# Constraints Generating Gauge Transformations and BRST

Given a gauge-invariant point particle action with first class primary constraints $$\phi_a$$ of the form ([1], eq. (2.36)) $$S = \int d \tau[p_I \dot{q}^I - u^a \phi_a]\tag{1}$$ we know immediately, since first class 'primary' constraints always generate gauge transformations, that $$\varepsilon^a \phi_a = i \theta c^a \phi_a = i \theta Q \tag{2}$$ generates gauge transformations (where I set $$\varepsilon^a = i \theta c^a$$ using anti-commuting parameters). Thus a gauge transformation reads as $$\delta q^I = \{q^I, \varepsilon^a \phi_a \} \ \ , \ \ \delta p_I = \{p_I , \varepsilon^a \phi_a \} \ \ , \ \ \delta \phi_a = \{\phi_a,\epsilon^b \phi_b \} = f_{ab}^c \epsilon^b \phi_c \ \ . \tag{3}$$ In order for this to be consistent, an infinitesimal gauge transformation on the gauge parameters $$u^a$$ turns out to be $$\delta u^c = \dot{\varepsilon}^c + f_{ab}^c \varepsilon^a u^b.\tag{4}$$

Let us quickly recall that Dirac's argument that the $$\phi_a$$ generate gauge transformations amounts to simply studying the change of a function $$F$$ w.r.t. time under $$H_T = u^a \phi_a$$ as $$\delta F = \{F,H_T\} dt = u^a dt \{F,\phi_a\} \tag{5}$$ noting that the result depends on an arbitrary function $$u^a$$, which can be changed e.g. into $$u^a = u'^a = (u^a - u'^a)$$ so that $$\delta F = u'^a dt \{F,\phi_a\} + (u^a - u'^a) dt \{F,\phi_a\} \tag{6}$$ is just a gauge-transformed version of $$\delta F$$.

If I now fix a partial gauge for the $$u^a$$ via a gauge-fixing function $$F^a(u) = u^a - u_0^a = 0$$ imposed via a Lagrange multiplier $$N_a$$, when I impose $$F^a(u) = 0$$ there may be non-trivial residual gauge transformation solutions of $$\delta F^a(u) = 0$$. However, the action with $$-N_a F^a$$ added (assuming $$\delta N_a = 0$$) is no longer gauge invariant, thus the gauge-fixed action no longer possess these residual gauge transformations, thus the theory as it stands is sick. I therefore need to add a term which cancels the $$- N_a \delta F^a$$ to restore these residual gauge transformations as symmetries. In path integral language this arises from FP ghost terms extending the action to \begin{align*} S &= \int d \tau[p_I \dot{q}^I - u^a \phi_a - N_a F^a + i b_c (\dot{c}^c + i f_{ab}^c c^a u^b)] \\ &= \int d \tau [p_I \dot{q}^I + i b_a \dot{c}^a - N_a F^a - u^a (\phi_a - i f_{ab}^c c^b b_c)] \\ &= \int d \tau [p_I \dot{q}^I + i b_a \dot{c}^a - N_a F^a - u^a \hat{\phi}_a].\tag{7} \end{align*} The vanishing of $$S$$ occurs when we define \begin{align*} \delta q^I &= \{q^I,i \theta c^a \phi_a \} \ , \ \delta p_I = \{p_I,i \theta c^a \phi_a \} \ , \ \delta \phi_a = \{\phi_a, i \theta c^b \phi_b \} = i \theta f_{ab}^c c^b \phi_c \ , \\ \delta F^a &= i \theta (\dot{c}^a + f_{bc}^a c^b u^c) \ , \ \delta N_a = 0 \ , \ \delta b_a = N_a \theta \ , \ \delta c^a = \frac{i}{2} \theta f_{bc}^a c^b c^c \ \ . \tag{8} \end{align*} where the definition of $$\delta b_a$$ is obvious based on wanting to cancel $$N_a \delta F^a$$, and the definition of $$\delta c^a$$ is suggested as necessary in order to cancel what remains. Note this is basically the exact motivation for the BRST transformations given at the start of Section 4.2 of [2].

We have now found $$H_T = u^a \hat{\phi}_a + N_a F^a$$ in terms of the new constraints $$\hat{\phi}_a = \phi_a - i f_{ab}^c c^b b_c \ \ \text{satisfying} \ \ \{\hat{\phi}_a,\hat{\phi}_b\} = f_{ab}^c \hat{\phi}_c \ \ , \ \ \text{and} \ \ F^a = u^a - u_0^a = 0 \ \ .\tag{9}$$ thus the $$\hat{\phi}_a$$ are first class primary amongst themselves. We added $$b_c (\dot{c}^c + i f_{ab}^c c^a u^b)$$ so that it would cancel the variation of $$N_a F^a$$ and such that the variation of all these terms would vanish.

Using the $$u^a$$ eom we can solve for $$N_a$$ thus integrating the $$N_a$$ out and fixing the gauge $$u^a = u^a_0$$, and we thus find $$S = \int d \tau [p_I \dot{q}^I + i b_a \dot{c}^a - u_0^a \hat{\phi}_a].\tag{10}$$ The transformations which ensure $$\delta S = 0$$ are now \begin{align*} \delta q^I &= \{q^I,i \theta c^a \phi_a \} \ , \ \delta p_I = \{p_I,i \theta c^a \phi_a \} \ , \ \delta \phi_a = \{\phi_a, i \theta c^b \phi_b \} = i \theta f_{ab}^c c^b \phi_c \ , \\ \delta b_a &= - \hat{\phi}_a \theta \ , \ \delta c^a = \frac{i}{2} \theta f_{bc}^a c^b c^c \ \ . \tag{11} \end{align*} Based on this alone, since first class primary constraints generate gauge transformations, you'd think that on simply writing $$\varepsilon^a \hat{\phi}_a = i \theta c^a \hat{\phi}_a = i \theta \hat{Q} \tag{12}$$ that the $$\hat{\phi}_a$$, i.e. $$\hat{Q}$$, generate(s) the above residual gauge transformations that ensure $$\delta S = 0$$ holds. Indeed, with $$H_T = u_0^a \hat{\phi}_a$$, Dirac's argument \begin{align*} \delta F &= u_0^a dt \{F,\hat{\phi}_a \} = u_0'^a dt \{F,\hat{\phi}_a\} + (u_0^a - u_0'^a) dt \{F,\hat{\phi}_a\} \tag{13} \end{align*} seems to go through directly, noting that the $$u_0^a$$ are only partially fixed up to solutions of $$\delta F^a = 0$$.

Thus, it is plausible to assume that the above symmetries in (8) and (11) are the precise gauge symmetries from (3) that we spoiled by adding $$N_a F^a$$ and then apparently restored by adding the ghost terms leading to (7), so that (10), where $$N_a$$ is integrated out, supposedly really just possesses the original gauge invariance (3) that (1) had, but they now just read as (11). The only difference between (11) and (3) is supposed to be that we can now consistently impose $$F^a = 0$$ without the remaining action losing the residual symmetries satisfying $$\delta F^a = 0$$, which (11) are supposed to be.

According to Dirac, we should thus be able to blindly copy the algorithm discussed above and should immediately find the correct gauge transformations in (11) to be generated by the new constraints in (10) \begin{align*} \delta q^I &=^? \{q^I,i \theta \hat{Q} \} \ , \ \delta p^I =^? \{p^I,i \theta \hat{Q} \} \ , \ \delta b_a =^? \{b_a,i \theta \hat{Q} \} \ , \ \delta c^a =^? \{c^a,i \theta \hat{Q} \} \ .\tag{14} \end{align*} Unfortunately this gives the wrong answer, we do not reproduce the variations of equation (11) that led to the total vanishing $$\delta S = 0$$ of $$S$$ in (10) (up to total derivatives that we ignore).

What we find is that the generator that reproduces the transformations of (11) is not $$\hat{Q}$$ it is $$Q = c^a (\phi_a - \frac{1}{2} i f_{ab}^c c^b b_c) = c^a \chi_a \tag{15}$$ where the correct gauge transformations reproducing (11) are given by \begin{align*} \delta q^I &= \{q^I,i \theta Q \} \ , \ \delta p^I = \{p^I,i \theta Q \} \ , \\ \delta \phi_a &= \{\phi_a,i \theta Q \} = i \theta f_{ab}^c c^b \phi_c \ , \ \delta \hat{\phi}_a = \{\hat{\phi}_a,i\theta Q \} = 0 \ , \\ \delta c^a &= \{c^a, i \theta Q \} = \frac{i}{2} \theta f_{bc}^a c^b c^c \ , \ \delta b_a = \{b_a,i \theta Q \} = - \hat{\phi}_a \theta \ \ . \tag{16} \end{align*} Comparing the transformations under $$Q$$ vs. $$\hat{Q}$$ we have \begin{align*} \delta q^I &= \{q^I,i \theta c^a \phi_a \} = \{q^I,i \theta Q \} = \{q^I, i \theta \hat{Q} \} \ , \\ \delta p_I &= \{p_I,i \theta c^a \phi_a \} = \{p_I,i \theta Q \} = \{p_I, i \theta \hat{Q} \} \ , \\ \delta \phi_a &= \{\phi_a, i \theta c^b \phi_b \} = i \theta f_{ab}^c c^b \phi_c = \{ \phi_a, i \theta Q \} = \{ \phi_a, i \theta \hat{Q} \} \ , \\ \delta b_a &= - \hat{\phi}_a \theta = \{b_a, i \theta Q \} \neq \{b_a,i \theta \hat{Q} \} = - \hat{\phi}_a \theta + i f_{ab}^c c^b b_c \ , \\ \delta c^a &= \frac{i}{2} \theta f_{bc}^a c^b c^c = \{c^a, i \theta Q \} \neq \{c^a,i\theta \hat{Q} \} = \frac{i}{2} \theta f_{bc}^a c^b c^c + \frac{i}{2} \theta f_{bc}^a c^b c^c \ \ . \tag{17} \end{align*} Note that the $$\chi_a$$ do not close, $$\{\chi_a,\chi_b\} \neq f_{ab}^c \chi_c$$, unlike the $$\hat{\phi}_a$$ which do close, $$\{\hat{\phi}_a,\hat{\phi}_b \} = f_{ab}^c \hat{\phi}_c$$.

Even $$H_T = u^a \hat{\phi}_a$$ can be written in terms of $$Q$$ via $$H_T = u^a \hat{\phi}_a = i \{u^a b_a , Q \} = u^a \frac{\delta b_a}{\delta \theta} . \tag{18}$$ The fact that $$i \theta Q = i \theta c^a \chi_a$$ generates the correct transformations, according to Dirac's logic, should immediately imply that $$\delta F = \{F,H_T \} =^? u_0^a \{F,\chi_a \} \tag{19}$$ should hold, however this is false we do not have $$H_T =^? u_0^a \chi_a$$ we have $$H_T = u_0^a \hat{\phi}_a$$ and $$\delta F = \{F,H_T \} = u_0^a \{F,\hat{\phi}_a \} = u_0^a \{F, \{b_a,Q \} \} \tag{20}$$ so that you'd think $$\delta c^a =^? \{c^a,i\theta c^b \hat{\phi}_b \}$$ was true, but this is false, instead $$\delta c^a = \{c^a,i \theta c^b \chi_b \}$$ is true. Examining (20), Dirac is telling us that the $$\{b_a,Q\} = \hat{\phi}_a$$ should be the correct generators of these gauge transformations, however we are actually finding that the $$Q$$ nested in $$\{b_a,Q\}$$ is the generator, it seems crazy and seems as though Dirac's statement is failing (assuming BRST transformations are the direct analog of residual gauge invariances).

The only thing left to do is to ask what happens to $$\delta S$$ when varied under the constraints $$\hat{\phi}_a$$ i.e. under $$\epsilon^a \hat{\phi}_a = i \theta c^a \hat{\phi}_a = i \theta \hat{Q}$$: \begin{align*} \delta q^I &= \{q^I, i \theta \hat{Q} \} = \{q^I, i \theta c^a \hat{\phi}_a \} = \{q^I,i \theta c^a \phi_a \} \ , \\ \delta p_I &= \{p_I, i \theta \hat{Q} \} = \{p_I , i \theta c^a \hat{\phi}_a \} = \{p_I,i \theta c^a \phi_a \} \ , \\ \delta \phi_a &= \{ \phi_a, i \theta \hat{Q} \} = i \theta f_{ab}^c c^b \phi_c = \{\phi_a , i \theta c^b \hat{\phi}_b \} = \{\phi_a, i \theta c^b \phi_b \} \ , \\ \delta \hat{\phi}_a & = \{\hat{\phi}_a,i \theta \hat{Q} \}= \{\hat{\phi}_a, i \theta c^b \hat{\phi}_b \} = i \theta f_{ab}^c c^b \hat{\phi}_c \ , \\ \delta b_a &= \{b_a,i \theta \hat{Q} \} = - \hat{\phi}_a \theta + i f_{ab}^c c^b b_c \ , \\ \delta c^a &= \{c^a,i\theta \hat{Q} \} = \frac{i}{2} \theta f_{bc}^a c^b c^c + \frac{i}{2} \theta f_{bc}^a c^b c^c \ \ . \tag{21} \end{align*} I am getting (ignoring total derivatives) $$\delta S = \int d \tau i \theta u_0^a c^b f_{ab}^c \hat{\phi}_c, \tag{22}$$ In other words, the variation of $$S$$ under the constraints does not completely vanish, however it varies into a linear combination of the constraints, thus it vanishes on the constraint surface.

According to definitions (17) and (18) of [3], if $$\delta B = \{B, i \theta c^a \hat{\phi}_a \} = i \theta u_0^i c^b f_{ab}^c \hat{\phi}_c \approx 0 \tag{23}$$ this implies that $$B$$ is gauge invariant. Since I am getting $$\delta S = \int d \tau i \theta u_0^a c^b f_{ab}^c \hat{\phi}_c$$, by this definition $$S$$ is gauge invariant (where I need to integrate $$\delta L$$ to eliminate total derivative terms).

By this definition $$S$$ is thus gauge-invariant under a canonical transformation generated by the constraints $$\hat{\phi}_a$$, it just doesn't vanish off the constraint surface, which suggests the existence of a distinct symmetry (BRST symmetry) for which the action will vanish on and off the constraint surface, without contradicting Dirac's theorem.

In fact, $$H_T = u_0^a \hat{\phi}_a$$ is their 'gauge invariant Hamiltonian' of their equation (19) since (their (19) reads as) $$\{H_t,\hat{\phi}_b \} = u_0^a \{\hat{\phi}_a,\hat{\phi}_b\} = u_0^a f_{ab}^c \hat{\phi}_c \tag{24}$$ holds, which implies the Hamiltonian is gauge invariant.

To Summarize: It appears that the transformations (21), applied to (10), are in fact the original gauge symmetries (3) of (1) that are generated by Dirac's procedure of treating the constraints in (10) as canonical transformations generating the gauge symmetry. These gauge symmetries were restored after we spoiled (3) by adding $$N_a F^a$$ to (1), and fixed things by adding the ghosts to form (7), and we just integrated $$N_a$$ out to consistently impose $$F^a = 0$$ leading to the action (10), which is invariant under the residual gauge symmetries (21) only on the constraint surface as (22) shows. Thus Dirac's procedure only preserves these residual gauge symmetries on the constraint surface, which is not a big deal. However the original gauge symmetries (3) of the original action (1) were symmetries of the action everywhere not just on the constraint surface, thus we should expect a residual symmetry that forces the action (10) to vanish everywhere, not just on the constraint surface, and the BRST transformations of (11), (15) and (16) give this residual gauge symmetry.

The question is:

• Is all this right?

• Does this suggest that Dirac's theorem is true in this case, that the action is gauge-invariant because it vanishes on the constraint surface under (21), and that BRST is simply a slight modification of residual gauge invariance that generalizes gauge-invariance to hold on and off the constraint surface? Thus even though $$i \theta \hat{Q}$$ is a direct gauge transformation generated by the constraints, because it only vanishes on the constraint surface, and we expect a residual gauge transformation should be a symmetry of the action on and off the constraint surface, a more general symmetry is needed, and the BRST transformation $$i \theta Q$$ encodes the fact that there is a symmetry of the action that holds both on and off the constraint surface?

• Summarizing everything: before you fix a gauge (eq (1)), the action vanishes under the gauge transformations (3) generated by (2) everywhere. Once you fix a gauge admitting residual gauge transformations and add ghost terms to restore gauge invariance of the gauge-fixed action (7), and integrate out the auxilliary fields thus imposing the gauge $$F^a = 0$$ in the action in (10), the gauge transformations in (21) that are generated by the total Hamiltonian in (10) only force $$\delta S$$ to vanish on the constraint surface as in (22). The BRST transformations of (15) to (16) are required to ensure the action vanishes on and off the constraint surface.

• I still find it jarring that BRST can encode residual gauge transformations yet not directly arise directly from the constraints $$\hat{\phi}_a$$ in the total Hamiltonian $$H_T = u_0^a \hat{\phi}_a$$. Any comments on this fact?

References:

1. Townsend, String Theory notes, p.62-64.
2. Polchinski, String Theory, Volume 1. Section 4.2.
3. Bastianelli, Constrained hamiltonian systems and relativistic particles

I) It seems that OP is already aware that

1. Gauge transformations are usually only defined in the original$$^1$$ field sector. In the extended field sector [which includes ghosts, Lagrange multipliers, and other auxiliary variables, etc.] the notion of gauge transformation is replaced by the notion of BRST transformation.

2. For non-abelian Hamiltonian gauge theories, the BRST charge is no longer just of the form 'ghost times first class constraint'. There are more terms. The next term is of the form 'ghost times ghost times ghost momentum times structure function'.

3. There is a famous factor $$\frac{1}{2}$$ in this second term. It is needed to ensure that the BRST charge squares to zero, i.e. is nilpotent. See e.g. Ref. 1.

II) In more detail: Let us start with the original action $$S_0~=~\int \! dt~L_0,\qquad L_0~=~p_i\dot{q}^i -H_0 -u^a\phi_a,\tag{A}$$ where $$\phi_a(q,p)$$ are independent non-abelian first-class constraints $$\{\phi_a,\phi_b\}~=~f_{ab}{}^c\phi_c.\tag{B}$$ Here the structure functions $$f_{ab}{}^c$$ are for simplicity assumed to be structure constants of a Lie algebra. [We will also assume for simplicity that all variables so far are Grassmann-even.]

III) Let us now use the Batalin-Fradkin-Vilkovisky (BFV) formalism [2]. The BFV formalism is needed if we want to have a fully Hamiltonian BRST formulation that includes gauge-fixing. The BFV formalism has an extended phase space as follows.

Min. sector Non-min. sector
Orig. sector Ghost Lagr. mult. Antighost
Position var. $$q^i$$ $$C^a$$ $$u^a$$ $$\bar{C}_a$$
Momentum var. $$p_i$$ $$\bar{P}_a$$ $$B_a$$ $$P^a$$
Grassmann par. $$0$$ $$1$$ $$0$$ $$1$$

The non-zero fundamental super Poisson brackets are $$\{q^i,p_j\}~=~\delta^i_j, \qquad\{C^a,\bar{P}_b\}~=~\delta^a_b,$$ $$\{u^a,B_b\}~=~\frac{1}{\epsilon}\delta^a_b,\qquad\{\bar{C}_a,P^b\}~=~\frac{1}{\epsilon}\delta^b_a.\tag{C}$$

The BRST transformation is $${\bf s}~=~\{Q,\cdot\},\tag{D}$$ where the BRST charge is $$Q~=~C^a\phi_a+\frac{1}{2}C^bC^af_{ab}{}^c\bar{P}_c + \epsilon B_a P^a, \qquad \{Q,Q\}~=~0.\tag{E}$$ The gauge-fixing fermion is $$\Psi~=~\bar{C}_a\left(\frac{\xi}{2}B^a +\chi^a+\epsilon\dot{u}^a\right)-u^a\bar{P}_a.\tag{F}$$ Here $$\epsilon,\xi$$ are gauge-fixing parameters, and $$\chi_a$$ are gauge-fixing functions. We have raised the index $$B^a:=\kappa^{ab}B_b$$ with the Killing metric of the Lie algebra.

The BFV action is $$S_{BFV}~=~\int \! dt~L_{BFV},\tag{G}$$ where$$^2$$ \begin{align} L_{BFV}~=~&p_i\dot{q}^i +C^a\dot{\bar{P}}_a -H_0 +\{\Psi,Q\}\cr ~\sim~&p_i\dot{q}^i +\bar{P}_a\left(\dot{C}^a-P^a\right) +\underbrace{\epsilon\left( B_a \dot{u}^a + \bar{C}_a \dot{P}^a\right)}_{\text{kinetic term}}\cr &-H_0 -u^a\breve{\phi}_a +\underbrace{\bar{C}_a\{\chi^a,\phi_b\}C^b}_{\text{FP term}} +\underbrace{B_a\left(\frac{\xi}{2}B^a +\chi^a\right)}_{\text{gauge-fixing term}},\end{align}\tag{H} where we have defined BRST-modified first-class constraints $$\breve{\phi}_a~=~{\bf s}\bar{P}_a~=~\phi_a +C^bf_{ba}{}^c\bar{P}_c,\tag{I}$$ that satisfy the same Lie algebra $$\{\breve{\phi}_a,\breve{\phi}_b\}~=~f_{ab}{}^c\breve{\phi}_c.\tag{J}$$ Eq. (I) is an example that the BRST construction modifies the gauge theory with ghost terms. [Although we are going to suppress it in this answer, there is strictly speaking a similar BRST-improvement of the Hamiltonian $$H_0$$ with ghost terms.] It is clear that the ghost modification in eq. (I) will not play a role when sandwiched between ghost-free states in the Hilbert space.

IV) After this review of the BFV construction let us try to address OP's question. It seems that OP around their eq. (10) effectively wants to replace the first-class constraint $$\phi_a$$ with $$\widetilde{\phi}_a~=~\phi_a +c^bf_{ba}{}^cb_c,\qquad \{c^a,b_b\}~=~\delta^a_b,\tag{K}$$ and add the corresponding kinetic term $$c^a\dot{b}_a$$ in a new original$$^1$$ action $$\widetilde{S}_0~=~\int \! dt~\widetilde{L}_0,\qquad\widetilde{L}_0~=~p_i\dot{q}^i +c^a\dot{b}_a -H_0 -u^a\widetilde{\phi}_a,\tag{L}$$ i.e. treat the Grassmann-odd $$c$$ and $$b$$ variables as part of the original$$^1$$ field sector, and treat $$\widetilde{\phi}_a$$ as a first class constraint in this super phase space.

To follow the BRST recipe in a consistent fashion, one should next introduce a new ghost, say $$C^a$$, for $$\widetilde{\phi}_a$$, and so forth. Long story short, it is clear that the new BFV action $$\widetilde{S}_{BFV}$$ is just the old BFV action (G) [with an extra kinetic term $$c^a\dot{b}_a$$] where $$\phi_a\to\widetilde{\phi}_a$$.

OP's $$b$$-$$c$$-system plays a role very similarly to the ghost-variables, and it decouples in a unitary gauge.

References:

1. J. Polchinski, String Theory Vol. 1, 1998; eq. (4.3.14).
2. M. Henneaux & C. Teitelboim, Quantization of Gauge Systems, 1994.

--

$$^1$$ The word original is meant relative to the pertinent BRST formulation.

$$^2$$ The $$\sim$$ symbol means equality modulo total time-derivative terms.

• May have found the answer, have re-written the post and hopefully clarified things. Commented Dec 10, 2023 at 20:34
• I think I made sense of things. I don't understand BFV, but I have an idea of how to. In Kaku's 'Strings, Conformal Fields, & Topology', Appendix I, he gives an intro to BFV, in part by saying we need 'ghosts for ghosts', w/ an example $S=-\frac{1}{12}\int d^4 x\sqrt{g}F_{\mu\nu\rho}F^{\mu\nu\rho}$ , $F_{\mu \nu \rho} = 3 \partial_{[\mu}A_{\nu\rho]}$, noting $\delta A_{\mu \nu} = 2 \partial_{[\mu} \theta_{\nu]}$ says the gauge parameter has ghosts, $\delta \theta_{\mu} = \partial_{\mu} \Lambda$. Retracing my post for it should lead to BFV without knowing it in advance, if you want a challenge. Commented Dec 11, 2023 at 7:10