The 1.5 Order Formalism does not Generate
a Valid BRS Transformation for Supergravity
John A<EMAIL_ADDRESS><EMAIL_ADDRESS>
University of Calgary
Calgary, Alberta, Canada

Abstract

The 1.5 order formalism (sometimes called a ‘trick’) is the cornerstone of
modern supergravity. In this paper, the free massive Wess Zumino theory is
used as a simple toy model to look at the BRS symmetry of the first, second
and 1.5 order formalisms. This easily shows that the 1.5 order formalism is
flawed for all theories. The 1.5 algebra naively appears to close. However,
when it is analyzed in detail, in a simple model, where easy calculations are
available, the 1.5 formalism always generates an invalid BRS operator, which
is not even nilpotent. This clearly is also the case for supergravity. It
follows that a revised and completed set of nilpotent first order supergravity
transformations is needed to properly understand 3+1 dimensional supergravity.
Such a set seems easy to write down, by simply adding two more auxiliary
fields so that the spin connection becomes part of a super–YM multiplet.

1. Introduction: In supergravity [1, 2, 3, 4, 5, 6] we start out with Einstein’s theory, but it needs to be formulated with a vierbein $e_{\mu a}$ and a spin connection $w_{\mu ab}$, instead of the metric tensor $g_{\mu\nu}$ and an affine connection $\Gamma_{\mu\nu\lambda}$. The vierbein and spin connection are needed so that the gravitino $\Psi_{\mu\alpha}$ can be coupled to gravity. All three of these fields are gauge fields, and their gauge transformations are of the form

$\delta e_{\mu}^{\;\;a}=\partial_{\mu}\xi^{a}+\cdots;\;\delta
w_{\mu}^{\;\;ab}=\partial_{\mu}\rho^{ab}+\cdots;\;\delta\Psi_{\mu\alpha}=\partial_{\mu}C_{\alpha}+\cdots$
(1)

where $\xi^{a},\rho^{ab},C_{\alpha}$ are the Faddeev-Popov ghosts of general
coordinate transformations, local Lorentz rotations, and local supersymmetry
transformations, respectively. We can take all these variations $\delta$ to be
Grassmann odd. That trick was invented by BRS and it facilitates the
verification of the closure of the algebra of these variations. So $e_{\mu
a},w_{\mu ab},C_{\alpha}$ are all Grassmann even and
$\Psi_{\mu\alpha},\xi^{a},\rho^{ab}$ are all Grassmann odd.

2. Supergravity: We will not need the actual details here of the supergravity theory or even its action ${\cal A}_{\rm Sugra}$. But it is useful to write down the action:

${\cal A}_{\rm Sugra}=\int
d^{4}x\left\\{eR[w,e]+\varepsilon_{\mu\nu\lambda\sigma}{\overline{\Psi}}^{\mu}\sigma^{\nu}D^{\lambda}[w]\Psi^{\sigma}-\frac{1}{3}B^{\mu}B_{\mu}+\frac{1}{3}M{\overline{M}}\right\\}$
(2)

Here $R[w,e]$ is Einstein’s curvature scalar as a function of the spin
connection and the vierbein, $e$ is the determinant of the vierbein, and
$D_{\mu}[w]$ is a covariant derivative formed using the spin connection. It
includes two auxiliary fields, $B^{\mu}$ and $M$, which enable the closure of
the variation of the gravitino.

3. First, second and 1.5 order formalisms for supergravity: In supergravity, it appears that one could choose to use the first order formalism, the second order formalism or the 1.5 order formalism. They differ by their treatment of the spin connection field $w_{\mu ab}$. This field $w_{\mu ab}$ seems to be an auxiliary in the theory, and so it can be removed in various ways, or left in place. The only formalism that is actually used at present is the 1.5 order formalism, because it is the simplest. However, as is shown here, it is also fundamentally flawed.

4. Organization of this Paper: This paper will discuss all three formalisms for supergravity in a general way, and then illustrate them in detail with the WZ model.

5. Results of this paper: It appears that, in the 1.5 order formalism, the form of all the transformations are known except $\Delta_{1}w_{\mu ab}$. It also appears that the transformation $\Delta_{1}w_{\mu ab}$ is not needed. However, as is shown in paragraph * below, the 1.5 formalism does not really generate a closed algebra. To corrrect this problem it is necessary to start from the first order formalism, and to find a new nilpotent transformation $\Delta_{1}w_{\mu ab}$ for the spin connection. But that is of course not the whole story, since that closure requires new auxiliary fields, and the transformations of the other fields and the action can be expected to change, to some extent, in order to accomodate the new transformations. The conclusion of this paper is that

  1. 1.

We do not yet know the transformation $\Delta_{1}w_{\mu ab}$ but there is an
obvious guess to make (see paragraph *), and

  2. 2.

We need the transformation $\Delta_{1}w_{\mu ab}$ to formulate any BRS
identity for supergravity.

There is a current candidate for a transformation $\Delta_{1.5}w_{\mu ab}$ in
the 1.5 order formulation, as we will describe below. But we will show that
this transformation $\Delta_{1.5}w_{\mu ab}$ has an incurable problem, and, as
a result,it follows that we do not yet have a correct formulation for
supergravity in 3+1 dimensions.

6. First Order Formalism for supergravity: In the first order formalism, the vierbein, the spin connection and the gravitino are all considered to be fundamental variables, and the equations of motion are obtained by varying each of them independently. It is called first order because it turns out that the equations of motion here for the fields $e_{\mu a}$ and $w_{\mu ab}$ are of first order in the derivatives.

7. The first order identity for supergravity might be expected to be of the form

${\cal M}_{\rm Sugra}=\int d^{4}x\left\\{\frac{\delta{\cal A}_{\rm
Sugra}}{\delta e_{\mu a}}\Delta_{1}e_{\mu a}+\frac{\delta{\cal A}_{\rm
Sugra}}{\delta\Psi_{\mu}}\Delta_{1}\Psi_{\mu}+\frac{\delta{\cal A}_{\rm
Sugra}}{\delta\Psi_{\mu}}\Delta_{1}{\overline{\Psi}}_{\mu}\right.$ (3)
$\left.+\frac{\delta{\cal A}_{\rm Sugra}}{\delta
M}\Delta_{1}M+\frac{\delta{\cal A}_{\rm
Sugra}}{\delta{\overline{M}}}\Delta_{1}{\overline{M}}+\frac{\delta{\cal
A}_{\rm Sugra}}{\delta B^{\mu}}\Delta_{1}B^{\mu}+\frac{\delta{\cal A}_{\rm
Sugra}}{\delta w_{\mu ab}}\Delta_{1}w_{\mu ab}\right\\}=0$ (4)

This identity would summarize all the invariances of the action. The
variations $\Delta e_{\mu a},\Delta\Psi_{\mu},\cdots$ are all complicated
local functions of the variables $e_{\mu a},\Psi_{\mu},\cdots$ and they also
are functions of the three ghosts of supergravity. These three ghosts
$\xi^{a},\rho^{ab},C_{\alpha}$ were mentioned in paragraph * above. To close
the algebra we also need the variations of the ghosts
$\Delta\xi^{a},\Delta\rho^{ab},\Delta C_{\alpha}$, which are also functions of
all the variables, including the ghosts themselves.

8. Closure of the algebra for the first order identity for supergravity would mean that the variations of the variations are all zero.

$\Delta_{1}^{2}e_{\mu
a}=0,\Delta_{1}^{2}\Psi_{\mu}=0,\Delta_{1}^{2}{\overline{\Psi}}_{\mu}=0,$ (5)
$\Delta_{1}^{2}B_{\mu}=0,\Delta_{1}^{2}M=0,\Delta_{1}^{2}{\overline{M}}=0,\Delta_{1}^{2}w_{\mu
ab}=0,\Delta_{1}^{2}\rho_{ab}=0,$ (6)
$\Delta_{1}^{2}\xi_{\mu}=0,\Delta_{1}^{2}C_{\alpha}=0,\Delta_{1}^{2}{\overline{C}}_{\dot{\alpha}}=0$
(7)

Again the operator $\Delta_{1}$ here is Grassmann odd. These transformations
are the starting point for the BRS cohomology [7, 8, 9, 10, 11, 12, 13, 14] of
supergravity. In this paper, we will not need the explicit forms of these
variations and we will not contemplate any other transformations, such as
those suggested in paragraph *.

9. Early in the history of supergravity, the paper [17] did contain some pieces of $\Delta_{1}w_{\mu ab}$. But that was before the auxiliary fields $B_{\mu}$ and $M$ had been discovered. Since then it appears that there has been little discussion of $\Delta_{1}w_{\mu ab}$, because the 1.5 order formalism has been used.

10. In the second order formalism, the idea is to replace the spin connection $w_{\mu ab}$ with its expression as a function of the vierbein and the gravitino, obtained by finding the equation of motion of the spin connection:

$\frac{\delta{\cal A}_{\rm Supergravity}}{\delta w_{\mu ab}}=0.$ (8)

Then the theory would be expressed in terms of the vierbein and the gravitino,
and one would proceed to deal with the equations of motion obtained by varying
each of those independently. The spin connection would play no further role in
the formalism. This is called second order because it turns out that the
equations of motion for the field $e_{\mu a}$ here are of second order in the
derivatives. For the second order formalism, we would expect to find an
identity of the form

${\cal M}_{\rm Second}=\int d^{4}x\left\\{\frac{\delta{\cal A}_{\rm
Second}}{\delta e_{\mu a}}\Delta_{2}e_{\mu a}+\frac{\delta{\cal A}_{\rm
Second}}{\delta\Psi_{\mu}}\Delta_{2}\Psi_{\mu}+\frac{\delta{\cal A}_{\rm
Second}}{\delta\Psi_{\mu}}\Delta_{2}{\overline{\Psi}}_{\mu}\right.$ (9)
$\left.+\frac{\delta{\cal A}_{\rm Second}}{\delta
M}\Delta_{2}M+\frac{\delta{\cal A}_{\rm
Second}}{\delta{\overline{M}}}\Delta_{2}{\overline{M}}+\frac{\delta{\cal
A}_{\rm Second}}{\delta B^{\mu}}\Delta_{2}B^{\mu}\right\\}=0$ (10)

Note that the term $\frac{\delta{\cal A}_{\rm Sugra}}{\delta w_{\mu
ab}}\Delta_{1}w_{\mu ab}$, which was present at the end of (4) has been
removed from (10). There is no term of the form $\frac{\delta{\cal A}_{\rm
Sugra}}{\delta w_{\mu ab}}\Delta_{2}w_{\mu ab}.$ This identity would summarize
all the invariances of the action. The action here would be obtained from the
action

${\cal A}_{\rm Second}={\cal A}_{\rm
Sugra}[e,\Psi,\overline{\Psi},w\rightarrow w[e,\Psi,{\overline{\Psi}}]]$ (11)

where the expression $w[e,\Psi,{\overline{\Psi}}]$ is obtained from the
solution of

$\frac{\delta{\cal A}_{\rm Sugra}}{\delta w_{\mu ab}}=0,$ (12)

which is linear in $w$.

11. By closure of the algebra here, for the second order formalism we again mean that the variations of the variations are all zero:

$\Delta_{2}^{2}e_{\mu
a}=0,\Delta_{2}^{2}\Psi_{\mu}=0,\Delta_{2}^{2}{\overline{\Psi}}_{\mu}=0,\Delta_{2}^{2}B_{\mu}=0,\Delta_{2}^{2}M=0,$
(13)
$\Delta_{2}^{2}{\overline{M}}=0,\Delta_{2}^{2}\rho_{ab}=0,\Delta_{2}^{2}\xi_{\mu}=0,\Delta_{2}^{2}C_{\alpha}=0,\Delta_{2}^{2}{\overline{C}}_{\dot{\alpha}}=0$
(14)

However the variations $\Delta_{2}$ here are clearly not the same as the
$\Delta_{1}$ were for the identity (4), because the field $w_{\mu ab}$ has
been eliminated, and so it has no variation. We shall return to the question
of the construction of these new variations $\Delta_{2}$ below. We will show
that if we had variations satisfying the equations in paragraph 7, then we
could construct solutions for $\Delta_{2}$ which satisfy $\Delta_{2}^{2}=0$.
However we would need to expand the equations (14) to include some sources. To
do that construction we would need to introduce the BRS formalism in the form
of the Master Equation.

12. The 1.5 order formalism tries to take advantage of the merits of both the first order formalism, and the second order formalism. The action is expressed as a function of the same variables as in the first order formulation (because that is the simplest formulation). All three variables are varied, as in the first order formalism. But then there is an effort to implement the equation of motion of the spin connection so that its independent variation does not complicate things.

${\cal M}_{\rm Sugra\;With\;1.5\;Trick}=\int d^{4}x\left\\{\frac{\delta{\cal
A}}{\delta e_{\mu a}}\Delta_{1.5}e_{\mu a}+\frac{\delta{\cal
A}}{\delta\Psi_{\mu}}\Delta_{1.5}\Psi_{\mu}+\frac{\delta{\cal
A}}{\delta\Psi_{\mu}}\Delta_{1.5}{\overline{\Psi}}_{\mu}\right.$ (15)
$\left.+\frac{\delta{\cal A}}{\delta M}\Delta_{1.5}M+\frac{\delta{\cal
A}}{\delta{\overline{M}}}\Delta_{1.5}{\overline{M}}+\frac{\delta{\cal
A}}{\delta B^{\mu}}\Delta_{1.5}B^{\mu}\right\\}=0$ (16)

The other attraction of the 1.5 order formalism is that it looks like it
avoids the problem of finding a form for $\Delta_{1}w_{\mu ab}$. But in fact
it does not succeed in that avoidance, as will be shown.

13. We would also expect to have the same form as for the second order formalism in (10):

$\Delta_{1.5}^{2}e_{\mu
a}=0,\Delta_{1.5}^{2}\Psi_{\mu}=0,\Delta_{2}^{2}{\overline{\Psi}}_{\mu}=0,\Delta_{1.5}^{2}B_{\mu}=0,\Delta_{1.5}^{2}M=0,$
(17)
$\Delta_{1.5}^{2}{\overline{M}}=0,\Delta_{1.5}^{2}\rho_{ab}=0,\Delta_{1.5}^{2}\xi_{\mu}=0,\Delta_{1.5}^{2}C_{\alpha}=0,\Delta_{1.5}^{2}{\overline{C}}_{\dot{\alpha}}=0$
(18)

We might ask at this point whether we expect these variations $\Delta_{1.5}$
to be the same as $\Delta_{1}$ or $\Delta_{2}$ or different from both of them,
and how we can find them. In fact this is not a real question, because the 1.5
formalism is not a valid formulation. An impatient reader here can look at
paragraph * to see how this happens.

14. The easy way to find $\Delta_{1.5}w_{\mu ab}$: It has been suggested in the literature that we can find the variation $\Delta_{1.5}w_{\mu ab}$ by using the equation of motion (8) to solve for $w_{\mu ab}$ as a function of $e_{\mu a}$ and $\Psi_{\mu}$, and then construct $\Delta_{1.5}w_{\mu ab}$ by using $\Delta_{1}e_{\mu a}$ and $\Delta_{1}\Psi_{\mu}$ in that expression to write down $\Delta_{1.5}w_{\mu ab}$.

15. This easy route to find $\Delta_{1.5}w_{\mu ab}$ does not work. We do not need to get into the terrific complexity of supergravity to show this. We can explain the problem using the first supersymmetric action that was discovered: the Wess-Zumino model. In that way we will see that if we use the equation of motion to write everything in terms of the fields other than $w_{\mu ab}$, then the nilpotence fails. As mentioned above, an impatient reader here can look at paragraph * to see how this happens.

16. The three kinds of order formalisms in the WZ multiplet: We can easily analyze the basic features of the three kinds of order formalism in a very much simpler model. Instead of dealing with the spin connection in supergravity we will deal with the auxiliary fields $F,{\overline{F}}$ in the WZ model. The same idea applies to any auxiliary field in any theory. For the WZ multiplet, we will start with the first order, then generate the second order, and then contemplate the 1.5 order. Then we will return to the situation in supergravity.

17. Here is the action for the WZ model, including a simple mass term of the chiral type. We use two component complex spinors, because the theory is naturally adapted to them.

${\cal A}_{\rm Fields}=\int
d^{4}x\left\\{\partial_{\mu}A\partial^{\mu}{\overline{A}}+\chi^{\alpha}\sigma^{\mu\dot{\beta}}_{\alpha}\partial_{\mu}{\overline{\chi}}_{\dot{\beta}}+F{\overline{F}}\right.$
(19)
$\left.+m\left(AF-\frac{1}{2}\chi^{\alpha}\chi_{\alpha}\right)+m\left({\overline{A}}{\overline{F}}-\frac{1}{2}{\overline{\chi}}^{\dot{\alpha}}{\overline{\chi}}_{\dot{\alpha}}\right)\right\\}$
(20)

This action obeys an identity that arises from supersymmetry. We define the
following variations:

$\Delta_{1}A=C^{\alpha}\chi_{\alpha};\;\Delta_{1}\chi_{\alpha}=\sigma^{\mu\dot{\beta}}_{\alpha}\partial_{\mu}A{\overline{C}}_{\dot{\beta}}+FC_{\alpha};\;\Delta_{1}F={\overline{C}}_{\dot{\beta}}\overline{\sigma}^{\mu\dot{\beta}}_{\alpha}\partial_{\mu}\chi^{\alpha}$
(21)
$\Delta_{1}{\overline{A}}={\overline{C}}^{\dot{\alpha}}{\overline{\chi}}_{\dot{\alpha}};\;\Delta_{1}{\overline{\chi}}_{\dot{\alpha}}=\overline{\sigma}^{\mu\beta}_{\dot{\alpha}}\partial_{\mu}{\overline{A}}C_{\beta}+{\overline{F}}C_{\dot{\alpha}};\;\Delta_{1}{\overline{F}}=C_{\beta}\sigma^{\mu\beta}_{\dot{\alpha}}\partial_{\mu}{\overline{\chi}}^{\dot{\alpha}}$
(22)

This is rigid supersymmetry, so we assume that the ghosts
$C_{\alpha},{\overline{C}}_{\dot{\alpha}}$ are constants, independent of the
coordinates, so that

$\Delta_{1}C_{\alpha}=\Delta_{1}{\overline{C}}_{\dot{\alpha}}=0;\;\partial_{\mu}C_{\alpha}=\partial_{\mu}{\overline{C}}_{\dot{\alpha}}=0$
(23)

18. The First Order Identity: The following is the identity in this theory that is generated by its supersymmetry:

${\cal M}_{\rm WZ}=\int d^{4}x\left\\{\frac{\delta{\cal A}}{\delta
A}\Delta_{1}A+\frac{\delta{\cal
A}}{\delta\chi_{\alpha}}\Delta_{1}\chi_{\alpha}+\frac{\delta{\cal A}}{\delta
F}\Delta_{1}F\right.$ $\left.\frac{\delta{\cal
A}}{\delta{\overline{A}}}\Delta_{1}{\overline{A}}+\frac{\delta{\cal
A}}{\delta{\overline{\chi}}_{\dot{\alpha}}}\Delta_{1}{\overline{\chi}}_{\dot{\alpha}}+\frac{\delta{\cal
A}}{\delta{\overline{F}}}\Delta_{1}{\overline{F}}\right\\}=0$ (24)

By saying that this is an identity, we mean that adding all the terms together
gives us exactly zero, without imposing any equation that is assumed to be
satisfied by the fields. Formula (24) is simply true, no matter what the
values of the fields are. This can be done explicitly in one page, and it
looks very much like paragraph * below.

19. A closely related identity is the nilpotence identity: Thus for example we have

$\Delta_{1}^{2}A=C^{\alpha}\Delta_{1}\chi_{\alpha}=(C^{\beta}\sigma^{\mu\dot{\gamma}}_{\beta}{\overline{C}}_{\dot{\gamma}})\partial_{\mu}A;\;\Delta_{1}^{2}\chi_{\alpha}=(C^{\beta}\sigma^{\mu\dot{\gamma}}_{\beta}{\overline{C}}_{\dot{\gamma}})\partial_{\mu}\chi_{\alpha}$
(25)

Similar results hold for all the fields in the formulae above. The operator
$\Delta_{1}^{2}$ gives rise to a derivative times the constant
$C^{\alpha}\sigma^{\mu\dot{\beta}}_{\alpha}{\overline{C}}_{\dot{\beta}}$. This
can be treated as zero, because if we take the derivative of the Lagrangian,
we get zero variation for the action, because the integral of a total
derivative is taken to be zero in quantum field theory.

20. The second order formalism for the WZ model: To generate this, we can start with the first order formalism above. The idea here is to first formulate the BRS ZJ Master Equation for the first order formalism. Then we will integrate the auxiliary fields $F,{\overline{F}}$ out of the theory. That will lead us to a Master Equation that does not refer to the auxiliary fields at all.

21. Formulating the Master Equation for the WZ model: The action we need to use is

${\cal A}_{\rm Total}={\cal A}_{\rm Fields}+{\cal A}_{\rm ZJ\;Sources}$ (26)

where we define the ZJ Source Action ${\cal A}_{\rm ZJ\;Sources}$:

${\cal A}_{\rm ZJ\;Sources}=\int
d^{4}x\left\\{\widetilde{A}\Delta_{1}A+\widetilde{{\overline{A}}}\Delta_{1}{\overline{A}}+\widetilde{\chi}_{\alpha}\Delta_{1}\chi^{\alpha}\right.$
(27)
$\left.+\widetilde{{\overline{\chi}}_{\dot{\alpha}}}\Delta_{1}{\overline{\chi}}^{\dot{\alpha}}+\widetilde{F}\Delta_{1}F+\widetilde{F}\Delta_{1}{\overline{F}}\right\\}$
(28)

Let us write this explicitly using the above expressions in (refdeltaF)
through (22):

${\cal A}_{\rm ZJ\;Sources}=\int
d^{4}x\left\\{\widetilde{A}C^{\alpha}\chi_{\alpha}+\widetilde{{\overline{A}}}{\overline{C}}^{\dot{\alpha}}{\overline{\chi}}_{\dot{\alpha}}+\widetilde{\chi}^{\alpha}\left(\sigma^{\mu\dot{\beta}}_{\alpha}\partial_{\mu}A{\overline{C}}_{\dot{\beta}}+FC_{\alpha}\right)\right.$
(29)
$\left.+\widetilde{{\overline{\chi}}^{\dot{\alpha}}}\left(\overline{\sigma}^{\mu\beta}_{\dot{\alpha}}\partial_{\mu}{\overline{A}}C_{\beta}+{\overline{F}}C_{\dot{\alpha}}\right)+\widetilde{F}{\overline{C}}_{\dot{\beta}}\sigma^{\mu\dot{\beta}}_{\alpha}\partial_{\mu}\chi^{\alpha}+\widetilde{F}C_{\beta}\overline{\sigma}^{\mu\beta}_{\dot{\alpha}}\partial_{\mu}{\overline{\chi}}^{\dot{\alpha}}\right\\}$
(30)

This will be the analogue of paragraph *, but for this much simpler theory.

22. Formulating the Master Equation in General for the First Order Formalism: This is a standard technique in BRS theories, but we want to emphasize the points that are crucial to show that the 1.5 order formalism does not work. It is easiest to formulate this in a general way. We suppose that there is a set of fields $\phi_{i}$ which may be physical fields, auxiliaries, ghosts or antighosts, and which may have even or odd Grassmann character. We suppose that there is a local action

${\cal A}[\phi]=\int d^{4}x{\cal L}[\phi]$ (31)

and a set of transformations $\Delta_{1}\phi^{i}$ which satisfy the equations

$\sum_{i}\int d^{4}x\;\frac{\delta{\cal
A}[\phi]}{\delta\phi^{i}}\Delta_{1}\phi^{i}=0;\;\Delta_{1}^{2}\phi^{i}=0$ (32)

We define the action by

${\cal A}_{\rm Integrand}={\cal A}[\phi,\widetilde{\phi}]+{\cal A}[{\rm
Field\;Sources}];\;cA[\phi,\widetilde{\phi}]={\cal A}[\phi]+{\cal A}[{\rm
ZJ\;Sources}]$ (33) ${\cal A}[{\rm ZJ\;Sources}]=\sum_{i}\int
d^{4}x\widetilde{\phi}_{i}\Delta_{1}\phi^{i};\;{\cal A}[{\rm
Field\;Sources}]=\sum_{i}\int d^{4}x\widehat{\phi}_{i}\phi^{i}$ (34)

23. The Master Equation: Now we consider the functional path integral:

$Z[\widetilde{\phi}_{i},\widehat{\phi}_{i}]=\Pi_{\rm i,x}\int
d\;{\rm\phi^{i}}(x)\;\;e^{\frac{i}{\hbar}{\cal A}_{\rm Integrand}}$ (35)

Then we see that there is an identity

$\sum_{i}\int d^{4}x\;\widehat{\phi}_{i}\frac{\delta
Z}{\delta\widetilde{\phi}_{i}}=\sum_{i}\int
d^{4}x\;\widehat{\phi}_{i}\Delta_{1}\phi^{i}=0$ (36)

This arises from a transformation

$\phi^{i}\rightarrow\phi^{i}+\epsilon\Delta_{1}\phi^{i}$ (37)

on the integration variables in (35) and using the equations (32) ($\epsilon$
is a Grassmann odd parameter to make the Grassmann characters match). Now we
define the 1PI generating functional ${\cal G}_{\rm Connected}$ which
generates connected diagrams by

$Z=e^{\frac{i}{\hbar}{\cal G}_{\rm Connected}}$ (38)

and then we define the 1PI vertex generating functional ${\cal G}[\phi]$,
which is a function of new classical variables $\phi^{i}$, by the Legendre
transform

${\cal G}[\widehat{\phi}]={\cal G}[\phi]-\sum_{i}\int
d^{4}x\phi^{i}\widehat{\phi}_{i}$ (39)

It follows that

$\widehat{\phi}_{i}=\frac{\delta{\cal G}[\phi]}{\delta\phi^{i}}$ (40)

Now we can rewrite the identity (36) in the form

${\cal M}=\sum_{i}^{N}\int d^{4}x\;\frac{\delta{\cal
G}}{\delta{\rm\phi}^{i}}\frac{\delta{\cal G}}{\delta\widetilde{\phi}_{i}}=0$
(41)

Here we are supposing that there are $N$ fields. We can expand in the
parameter $\hbar$ and the first term is the action

${\cal G}=cA[\phi,\widetilde{\phi}]+O(\hbar)$ (42)

So we should find that

${\cal M}[{\cal A}[\phi,\widetilde{\phi}]]=\sum_{i}^{N}\int
d^{4}x\;\frac{\delta{\cal
A}[\phi,\widetilde{\phi}]}{\delta{\rm\phi}^{i}}\frac{\delta{\cal
A}[\phi,\widetilde{\phi}]}{\delta\widetilde{\phi}_{i}}=0$ (43)

The identity (43) is the same as (24), for the case of the WZ multiplet.

24. The consequence of the above is to show that for any theory, including supergravity, in order to formulate the BRS Master Equation for that theory, we need to have the theory satisfy two requirements:

  1. 1.

The action must be invariant.

  2. 2.

The variations must be nilpotent.

We have stated these requirements for the WZ model in the first order
formalism in paragraphs * and *. Also, we have assumed these requirements, in
the first order formalism, in paragraphs * and *, for supergravity.

25. We will now show how to derive the second order formalism from the first order formalism for the WZ model: This uses the BRS Master Equation in a fundamental way. We expect all theories and all integrations of all auxiliary fields to behave in the same way, so in particular this should be an exact analogue of what is needed to complete the second order formalism for supergravity discussed in paragraphs * and *.

26. How to Integrate an Auxiliary out of the Action: All we need to do to accomplish this is to drop the ZJ sources and the Field Sources for the auxiliary, and then to perform the integration of the auxiliary explicitly in the path integral. This amounts to doing the path integral:

$Z=\Pi_{\rm x}\int dF(x)d{\overline{F}}(x)\;\;e^{\frac{i}{\hbar}\int d^{4}x\;\
\left\\{-F+{\overline{m}}{\overline{A}}+\widetilde{\chi}^{\alpha}C_{\alpha}\right\\}\left\\{-{\overline{F}}+mA+\widetilde{{\overline{\chi}}}^{\dot{\alpha}}{\overline{C}}_{\dot{\alpha}}\right\\}}$
(44) $e^{-\frac{i}{\hbar}\int d^{4}x\;\
\left\\{{\overline{m}}{\overline{A}}+\widetilde{\chi}^{\alpha}C_{\alpha}\right\\}\left\\{mA+\widetilde{{\overline{\chi}}}^{\dot{\alpha}}{\overline{C}}_{\dot{\alpha}}\right\\}}=e^{-\frac{i}{\hbar}\int
d^{4}x\;\left\\{{\overline{m}}{\overline{A}}+\widetilde{\chi}^{\alpha}C_{\alpha}\right\\}\left\\{mA+\widetilde{{\overline{\chi}}}^{\dot{\alpha}}{\overline{C}}_{\dot{\alpha}}\right\\}}$
(45)

where we shift $F$ in the first term to do the integral, and we absorb an
irrelevant constant. All the terms in the action that contain the auxiliaries
$F$ or ${\overline{F}}$ are dropped from the action at the same time.

27. Starting with the action 26, the total equations of motion of F, used above, are

$\frac{\delta{\cal A}_{\rm Total}}{\delta
F^{i}}=-{\overline{F}}+mA+\widetilde{{\overline{\chi}}}^{\dot{\alpha}}{\overline{C}}_{\dot{\alpha}};\;\frac{\delta{\cal
A}_{\rm
Total}}{\delta{\overline{F}}}=-F+{\overline{m}}{\overline{A}}+\widetilde{\chi}^{\alpha}C_{\alpha}$
(46)

Let us suppose that we put the action into a path integral and integrate the
fields $F,{\overline{F}}$. We get the new action

${\cal A}_{\rm New}={\cal A}_{\rm Total\;with\;F\rightarrow
0,\widetilde{F}\rightarrow 0,\widehat{F}\rightarrow 0}+{\cal A}_{\rm
From\;integration\;of\;F}$ (47)

where

${\cal A}_{\rm Total\;with\;F\rightarrow 0,\widetilde{F}\rightarrow
0,\widehat{F}\rightarrow 0}$ (48) $=\int
d^{4}x\left\\{\partial_{\mu}A\partial_{\mu}{\overline{A}}+\chi^{\alpha}\sigma^{\mu\dot{\beta}}_{\alpha}{\overline{\chi}}_{\dot{\beta}}+m\left(-\chi^{\alpha}\chi_{\alpha}\right)+m\left(-{\overline{\chi}}^{\dot{\alpha}}{\overline{\chi}}_{\dot{\alpha}}\right)\right\\}$
(49) $+\int
d^{4}x\left\\{\widetilde{A}C^{\alpha}\chi_{\alpha}+\widetilde{{\overline{A}}}{\overline{C}}^{\dot{\alpha}}{\overline{\chi}}_{\dot{\alpha}}+\widetilde{\chi}^{\alpha}\left[\sigma^{\mu\dot{\beta}}_{\alpha}\partial_{\mu}A{\overline{C}}_{\dot{\beta}}\right]+\widetilde{{\overline{\chi}}^{\dot{\alpha}}}\left[\overline{\sigma}^{\mu\beta}_{\dot{\alpha}}\partial_{\mu}{\overline{A}}C_{\beta}\right]\right\\}$
(50) ${\cal A}_{\rm From\;integration\;of\;F}=-\int
d^{4}x\left\\{mA+\widetilde{\chi}^{\alpha}C_{\alpha}\right\\}\left\\{m{\overline{A}}+\widetilde{{\overline{\chi}}}^{\dot{\alpha}}{\overline{C}}_{\dot{\alpha}}\right\\}$
(51)

and we expect the new identity:

${\cal M}_{\rm New}=\int d^{4}x\left\\{\frac{\delta{\cal A}_{\rm New}}{\delta
A}\frac{\delta{\cal A}_{\rm New}}{\delta\widetilde{A}}+\frac{\delta{\cal
A}_{\rm New}}{\delta\chi_{\alpha}}\frac{\delta{\cal A}_{\rm
New}}{\delta\widetilde{\chi}^{\alpha}}\right.$ (52) $\left.+\frac{\delta{\cal
A}_{\rm New}}{\delta{\cal A}_{\rm New}}\frac{\delta{\cal A}_{\rm
New}}{\delta\widetilde{{\overline{A}}}}+\frac{\delta{\cal A}_{\rm
New}}{\delta{\overline{\chi}}_{\dot{\alpha}}}\frac{\delta{\cal A}_{\rm
New}}{\delta\widetilde{{\overline{\chi}}}^{\dot{\alpha}}}\right\\}=0$ (53)

28. Let us verify that this is an identity, and observe how it works: We have

$\int d^{4}x\frac{\delta{\cal A}_{\rm New}}{\delta A}\frac{\delta{\cal A}_{\rm
New}}{\delta\widetilde{A}}=\int
d^{4}x\left\\{-\Box{\overline{A}}-\partial_{\mu}\widetilde{\chi}_{\alpha}\sigma^{\mu\dot{\beta}}_{\alpha}{\overline{C}}_{\dot{\beta}}-m\left({\overline{m}}{\overline{A}}+\widetilde{{\overline{\chi}}}^{\dot{\alpha}}{\overline{C}}_{\dot{\alpha}}\right)\right\\}\left\\{C^{\alpha}\chi_{\alpha}\right\\}$
(54) $\int d^{4}x\frac{\delta{\cal A}_{\rm
New}}{\delta\chi_{\alpha}}\frac{\delta{\cal A}_{\rm
New}}{\delta\widetilde{\chi}^{\alpha}}=\int
d^{4}x\left\\{\sigma^{\nu\alpha\dot{\gamma}}\partial_{\nu}{\overline{\chi}}_{\dot{\gamma}}+m\chi^{\alpha}\right\\}\left\\{\sigma^{\mu\dot{\beta}}_{\alpha}\partial_{\mu}A{\overline{C}}_{\dot{\beta}}+{\overline{m}}{\overline{A}}C_{\alpha}+\widetilde{{\overline{\chi}}}^{\dot{\alpha}}{\overline{C}}_{\dot{\alpha}}C_{\alpha}\right\\}$
(55) $\int d^{4}x\frac{\delta{\cal A}_{\rm
New}}{\delta{\overline{A}}}\frac{\delta{\cal A}_{\rm
New}}{\delta\widetilde{{\overline{A}}}}=\int d^{4}x\left\\{-\Box
A-\partial_{\mu}\widetilde{{\overline{\chi}}}_{\dot{\alpha}}\overline{\sigma}^{\mu\beta}_{\dot{\alpha}}C_{\beta}-{\overline{m}}\left(mA+\widetilde{\chi}^{\alpha}C_{\alpha}\right)\right\\}\left\\{{\overline{C}}^{\dot{\alpha}}{\overline{\chi}}_{\dot{\alpha}}\right\\}$
(56) $\int d^{4}x\frac{\delta{\cal A}_{\rm
New}}{\delta{\overline{\chi}}_{\dot{\alpha}}}\frac{\delta{\cal A}_{\rm
New}}{\delta\widetilde{{\overline{\chi}}}^{\dot{\alpha}}}=\int
d^{4}x\left\\{\overline{\sigma}^{\nu\dot{\alpha}\gamma}\partial_{\nu}\chi_{\gamma}+{\overline{m}}{\overline{\chi}}^{\dot{\alpha}}\right\\}\left\\{\overline{\sigma}^{\mu\beta}_{\dot{\alpha}}\partial_{\mu}{\overline{A}}C_{\beta}+mA{\overline{C}}_{\dot{\alpha}}+\widetilde{\chi}^{\alpha}C_{\alpha}{\overline{C}}_{\dot{\alpha}}\right\\}$
(57)

It is a simple matter to pick out the cancelling pairs in the above set of
objects. One gets a total of zero, thus verifying (53).

29. Also we now have nilpotence. But the nilpotence involves both the variations of the fields and the variation of the ZJ sources. Thus we have, for example,

$\Delta{\chi}_{\alpha}=\frac{\delta{\cal A}_{\rm
New}}{\delta\widetilde{\chi}^{\alpha}}=\left\\{\sigma^{\mu\dot{\beta}}_{\alpha}\partial_{\mu}A{\overline{C}}_{\dot{\beta}}+{\overline{m}}{\overline{A}}C_{\alpha}+\widetilde{{\overline{\chi}}}^{\dot{\alpha}}{\overline{C}}_{\dot{\alpha}}C_{\alpha}\right\\}$
(58)

and

$\Delta\widetilde{A}=\frac{\delta{\cal A}_{\rm New}}{\delta
A}=\left[\Box{\overline{A}}-\partial_{\mu}\widetilde{\chi}_{\alpha}\sigma^{\mu\dot{\beta}}_{\alpha}{\overline{C}}_{\dot{\beta}}-m\left({\overline{m}}{\overline{A}}+\widetilde{{\overline{\chi}}}^{\dot{\alpha}}{\overline{C}}_{\dot{\alpha}}\right)\right]$
(59)

and we can derive nilpotence equations quite easily for all the fields and all
the sources. For example we can show that:

$\Delta^{2}\chi_{\alpha}={\overline{C}}_{\dot{\beta}}\sigma^{\mu\dot{\beta}}_{\alpha}C^{\alpha}\partial_{\mu}\chi_{\alpha};\;\Delta^{2}\widetilde{A}={\overline{C}}_{\dot{\beta}}\sigma^{\mu\dot{\beta}}_{\alpha}C^{\alpha}\partial_{\mu}\widetilde{A}$
(60)

30. So, using the ZJ sources, and the Master equation for the first order formalism, we have derived the second order formalism by integrating the auxiliary F using the path integral.

31. Using the path integral, one can derive the BRS identity quite easily in the form (53) for the one particle irreducible vertices. But this requires that we start with invariance and nilpotence for all the fields, including the ones we are going to integrate out of the theory ($F,{\overline{F}}$ in this case, and the three auxiliaries $M,B_{\mu},w_{\mu ab}$ for supergravity).

32. The following is the 1.5 order formalism for the WZ theory: It is not an identity. The following is exactly the same as what is being done in supergravity with the 1.5 trick as described in paragraph *.

${\cal M}_{\rm WZ\;With\;1.5\;Trick}=\int d^{4}x\left\\{\frac{\delta{\cal
A}}{\delta A}\Delta A+\frac{\delta{\cal
A}}{\delta{\overline{A}}}\Delta{\overline{A}}+\frac{\delta{\cal
A}}{\delta\chi_{\alpha}}\Delta\chi_{\alpha}+\frac{\delta{\cal
A}}{\delta{\overline{\chi}}_{\dot{\alpha}}}\Delta{\overline{\chi}}_{\dot{\alpha}}\right\\}=0$
(61)

If we evaluate the expression (61), we do not get zero. We can say that (61)
is true, if, and only if, we use the equation of motion of F and
${\overline{F}}$. The missing tems are

$\int d^{4}x\frac{\delta{\cal A}_{\rm New}}{\delta F}\Delta F=\int
d^{4}x\left\\{{\overline{F}}+mA\right\\}\left\\{{\overline{C}}_{\dot{\beta}}\sigma^{\mu\dot{\beta}}_{\alpha}\partial_{\mu}\chi^{\alpha}\right\\}$
(62) $\int d^{4}x\frac{\delta{\cal A}_{\rm
New}}{\delta{\overline{F}}}\Delta{\overline{F}}=\int
d^{4}x\left\\{F+m{\overline{A}}\right\\}\left\\{C_{\beta}\overline{\sigma}^{\mu\beta}_{\dot{\alpha}}\partial_{\mu}{\overline{\chi}}^{\dot{\alpha}}\right\\}$
(63)

and, indeed, they vanish by the equations of motion of $F$ and
${\overline{F}}$.

33. The value of the BRS formalism occurs when there is an identity, not when there is an identity which also requires the use of the equations of motion in addition. If we are content to use the equations of motion we could actually say that each individual term is zero:

$\frac{\delta{\cal A}}{\delta A}=\frac{\delta{\cal
A}}{\delta{\overline{A}}}=\frac{\delta{\cal
A}}{\delta\chi_{\alpha}}=\frac{\delta{\cal
A}}{\delta{\overline{\chi}}_{\dot{\alpha}}}=\frac{\delta{\cal A}}{\delta
F}=\frac{\delta{\cal A}}{\delta{\overline{F}}}$ (64)

In that case we would not need any of the $\Delta$ variations at all. Now one
might argue that $F$ and ${\overline{F}}$ are auxiliary fields, and so they
are special, so that nothing goes wrong if we use the 1.5 order formalism for
auxiliary fields. But that is not true. Something very serious does go wrong,
as we now show.

34. The Wrong Way to Find $\Delta w_{\mu ab}$: As mentioned above in paragraph *, sometimes we are told to find $\Delta w_{\mu ab}$ by applying the variations of the other fields to its form as a function of the other fields, found by using the equation of motion. The claim here is that that idea does not work. To show it does not work, let us try the same idea here, for the simple case of the WZ model, except that here we are going to use the idea to generate $\delta F$:

  1. 1.

If one adds all the terms in (24) or in (53), one gets identically zero–there
is no need to assume an equation of motion. This was shown explicitly in
detail in paragraph *.

  2. 2.

This is not true for the expression (61). In fact the equations that need to
be used there, to get zero in (61), are:

$\frac{\delta{\cal A}}{\delta F}=-{\overline{F}}+mA=0;\;\frac{\delta{\cal
A}}{\delta F}=-F+m{\overline{A}}=0$ (65)

  3. 3.

Using the philosophy of the 1.5 formalism, we need to assume that the equation
of motion of the auxiliary can be used to prove (61), and this means here that

$F=m{\overline{A}}$ (66)

  4. 4.

As was mentioned above, we are told in the literature that we can take this
equation to get the variation of F, by using the known variations of the other
fields. That would mean that we should take

$\Delta
A=C^{\alpha}\chi_{\alpha};\;\Delta\chi_{\alpha}=\sigma^{\mu\dot{\beta}}_{\alpha}\partial_{\mu}A{\overline{C}}_{\dot{\beta}}+FC_{\alpha}\longrightarrow\sigma^{\mu\dot{\beta}}_{\alpha}\partial_{\mu}A{\overline{C}}_{\dot{\beta}}+m{\overline{A}}C_{\alpha}$
(67) $\Delta F\longrightarrow
m\Delta{\overline{A}}=m{\overline{C}}^{\dot{\alpha}}{\overline{\chi}}_{\dot{\alpha}}$
(68)

  5. 5.

Is that actually correct? Now of course the field $F$ does not actually appear
in the final result. But there are two fields (and their complex conjugates)
that do appear. Let us look at the square $\Delta^{2}$ of their variations, to
see if the algebra closes:

    1. (a)

From the above we get:

$\Delta^{2}A=C^{\alpha}(\sigma^{\mu\dot{\beta}}_{\alpha}\partial_{\mu}A{\overline{C}}_{\dot{\beta}}+m{\overline{A}}C_{\alpha})=(C^{\alpha}\sigma^{\mu\dot{\beta}}_{\alpha}{\overline{C}}_{\dot{\beta}})\partial_{\mu}A$
(69)

So that has not changed. This expression is nilpotent.

    2. (b)

However

$\Delta^{2}\chi_{\alpha}=\sigma^{\mu\dot{\beta}}_{\alpha}\partial_{\mu}\Delta
A{\overline{C}}_{\dot{\beta}}+m\Delta{\overline{A}}C_{\alpha}=\sigma^{\mu\dot{\beta}}_{\alpha}\partial_{\mu}(C^{\beta}\chi_{\beta}){\overline{C}}_{\dot{\beta}}+m({\overline{C}}^{\dot{\alpha}}{\overline{\chi}}_{\dot{\alpha}})C_{\alpha}$
(70)

    3. (c)

Here we expect, and we need, to get

$\Delta^{2}\chi_{\alpha}=(C^{\beta}\sigma^{\mu\dot{\gamma}}_{\beta}{\overline{C}}_{\dot{\gamma}})\partial_{\mu}\chi_{\alpha}$
(71)

    4. (d)

Equation (70) is not the derivative form (71). What has gone wrong?

    5. (e)

What is happening of course, is that instead of using the original result for
$\Delta F$, which was

$\Delta
F={\overline{C}}_{\dot{\beta}}\sigma^{\mu\dot{\beta}}_{\alpha}\partial_{\mu}\chi^{\alpha}$
(72)

we are using the new result for $\Delta F$

$\Delta F=m{\overline{C}}^{\dot{\alpha}}{\overline{\chi}}_{\dot{\alpha}},$
(73)

and they are not the same. So if we find $\Delta F$ by evaluating the
variation of its value using the equation of motion of $F$, we lose the
closure of the algebra. This then spoils the nilpotence that would be present
in (71) and yields (70) instead. The correct result (71) requires (72) rather
than (73).

    6. (f)

There is a loss of information when we abandon the original variation in favor
of the new variation. To preserve the closure of the algebra we need to go
through the work of getting the proper second order formulation, as we did
above to get equation (60). And to do that we need to start with a correct
formulation of the first order formalism, as we did above to establish the
master equation for the WZ model.

  6. 6.

We would find a similar result for supergravity, if we substitute the
auxiliaries $F\rightarrow w_{\mu ab}$ appropriately.

  7. 7.

So we can expect to find that the algebra does not close for supergravity when
we use the 1.5 order formalism. We need to start with a correct formulation of
the first order formalism, and then get the second order formulation using the
master equation for supergravity too.

  8. 8.

Of course this is a huge calculation in supergravity, though it is a simple
one in the WZ model.

35. This is not at all surprising: From the general form of the Master Equation in (41) we can see that there is a $\delta$ that is nilpotent of the form:

$\delta=\int d^{4}x\;\sum_{i=1}^{N}\left\\{\frac{\delta{\cal
G}}{\delta{\rm\phi}^{i}}\frac{\delta}{\delta\widetilde{\phi}_{i}}+\frac{\delta{\cal
G}}{\delta\widetilde{\phi}_{i}}\frac{\delta}{\delta{\rm\phi}^{i}}\right\\}$
(74)

and it follows from (41) that

$\delta^{2}=0$ (75)

These are both identities, not merely equations that are true assuming some
equations of motion. However if we remove a term, which is what happens in the
1.5 formalism, we get:

${\cal M}_{\rm Term\;Missing}=\sum_{i}^{N-1}\int d^{4}x\;\frac{\delta{\cal
G}}{\delta{\rm\phi}^{i}}\frac{\delta{\cal G}}{\delta\widetilde{\phi}_{i}}\neq
0;\;$ (76) $\delta_{\rm Term\;Missing}=\int
d^{4}x\;\sum_{i=1}^{N-1}\left\\{\frac{\delta{\cal
G}}{\delta{\rm\phi}^{i}}\frac{\delta}{\delta\widetilde{\phi}_{i}}+\frac{\delta{\cal
G}}{\delta\widetilde{\phi}_{i}}\frac{\delta}{\delta{\rm\phi}^{i}}\right\\}\;{\rm
Not\;nilpotent}$ (77)

In this case we can be sure that we do not get zero for the first and we do
not get nilpotence for the second. We saw explicitly above that the nilpotence
does not arise even if we substitute the value of the field equation for the
field.

36. BRS Cohomology and Supergravity: This problem stands in the way of computing the BRS cohomology of supergravity in a convincing way. For me, this problem arose from [15], when I gradually realized that I could not find out how to write down the nilpotent transformations. The methods in [16] require identities for the nilpotence. However as shown above in paragraph *, using the 1.5 formalism, one gets BRS operators that are not nilpotent at all, not even when using the equations of motion.

37. It seems rather easy to close the first order algebra here. As is well known, the Einstein curvature scalar $R$ looks like a YM curvature made from the spin connection, contracted with vierbeins. So the way to make this supersymmetric is evident. We add a new auxiliary spinor field $\Omega^{ab\alpha},\overline{\Omega}^{ab\dot{\alpha}}$ and a new auxiliary scalar $D^{ab}$, and with $w_{\mu}^{ab}$ and the ghost $\rho^{ab}$, these make a full super–YM multiplet. This multiplet is analogous to the usual super–YM multiplet, except that the gauge group is the non-compact Lorentz group. The details of the transformations need careful work. This new multiplet raises many issues. One of the problems is that this makes it very tempting to add the usual quadratic super-Yang Mills type action for the new multiplet, but that would mean that the new super–YM multiplet no longer consists of auxiliary fields. They would become propagating and dynamical.

Acknowledgments

I thank Doug Baxter, Carlo Becchi, Friedemann Brandt, James Dodd, Mike Duff,
Pierre Ramond, Peter Scharbach, Kelly Stelle, J.C. Taylor, and Peter West for
recent stimulating correspondence and conversations.

## References

  * [1] D. Z. Freedman and A. Van Proeyen, “Supergravity”, Cambridge, UK (2012): ISBN: 9781139368063 (eBook), 9780521194013 (Print) (Cambridge University Press).
  * [2] Many of the original papers on SUSY and Supergravity are collected in “Supersymmetry”, Vols. 1 and 2, ed. Sergio Ferrara, North Holland, World Scientific, (1987).
  * [3] P. C. West, “Introduction to supersymmetry and supergravity,” Singapore, Singapore: World Scientific (1990) 425 p
  * [4] Brandt, Friedemann “Local BRST Cohomology in Minimal D=4, N=1 Supergravity”, Annals of Physics 259, pp 357–416 (1997)
  * [5] K. S. Stelle and P. C. West, “Matter Coupling and BRS Transformations With Auxiliary Fields in Supergravity,” Nucl. Phys. B 140, 285-293 (1978) doi:10.1016/0550-3213(78)90536-9
  * [6] Steven Weinberg: “The Quantum Theory of Fields” Volume 3, Cambridge University Press, ISBN 052155002.
  * [7] Steven Weinberg: “The Quantum Theory of Fields” Volume 2, Cambridge University Press.
  * [8] J. C. Taylor, “Gauge Theories of Weak Interactions,” Cambridge 1976, 167p
  * [9] The Master Equation goes back to [10] and Zinn-Justin’s early contribution is set out in a later textbook [11]. An early and pithy introduction was in [8]. A bit of history and Zinn Justin’s involvement is in [12]. A more recent treatment is in [7].
  * [10] C. Becchi, A. Rouet and R. Stora, “Renormalization of Gauge Theories,” Annals Phys. 98, 287 (1976).
  * [11] J. Zinn-Justin, “Quantum Field Theory and Critical Phenomena”, Oxford Science Publications, Reprinted 1990.
  * [12] A summary and some history can be found in J. Zinn-Justin, “From Slavnov-Taylor identities to the ZJ equation,” Proc. Steklov Inst. Math. 272, 288 (2011).
  * [13] C. Becchi, “Slavnov–Taylor and Ward identities in the electroweak theory,” Theor. Math. hys. 182, no. 1, 52 (2015) [Teor. Mat. Fiz. 182, no. 1, 65 (2014)] doi:10.1007/s11232-015-0244-8 [arXiv:1407.3960 [hep-th]].
  * [14] C. Becchi, “Becchi–Rouet–Stora–Tyutin symmetry”, http://www. scholarpedia .org/article/Becchi-Rouet-Stora-Tyutin symmetry
  * [15] John A. Dixon, “A Candidate for a Supergravity Anomaly”, arXiv 2110.06213. This is a rambling paper containing guesses and conjectures. They cannot be resolved unless the BRS transformations for supergravity can be found first.
  * [16] Ibid, “Calculation of BRS cohomology with spectral sequences”, Commun. Math. Phys., 139 pp 495–526 (1991)
  * [17] S. Deser and B. Zumino, “Consistent Supergravity”, Phys. Lett. B 62, 335 (1976)

JB277 The1.5OrderisFlawed