Prepotentials for linearized supergravity

The Einstein tensor is obtained by taking a curl of $\phi$ on both groups of indices,

It is invariant under the gauge symmetries parametrized by M and A. It is identically divergenceless (on both groups of indices) and also has the $(d-2,1)$ Young symmetry. The converse of these properties is also true and is easily proven using the generalized Poincaré lemma of [41] for mixed symmetry fields:

• The condition $G[\phi] = 0$ implies that the field is pure gauge,
  
  for some A and M with the appropriate symmetry.
• Any divergenceless tensor of $(d-2,1)$ symmetry is the Einstein tensor of some $(d-2,1)$ field,
  
  (The divergencelessness of T in the j  index follows from the cyclic identity $T_{[i_1 \dots i_{d-2} j]} = 0$.)

The Einstein tensor is not invariant under the Weyl symmetries parametrized by B, under which it transforms as

For the trace , this implies

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:vartraceGphi} \delta G^{i_2 \dots i_{d-2}} [\phi] = \partial^k \partial_m B^{l_2 \dots l_{d-2} } \varepsilon^{mnj i_2 \dots i_{d-2}} \varepsilon_{jknl_2 \dots l_{d-2}} = 2(d-2)! \partial^k \partial_m B^{l_2 \dots l_{d-2} } \delta^{m i_2 \dots i_{d-2}}_{k l_2 \dots l_{d-2}}. \nonumber \end{align} \tag{ 2.7 }$$

From the Einstein tensor and its trace, one can define the Schouten tensor

which has the important property of transforming simply as

under a Weyl transformation, as follows from (2.6) and (2.7).

The relation between the Einstein and the Schouten can be inverted to

where is the trace of the Schouten tensor. The divergencelessness of G is then equivalent to

It also implies the identity

because G is divergenceless on its last index.

The Cotton tensor is defined as

Because of the transformation law (2.9) of the Schouten tensor, it is invariant under the full gauge and Weyl transformations (2.2). Moreover, $D[\phi] = 0$ implies that the field takes the form (2.2) (see appendix A). It also satisfies the following properties:

• It has the $(d-2,d-2)$ Young symmetry,
  $$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:cyclicDphi} D_{[i_1 \dots i_{d-2} \, j_1]\,j_2 \dots j_{d-2}}[\phi] = 0, \nonumber \end{align} \tag{ 2.14 }$$
• It is divergenceless,
  $$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \partial^{i_1}D_{i_1 \dots i_{d-2}\, j_1 \dots j_{d-2}}[\phi] = 0, \nonumber \end{align} \tag{ 2.15 }$$
• Its complete trace vanishes,
  

The first of these properties is equivalent to the identity (2.11). The second is evident from the definition of D; divergencelessness on the second group of indices then follows from the cyclic identity (2.14). The last property follows from the cyclic identity for the Schouten tensor. Conversely, any tensor that satisfies these three properties must be the Cotton tensor of some $(d-2,1)$ field.

Note that the transformation property (2.9) of the Schouten implies that the tensor

is also invariant under Weyl transformations of $\phi$. (With respect to definition (2.13), $D'$ is obtained by taking the curl of S on the other group of indices.) Therefore, it must be a function of the Cotton tensor. Indeed, using the cyclic identity $S_{[i_1 \dots i_{d-2} | j]} = 0$, one finds

The Einstein tensor of P is defined as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle G_{ij}[P] = \varepsilon_{ikm_1 \dots m_{d-2}} \varepsilon_{jln_1 \dots n_{d-2}} \partial^k \partial^l P^{m_1 \dots m_{d-2} \, n_1 \dots n_{d-2} } . \nonumber \end{align} \tag{ 2.21 }$$

It is invariant under the $\alpha$ gauge symmetries. It is also symmetric and divergenceless. Conversely, the condition G_ij[P]  =  0 implies that P is pure gauge, and any symmetric divergenceless tensor is the Einstein tensor of some field P with the $(d-2,d-2)$ symmetry. These two theorems are easily proven using the Poincaré lemma of [39, 40] for rectangular Young tableaux.

Under a Weyl transformation, we have

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\lap}{\,\Delta} \newcommand{\re}{{\rm Re}} \displaystyle \delta G_{ij}[P] = (d-2)! (\delta_{ij} \lap \xi - \partial_i \partial_j \xi) \nonumber \end{align} \tag{ 2.22 }$$

and $\newcommand{\re}{{\rm Re}} \newcommand{\lap}{\,\Delta} \delta G[P] = (d-1)! \lap \xi$ for the trace . The Schouten tensor is then defined as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle S_{ij}[P] = G_{ij}[P] - \frac{1}{d-1} \, \delta_{ij}\, G[P]. \nonumber \end{align} \tag{ 2.23 }$$

It transforms in a simple way under Weyl transformations,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:deltaSP} \delta S_{ij}[P] = -(d-2)! \,\partial_i \partial_j \xi . \nonumber \end{align} \tag{ 2.24 }$$

It is also symmetric. One can invert the relation between the Schouten and the Einstein as $G_{ij} = S_{ij} - \delta_{ij} S$. Therefore, the Schouten tensor satisfies

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:SPdiv} \partial^i S_{ij} = \partial_j S \nonumber \end{align} \tag{ 2.25 }$$

because G_ij is divergenceless.

The Cotton tensor is then defined as

It is invariant under all the gauge symmetries (2.20) as a consequence of (2.24), and the condition D[P]  =  0 implies that P is of the form (2.20). It is a tensor of $(d-2,1)$ mixed symmetry,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:DPsym} D_{i_1 \dots i_{d-2} \, j } = D_{[i_1 \dots i_{d-2}] \, j }, \quad D_{[i_1 \dots i_{d-2} \, j]} = 0. \nonumber \end{align} \tag{ 2.27 }$$

The second of these equations is equivalent to (2.25). Moreover, because the Schouten is symmmetric, the Cotton is traceless,

Lastly, the Cotton is also identically divergenceless on the first group of indices,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \partial^{i_1} D_{i_1 \dots i_{d-2} \, j } = 0 . \nonumber \end{align} \tag{ 2.29 }$$

The divergencelessness on the last index, $\partial^{\,j} D_{i_1 \dots i_{d-2} \, j } = 0$, is then a consequence of this and the second of (2.27). Conversely, any $(d-2,1)$ field that is traceless and divergenceless is the Cotton tensor of some P.

The invariant tensor for the $\newcommand{\e}{{\rm e}} \eta$ transformations is of course the curl

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle G_i[\chi] = \varepsilon_{ijk_1\dots k_{d-2}} \partial^{\,j} \chi^{k_1 \dots k_{d-2}} , \nonumber \end{align} \tag{ 2.31 }$$

which we also call 'Einstein tensor' by analogy with the previous cases. It satisfies the following properties, which are easily proved using the usual Poincaré lemma with a spectator spinor index:

• It is invariant under the $\newcommand{\e}{{\rm e}} \eta$ gauge transformations. Conversely, $G_i[\chi] = 0$ implies that $\chi_{i_1 \dots i_{d-2}}$ is pure gauge,
  $$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle G_i[\chi] = 0 \quad \Leftrightarrow\quad \chi_{i_1 \dots i_{d-2}} = (d-2) \partial_{[i_1} \eta_{i_2 \dots i_{d-2}]} \;{{\rm ~for~some~}} \eta. \nonumber \end{align} \tag{ 2.32 }$$
• It is identically divergenceless, $\partial^i G_i = 0$. Conversely, any divergenceless vector-spinor is the Einstein tensor of some antisymmetric tensor-spinor $\chi_{i_1 \dots i_{d-2}}$,
  $$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \partial^i T_i = 0 \quad \Leftrightarrow \quad T_i = G_i[\chi] \;{{\rm ~for~some~}} \chi. \nonumber \end{align} \tag{ 2.33 }$$

The Einstein tensor is not invariant under Weyl transformations. We have

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \delta G^i[\chi] = \varepsilon^{ijl_1 \dots l_{d-2}} \gamma_{l_1\dots l_2}\partial_j \rho = i^{m+1} (d-2)!\, \gamma^{ij} \gamma_0 \hat{\gamma}\, \partial_j \rho \nonumber \end{align} \tag{ 2.34 }$$

where $m = \lfloor (d+1)/2 \rfloor$ and the dimension-dependent matrix $\hat{\gamma}$ is defined in appendix E.

The Schouten tensor is then defined as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle S_i[\chi] = \frac{1}{d-1} \left( \gamma_{ij} - (d-2) \delta_{ij} \right) G^{\,j}[\chi]. \nonumber \end{align} \tag{ 2.35 }$$

Using the gamma matrix identity (E.4), one can see that it transforms simply as a total derivative under Weyl rescalings,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \delta S_i [\chi] = \partial_i \nu, \qquad \nu = i^{m+1} (d-2)! \gamma_0 \hat{\gamma}\, \rho . \nonumber \end{align} \tag{ 2.36 }$$

The Einstein can be written in terms of the Schouten as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:GofSchi} G_i[\chi] = \gamma_{ij} S^{\,j}[\chi] , \nonumber \end{align} \tag{ 2.37 }$$

which implies that the Schouten identically satisfies

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:dSfermionic} \gamma_{ij} \partial^i S^{\,j}[\chi] = 0 . \nonumber \end{align} \tag{ 2.38 }$$

The invariant tensor for Weyl transformations is the Cotton tensor

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:defcottonchi} D_{i_1 \dots i_{d-2}}[\chi] = \varepsilon_{i_1\dots i_{d-2}jk} \partial^{\,j} S^k[\chi]. \nonumber \end{align} \tag{ 2.39 }$$

It is identically divergenceless. Its complete gamma-trace also vanishes,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:cottongammatrace} \gamma^{i_1 \dots i_{d-2}} D_{i_1 \dots i_{d-2}} = 0, \nonumber \end{align} \tag{ 2.40 }$$

as follows from (2.38) and identity (E.11). Conversely, $D[\chi] = 0$ implies that $\chi$ takes the form (2.30), and any divergenceless, rank d  −  2 antisymmetric tensor-spinor satisfying the complete gamma-tracelessness condition (2.40) is the Cotton tensor of some $\chi$. The proof of those properties is done in appendix A.

First, it follows from (5.19) that the variation of the Cotton tensor of $\chi$ is

where we used the relation (3.13) between the Riemann of $h[\phi]$ and the Cotton of $\phi$. Writing both sides as the curl of the respective Schouten tensors, this gives

up to a total derivative that can cancelled by adding the appropriate Weyl transformation to $\delta_\phi \chi$. (This is more convenient than (5.19) because it involves directly the Schouten tensors on both sides.) We now write this in terms of the Einstein tensors: this gives

The computation uses relations (2.37), (2.8) and the gamma matrix identity (E.5). From there, the variation $\delta_\phi \chi$ is found by writing the right-hand side as a curl: this yields expression (5.5).

Using the identity $\gamma_{ij} S^{\,j}[\chi] = G_i[\chi]$, we must have for the Einstein tensor

The variations $\delta_P \chi$ can the be identified up to a gauge transformation by writing the right-hand sides as a curl. Using the identity and the fact that S_ij[P] is symmetric, we get

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle = \varepsilon_{ikl_1\dots k_{d-2}} \partial^k \left( - \frac{1}{2}\varepsilon_{jpq_1 \dots q_{d-2}} \partial^{\,p} P^{l_1 \dots l_{d-2} q_1 \dots q_{d-2}} \gamma^{\,j} \gamma^0 \epsilon \right) \nonumber \end{align} \tag{ 5.26 }$$

from which we deduce the expression (5.6).

We want to prove the implication

(The opposite implication is true by construction.) The proof goes in two steps:

• 1.  
  First, D  =  0 implies that the curl of the Schouten tensor on both groups of indices vanishes,

  

  Indeed, the second of those equations follows from the definition of the Cotton tensor, and the first from the relation (2.18) between this curl and the trace of the Cotton. The generalized Poincaré lemma of [41] then implies that S is itself a double curl, i.e.

  

  for some antisymmetric B. This is precisely the variation of the Schouten induced by a Weyl transformation of $\phi$.
• 2.  
  Then, using the inversion relations between the Einstein and Cotton tensors, this implies that

  

  where $\delta B$ stands for the tensor $\delta^{[i_1}_j B^{i_2 \dots i_{d-2}]}$. The Poincaré lemma of [41] (property (2.4) of the Einstein tensor) now implies that $\phi - \delta B$ is pure gauge, which is what we wanted to prove.

The implication to be proven is now

The proof goes as before:

• 1.  
  First, D  =  0 implies that the curl of S_ij[P] vanishes,

  $$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \partial_{[i} S_{j]k} = 0. \nonumber \end{align} \tag{ A.6 }$$

  The Poincaré lemma of [39, 40] applied to symmetric tensors implies then that

  $$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\pd}{\partial} \displaystyle S_{ij}[P] = -(d-2)! \pd_i \pd_j \xi \nonumber \end{align} \tag{ A.7 }$$

  for some $\xi$, which is again the form induced by a Weyl transformation of $\xi$.
• 2.  
  Then, this implies that the Einstein tensor of $P - \delta^{d-2} \xi$ vanishes, where $\delta^{d-2} \xi$ stands for $\delta^{i_1 \dots i_{d-2}}_{j_1 \dots j_{d-2}} \,\xi$. The combination $P - \delta^{d-2} \xi$ is therefore pure gauge, which proves the proposition.

We now should prove

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle D_{i_1 \dots i_{d-2}}[\chi] = 0 \;\Rightarrow\; \chi_{i_1 \dots i_{d-2}} = {{\rm (gauge)}} + \gamma_{i_1 \dots i_{d-2}} \rho \;~{{\rm ~for~~some~}}~ \rho. \nonumber \end{align} \tag{ A.8 }$$

We take the same steps:

• 1.  
  First, D  =  0 is equivalent to $\partial_{[i} S_{j]}[\chi] = 0$, which by the usual Poincaréé lemma with a spectator spinorial index implies

  $$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\pd}{\partial} \displaystyle S_i[\chi] = \pd_i \nu \nonumber \end{align} \tag{ A.9 }$$

  for some $\nu$ that can always be written as $\nu = i^{m+1} (d-2)! \gamma_0 \hat{\gamma}\, \rho$. This is the form of the Schouten induced by a Weyl transformation of $\chi$.
• 2.  
  This implies that the Einstein tensor of $\chi_{i_1 \dots i_{d-2}} - \gamma_{i_1 \dots i_{d-2}} \rho$ is zero. That combination is therefore pure gauge, which is what had to be proven.

We need to prove that, for any $(d-2, d-2)$ tensor $T_{i_1 \dots i_{d-2} j_1 \dots j_{d-2}}$ that is divergenceless and completely traceless, there exists a $(d-2,1)$ field such that $T = D[\phi]$. Moreover, $\phi$ is determined from this condition up to gauge and Weyl transformations of the form (2.2).

• 1.  
  First, the divergencelessness of T and the usual Poincaré lemma implies that T can be written as the curl of some other tensor U,

  

  The tensor U is determined up to , where $V$ is totally antisymmetric.
• 2.  
  At this stage, U is not of irreducible Young symmetry $(d-2,1)$, but could have a completely antisymmetric component. However, the condition that T is completely traceless implies that this component satisfies $\newcommand{\pd}{\partial} \pd_{[k} U_{j_1 \dots j_{d-2} l]} = 0$, i.e. is of the form $\newcommand{\pd}{\partial} U_{[\,j_1 \dots j_{d-2} l]} = \pd_{[l} W_{j_1 \dots j_{d-2}]}$ for some antisymmetric W. Now, the ambiguity in U described above can be used to precisely cancel this contribution. We can therefore assume the U has the $(d-2,1)$ symmetry from now on.
• 3.  
  We now use the fact that T has the $(d-2,d-2)$ symmetry. This implies that

  $$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \varepsilon_{i_1 \dots i_{d-2} j_1 m} T^{i_1 \dots i_{d-2} j_1 \dots j_{d-2}} = 0 \nonumber \end{align} \tag{ A.11 }$$

  which, using the expression (A.10), is equivalent to the differential identity

  

  on U. We now define another tensor X of $(d-2,1)$ symmetry by the equation

  

  The identity (A.12) is then equivalent to the fact that X is divergenceless.
• 4.  
  We are now ready to conclude. Because X is a divergenceless $(d-2,1)$ tensor, it must be the Einstein tensor of some $(d-2,1)$ field $\phi$, $X = G[\phi]$. The relation between U and X then implies that U is the Schouten tensor of $\phi$, $U = S[\phi]$, and equation (A.10) gives at last $T = D[\phi]$, which was to be proven.
• 5.  
  The ambiguity in $\phi$ is easily determined from the first property of the Cotton tensor. Indeed, by linearity, the condition $D[\phi'] = D[\phi]$ is equivalent to $D[\phi'-\phi] = 0$. The first property of the Cotton then implies that $\phi' - \phi$ takes the form 'Weyl  +  gauge transformation'.

For the $(d-2,d-2)$ field, we now prove that, for any divergenceless, traceless $(d-2, 1)$ tensor $T_{i_1 \dots i_{d-2} j}$, there exists a $(d-2,d-2)$ field $P_{i_1 \dots i_{d-2} j_1 \dots j_{d-2}}$ such that T  =  D[P]. This field P is determined from this condition up to a gauge and Weyl transformation of the form (2.20). The proof goes as before:

• 1.  
  First, the divergencelessness of T implies that it can be written as a curl,

  

  The tensor U is determined up to .
• 2.  
  The tensor U is not necessarily symmetric. However, the condition that T is traceless implies that the antisymmetric component satisfies $\newcommand{\pd}{\partial} \pd_{[k} U_{jl]} = 0$, i.e. is of the form $\newcommand{\pd}{\partial} U_{[\,jl]} = \pd_{[l} W_{j]}$ for some vector W. By using $V$, we can precisely cancel this contribution and we can assume that U is symmetric.
• 3.  
  We now use the fact that T satisfies $T_{[i_1 \dots i_{d-2}j]}=0$. This is equivalent to

  

  Defining the symmetric tensor X by

  

  that identity is equivalent to the fact that X is divergenceless.
• 4.  
  We can now conclude: because X is a symmetric divergenceless tensor, it must be the Einstein tensor of some $(d-2,d-2)$ field P, X  =  G[P]. The relation between U and X then implies that U is the Schouten tensor of P, U  =  S[P], and equation (A.14) gives $T = D[\phi]$.
• 5.  
  The ambiguity in P is given by the first property of the Cotton tensor. Indeed, the condition $D[P'] = D[P]$ is equivalent to $D[P'-P] = 0$, which shows that $P'$ and P differ by a gauge and Weyl transformation.

We fnish this appendix by proving the analogous property for the fermionic antisymmetric field, namely: For any divergenceless, completely gamma-traceless rank d  −  2 tensor-spinor $T_{i_1 \dots i_{d-2}}$, there exists an antisymmetric tensor-spinor $\chi_{i_1 \dots i_{d-2}}$ such that $T = D[\chi]$. Moreover, $\chi$ is determined from this condition up gauge and Weyl transformations of the form (2.30). We follow the same steps as before, but without the complications of Young symmetries

• 1.  
  The divergencelessness of T implies that it can be written as

  $$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:TofUchi} T^{i_1 \dots i_{d-2}} = \varepsilon^{i_1 \dots i_{d-2}kl} \partial_k U_l, \nonumber \end{align} \tag{ A.17 }$$

  where U is determined up to a total derivative $\newcommand{\pd}{\partial} \pd_l V$ for some spinor field $V$.
• 2.  
  Using the gamma matrix identity (E.11), the complete gamma-tracelessness

  $$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \gamma^{i_1 \dots i_{d-2}} T_{i_1 \dots i_{d-2}} = 0 \nonumber \end{align} \tag{ A.18 }$$

  of T is equivalent to the differential equation $\newcommand{\pd}{\partial} \gamma_{ij} \pd^i U^{\,j} = 0$, which is in turn equivalent to $\newcommand{\pd}{\partial} \pd^i X_i = 0$ if we define $X_i = \gamma_{ij} S^{\,j}$.
• 3.  
  Because X is a divergenceless vector-spinor, it is the Einstein tensor of some antisymmetric tensor-spinor $\chi$, $X = G[\chi]$. As before, the relation between U and X then implies that U is the Schouten tensor of $\chi$, U  =  S[P], and the relation between T and U then gives $T = D[\phi]$.
• 4.  
  The ambiguity in $\chi$ is given by the first property of the Cotton tensor: $D[\chi'] = D[\chi]$ is equivalent to $D[\chi'-\chi] = 0$, which shows that $\chi'$ and $\chi$ differ by a transformation of the form (2.30).

We reduce from D  +  1  =  d  +  2 to D  =  d  +  1 dimensions and write the extra coordinate as z. The two higher-dimensional potentials $\hat{A}$ and $\hat{Z}$ split as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \hat{A}_I \;\rightarrow\; \hat{A}_i,\; \hat{A}_z, \quad \hat{Z}_{I_1 \dots I_{d-1}} \;\rightarrow\; \hat{Z}_{i_1 \dots i_{d-1}},\; \hat{Z}_{i_1 \dots i_{d-2} z}, \nonumber \end{align} \tag{ B.10 }$$

with gauge transformations

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \delta \hat{A}_i = \partial_i \Lambda, \qquad \delta \hat{Z}_{i_1 \dots i_{d-1}} = (d-1) \partial_{[i_1} \tilde{\Lambda}_{i_2 \dots i_{d-1}]}, \nonumber \end{align} \tag{ B.11 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \delta \hat{A}_z = 0, \qquad \delta \hat{Z}_{i_1 \dots i_{d-2} z} = (d-2) \partial_{[i_1} \tilde{\Lambda}_{i_2 \dots i_{d-2}] z} . \nonumber \end{align} \tag{ B.12 }$$

The potentials $\hat{A}_i$ and $\hat{Z}_{i_1 \dots i_{d-2} z}$ are those of a vector field in D dimensions, and the other two correspond to a free scalar field. We therefore write

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \hat{A}_i = A_i, \quad \hat{Z}_{i_1 \dots i_{d-2}z} = Z_{i_1 \dots i_{d-2}}, \quad \hat{A}_z = \varphi, \quad \hat{Z}_{i_1 \dots i_{d-1}} = (-1)^d (d-1)!\, \omega_{i_1 \dots i_{d-1}} . \nonumber \end{align} \tag{ B.13 }$$

The magnetic fields of $\hat{A}$ and $\hat{Z}$ then reduce as

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\cB}{\mathcal{B}} \displaystyle \cB_{i_1 \dots i_{d-1}}[\hat{A}] = \varepsilon_{i_1 \dots i_{d-1} j} \partial^{\,j} \varphi, \quad \cB_{i_1 \dots i_{d-2} z}[\hat{A}] = \cB_{i_1 \dots i_{d-2}}[A] \nonumber \end{align} \tag{ B.14 }$$

and

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\cB}{\mathcal{B}} \displaystyle \cB_i[\hat{Z}] = \cB_i[Z], \quad \cB_z[\hat{Z}] = \varepsilon_{i j_1 \dots j_{d-1}} \partial^i \omega^{\,j_1 \dots j_{d-1}} = \pi[\omega] \nonumber \end{align} \tag{ B.15 }$$

respectively. The action then becomes

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\cB}{\mathcal{B}} \displaystyle S[A,Z,\varphi,\omega] = \int {\rm d}t ~{\rm d}^{d}x \,\Big( &\cB_i[Z] \dot{A}^i - \frac{1}{2} \cB_i[Z] \cB^i[Z] - \frac{1}{2(d-2)!} \cB_{i_1 \dots i_{d-2}}[A]\cB^{i_1 \dots i_{d-2}}[A] \nonumber \\ & + \pi[\omega] \dot{\varphi}- \frac{1}{2} \pi[\omega]^2 - \frac{1}{2} \partial_i \varphi \, \partial^i \varphi \Big), \nonumber \end{align} \tag{ B.16 }$$

which is indeed the sum of a free vector action and a free scalar action. Remark that the momentum $\pi$ is written as the curl of a rank $(d-1)$ antisymmetric tensor in this reduction. This is always allowed and does not restrict $\pi$ (this is also discussed in [27]).

We take the higher-dimensional graviton as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:ansatz} \hat{h}_{\mu\nu} = h_{\mu\nu} + 2 \alpha \,\varphi \,\eta_{\mu\nu}, \quad \hat{h}_{\mu z} = A_\mu, \quad \hat{h}_{zz} = 2 \beta \,\varphi \nonumber \end{align} \tag{ C.1 }$$

for some constants $\alpha$, $\beta$. This ansatz can be inverted to

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle h_{\mu\nu} = \hat{h}_{\mu\nu} - \frac{\alpha}{\beta} \hat{h}_{zz} \eta_{\mu\nu}, \quad A_{\mu} = \hat{h}_{\mu \,z}, \quad \varphi = \frac{1}{2\beta} \hat{h}_{zz} . \nonumber \end{align} \tag{ C.2 }$$

Therefore, it is a good parametrization of the higher-dimensional metric as long as $\beta \neq 0$.

Then, the Lagrangian (3.1) reduces to the sum of the free Lagrangians for $h_{\mu\nu}$, $A_\mu$ and $\varphi$:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{Lreduced} \mathcal{L}^{(D+1)}_{\rm PF}[\hat{h}] = \mathcal{L}^{(D)}_{\rm PF}[h] - \frac{1}{4} F_{\mu\nu} F^{\mu\nu} - \frac{1}{2} \partial_\mu \varphi \partial^\mu \varphi , \nonumber \end{align} \tag{ C.3 }$$

provided the constants $\alpha$, $\beta$ are given by

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \alpha^2 = \frac{1}{2(D-2)(D-1)}, \quad \beta = - \alpha (D-2). \nonumber \end{align} \tag{ C.4 }$$

This is of course consistent with the usual KK ansatz at the non-linear level (see for example [25]).

From equation (C.1), we get the reduction formulas

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \hat{h}_{ij} = h_{ij} + 2 \alpha \,\varphi \,\delta_{ij},\qquad \hat{h}_{i z} = A_i, \qquad \hat{h}_{zz} = 2 \beta \varphi, \nonumber \end{align} \tag{ C.5 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \hat{\pi}^{ij} = \pi^{ij}, \hskip 58 pt \hat{\pi}^{iz} = \frac{1}{2} \pi^i, \qquad \hat{\pi}^{zz} = \frac{1}{2\beta}( \pi - 2 \alpha \, \delta_{ij} \pi^{ij}) \nonumber \end{align} \tag{ C.6 }$$

with inverse relations

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle h_{ij} = \hat{h}_{ij} - \frac{\alpha}{\beta} \delta_{ij} \hat{h}_{zz}, \qquad A_i = \hat{h}_{i\,z},\qquad \varphi = \frac{1}{2\beta} \hat{h}_{zz} ,\label{eq:ansatzh} \nonumber \end{align} \tag{ C.7 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \pi^{ij} = \hat{\pi}^{ij}, \hskip 58 pt \pi^i = 2 \hat{\pi}^{iz}, \qquad \pi = 2 \beta \hat{\pi}^{zz} + 2 \alpha \delta_{ij} \hat{\pi}^{ij} .\label{eq:ansatzpi} \nonumber \end{align} \tag{ C.8 }$$

Plugging this in the higher-dimensional Hamiltonian action (3.10), direct computation shows that it indeed reduces to the sum of the Hamiltonian actions (B.1), (B.3) and (3.10) for a scalar field, a vector field and linearized gravity in D  =  d  +  1 dimensions.

The prepotential $\Phi_{I_1 \dots I_{d-1}J}$ splits into four pieces,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:phired} \Phi_{i_1 \dots i_{d-1}j}, \quad \Phi_{z i_1 \dots i_{d-2} j}, \quad \Phi_{i_1 \dots i_{d-1} z}, \quad \Phi_{i_1 \dots i_{d-2} zz}. \nonumber \end{align} \tag{ C.9 }$$

It is convenient to do the following field redefinition:

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle Q_{i_1 \dots i_{d-1}j} = \Phi_{i_1 \dots i_{d-1}j} - (d-1) \Phi_{[i_1 \dots i_{d-2} |zz|} \delta_{i_{d-1}]\,j}, \nonumber \end{align} \tag{ C.10 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle M_{i_1 \dots i_{d-2}j} = \mathbb{P}_{(d-2,1)}(\Phi_{z i_1 \dots i_{d-2} j}) = \Phi_{z i_1 \dots i_{d-2} j} - \Phi_{z[i_1 \dots i_{d-2} j]}, \nonumber \end{align} \tag{ C.11 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle A_{i_1 \dots i_{d-1}} = \Phi_{z[i_1 \dots i_{d-1}]}, \nonumber \end{align} \tag{ C.12 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle B_{i_1 \dots i_{d-2}} = \Phi_{i_1 \dots i_{d-2} zz} . \nonumber \end{align} \tag{ C.13 }$$

This change of variables is invertible: explicitely, one can express the quantities in (C.9) as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Phi_{i_1 \dots i_{d-1}j} = Q_{i_1 \dots i_{d-1}j} + (d-1) B_{[i_1 \dots i_{d-2}} \delta_{i_{d-1}]\,j}, \nonumber \end{align} \tag{ C.14 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Phi_{z i_1 \dots i_{d-2} j} = M_{i_1 \dots i_{d-2}j} + A_{i_1 \dots i_{d-2}j}, \label{eq:splitphiz} \nonumber \end{align} \tag{ C.15 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Phi_{i_1 \dots i_{d-1} z} = (-1)^{d-1} (d-1) A_{i_1 \dots i_{d-1}}, \nonumber \end{align} \tag{ C.16 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Phi_{i_1 \dots i_{d-2} zz} = B_{i_1 \dots i_{d-2}}, \nonumber \end{align} \tag{ C.17 }$$

where we used the identity

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Phi_{z[i_1 \dots i_{d-2}j]} = \frac{(-1)^d}{d-1} \Phi_{i_1 \dots i_{d-2} j z}, \nonumber \end{align} \tag{ C.18 }$$

which follows from $\Phi_{[i_1 \dots i_{d-2}jz]}=0$. The new variables have the irreducible Young symmetry suggested by their indices: Q has the $(d-1,1)$ symmetry, M the $(d-2,1)$ symmetry, and A and B are totally antisymmetric. Equation (C.15) corresponds to the split of $\Phi_{z i_1 \dots i_{d-2} j}$ into its irreducible components.

The reduction of the gauge symmetries of the higher-dimensional prepotential are give the usual gauge symmetries for Q, M, A, B according to their Young symmetry. For the reduction of the local Weyl transformations, one finds that

• Q has no Weyl transformation (this motivates its definition);
• M has the same Weyl symmetries (2.2) as the first prepotential of linearized gravity;
• A has no Weyl symmetry;
• B can be set to zero.

The link between these fields and the appropriate prepotentials for the lower-dimensional fields is done by using equation (C.7). This gives

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\pd}{\partial} \displaystyle \varphi = \frac{1}{2\beta} (-1)^{d-1} (d-1) \,\varepsilon_{j_1 \dots j_{d-1}j} \pd^{\,j} A^{i_1 \dots i_{d-1}} . \nonumber \end{align} \tag{ C.21 }$$

Therefore, we identify the lower-dimensional prepotential $\phi$ as

The prepotential $P_{I_1 \dots I_{d-1}J_1 \dots J_{d-1}}$ splits into three pieces,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:pred} P_{i_1 \dots i_{d-1}j_1 \dots j_{d-1}}, \quad P_{i_1 \dots i_{d-1}j_1 \dots j_{d-2} z}, \quad P_{i_1 \dots i_{d-2} z j_1 \dots j_{d-2} z}. \nonumber \end{align} \tag{ C.23 }$$

As before, it is useful to introduce a combination that has no Weyl symmetry,

The lower-dimensional momenta are then

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\pd}{\partial} \displaystyle \pi^{ij} &= G_{ij}[P] \nonumber \\ &= (d-1)^2 \varepsilon^{ikm_1 \dots m_{d-2}}\varepsilon^{\,jln_1 \dots n_{d-2}} \pd_k \pd_l P_{m_1 \dots m_{d-2}zn_1 \dots n_{d-2}z} \nonumber \end{align} \tag{ C.25 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\pd}{\partial} \displaystyle \pi &= 2 \beta G^{zz}[P] + 2 \alpha \delta_{ij} G^{ij}[P] \nonumber \\ &= 2 \beta \varepsilon^{k m_1 \dots m_{d-1}}\varepsilon^{\,j n_1 \dots n_{d-1}} \pd_k \pd_l T_{m_1 \dots m_{d-1} n_1 \dots n_{d-1}} . \nonumber \end{align} \tag{ C.27 }$$

And we identify the potentials as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle P_{m_1 \dots m_{d-2}n_1 \dots n_{d-2}} = (d-1)^2 P_{m_1 \dots m_{d-2}zn_1 \dots n_{d-2}z} , \nonumber \end{align} \tag{ C.28 }$$

It is useful to first write the reduction in covariant form. We have the three terms

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle -\bar{\Psi}_{\hat{\mu}} \hat{\gamma}^{\hat{\mu}\hat{\nu}\hat{\rho}} \partial_{\hat{\nu}} \Psi_{\hat{\rho}} &= - \bar{\Psi}_z \hat{\gamma}^z \hat{\gamma}^{\mu\nu} \partial_\mu \Psi_\nu - \bar{\Psi}_\mu \hat{\gamma}^{\mu\nu} \hat{\gamma}^z \partial_\nu \Psi_z - \bar{\Psi}_\mu \hat{\gamma}^{\mu\nu\rho} \partial_\nu \Psi_\rho \nonumber \\ &= \bar{\eta} \hat{\gamma}^{\mu\nu} \partial_\mu \Psi_\nu - \bar{\Psi}_\mu \hat{\gamma}^{\mu\nu} \partial_\nu \eta - \bar{\Psi}_\mu \hat{\gamma}^{\mu\nu\rho} \partial_\nu \Psi_\rho, \nonumber \end{align} \tag{ C.30 }$$

where in the second line we defined $\newcommand{\e}{{\rm e}} \eta = \hat{\gamma}^z \Psi_z$. (This definition implies $\newcommand{\e}{{\rm e}} \bar{\eta} = - \bar{\Psi}_z \hat{\gamma}^z$.) We now take the ansatz

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \eta = a \lambda, \quad \Psi_\mu = \psi_\mu + b \hat{\gamma}_\mu \lambda, \nonumber \end{align} \tag{ C.31 }$$

where a and b are constants. Requiring that the cross-terms between $\lambda$ and $\psi$ vanish first gives $a=-b(D-2)$. The constant b then gives the normalization of the part containing $\lambda$: we choose b²(D  −  1)(D  −  2)  =  1 for convenience. This gives the constants a, b as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle a = -\sqrt{\frac{D-2}{D-1}}, \qquad b = \frac{1}{\sqrt{(D-1)(D-2)}} . \nonumber \end{align} \tag{ C.32 }$$

The action for the gravitino then reduces to

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\dx}[1][D]{\! {\rm d}^{#1}\! x} \displaystyle \label{eq:actiongravitinored} S[\lambda, \psi] = - \int \dx \left( \bar{\lambda} \hat{\gamma}^\mu \partial_\mu \lambda + \bar{\psi}_\mu \hat{\gamma}^{\mu\nu\rho} \partial_\nu \psi_\rho \right). \nonumber \end{align} \tag{ C.33 }$$

The ansatz above can be inverted to

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:gravitinored} \lambda = - \sqrt{\frac{D-1}{D-2}}\,\hat{\gamma}^z \Psi_z, \quad \psi_\mu = \Psi_\mu + \frac{1}{D-2}\, \hat{\gamma}_\mu \hat{\gamma}^z \Psi_z . \nonumber \end{align} \tag{ C.34 }$$

Note that the action (C.33) is not quite the action for a free Dirac field and a free gravitino in D dimensions, since it still contains the higher-dimensional gamma matrices $\hat{\gamma}_\mu$. To go further, we must specify the link between gamma matrices in D  +  1 and D dimensions.

We now set up our choice of gamma matrix representations. Starting from a given representation (that we do not need to specify) in even dimension D  =  2m, we go up in dimensions and give representations in dimensions 2m  +  1 and 2m  +  2.

In odd dimensions D  =  2m  +  1, we can take the first 2m matrices to be exactly those in dimension D  =  2m, and the last one to be $\gamma_*$. Indeed, because of the first two equations of (E.8), the set

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle (\hat{\gamma}_\mu) = ( \gamma_0, \gamma_1, \dots, \gamma_{2m-1}, \gamma_{2m} = \gamma_*) \nonumber \end{align} \tag{ C.35 }$$

satisfies the defining property (E.1) in dimension 2m  +  1. (Note that we could also take $\gamma_{2m} = - \gamma_*$: this gives an equivalent representation.)

In dimension D  =  2m  +  2, one can take the gamma matrices in the block form

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:explicitgamma} \hat{\gamma}_\mu &= \sigma_1 \otimes \gamma_\mu = \left(\begin{array}{@{}cc@{}} 0 & \gamma_\mu \nonumber \\ \gamma_\mu & 0 \end{array}\right) \quad (\mu = 0, \dots, D-2), \nonumber \\ \hat{\gamma}_{D-1} &= \sigma_2 \otimes I= \left(\begin{array}{@{}cc@{}} 0 & - iI \nonumber \\ iI & 0 \end{array}\right). \nonumber \end{align} \tag{ C.36 }$$

The first are given in terms of the D  −  1  =  2m  +  1 gamma matrices (in particular, $\gamma_{2m} = (-i){}^{m+1} \gamma_0 \gamma_1 \dots \gamma_{2m-1}$ is the chirality matrix in D  −  2  =  2m dimensions). In this representation, the $\hat{\gamma}_*$ matrix is diagonal,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \hat{\gamma}_* = \left(\begin{array}{@{}cc@{}} I & 0 \nonumber \\ 0 & -I \end{array}\right) . \nonumber \end{align} \tag{ C.37 }$$

Covariant formulation.

The gamma matrices in D  +  1 and D dimensions have the same size, and so do the spinors. The lower-dimensional fields are, using $\hat{\gamma}_\mu = \gamma_\mu$ and $\hat{\gamma}_z = \gamma_*$,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:gravitinoredodd} \lambda = - \sqrt{\frac{D-1}{D-2}}\,\gamma_* \Psi_z, \quad \psi_\mu = \Psi_\mu + \frac{1}{D-2}\, \gamma_\mu \gamma_* \Psi_z . \nonumber \end{align} \tag{ C.38 }$$

The action (C.33) is just the sum of the free action for a Dirac field $\lambda$ and a Rarita–Schwinger field $\psi_\mu$ in D dimensions. Note also that redefinitions of the fields by a phase may be required to accommodate for (Majorana or symplectic-Majorana) reality conditions in D dimensions; this poses no difficulty and will not be done here.

Prepotential formulation.

The prepotential reduces to two pieces,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle X_{I_1 \dots I_{d-1}} \;\rightarrow\; X_{i_1 \dots i_{d-1}},\; X_{i_1 \dots i_{d-2} z} , \nonumber \end{align} \tag{ C.39 }$$

with the associated gauge and Weyl transformations

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \delta X_{i_1 \dots i_{d-1}} = (d-1) \partial_{[i_1} \Lambda_{i_2 \dots i_{d-1}]} + \gamma_{i_1 \dots i_{d-1}} W \nonumber \end{align} \tag{ C.40 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \delta X_{i_1 \dots i_{d-2} z} = (d-2) \partial_{[i_1} \Lambda_{i_2 \dots i_{d-2}] z} + \gamma_{i_1 \dots i_{d-2}} \gamma_* W \label{eq:chigaugered}. \nonumber \end{align} \tag{ C.41 }$$

To identify those prepotentials with those of the D-dimensional fields, one must compare with (C.38). For that, one first needs the reduction of the Schouten tensor of X (indeed, the higher-dimensional gravitino is determined by X through the equation $\Psi_I = S_I[X]$). One finds

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle S_i[X] = \frac{1}{d} \left( \gamma_{ij} - (d-1) \delta_{ij} \right) G^{\,j}[X] + \gamma_i \gamma_* G_z[X], \nonumber \end{align} \tag{ C.42 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle S_z[X] = - \frac{1}{d} \left( \gamma_i \gamma_* G^i[X] + (d-1) G_z[X] \right) \nonumber \end{align} \tag{ C.43 }$$

in terms of the Einstein tensor. The lower-dimensional fields are then

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \psi_i = S_i[X] + \frac{1}{d-1} \gamma_i \gamma_* S_z[X] = \frac{1}{d-1} \left( \gamma_{ij} - (d-2) \delta_{ij} \right) G^{\,j}[X] \label{eq:psiofX} \nonumber \end{align} \tag{ C.44 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \lambda = - \sqrt{\frac{d}{d-1}} \gamma_* S_z[X] = \frac{1}{\sqrt{d(d-1)}} \left( -\gamma_i G^i[X] + (d-1) \gamma_* G_z[X] \right) . \nonumber \end{align} \tag{ C.45 }$$

The Einstein itself reduces as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle G_i[X] = (d-1) \varepsilon_{ij k_1 \dots k_{d-2}} \partial^{\,j} X^{k_1 \dots k_{d-2} z}, \quad G_z[X] = \varepsilon_{i j_1 \dots j_{d-1}} \partial^i X^{\,j_1 \dots j_{d-1}}. \nonumber \end{align} \tag{ C.46 }$$

Therefore, (C.44) is exactly the formula $\psi_i = S_i[\chi]$ relating the lower-dimensional gravitino to its prepotential, that we identify as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \chi_{i_1 \dots i_{d-2}} = (d-1) X_{i_1 \dots i_{d-2} z} \nonumber \end{align} \tag{ C.47 }$$

(the gauge and Weyl symmetries also match, see (C.41)). For the Dirac field $\lambda$, one finds

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \lambda = \varepsilon_{i_1 \dots i_d} \partial^{i_1} \zeta^{i_2 \dots i_d}, \quad \zeta_{i_1 \dots i_{d-1}} = \sqrt{\frac{d-1}{d}} \left( \gamma_* X_{i_1 \dots i_{d-1}} + \gamma_{[i_1} X_{i_2 \dots i_{d-1}]z} \right) . \nonumber \end{align} \tag{ C.48 }$$

The gauge transformation of the right-hand side is easily checked to give $\delta \lambda = 0$, as it should. Note also that the change of variables from the two X's to the pair $(\chi, \zeta)$ is invertible.

This identification of the fields now implies the reduction of the action (4.11) as above.

Covariant formulation.

This case is a bit more involved, since the gamma matrices in even dimension D  +  1 now have twice the size of those in 1D below. Accordingly, we write the higher-dimensional gravitino in the block form

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\col}[2]{\left(\begin{array}{c} #1 \nonumber \\ #2 \end{array}\right)} \displaystyle \Psi_{\mu} = \col{\Psi^+_\mu}{\Psi^-_\mu} . \nonumber \end{align} \tag{ C.49 }$$

Note that a chirality condition $\Gamma_* \Psi_\mu = \pm \Psi_\mu$ on the higher-dimensional gravitino corresponds to keeping only the component $\Psi^\pm$ in this decomposition.

According to (C.34), the D-dimensional fields are then

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\col}[2]{\left(\begin{array}{c} #1 \nonumber \\ #2 \end{array}\right)} \displaystyle \label{eq:psipm} \psi_\mu = \col{\psi^+_\mu}{\psi^-_\mu}, \quad \psi^\pm_\mu = \Psi^\pm_\mu \pm \frac{\rm i}{D-2} \, \gamma_\mu \Psi^\pm_z \nonumber \end{align} \tag{ C.50 }$$

and

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\col}[2]{\left(\begin{array}{c} #1 \nonumber \\ #2 \end{array}\right)} \displaystyle \label{eq:lambdapm} \lambda = \col{\lambda^+}{\lambda^-}, \quad \lambda^\pm = \pm {\rm i} \sqrt{\frac{D-1}{D-2}} \Psi^\mp_z. \nonumber \end{align} \tag{ C.51 }$$

The action (C.33) then reduces to the sum of two Dirac and two gravitino actions,

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\dx}[1][D]{\! {\rm d}^{#1}\! x} \displaystyle S[\lambda^\pm, \psi^\pm] = - \int \dx \left( \bar{\lambda}^+ \gamma^\mu \partial_\mu \lambda^+ + \bar{\lambda}^- \gamma^\mu \partial_\mu \lambda^- + \bar{\psi}^+_\mu \gamma^{\mu\nu\rho} \partial_\nu \psi^+_\rho + \bar{\psi}^-_\mu \gamma^{\mu\nu\rho} \partial_\nu \psi^-_\rho \right) . \nonumber \end{align} \tag{ C.52 }$$

Again, phases can be incorporated in the various fields to accommodate for reality conditions in D dimensions.

Prepotential formulation.

We write the prepotential for $\Psi_\mu$ in the block form

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\col}[2]{\left(\begin{array}{c} #1 \nonumber \\ #2 \end{array}\right)} \displaystyle \label{eq:chiblock} X_{I_1 \dots I_{d-1}} = \col{X^+_{I_1 \dots I_{d-1}}}{X^-_{I_1 \dots I_{d-1}}} . \nonumber \end{align} \tag{ C.53 }$$

One has therefore four fields after reduction,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle X^\pm_{I_1 \dots I_{d-1}} \;\rightarrow\; X^\pm_{i_1 \dots i_{d-1}},\; X^\pm_{i_1 \dots i_{d-2} z} , \nonumber \end{align} \tag{ C.54 }$$

with the associated gauge and Weyl transformations

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \delta X^\pm_{i_1 \dots i_{d-1}} = (d-1) \partial_{[i_1} \Lambda^\pm_{i_2 \dots i_{d-1}]} + \gamma_{i_1 \dots i_{d-1}} W^\mp \nonumber \end{align} \tag{ C.55 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \delta X^\pm_{i_1 \dots i_{d-2} z} = (d-2) \partial_{[i_1} \Lambda^\pm_{i_2 \dots i_{d-2}] z} \mp {\rm i} \gamma_{i_1 \dots i_{d-2}} W^\mp \label{eq:psigaugereduced}, \nonumber \end{align} \tag{ C.56 }$$

where $\Lambda^\pm$ and $W^\pm$ are defined from the higher-dimensional parameters by a block form analogous to (C.53).

To identify those prepotentials with those of D-dimensional fields, one must compare with equations (C.50) and (C.51). For that, one first needs the reduction of the Schouten tensor of X (indeed, the higher-dimensional gravitino is determined by X through the equation $\Psi_I = S_I[X]$). One finds

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle S_i^\pm[X] = \frac{1}{d} \left( \gamma_{ij} - (d-1) \delta_{ij} \right) G^{\pm j}[X] \pm \frac{i}{d} \gamma_i G^\pm_z[X], \nonumber \end{align} \tag{ C.57 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle S^\pm_z[X] = \frac{1}{d} \left( \mp {\rm i} \gamma_i G^{\pm i}[X] - (d-1) G^\pm_z[X] \right) \nonumber \end{align} \tag{ C.58 }$$

in terms of the Einstein tensor. The lower-dimensional fields are therefore

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \psi_i^\pm = S^\pm_i[X] \pm \frac{i}{d-1} \gamma_i S^\pm_z[X] = \frac{1}{d-1} \left( \gamma_{ij} - (d-2) \delta_{ij} \right) G^{\pm j}[X] \label{eq:psipmofX} \nonumber \end{align} \tag{ C.59 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \lambda^\pm = \pm {\rm i} \sqrt{\frac{d}{d-1}} S^\mp_z[X] = \frac{1}{\sqrt{d(d-1)}} \left( - \gamma_i G^{\mp i}[X] \mp {\rm i} (d-1) G^\mp_z[X] \right) . \nonumber \end{align} \tag{ C.60 }$$

The Einstein itself reduces as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle G^\pm_i[X] = (d-1) \varepsilon_{ij k_1 \dots k_{d-2}} \partial^{\,j} X^{\pm k_1 \dots k_{d-2} z}, \quad G^\pm_z[X] = \varepsilon_{i j_1 \dots j_{d-1}} \partial^i X^{\pm j_1 \dots j_{d-1}}. \nonumber \end{align} \tag{ C.61 }$$

Therefore, (C.59) is exactly the formula $\psi^\pm_i = S_i[\chi^\pm]$ relating the lower-dimensional gravitinos to their prepotentials

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \chi^\pm_{i_1 \dots i_{d-2}} = (d-1) X^\pm_{i_1 \dots i_{d-2} z} \nonumber \end{align} \tag{ C.62 }$$

(the gauge and Weyl symmetries also match with the appropriate identifications, see (C.56)). For the Dirac fields $\lambda^\pm$, one finds

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \lambda^\pm = \varepsilon_{i_1 \dots i_d} \partial^{i_1} \zeta^{\pm i_2 \dots i_d}, \quad \zeta^{\pm}_{i_1 \dots i_{d-1}} = \sqrt{\frac{d-1}{d}} \left( \mp {\rm i} X^\mp_{i_1 \dots i_{d-1}} + \gamma_{[i_1} X^\mp_{i_2 \dots i_{d-1}]z} \right) . \nonumber \end{align} \tag{ C.63 }$$

The action yielding the equations of motion (D.3) was first written in [34], and it involves prepotentials in an essential way. The prepotential is a $(2,2)$ tensor Z_IJKL, determined up to the local gauge and Weyl transformations

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:6d22gauge} \delta Z_{IJKL} = \xi_{IJ[K,L]} + \xi_{KL[I,J]} + \delta_{[I[K} \lambda_{J]L]}, \nonumber \end{align} \tag{ D.4 }$$

where $\xi_{IJK}$ is a $(2,1)$ tensor parametrizing the gauge transformations and $\lambda_{IJ}$ is a symmetric tensor parametrizing the Weyl rescalings.

The Einstein, Schouten and Cotton tensors of Z are

where the traces of the Einstein tensor are defined by and . The action is

The prepotential Z_IJKL and its gauge parameters split into several pieces,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle Z_{IJKL} \;\longrightarrow\; Z_{ijkl},\; Z_{ij k5},\; Z_{i5j5}, \nonumber \end{align} \tag{ D.9 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \xi_{IJK} \;\longrightarrow\; \xi_{ijk},\; \xi_{i5j},\; \xi_{ij5},\; \xi_{i55}, \nonumber \end{align} \tag{ D.10 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \lambda_{IJ} \;\longrightarrow\; \lambda_{ij},\; \lambda_{i5},\; \lambda_{55}. \nonumber \end{align} \tag{ D.11 }$$

The transformation of Z_i5j 5 under Weyl rescalings is

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \delta Z_{i5j5} = \frac{1}{4} ( \lambda_{ij} + \delta_{ij} \lambda_{55}) . \nonumber \end{align} \tag{ D.12 }$$

Therefore, one can gauge away Z_i5j 5 by a Weyl transformation. We will set Z_i5j 5  =  0 from now on. To preserve the condition, one must then set $\xi_{i55} = 0$ and $\lambda_{ij} = - \delta_{ij} \lambda_{55}$. Also, we can split $\xi_{i5j} = - \xi_{5ij}$ into its symmetric and antisymmetric parts,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \xi_{5ij} = s_{ij} + a_{ij}, \qquad s_{ij} = \xi_{5(ij)}, \qquad a_{ij} = \xi_{5[ij]} . \nonumber \end{align} \tag{ D.13 }$$

The cyclic identity $3\,\xi_{[IJK]} = \xi_{IJK} + \xi_{JKI} + \xi_{KIJ} = 0$ implies that the $\xi_{ij5}$ component is not independent: $\xi_{ij5} = - 2 a_{ij}$. The gauge transformations of the remaining fields are then

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \delta Z_{ijkl} = \xi_{ij[k,l]} + \xi_{kl[i,j]} - \frac{1}{2} (\delta_{ik} \delta_{jl} - \delta_{il} \delta_{jk}) \lambda_{55} \nonumber \end{align} \tag{ D.14 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \delta Z_{ijk5} = \partial_k a_{ij} + \partial_{[i} s_{j]k} - \partial_{[i} a_{j]k} + \frac{1}{2} \delta_{k[i} \lambda_{j]5}. \nonumber \end{align} \tag{ D.15 }$$

Those are exactly the gauge transformations (2.2) and (2.20) for the prepotentials of linearized gravity in 5D, provided we identify the fields and gauge parameters as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle P_{ijkl} = Z_{ijkl}, \quad \phi_{ijk} = Z_{ijk5}, \nonumber \end{align} \tag{ D.16 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \alpha_{ijk} = \xi_{ijk}, \quad A_{ij} = a_{ij}, \quad M_{ij} = s_{ij}, \quad \xi = -\lambda_{55}, \quad B_i = \frac{1}{2}\lambda_{i5} . \nonumber \end{align} \tag{ D.17 }$$

The Einstein, Schouten and Cotton tensors of Z_IJKL reduce as follows:

• Einstein:
  $$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle G_{ijkl}[Z] = 0, \quad G_{ijk5}[Z] = - \frac{1}{18} G_{ijk}[\phi], \quad G_{i5j5}[Z] = \frac{1}{3!^2} G_{ij}[P], \nonumber \end{align} \tag{ D.18 }$$
• Schouten:
  $$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle S_{ijkl}[Z] = - \frac{1}{18} \delta_{[i[k} S_{l]\,j]}[P], \quad S_{ijk5}[Z] = - \frac{1}{18} S_{ijk}[\phi], \quad S_{i5j5}[Z] = \frac{1}{2.3!^2} S_{ij}[P], \nonumber \end{align} \tag{ D.19 }$$
• Cotton:
  

where the right-hand sides are given by the corresponding tensors of sections 2.1 and 2.2 for the graviton prepotentials.

We can now use those formulas to reduce the 6D action (D.8). This gives

This is the action of section 3.4, up to the redefinitions $P_{ijkl} = 12 \sqrt{3} \,P'_{ijkl}$ and $\phi_{ijk} = 3 \sqrt{3} \,\phi'_{ijk}$. This result was annoucend in [34]⁹ but is made more transparent by using the appropriate Cotton tensors in 5D.

In [35], it was shown that the equation of motion (D.23) is equivalent to the self-duality equation $H = \star H$ on the field strength, i.e.

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle H_{\mu\nu\rho} = \frac{1}{3!} \varepsilon_{\mu\nu\rho\sigma\tau\lambda} H^{\sigma\tau\lambda}, \nonumber \end{align} \tag{ D.24 }$$

supplemented by the purely spatial constraint

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Gamma^{IABC} H_{ABC} = 0. \nonumber \end{align} \tag{ D.25 }$$

Again, this constraint can be solved by the use of prepotentials; one finds that the appropriate prepotential is an antisymmetric left-handed spinor X_IJ, i.e.

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle X_{IJ} = X_{[IJ]}, \quad \Gamma_* X_{IJ} = X_{IJ} , \nonumber \end{align} \tag{ D.26 }$$

just like the spatial components $\Psi_{IJ}$ themselves. It is determined by $\Psi_{IJ}$ up to a gauge and Weyl transformation

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{eq:GaugeChi} \delta X_{IJ} = \partial_{[I} \Lambda_{J]} + \Gamma_{[I} W_{J]} , \nonumber \end{align} \tag{ D.27 }$$

where the gauge parameters satisfy $\Gamma_* \Lambda_I = \Lambda_I$ and $\Gamma_* W_I = - W_I$ in order to preserve the chirality condition $\Gamma_* X_{IJ} = X_{IJ}$. The Einstein, Schouten and Cotton tensors of X are given by

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle G_{IJ}[X] = \varepsilon_{IJKLM} \partial^K X^{LM}, \nonumber \end{align} \tag{ D.28 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle D_{IJ}[X] = \varepsilon_{IJKLM} \partial^K S^{LM}[X]. \nonumber \end{align} \tag{ D.30 }$$

The spatial components $\Psi_{IJ}$ are given in terms of the prepotential by $\Psi_{IJ} = S_{IJ}[X]$. The action for $\Psi_{\mu\nu}$ can then be written as

$$\begin{align} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \displaystyle \label{eq:exoticgravitino} S[X] = -2{\rm i} \int \!{\rm d}t \,{\rm d}^5\!x\, X^\dagger_{IJ} \left( \dot{D}^{IJ}[X] - \frac{1}{2} \varepsilon^{IJKLM} \partial_K D_{LM}[X] \right). \nonumber \end{align} \tag{ D.31 }$$

For the dimensional reduction, we use the form (C.36) of the 6D gamma matrices. In particular, the block-diagonal form of $\Gamma_*$ implies that the prepotential X_IJ only has components in the first block,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle X_{IJ} = \left(\begin{array}{@{}c@{}} \hat{\chi}_{IJ} \nonumber \\ 0 \end{array}\right). \nonumber \end{align} \tag{ D.32 }$$

The field $\hat{\chi}_{IJ}$ then splits into two parts, $\hat{\chi}_{ij}$ and $\hat{\chi}_{i5}$. From (D.27), we find their gauge transformations

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \delta \hat{\chi}_{ij} = \partial_{[i} \hat{\eta}_{j]} + \gamma_{[i} \hat{\rho}_{j]}, \quad \delta \hat{\chi}_{i5} = \frac{1}{2} (\partial_i \hat{\eta}_5 + \gamma_i \hat{\rho}_5 + i \hat{\rho}_i), \nonumber \end{align} \tag{ D.33 }$$

where $\newcommand{\e}{{\rm e}} \hat{\eta}_I$ and $\hat{\rho}_I$ are given from the 6D gauge parameters by the block form

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Lambda_{I} = \left(\begin{array}{@{}c@{}} \hat{\eta}_{I} \nonumber \\ 0 \end{array}\right),\qquad W_{I} = \left(\begin{array}{@{}c@{}} 0 \nonumber \\ \hat{\rho}_{I} \end{array}\right). \nonumber \end{align} \tag{ D.34 }$$

Using $\hat{\rho}_i$, the field $\hat{\chi}_{i5}$ can be set to zero. The residual gauge transformations must satisfy $\delta \hat{\chi}_{i5} = 0$ to respect this choice; this imposes $\newcommand{\e}{{\rm e}} \hat{\rho}_i = {\rm i} (\partial_i \hat{\eta}_5 + \gamma_i \hat{\rho}_5)$. The gauge transformations of $\hat{\chi}_{ij}$ are then

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \delta \hat{\chi}_{ij} = 2 \partial_{[i} \eta_{j]} + \gamma_{ij} \rho, \qquad {{\rm with}} \quad \eta_j = \frac{1}{2} \left( \hat{\eta}_j - {\rm i} \gamma_j \hat{\eta}_5\right), \quad \rho = {\rm i} \hat{\rho}_5 . \nonumber \end{align} \tag{ D.35 }$$

They have exactly the form (2.30) of the gauge transformations of the prepotential for the 5D gravitino.

The various curvature tensors of X_IJ reduce as follows:

• Einstein:
  $$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \hat{G}_{ij} = 0, \quad \hat{G}_{i5} = - G_i[\hat{\chi}] , \nonumber \end{align} \tag{ D.36 }$$
• Schouten:
  $$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \hat{S}_{ij} = {\rm i} \gamma_{[i} S_{j]}[\hat{\chi}], \quad \hat{S}_{i5} = \frac{1}{2} S_i[\hat{\chi}] , \nonumber \end{align} \tag{ D.37 }$$
• Cotton:
  $$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \hat{D}_{ij} = D_{ij}[\hat{\chi}], \quad \hat{D}_{i5} = {\rm i} \gamma^k D_{ik}[\hat{\chi}] , \nonumber \end{align} \tag{ D.38 }$$

where the left-hand sides are defined by the block forms

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle G_{IJ}[X] = \left(\begin{array}{@{}c@{}} \hat{G}_{IJ} \nonumber \\ 0 \end{array}\right), \quad S_{IJ}[X] = \left(\begin{array}{@{}c@{}} \hat{S}_{IJ} \nonumber \\ 0 \end{array}\right), \quad D_{IJ}[X] = \left(\begin{array}{@{}c@{}} \hat{D}_{IJ} \nonumber \\ 0 \end{array}\right), \nonumber \end{align} \tag{ D.39 }$$

and the right-hand sides are given by the appropriate tensors of section 2.3.

Using the above formulas, the action (D.31) reduces to

$$\begin{align} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \displaystyle S[\chi] = -2{\rm i} \int \!{\rm d}t \,{\rm d}^4\!x\, \hat{\chi}^\dagger_{ij} \left( \dot{D}^{ij}[\hat{\chi}] - {\rm i} \varepsilon^{ijkl} \gamma^m \partial_k D_{lm}[\hat{\chi}] \right), \nonumber \end{align} \tag{ D.40 }$$

which is exactly the action (4.11) for the gravitino in 5D in the prepotential formalism, up to the redefinition $\hat{\chi}_{ij} = \chi_{ij} / \sqrt{2}$. This result was announced in [35].

The sum of actions (D.8) and (D.31) is invariant under a supersymmetry transformation mixing the exotic graviton and gravitino, written explicitely in [35]. Direct reduction of that variation does not reproduce the supersymmetry variations of section 5: this is because the formula written in [35] does not preserve the gauge condition $\hat{\chi}_{i5} = 0$ in 5D. However, a pure gauge term may be added to the variation $\delta X_{IJ}$ in 6D to ensure $\delta \hat{\chi}_{i5} = 0$ upon reduction to 5D. Once this is done, we recover correctly the formulas of this paper for the supersymmetry variations of the prepotentials in 5D¹⁰.

The action for this field was first written in [3]; it is expressed in terms of the spatial components $\hat{A}_{IJ}$ only and reads

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\cB}{\mathcal{B}} \displaystyle S[\hat{A}] = \frac{1}{2} \int {\rm d}t ~{\rm d}^{5}x \left( \dot{\hat{A}}_{IJ} \cB^{IJ}[\hat{A}] - \cB_{IJ}[\hat{A}] \cB^{IJ}[\hat{A}] \right), \nonumber \end{align} \tag{ D.42 }$$

where $\newcommand{\cB}{\mathcal{B}} \cB^{IJ}[\hat{A}]$ is the magnetic field

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\cB}{\mathcal{B}} \displaystyle \cB_{IJ}[\hat{A}] = \frac{1}{2} \varepsilon_{IJKLM} \partial^K \hat{A}^{LM} . \nonumber \end{align} \tag{ D.43 }$$

The field $\hat{A}_{IJ}$ reduces to two pieces,

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \hat{A}_{IJ} \;\longrightarrow\; \hat{A}_{ij},\; \hat{A}_{i5} . \nonumber \end{align} \tag{ D.44 }$$

Their gauge transformations are $\delta \hat{A}_{ij} = 2\partial_{[i} \lambda_{j]}$ and $\delta \hat{A}_{i5} = \partial_i \lambda_5$. Those are exactly the potentials and gauge transformations of the two-potential formulation of a free Maxwell field in 5D (see [27] and appendix B.3). We therefore write

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \hat{A}_{ij} = Z_{ij}, \quad \hat{A}_{i5} = A_i \nonumber \end{align} \tag{ D.45 }$$

from now on.

The magnetic field splits as

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\cB}{\mathcal{B}} \displaystyle \cB_{ij}[\hat{A}] = \cB_{ij}[A], \quad \cB_{i5}[\hat{A}] = - \cB_i[Z], \nonumber \end{align} \tag{ D.46 }$$

where the right-hand sides are given by the 5D magnetic fields defined in appendix B.3.

The action the becomes

$$\begin{align} \newcommand{\e}{{\rm e}} \newcommand{\cB}{\mathcal{B}} \displaystyle S[A,Z] = \frac{1}{2} \int {\rm d}t ~{\rm d}^{4}x \left( \dot{Z}_{ij} \cB^{ij}[A] - 2 \dot{A}_i \cB^i[Z] - \cB_{ij}[A] \cB^{ij}[A] - 2 \cB_i[Z] \cB^i[Z] \right). \nonumber \end{align} \tag{ D.47 }$$

Up to the rescalings $A_i = A'_i/\sqrt{2}$, $Z_{ij} = - Z'_{ij}/\sqrt{2}$, this is the two-potential action for a free vector field.