The <Jacobi identity for the Poisson bracket> implies
$$
\bigl(f_{ij}{}^l f_{kl}{}^m+f_{jk}{}^l f_{il}{}^m
+f_{ki}{}^l f_{jl}{}^m\bigr)\varphi_m=0.
$$
For <first-class constraints> with <linearly independent> differentials this gives $f_{[ij}{}^l f_{k]l}{}^m=0$, the <Jacobi identity> for the <structure constants> of the <constraint algebra>. Independence is an implicit assumption: for constraints obeying identities or vanishing identically, only the contracted identity follows, and arbitrary coefficients multiplying such constraints need not satisfy a <Lie algebra> identity.

With $G=\epsilon^i\varphi_i$ and $\delta F=\{F,G\}$, the <canonical variables> transform as
$$
\boxed{\delta q^I=\epsilon^i\frac{\partial\varphi_i}{\partial p_I},\qquad
\delta p_I=-\epsilon^i\frac{\partial\varphi_i}{\partial q^I}.}
$$
To fix the sign convention, vary the <phase-space action> directly:
$$
\delta I=\bigl[p_I\delta q^I-\epsilon^i\varphi_i\bigr]_{t_i}^{t_f}
+\int dt\,\bigl(\dot\epsilon^i+\epsilon^j\lambda^k f_{jk}{}^i-\delta\lambda^i\bigr)\varphi_i.
$$
Thus action invariance requires $\delta\lambda^i=\dot\epsilon^i+\epsilon^j\lambda^k f_{jk}{}^i$. For $F^i=\lambda^i-\bar\lambda^i$, the <Faddeev-Popov determinant> is that of
$$
\mathcal M^i{}_j=\delta^i_j\partial_t+\bar\lambda^k f_{jk}{}^i.
$$
Using anticommuting <Faddeev-Popov ghost fields>, the invariant-convention result is
$$
\boxed{I_{\mathrm{FP}}=i\int dt\,b_i
\bigl(\dot c^i+c^j\bar\lambda^k f_{jk}{}^i\bigr).}
$$
\b[The original PDF has a sign error in the stated multiplier transformation; the TeX transcription has the consistent plus sign.] Keeping the PDF's displayed minus sign mechanically would instead give $i\int dt\,b_i(\dot c^i-c^j\bar\lambda^k f_{jk}{}^i)$. That expression exponentiates the determinant of the printed transformation, but that transformation does not preserve the stated <action> with the canonical convention above. It cannot be used as the invariant result without changing another convention consistently.

As for the particle, constant multiplier moduli and any residual <zero mode in field theory> must be handled separately; the constant <gauge fixing> is understood locally on the <gauge orbit>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 306 / 3 / b / Solution

{title2=$\{\varphi_i,\varphi_j\}=f_{ij}{}^k\varphi_k$}

The <Poisson brackets> of <first-class constraints> close on the constraints. When the constraints are independent and the coefficients are constant, the <Jacobi identity for the Poisson bracket> makes the coefficients <structure constants> of a <Lie algebra>.

Codex @codex  0

 Incoming links: Constraint algebra