A Unified Approach to Standard and Exotic Dualizations Through Graded Geometry

Abstract

Gauge theories can often be formulated in different but physically equivalent ways, a concept referred to as duality. Using a formalism based on graded geometry, we provide a unified treatment of all parent theories for different types of standard and exotic dualizations. Our approach is based on treating tensor fields as functions of a certain degree on graded supermanifolds equipped with a suitable number of odd coordinates. We present a universal two-parameter first order action for standard and exotic electric/magnetic dualizations and prove in full generality that it yields two dual second order theories with the desired field content and dynamics. Upon choice of parameters, the parent theory reproduces (i) the standard and exotic duals for p-forms and (ii) the standard and double duals for (p, 1) bipartite tensor fields, such as the linearized graviton and the Curtright field. Moreover, we discuss how deformations related to codimension-1 branes are included in the parent theory.

Appendices

Useful Identities and Additional Proofs

In this appendix we collect some formulas used in the main text. First, we present some relations among the maps defined in Sect. 2, which appear in Refs. [2, 29, 30] in a slightly different setting. Although graded geometry is not used in those works, the following formulas still hold for maps between bipartite tensors, since in local coordinates our definitions are completely equivalent to the ones proven in the aforementioned papers. Alternatively, they may be easily proven using (2.32) and the following definitions of “number operators”

$$\begin{aligned} \hat{p}=\theta ^i\bar{\theta }_i \quad \text {and} \quad \hat{q}=\chi ^i\bar{\chi }_i, \end{aligned}$$

(A.1)

which count the degrees p and q respectively of a bipartite tensor of type (p,q). Then the set of six operators $(\eta , \text {tr},\sigma ,\widetilde{\sigma },\hat{p},\hat{q})$ satisfy the commutation relations

$$\begin{aligned}&[\sigma ,\widetilde{\sigma }]=\hat{p}-\hat{q},\quad [\text {tr},\eta ]=D-\hat{p}-\hat{q}, \quad [\hat{p},\sigma ]=-[\hat{q},\sigma ]=\sigma ,\nonumber \\&[\hat{p},\widetilde{\sigma }]=-[\hat{q},\widetilde{\sigma }]=-\widetilde{\sigma },\quad [\hat{p},\text {tr}]=[\hat{q},\text {tr}]=-\text {tr},\quad [\hat{p},\eta ]=[{\hat{q}},\eta ]=\eta ,\nonumber \\&[\sigma ,\eta ]=[\widetilde{\sigma },\eta ]=[\sigma ,\text {tr}]=[\widetilde{\sigma },\text {tr}]=0. \end{aligned}$$

(A.2)

We note in passing that these six maps and their commutation relations may be related to the six generators $J_{ab}, a,b=1,2,3,4$ of the Lie algebra $\mathfrak {so}(4)$, see [46, 47], as

$$\begin{aligned} J_{13}= & {} -{\hat{p}},\quad J_{24}=-{\hat{q}},\nonumber \\ J_{12}= & {} -\frac{i}{2}(\sigma -\widetilde{\sigma }+\eta -\text {tr}), \quad J_{14}=-\frac{1}{2}(\sigma +\widetilde{\sigma }-\eta -\text {tr}),\nonumber \\ J_{23}= & {} -\frac{1}{2}(\eta +\text {tr}+\sigma +\widetilde{\sigma }), \quad J_{34}=-\frac{i}{2}(\eta -\text {tr}-\sigma +\widetilde{\sigma }), \end{aligned}$$

(A.3)

where

$$\begin{aligned}{}[J_{ab},J_{cd}]=4i\delta _{[a[c}J_{b]d]}. \end{aligned}$$

(A.4)

This $\mathfrak {so}(4)$ is generated by the Poisson algebra of the associated Poisson bracket that can be defined for the graded coordinates and momenta of the graded phase space mentioned in Sect. 2.1. Identities containing (co-)differentials may be worked out in the same way. The ones we use in this paper are summarized as

$$\begin{aligned}&\mathrm {d}\,\sigma +\sigma \,\mathrm {d}=0, \end{aligned}$$

(A.5a)

$$\begin{aligned}&\mathrm {d}\,\eta +\eta \,\mathrm {d}=0, \end{aligned}$$

(A.5b)

$$\begin{aligned}&\mathrm {d}^{\dagger }\,\text {tr}+\text {tr}\,\mathrm {d}^{\dagger }=0, \end{aligned}$$

(A.5c)

$$\begin{aligned}&\mathrm {d}\,{\widetilde{\sigma }}^{n}+(-1)^{n+1}{\widetilde{\sigma }}^n\,\mathrm {d}=-n\, {\widetilde{\mathrm {d}}}\,{\widetilde{\sigma }}^{n-1}, \end{aligned}$$

(A.5d)

$$\begin{aligned}&\mathrm {d}\,\text {tr}^n+(-1)^{n+1}\text {tr}^n\,\mathrm {d}=n\,{\widetilde{\mathrm {d}}}^{\dagger }\,\text {tr}^{n-1}, \end{aligned}$$

(A.5e)

$$\begin{aligned}&\mathrm {d}^{\dagger }\,\eta ^{n}+(-1)^{n+1}\eta ^{n}\,\mathrm {d}^{\dagger }=n\,{\widetilde{\mathrm {d}}}\,\eta ^{n-1}. \end{aligned}$$

(A.5f)

Note that there are also the corresponding identities for transposed maps. It is worth noting that the quadruple $(\mathrm {d},\widetilde{\mathrm {d}},\mathrm {d}^{\dagger },\widetilde{\mathrm {d}}^{\dagger })$ may be related to four supercharges $Q_{a}$ as

$$\begin{aligned} Q_1\equiv & {} \frac{-i}{\sqrt{2}}(\mathrm {d}+\mathrm {d}^\dagger ),\quad Q_2\equiv \frac{-i}{\sqrt{2}}(\widetilde{\mathrm {d}}+\widetilde{\mathrm {d}}^\dagger ),\quad Q_3\equiv \frac{1}{\sqrt{2}}(\mathrm {d}-\mathrm {d}^\dagger ),\nonumber \\ Q_4\equiv & {} \frac{1}{\sqrt{2}}(\widetilde{\mathrm {d}}-\widetilde{\mathrm {d}}^\dagger ), \end{aligned}$$

(A.6)

satisfying

$$\begin{aligned} Q_{a}^2=H \quad \text {and}\quad [Q_{a},H]=0, \end{aligned}$$

(A.7)

where H is the bosonic operator $-\frac{\Box }{2}$ (the Hamiltonian) [47]. In other words they generate a ${{\mathcal {N}}}=4$ supersymmetry algebra and the $\mathfrak {so}(4)$ algebra is its R-symmetry, as can be seen through (A.5).¹⁸ Here we do not utilize further the relation to supersymmetric mechanics, but only use the full set of identities (A.2) and (A.5).

Moreover, a number of integral identities hold for all bipartite tensor fields with degree as indicated. One of them is

$$\begin{aligned} \int _{\theta ,\chi }\omega _{p,q}\, \xi _{D-p,D-q}=-\int _{\theta ,\chi }*\omega _{p,q}*\xi _{D-p,D-q}=-\int _{\theta ,\chi }{\widetilde{*}}\,\omega _{p,q}\,{\widetilde{*}}\,\xi _{D-p,D-q},\nonumber \\ \end{aligned}$$

(A.8)

which can be easily proven using the definitions (2.14) and (2.15). A second identity we use appears in [30] and reads as

$$\begin{aligned} \int _{\theta ,\chi }\omega _{p,q}*{\widetilde{*}}\, \eta \, \xi _{p-1,q-1}=\int _{\theta ,\chi }\text {tr}\,\omega _{p,q}*{\widetilde{*}}\,\xi _{p-1,q-1}. \end{aligned}$$

(A.9)

Finally, the identities

$$\begin{aligned} \int _{\theta ,\chi }{\widetilde{\sigma }}^n\sigma ^n*\omega _{p,q}\,{\widetilde{*}}\, \xi _{p,q}= & {} (-1)^{p(D-p)}\int _{\theta ,\chi }\eta ^n\,\text {tr}^n\omega _{p,q}*{\widetilde{*}}\,\xi _{p,q},\nonumber \\ \int _{\theta ,\chi }\sigma ^n{\widetilde{\sigma }}^n\,{\widetilde{*}}\,\omega _{p,q}*\xi _{p,q}= & {} (-1)^{q(D-q)}\int _{\theta ,\chi }\eta ^n\,\text {tr}^n\omega _{p,q}*{\widetilde{*}}\,\xi _{p,q}, \end{aligned}$$

(A.10)

also hold for any n. We will now prove the first of these identities for $n=1$; the other cases, as well as the second identity, can be proven in exactly the same manner. Using the definitions of the maps appearing in the left hand side, we find

$$\begin{aligned} \int _{\theta ,\chi }{\widetilde{\sigma }}\,\sigma *\omega _{p,q}\,{\widetilde{*}}\, \xi _{p,q}&=(-1)^{qD+p+1}\int _{\theta ,\chi }{\widetilde{*}} (\text {tr}*{\widetilde{*}}\,\text {tr}\,\omega _{p,q})\, {\widetilde{*}}\,\xi _{p,q}\nonumber \\&\overset{(\mathrm{A.8})}{=}(-1)^{qD+p}\int _{\theta ,\chi } \text {tr}(*{\widetilde{*}}\,\text {tr}\,\omega _{p,q})\,\xi _{p,q}\nonumber \\&=(-1)^{pD+q}\int _{\theta ,\chi }\text {tr}(*{\widetilde{*}}\, \text {tr}\,\omega _{p,q})*{\widetilde{*}}(*{\widetilde{*}}\,\xi _{p,q})\nonumber \\&\overset{(\mathrm{A.9})}{=}(-1)^{pD+q}\int _{\theta ,\chi } *{\widetilde{*}}\,\text {tr}\,\omega _{p,q}*{\widetilde{*}} (\eta *{\widetilde{*}}\,\xi _{p,q})\nonumber \\&\overset{(\mathrm{A.8})}{=}(-1)^{pD+q}\int _{\theta ,\chi } \text {tr}\,\omega _{p,q}\,\eta *{\widetilde{*}}\,\xi _{p,q}\nonumber \\&=(-1)^{p(D-p)}\int _{\theta ,\chi }\eta \,\text {tr}\,\omega _{p,q} *{\widetilde{*}}\,\xi _{p,q}, \end{aligned}$$

(A.11)

which is indeed the first identity in (A.9) for $n=1$.

Cyclicity of the integral for the $\star$operator. Now we proceed in proving that

$$\begin{aligned} \int _{\theta ,\chi }\omega _{p,q}\star \xi _{p,q}=\int _{\theta ,\chi }\xi _{p,q}\star \omega _{p,q}. \end{aligned}$$

(A.12)

First, we note that integrals of this form are obviously invariant under the transposition operator $\top$ (tilde operation) since they correspond to spacetime scalars. Thus, we have

$$\begin{aligned} \int _{\theta ,\chi }\omega _{p,q}\star \xi _{p,q}&=\left( \int _{\theta ,\chi }\omega _{p,q}\star \xi _{p,q}\right) ^\top = \int _{\theta ,\chi }\omega ^\top _{q,p}(\star \,\xi _{p,q})^\top \nonumber \\&\overset{\eta ^\top =\,\eta }{=}\int _{\theta ,\chi }\omega ^\top _{q,p}\frac{1}{(D-p-q)!}\,\eta ^{D-p-q}(\xi _{p,q}^\top )^\top \nonumber \\&\overset{(\xi ^\top )^\top =\,\xi }{=}\int _{\theta ,\chi }\xi _{p,q}\frac{1}{(D-p-q)!}\,\eta ^{D-p-q}\omega _{q,p}^\top =\int _{\theta ,\chi }\xi _{p,q}\star \omega _{p,q}, \end{aligned}$$

(A.13)

which proves (A.12).

Proof of Lemma 2.59

We give a proof that is valid for an arbitrary smooth manifold and repeatedly uses the ordinary Poincaré lemma, which holds in either of the odd variables $\theta ^i$ and $\chi ^i$ separately. On a contractible patch, $\mathrm {d}{\widetilde{\mathrm {d}}} \,\xi _{p,q} = \mathrm {d}({\widetilde{\mathrm {d}}} \,\xi _{p,q}) = 0$ implies

$$\begin{aligned} {\widetilde{\mathrm {d}}}\, \xi _{p,q} = \mathrm {d}\,\xi _{p-1, q+1}, \end{aligned}$$

(A.14)

for some new mixed symmetry tensor $\xi _{p-1, q+1}$. Since $({\widetilde{\mathrm {d}}})^2 = 0$, this in turn implies $\mathrm {d}{\widetilde{\mathrm {d}}} \,\xi _{p-1, q+1} = 0$. Vice versa, $\mathrm {d}{\widetilde{\mathrm {d}}} \,\xi _{p-1, q+1} = 0$ implies (A.14) and $\mathrm {d}{\widetilde{\mathrm {d}}} \,\xi _{p,q}=0$. We can use this chain of relations with increasing and decreasing degrees in the odd variables to prove the lemma by induction. With the conventions explained above, let us start with the induction hypothesis

$$\begin{aligned} \xi _{p-1,q+1} = \mathrm {d}\kappa _{p-2,q+1} + {\widetilde{\mathrm {d}}} \kappa _{p-1,q} + c_{i_1 \ldots i_p k_0 \ldots k_q} \theta ^{i_1} \ldots \theta ^{i_{p-1}} x^{i_p} \chi ^{k_0} \ldots \chi ^{k_q}. \end{aligned}$$

(A.15)

Using (A.14), this implies

$$\begin{aligned} {\widetilde{\mathrm {d}}}\, \xi _{p,q} = 0 + \mathrm {d}{\widetilde{\mathrm {d}}} \kappa _{p-1,q} + c_{i_1 \ldots i_p k_0 \ldots k_q} \theta ^{i_1} \ldots \theta ^{i_{p}} \chi ^{k_0} \ldots \chi ^{k_q}, \end{aligned}$$

(A.16)

which can be rewritten to obtain

$$\begin{aligned} {\widetilde{\mathrm {d}}}(\xi _{p,q} - \mathrm {d}\kappa _{p-1,q} - c_{i_1 \ldots i_p k_0 \ldots k_q} \theta ^{i_1} \ldots \theta ^{i_{p}} x^{k_0} \chi ^{k_1} \ldots \chi ^{k_q}) = 0, \end{aligned}$$

(A.17)

and hence

$$\begin{aligned} \xi _{p,q} - \mathrm {d}\kappa _{p-1,q} - c_{i_1 \ldots i_p k_0 k_1\ldots k_q} \theta ^{i_1} \ldots \theta ^{i_p} x^{k_0} \chi ^{k_1} \ldots \chi ^{k_q} = {\widetilde{\mathrm {d}}}\, \kappa _{p,q-1} , \end{aligned}$$

(A.18)

i.e. (2.60) for some mixed symmetry tensor $\kappa _{p,q-1}$.

The suitable choice of induction anchor depends on the value of $p+q$:

For $p+q \ge D$ we can use the chain of relations until we reach $\mathrm {d}{\widetilde{\mathrm {d}}}\, \xi _{p+q-D, D}=0$. Clearly, ${\widetilde{\mathrm {d}}} (\xi _{p+q-D, D} - \mathrm {d}\kappa _{p+q-D-1,D}) = 0$ (both terms are of maximal degree in $\chi ^k$) and hence $\xi _{p+q-D, D} = \mathrm {d}\kappa _{p+q-D-1,D} + {\widetilde{\mathrm {d}}} \kappa _{p+q-D, D-1}$. (The first term can in fact be absorbed in the second one, but we write the expression in this way to match the lemma that we would like to prove.) Starting from this anchor, we can use the induction step by step to prove the lemma for the case $p+q \ge D$ without c-term, as desired.

For $p+q < D$ we can use the chain of relations until we reach $\mathrm {d}{\widetilde{\mathrm {d}}}\, \xi _{0,p+q}=0$. This implies ${\widetilde{\mathrm {d}}}\, \xi _{0,p+q} = \text {const.} =: c_{k_0 \ldots k_{p+q}} \chi ^{k_0} \ldots \chi ^{k_{p+q}}$, where $c_{k_0 \ldots k_{p+q}}$ is a totally antisymmetric constant tensor, and rearranging slightly

$$\begin{aligned} {\widetilde{\mathrm {d}}}(\xi _{0,p+q} - c_{k_0 k_1 \ldots k_{p+q}} x^{k_0} \chi ^{k_1} \ldots \chi ^{k_{p+q}}) =0 , \end{aligned}$$

(A.19)

hence, using Poincaré lemma in $\chi ^k$

$$\begin{aligned} \xi _{0,p+q} = {\widetilde{\mathrm {d}}} \kappa _{0,p+q-1} + c_{k_0 k_1 \ldots k_{p+q}} x^{k_0} \chi ^{k_1} \ldots \chi ^{k_{p+q}} . \end{aligned}$$

(A.20)

This is the desired anchor for the proof by induction in the case $p+q < D$. $\quad \square$

Proof of Eq. (3.6).

Using the map identities (A.5a) and (A.5d), we compute

$$\begin{aligned} \mathrm {d}\,\sigma ^n\,{\widetilde{\sigma }}^n= & {} (-1)^n\,\sigma ^n\,\mathrm {d}\,{\widetilde{\sigma }}^n \nonumber \\= & {} (-1)^n\,\sigma ^n\left( -(-1)^{n+1}\,{\widetilde{\sigma }}^n\,\mathrm {d}-n\,{\widetilde{\mathrm {d}}}\,{\widetilde{\sigma }}^{n-1}\right) \nonumber \\= & {} \sigma ^n\,{\widetilde{\sigma }}^{n}\,\mathrm {d}+(-1)^{n+1}n\,\sigma ^n\,{\widetilde{\mathrm {d}}}\,{\widetilde{\sigma }}^{n-1} \nonumber \\= & {} \sigma ^n\,{\widetilde{\sigma }}^{n}\,\mathrm {d}+(-1)^{n+1}n\left( (-1)^n{\widetilde{\mathrm {d}}}\,\sigma ^n+(-1)^nn\,\mathrm {d}\,\sigma ^{n-1}\right) {\widetilde{\sigma }}^{n-1} \nonumber \\= & {} \sigma ^n\,{\widetilde{\sigma }}^{n}\,\mathrm {d}+{\widetilde{\mathrm {d}}} Y-n^2\mathrm {d}\,\sigma ^{n-1}\,{\widetilde{\sigma }}^{n-1}, \end{aligned}$$

(A.21)

which directly leads to the recursive formula

$$\begin{aligned} \mathrm {d}\left( \sigma ^n\,{\widetilde{\sigma }}^n+n^2\,\sigma ^{n-1}\,{\widetilde{\sigma }}^{n-1}\right) \omega =\sigma ^n \,{\widetilde{\sigma }}^n\mathrm {d}\omega +{\widetilde{\mathrm {d}}} Y, \end{aligned}$$

as desired. Here $Y=-n\,\sigma ^n\,{\widetilde{\sigma }}^{n-1}\omega$, although it is unnecessary to present explicitly such quantities for our purposes, since they always do not influence the results. $\quad \square$

Proof of Eq. (3.7)

First we define for brevity

$$\begin{aligned} P(n):=\mathrm {d}\,\sigma ^n\,{\widetilde{\sigma }}^n \quad \text {and} \quad Q(n)=\sigma ^n\,{\widetilde{\sigma }}^n\,\mathrm {d}. \end{aligned}$$

(A.22)

Then we write (3.6) for every n as follows

$$\begin{aligned} P(n)+n^2P(n-1)= & {} Q(n)+{\widetilde{\mathrm {d}}} Y_1 \\ -\,n^2\left( P(n-1)+(n-1)^2P(n-2)\right)= & {} -\,n^2Q(n-1)+{\widetilde{\mathrm {d}}} Y_2 \\ +\,n^2(n-1)^2\left( P(n-2)+(n-2)^2P(n-3)\right)= & {} +\,n^2(n-1)^2Q(n-2)+{\widetilde{\mathrm {d}}} Y_3 \\ \vdots \qquad \qquad \qquad= & {} \qquad \qquad \qquad \vdots \\ +\,(-1)^{n-1}n^2(n-1)^2\dots 2^2\left( P(1)+P(0)\right)= & {} +\,(-1)^{n-1}n^2(n-1)^2\dots 2^2 Q(1)+{\widetilde{\mathrm {d}}} Y_{n}\\ +\,(-1)^n(n!)^2 \,P(0)= & {} +\,(-1)^n(n!)^2 \, Q(0)+{\widetilde{\mathrm {d}}} Y_{n+1}. \end{aligned}$$

Summing up these $n+1$ equations, the right hand side yields the sought-after P(n), which turns out to be

$$\begin{aligned} P(n)=Q(n)+\sum _{k=1}^n(-1)^k\prod _{m=0}^{k-1}(n-m)^2\,Q(n-k)+{\widetilde{\mathrm {d}}} Y', \end{aligned}$$

(A.23)

for some calculable $Y'$, which is precisely the relation (3.7). $\quad \square$

Proof of Theorem 3.29

In this second appendix, we provide the missing technical details for the proof of Theorem 3.29. The part that refers to the Proposition 3.15 has already been proven in the main text in full detail. Hence we concentrate in showing how the correct field equations and degrees of freedom for the dual field are obtained in each of the remaining three out of four domains of p and q values of the two-parameter parent Lagrangian. (Recall that the simplest case of the first domain was already fully proven in the main text.)

In what follows, we vary the parent Lagrangian (3.1) w.r.t. the field $F_{p,q}$. This variation is not trivial; we first show that

$$\begin{aligned} \int _{\theta ,\chi }\delta \left( F\star \mathcal {O}F\right) =\int _{\theta ,\chi }\delta F\star \mathcal {O}F+\int _{\theta ,\chi }F\star \mathcal {O}\delta F=2\int _{\theta ,\chi }\delta F\star \mathcal {O}F \end{aligned}$$

(B.1)

for any $F_{p,q}$ and $\mathcal {O}^{(p,q)}$ given by (3.8). Acting on any bipartite tensor of type (p, q), the general identities

$$\begin{aligned} \text {tr}^n*=(-1)^{n(p+1)}*\sigma ^n,\qquad \text {tr}^n{\widetilde{*}} =(-1)^{n(q+1)}\,{\widetilde{*}}\,{\widetilde{\sigma }}^n \end{aligned}$$

(B.2)

follow directly from definition (2.272.28) and are valid for every $n\ge 1$. Since the form of $\mathcal {O}$ depends on whether $p\ge q+1$ or $p<q+1$, (B.1) must be proven in both cases. We refrain from writing the proof for the second case, since all necessary steps and technical manipulations are completely analogous to the first one.

The first step is to observe that $\star \,\mathcal {O}=\mathcal {O}\,\star$ by definition, since both $\sigma$ and ${\widetilde{\sigma }}$ maps contained in $\mathcal {O}$ commute with both $\eta$ and $\text {tr}$ maps contained in $\star$. Thus, for $p\ge q+1$ one computes

$$\begin{aligned} \int _{\theta ,\chi }F\star \mathcal {O}\delta F&=\int _{\theta ,\chi }F\mathcal {O}(\star \delta F) =\int _{\theta ,\chi }\mathcal {O}\left[ *(*\star \delta F)\right] {\widetilde{*}} ({\widetilde{*}} F)\nonumber \\&\overset{(3.8)}{=}\int _{\theta ,\chi }\delta F\star F+\sum _{n=1}^qc_n \int _{\theta ,\chi }{\widetilde{\sigma }}^n\sigma ^n*(*\star \delta F){\widetilde{*}}({\widetilde{*}} F)\nonumber \\&\overset{(\mathrm{A.10})}{=}\int _{\theta ,\chi }\delta F\star F+(-1)^{p(D-p)} \sum _{n=1}^qc_n\int _{\theta ,\chi }\eta ^n\text {tr}^n(*\star \delta F)*{\widetilde{*}}({\widetilde{*}} F)\nonumber \\&\overset{(\mathrm{A.9})}{=}\int _{\theta ,\chi }\delta F\star F+(-1)^{p(D-p)}\sum _{n=1}^qc_n\int _{\theta ,\chi }*\star \delta F*{\widetilde{*}} (\eta ^n\text {tr}^n{\widetilde{*}} F). \end{aligned}$$

(B.3)

Using the second identity in (B.2), (A.8) and (A.9) one gets

$$\begin{aligned} \int _{\theta ,\chi }F\star \mathcal {O}\delta F=\int _{\theta ,\chi }\delta F\star F+(-1)^{q(D-q)}\sum _{n=1}^q c_n(-1)^{np}\int _{\theta ,\chi }{\widetilde{*}}\star \delta F*{\widetilde{*}} (\text {tr}^n*{\widetilde{\sigma }}^nF). \end{aligned}$$

Using now the first identity in (B.2), (A.8) and the integral cyclicity (A.12) we have

$$\begin{aligned} \int _{\theta ,\chi }F\star \mathcal {O}\delta F= & {} \int _{\theta ,\chi }\delta F\star F+\sum _{n=1}^qc_n\int _{\theta ,\chi }\sigma ^n{\widetilde{\sigma }}^nF\star \delta F\nonumber \\= & {} \int _{\theta ,\chi }\mathcal {O}F\star \delta F=\int _{\theta ,\chi }\delta F\star \mathcal {O}F, \end{aligned}$$

(B.4)

which proves (B.1) for the $p\ge q+1$ case. We are now ready to detail the proof of Theorem (3.29) for the rest of parameter domains.

The second domain. The second domain is the one for which $\{p\in [2,D-2], q=1\}$ and it corresponds to the standard dualization of generalized graviton fields. In this case, the parent Lagrangian reads as

$$\begin{aligned} \mathcal {L}_{\text {P}}=\int _{\theta ,\chi }F_{p,1}\star \mathcal {O}^{(p,1)}F_{p,1}+\int _{\theta ,\chi }\mathrm {d}F_{p,1}*{\widetilde{*}}\,\lambda _{p+1,1}~; \end{aligned}$$

(B.5)

varying w.r.t. to F and taking into account (B.1), the following duality relation is readily obtained

$$\begin{aligned} \star \mathcal {O}F=\frac{(-1)^{p+1}}{2}\,{\widetilde{*}}\,\mathrm {d}*\lambda . \end{aligned}$$

(B.6)

Firstly, we will show that the on-shell version of this relation leads to an equation of the form (3.21). More precisely, we will assume the equation $\mathrm {d}F=0$, which comes from varying (B.5) w.r.t. $\lambda$. This is equivalent to demanding that locally $F=\mathrm {d}\omega$, for a reducible field $\omega _{p-1,1}$. According to the defining property (3.3) of $\mathcal {O}$, we get

$$\begin{aligned} \star \mathrm {d}[\omega ]=\frac{(-1)^{p+1}}{2}\,\widetilde{*}\,\mathrm {d}*\lambda -\star \,\widetilde{\mathrm {d}}X, \end{aligned}$$

(B.7)

where $[\omega ]:=\omega _{[p-1,1]}$ and $X:=X_p$ is an arbitrary p-form. Using (2.20), the above relation takes the form

$$\begin{aligned} \mathrm {d}[\omega ]-\eta \,\text {tr}\mathrm {d}[\omega ]=\frac{(-1)^{\epsilon (p,1)+1}}{2}\,\mathrm {d}^\dagger \lambda -\widetilde{\mathrm {d}}X+\eta \,\text {tr}\widetilde{\mathrm {d}}\,X . \end{aligned}$$

(B.8)

Tracing both sides of this relation will give

$$\begin{aligned} \text {tr}\mathrm {d}[\omega ]=\frac{(-1)^{\epsilon (p,1)}}{2(D-p)}\,\text {tr}\mathrm {d}^\dagger \lambda -\text {tr}\widetilde{\mathrm {d}}X \end{aligned}$$

(B.9)

and, by substituting this back into (B.8), one finds

$$\begin{aligned} \mathrm {d}[\omega ]=\frac{(-1)^{\epsilon (p,1)+1}}{2}\left( \mathbb {I}-\frac{1}{D-p}\,\eta \,\text {tr}\right) \mathrm {d}^\dagger \lambda -\widetilde{\mathrm {d}}X. \end{aligned}$$

(B.10)

This is the on-shell duality relation. Decomposing $\lambda$ as in (3.20) and acting on (B.10) with $\widetilde{\mathrm {d}}$, a little algebra leads to the relation (3.21).

Let us now proceed in finding the dual action. Applying (2.20) to ${{\mathcal {O}}}F$, relation (B.6) may be rewritten as

$$\begin{aligned} \mathcal {O}(F-\eta \,\text {tr}F)=\frac{(-1)^{\epsilon (p,1)+1}}{2}\,\mathrm {d}^\dagger \lambda , \end{aligned}$$

(B.11)

where we used that $\mathcal {O}$ contains only the maps $\sigma$ and ${\widetilde{\sigma }}$, which commute with both $\eta$ and $\text {tr}$ by virtue of identities (A.2). Next, the duality relation should be solved for F, which requires the inverse operator $\mathcal {O}^{-1}$, namely the one that satisfies $\mathcal {O}^{-1}\mathcal {O}=\mathcal {O}\mathcal {O}^{-1}=\mathbb {I}$ on any bipartite tensor. Rather than proving the invertibility of ${{\mathcal {O}}}$ abstractly, we construct its inverse using direct methods. In the present case, $p\ge q+1=2$ and one finds

$$\begin{aligned} (\mathcal {O}^{(p,1)})^{-1}=\mathbb {I}-{\widetilde{\sigma }}\,\sigma . \end{aligned}$$

(B.12)

Acting with it on both sides of (B.11) one gets

$$\begin{aligned} F-\eta \,\text {tr}F=\frac{(-1)^{\epsilon (p,1)+1}}{2}(\mathbb {I}-{\widetilde{\sigma }}\sigma )\,\mathrm {d}^\dagger \lambda . \end{aligned}$$

(B.13)

Next, taking the trace of (B.13) and using the second commutation relation in (A.2), one can determine $\text {tr}F_{p,1}$ in terms of $\lambda _{p+1,1}$. Inserting it back in (B.13) one finds

$$\begin{aligned} F=\frac{(-1)^{\epsilon (p,1)+1}}{2}\left( \mathbb {I}-{\widetilde{\sigma }}\sigma -\frac{1}{D-p}\,\eta \,\text {tr}\right) \mathrm {d}^\dagger \lambda . \end{aligned}$$

(B.14)

Substituting (B.6) and (B.14) into (B.5), the dual Lagrangian (3.19) in terms of $\lambda$ is obtained as

$$\begin{aligned} \mathcal {L}_{\text {dual}}(\lambda )=\frac{(-1)^{\epsilon (p,1)+1}}{4}\int _{\theta ,\chi }\left( \mathbb {I}-{\widetilde{\sigma }}\sigma -\frac{1}{D-p}\,\eta \,\text {tr}\right) \mathrm {d}^\dagger \lambda *{\widetilde{*}}\,\mathrm {d}^\dagger \lambda . \end{aligned}$$

(B.15)

Now we decompose the Lagrange multiplier as in (3.20) and introduce the two independent fields ${\widehat{\omega }}_{[D-p-1,1]}$ and $\mathring{\lambda }_{p,0}$. Using the second of (A.2), (A.5d) and (A.5f) one can easily show that the first factor in the above integrand only depends on ${\widehat{\omega }}$, i.e.

$$\begin{aligned} \mathcal {L}_{\text {dual}}(\widehat{\omega },\mathring{\lambda })= & {} \frac{(-1)^{\epsilon (p,1)+1}}{4}\int _{\theta ,\chi }\left\{ \left( \mathbb {I}-{\widetilde{\sigma }}\sigma -\frac{1}{D-p}\,\eta \,\text {tr}\right) \mathrm {d}^\dagger *\widehat{\omega }\right\} *{\widetilde{*}}\,\mathrm {d}^\dagger (*\,\widehat{\omega }+\eta \mathring{\lambda })\nonumber \\= & {} \frac{(-1)^{\epsilon (p,1)+1}}{4}\int _{\theta ,\chi }\left\{ \left( \mathbb {I}-{\widetilde{\sigma }}\sigma \right) \mathrm {d}^\dagger *\widehat{\omega }\right\} *{\widetilde{*}}\,\mathrm {d}^\dagger (*\,\widehat{\omega }+\eta \mathring{\lambda }), \end{aligned}$$

(B.16)

where in the second line we used identity (A.5c) and that $\text {tr}*\widehat{\omega }_{[D-p-1,1]}=0$ due to the tracelessness of $\widehat{\lambda }$. Finally, using the integral identity (A.9), the identities (A.5e)–(A.5f) and the definitions (2.272.28)–(2.252.26), one finds that the $\mathring{\lambda }$-dependence in the dual Lagrangian cancels algebraically. The result is then

$$\begin{aligned} \mathcal {L}_{\text {dual}}(\widehat{\omega })&=\frac{(-1)^{\epsilon (p,1)+p(D-p)}}{4}\int _{\theta ,\chi }\left( \mathbb {I}-{\widetilde{\sigma }}\sigma \right) *\mathrm {d}\,\widehat{\omega }\,{\widetilde{*}}\,\mathrm {d}\,\widehat{\omega }\nonumber \\&=\frac{(-1)^{\epsilon (p,1)}}{4}\int _{\theta ,\chi }\mathrm {d}\,\widehat{\omega }*{\widetilde{*}}\left( \mathrm {d}\,\widehat{\omega }-\eta \,\text {tr}\,\mathrm {d}\,\widehat{\omega }\right) =\frac{1}{4}\int _{\theta ,\chi }\mathrm {d}\,\widehat{\omega }\star \mathrm {d}\,\widehat{\omega }, \end{aligned}$$

(B.17)

where the three integral identities (A.8), (A.9) and (A.10) were used. Finally, we note that the original field $\omega _{[p-1,1]}$ and the dual field ${\widehat{\omega }}_{[D-p-1,1]}$ have the same number of components as $SO(D-2)$ representations, since

$$\begin{aligned}&\left( {\begin{array}{c}D-2\\ p-1\end{array}}\right) \left( {\begin{array}{c}D-1\\ 1\end{array}}\right) (1-\frac{1}{p})-\left( {\begin{array}{c}D-2\\ p-2\end{array}}\right) \nonumber \\&\quad =\left( {\begin{array}{c}D-2\\ D-p-1\end{array}}\right) \left( {\begin{array}{c}D-1\\ 1\end{array}}\right) (1-\frac{1}{D-p})-\left( {\begin{array}{c}D-2\\ D-p-2\end{array}}\right) .\nonumber \\ \end{aligned}$$

(B.18)

This implies directly that in the physical gauge the number of field equations for the two fields is the same. As discussed in the main text, this establishes Theorem 3.29 for this domain of parameters.

The third domain. Now $\{p=1, q\in [1,D-2]\}$ and this domain corresponds to the exotic dualization of a q-form field. The parent Lagrangian reads in this case as

$$\begin{aligned} \mathcal {L}_{\text {P}}=\int _{\theta ,\chi }F_{1,q}\star F_{1,q}+\int _{\theta ,\chi }\mathrm {d}F_{1,q}*{\widetilde{*}}\lambda _{2,q} \end{aligned}$$

(B.19)

and the duality relation obtained from the variation w.r.t. F can be brought to the form

$$\begin{aligned} F=\frac{(-1)^{\epsilon }}{2}\left( \mathbb {I}-\frac{1}{D-q}\,\eta \,\text {tr}\right) \mathrm {d}^\dagger \lambda . \end{aligned}$$

(B.20)

One can easily prove that the on-shell version of this relation leads to (3.25). First, we decompose $\lambda$ as in (3.24) and consider that locally $F=\mathrm {d}\omega$, $\omega _{0,q}$ being a q-form. Acting with $\widetilde{\mathrm {d}}$ on the resulting relation will give

$$\begin{aligned} 2(-1)^{\epsilon (1,q)}\widetilde{\mathrm {d}}\,\mathrm {d}\,\omega =\widetilde{\mathrm {d}}\mathrm {d}^\dagger *{\widehat{\omega }}+\widetilde{\mathrm {d}}\mathrm {d}^\dagger \eta \mathring{\lambda }+\frac{1}{D-q}\,\widetilde{\mathrm {d}}\,\eta \,\mathrm {d}^\dagger \text {tr}\,\eta \mathring{\lambda }. \end{aligned}$$

(B.21)

The $\mathring{\lambda }$-dependent terms cancel each other, as can be seen by using suitable identities from Appendix A. Thus, the on-shell duality relation reads as

$$\begin{aligned} \widetilde{\mathrm {d}}\,\mathrm {d}\,\omega =\frac{(-1)^{qD}}{2}\,*\,\widetilde{\mathrm {d}}\,\mathrm {d}\,{\widehat{\omega }}, \end{aligned}$$

(B.22)

which is precisely (3.25).

The dual Lagrangian (3.23) in terms of $\lambda$ is obtained by substitution of F,

$$\begin{aligned} \mathcal {L}_{\text {dual}}[\lambda ]=\frac{3(-1)^{\epsilon }}{8}\int _{\theta ,\chi }\left( \mathbb {I}-\frac{1}{D-q}\eta \,\text {tr}\right) \mathrm {d}^\dagger \lambda *{\widetilde{*}}\,\mathrm {d}^\dagger \lambda . \end{aligned}$$

(B.23)

In contrast to the previous case, this Lagrangian depends explicitly on the trace part $\mathring{\lambda }_{1,q-1}$ of the Lagrange multiplier. However, one should obtain equations of motion only for the dual field $\widehat{\omega }_{[D-2,q]}$ such that the duality of the two theories is justified, as explained in the main text. Varying (B.23) w.r.t. $\lambda$ one obtains

$$\begin{aligned} \mathrm {d}\left( \mathbb {I}-\frac{1}{D-q}\,\eta \,\text {tr}\right) \mathrm {d}^{\dagger }\lambda =0. \end{aligned}$$

(B.24)

After the decomposition (3.24), and using the fact that $\text {tr}*\mathrm {d}\widehat{\omega }_{[D-2,q]}\propto *\mathrm {d}\sigma \, \widehat{\omega }_{[D-2,q]}=0$, which follows from (2.272.28) and (A.5a), one obtains

$$\begin{aligned} \mathrm {d}*\mathrm {d}{\widehat{\omega }}-\mathrm {d}\mathrm {d}^\dagger \eta \mathring{\lambda }+\frac{1}{D-q}\mathrm {d}\eta \,\text {tr}\,\mathrm {d}^\dagger \eta \mathring{\lambda }=0. \end{aligned}$$

(B.25)

Acting now with $\text {tr}^q*$ on both sides, it is observed that the second and third terms vanish due to the first identity in (B.2). Then the second term in (B.25) is proportional to $\mathrm {d}\sigma ^q\mathrm {d}^\dagger \eta \mathring{\lambda }_{1,q-1}$, which vanishes as may be shown using the identities (A.5f), (A.5d) and (A.2). For the third term in (B.25), one can see that it is proportional to $\sigma ^q\,\text {tr}\,\mathrm {d}^\dagger \eta \mathring{\lambda }_{1,q-1}$ by using the identities (A.5a) and (A.2). Again, this expression is zero.

Thus, the resulting equations of motion for the dual field ${\widehat{\omega }}$ become

$$\begin{aligned} \text {tr}^{q}\,\mathrm {d}^\dagger \mathrm {d}\,\widehat{\omega }_{[D-2,q]}=0 \end{aligned}$$

(B.26)

and we will now show their equivalence to (3.28). Indeed,

(B.27)

where we also used that ${\widetilde{\mathrm {d}}}\text {tr}^{q+1}\mathrm {d}\widehat{\omega }_{[D-2,q]}=0$. The effect of the q traces is that after complete gauge fixing, the number of independent field equation becomes simply $\left( {\begin{array}{c}D-2\\ D-q-2\end{array}}\right)$, which is equal to the ones for a q-form, as required. As explained in the main text, this proves Theorem 3.29 for these values of parameters.

The fourth domain. For the fourth domain $\{p=2, q\in [2,D-3]\}$, the parent Lagrangian reads as

$$\begin{aligned} \mathcal {L}_{\text {P}}=\int _{\theta ,\chi }F_{2,q}\star \mathcal {O}^{(2,q)}F_{2,q}+\int _{\theta ,\chi }\mathrm {d}F_{2,q}*{\widetilde{*}}\lambda _{3,q} \end{aligned}$$

(B.28)

and the duality relation after varying w.r.t. to F is

$$\begin{aligned} \begin{aligned} \star \mathcal {O}F=-\frac{1}{2}{\widetilde{*}}\,\mathrm {d}*\lambda . \end{aligned} \end{aligned}$$

(B.29)

Using (2.20) we can expand $\star \,\mathcal {O}F=(-1)^{\epsilon }*{\widetilde{*}}(\mathcal {O}F-\eta \,\text {tr}\,\mathcal {O}F+\frac{1}{4}\,\eta ^2\,\text {tr}^2\,\mathcal {O}F)$ and rewrite the above relation as

$$\begin{aligned} \mathcal {O}\left( F-\eta \,\text {tr}\,F+\frac{1}{4}\,\eta ^2\,\text {tr}^2\,F\right) =\frac{(-1)^{\epsilon +1}}{2}\,\mathrm {d}^\dagger \lambda , \end{aligned}$$

(B.30)

since $\mathcal {O}$ commutes both with $\eta$ and $\text {tr}$. Now one has to determine the inverse $\mathcal {O}^{-1}_{(2,q)}$ of $\mathcal {O}^{(2,q)}$. In the present case we have $p=2<q+1$ and we find

$$\begin{aligned} \mathcal {O}^{-1}_{(2,q)}=b_1\,\mathbb {I}+b_2\,\sigma \,{\widetilde{\sigma }}+b_3\,\sigma ^2\,{\widetilde{\sigma }}^2, \end{aligned}$$

(B.31)

with coefficients given by

$$\begin{aligned} b_1\equiv \frac{q+1}{q+2},\qquad b_2\equiv \frac{q+1}{2(q+2)},\qquad b_3\equiv -\frac{q+1}{2q(q+2)}. \end{aligned}$$

(B.32)

Acting with this $\mathcal {O}^{-1}$ on both sides of (B.30) yields

$$\begin{aligned} F-\eta \,\text {tr}F+\frac{1}{4}\,\eta ^2\,\text {tr}^2F=\frac{(-1)^{\epsilon +1}}{2}\,\mathcal {O}^{-1}\mathrm {d}^\dagger \lambda . \end{aligned}$$

(B.33)

One can then act with $\text {tr}^2$ on both sides of (B.33) and repeatedly use the second relation in (A.2) to find

$$\begin{aligned} \text {tr}^2F=\frac{(-1)^{\epsilon +1}}{(D-q)(D-q-1)}\,\mathcal {O}^{-1}\text {tr}^2\,\mathrm {d}^\dagger \lambda . \end{aligned}$$

(B.34)

After substituting (B.34) back into (B.33), one can then act with $\text {tr}$ on both sides of the resulting equation. Using again the second in (A.2), one finds

$$\begin{aligned} \text {tr}F=\frac{(-1)^{\epsilon }}{2(D-q-1)}\,\mathcal {O}^{-1}\left( \mathbb {I}-\frac{1}{D-q}\,\eta \,\text {tr}\right) \text {tr}\,\mathrm {d}^\dagger \lambda . \end{aligned}$$

(B.35)

Finally, plugging both (B.34) and (B.35) into (B.33) yields

$$\begin{aligned} F=\frac{(-1)^{\epsilon +1}}{2}\,\mathcal {O}^{-1}\left( \mathbb {I}-\frac{1}{D-q-1}\,\eta \,\text {tr}-\frac{1}{2(D-q)(D-q-1)}\,\eta ^2\,\text {tr}^2\right) \mathrm {d}^\dagger \lambda .\nonumber \\ \end{aligned}$$

(B.36)

The dual Lagrangian in terms of $\lambda$ is then readily obtained by substituting (B.29) and (B.36) into (B.28). In the standard spirit of the dualization procedure, one can then decompose the Lagrange multiplier into traceless and trace parts

$$\begin{aligned} \lambda _{3,q}={\widehat{\lambda }}_{3,q}+\eta \mathring{\lambda }_{2,q-1} \end{aligned}$$

(B.37)

and define the GL(D)-irreducible dual field $\widehat{\omega }_{[D-3,q]}$ to be the partial Hodge dual of the traceless part $\widehat{\lambda }_{3,q}$, i.e. $\widehat{\lambda }=*\,\widehat{\omega }$. Not surprisingly, the $\mathring{\lambda }$-dependence in the dual Lagrangian cannot be algebraically canceled—c.f. Ref. [4], where this is discussed for the special case of $q=D-3$. This is a common characteristic between the double standard dualization of generalized graviton fields and the exotic dualization of differential forms, since this was also the case in our previous analysis for the third domain of parameters. The logic remains the same, in that the Lagrangian contains additional off shell fields, whose elimination on shell leads to equations of motion for the dual field only.

The equations of motion obtained by variation of the dual Lagrangian w.r.t. $\lambda$ read as

$$\begin{aligned} \begin{aligned} \mathrm {d}\mathcal {O}^{-1}\left( \mathbb {I}-\frac{1}{D-q-1}\,\eta \ \text{ tr }-\frac{1}{2(D-q)(D-q-1)}\,\eta ^2\text{ tr }^2\right) \mathrm {d}^\dagger \lambda =0. \end{aligned} \end{aligned}$$

(B.38)

Acting now on both sides with $\text {tr}^{q-1}*$ and using (B.2) and (A.5a), the above equation can be rewritten as

$$\begin{aligned} \mathrm {d}\sigma ^{q-1}\,\mathcal {O}^{-1}\left( \mathbb {I}-\frac{1}{D-q-1}\,\eta \,\text {tr}-\frac{1}{2(D-q)(D-q-1)}\,\eta ^2\,\text {tr}^2\right) \mathrm {d}^\dagger \lambda =0.\nonumber \\ \end{aligned}$$

(B.39)

Using the decomposition (B.37) and the map identities (A.2), (A.5e) and (A.5f), we then compute

$$\begin{aligned} \eta \,\text {tr}\,\mathrm {d}^\dagger \lambda =-\eta \,{\widetilde{\mathrm {d}}}\,\text {tr}\mathring{\lambda }-(D-q-1)\,\eta \,\mathrm {d}^\dagger \mathring{\lambda }-\eta ^2\,\text {tr}\,\mathrm {d}^\dagger \mathring{\lambda } \end{aligned}$$

(B.40)

and

$$\begin{aligned} \eta ^2\,\text {tr}^2\,\mathrm {d}^\dagger \lambda =\eta ^2\,\text {tr}^2\,{\widetilde{\mathrm {d}}}\mathring{\lambda }-2(D-q+1)\,\eta ^2\,\text {tr}\,\mathrm {d}^\dagger \mathring{\lambda }. \end{aligned}$$

(B.41)

Plugging these expressions into (B.39) we obtain

$$\begin{aligned} \mathrm {d}\sigma ^{q-1}\mathcal {O}^{-1}\mathrm {d}^\dagger {\hat{\lambda }}+\mathrm {d}\sigma ^{q-1}\mathcal {O}^{-1}{\widetilde{\mathrm {d}}}\mathring{\lambda }=0, \end{aligned}$$

(B.42)

since, using map identities from Appendix A and the inverse operator $\mathcal {O}^{-1}$ in (B.31), all additional terms

$$\begin{aligned} \mathrm {d}\sigma ^{q-1}\mathcal {O}^{-1}\eta \,{\widetilde{\mathrm {d}}}\text {tr}\mathring{\lambda },\,\,\,\,\,\mathrm {d}\sigma ^{q-1} \mathcal {O}^{-1}\eta ^2\text {tr}^2{\widetilde{\mathrm {d}}}\mathring{\lambda }\qquad \text {and}\qquad \mathrm {d}\sigma ^{q-1}\mathcal {O}^{-1}\eta ^2\text {tr}\mathrm {d}^\dagger \mathring{\lambda } \end{aligned}$$

(B.43)

vanish identically. Using now the identity (A.5d) and the nilpotency property of the exterior derivatives, one can write (B.42) as

$$\begin{aligned} \mathrm {d}\sigma ^{q-1}\mathcal {O}^{-1}\mathrm {d}^\dagger {\widehat{\lambda }}=(-1)^q\mathrm {d}{\widetilde{\mathrm {d}}}\left( b_1\sigma ^{q-1}\mathring{\lambda }+b_2\,\sigma ^q{\widetilde{\sigma }}\mathring{\lambda }+b_3\,\sigma ^{q+1}{\widetilde{\sigma }}^2\mathring{\lambda }\right) \end{aligned}$$

(B.44)

for $b_1,b_2$ and $b_3$ given by (B.32). Then, we can use the first map identity (A.2) successively to obtain

$$\begin{aligned} \sigma ^q\,{\widetilde{\sigma }}\mathring{\lambda }=2q\,\sigma ^{q-1}\mathring{\lambda }\qquad \text {and}\qquad \sigma ^{q+1}{\widetilde{\sigma }}^2\mathring{\lambda }=2q(q+1)\sigma ^{q-1}\mathring{\lambda }, \end{aligned}$$

(B.45)

which can then be used to transform (B.44) into

$$\begin{aligned} \mathrm {d}\sigma ^{q-1}\mathcal {O}^{-1}\mathrm {d}^\dagger {\widehat{\lambda }}=(-1)^q\mathrm {d}{\widetilde{\mathrm {d}}}\left( b_1+2qb_2+2q(q+1)b_3\right) \sigma ^{q-1}\mathring{\lambda }. \end{aligned}$$

(B.46)

However, the b-coefficients given by (B.32) satisfy the algebraic equation

$$\begin{aligned} b_1+2q\,b_2+2q(q+1)\,b_3=0 \end{aligned}$$

(B.47)

and therefore the $\mathring{\lambda }$-dependence in the equations of motion cancels completely. Thus we are left with

$$\begin{aligned} \mathrm {d}\sigma ^{q-1}\mathcal {O}^{-1}\mathrm {d}^\dagger {\widehat{\lambda }}=0. \end{aligned}$$

(B.48)

As before, we can make a successive use of the first map identity in (A.2) again to find

$$\begin{aligned} \begin{aligned} \sigma ^q\,{\widetilde{\sigma }}\,\mathrm {d}^\dagger \widehat{\lambda }=q\,\sigma ^{q-1}\mathrm {d}^\dagger {\widehat{\lambda }}\qquad \text{ and }\qquad \sigma ^{q+1}{\widetilde{\sigma }}^2\mathrm {d}^\dagger {\widehat{\lambda }}=0. \end{aligned}\end{aligned}$$

(B.49)

Using these expressions, one can rewrite (B.48) as

$$\begin{aligned} (b_1+q\,b_2)\,\mathrm {d}\sigma ^{q-1}\mathrm {d}^\dagger {\widehat{\lambda }}=0\qquad \Leftrightarrow \qquad \mathrm {d}\sigma ^{q-1}\mathrm {d}^\dagger {\widehat{\lambda }}=0. \end{aligned}$$

(B.50)

These correspond exactly to the expected field equations

$$\begin{aligned} \text {tr}^{q-1}\mathrm {d}^\dagger \mathrm {d}{\widehat{\omega }}=0\qquad \Leftrightarrow \qquad \text {tr}^q\mathrm {d}{\widetilde{\mathrm {d}}}{\widehat{\omega }}=0 \end{aligned}$$

(B.51)

for the GL(D)-irreducible dual field ${\widehat{\omega }}_{[D-3,q]}$ defined by ${\widehat{\lambda }}\equiv *\,{\widehat{\omega }}$. Note that the above equation correspondence can be proven in the same way as (B.27).

Just as in the case of exotic dualization of q-forms (third domain), here one can again define a Riemann-like tensor $R_{[D-2,q+1]}:= \mathrm {d}{\widetilde{\mathrm {d}}}\,{\widehat{\omega }}_{[D-3,q]}$, as in the main text. The equations of motion $\text {tr}^{n}R=0$ will now imply the vanishing of the whole tensor R for any $n<q$, thus the correct equations for the gauge field ${\widehat{\omega }}$ will be $\text {tr}^{q}R=0$. These are exactly the equations (B.51) found above. Obviously, the effect of taking the q traces is that the number of independent field equations for the dual field after complete gauge fixing is the same as the ones for the original irreducible field $\omega _{[1,q]}$, as required.