F}$ is the homogenuous part. Therefore $\bar{F}$ is an integral of motion, and will not induce a gauge transformation, which we should be able to prescribe at any time $t$ independently of all other times. So a gauge transformation corresponds to $F=f^m \phi_m$. $$\delta q^n = \frac{\partial F}{\partial p_n} = \{q^n,F\} = \{q^n,f^m\phi_m\}\,.$$ So the gauge transformation is generated by first-class constraints.

References: Henneaux, Teitelboim. Quantization of Gauge Systems. Chapter 3, Ex. (3.24)