A. No-go result for $I=5$ in $a^{\mathrm {int}}$

In agreement with (86), the general solution to the equation $sa^{ \mathrm{int}}=\partial ^{\mu }m_{\mu }^{\mathrm{int}}$ can be chosen to stop at antighost number $I=5$

$$a^{\mathrm{int}}=a_{0}^{\mathrm{int}}+a_{1}^{\mathrm{int}}+a_{2}^{\mathrm{int }}+a_{3}^{\mathrm{int}}+a_{4}^{\mathrm{int}}+a_{5}^{\mathrm{int}},$$ (220)

where the components on the right-hand side of (181) are subject to the equations (68) and (66)-(67) for $I=5$.

The piece $a_{5}^{\mathrm{int}}$ as solution to equation (68) for $I=5$ has the general form expressed by (75) for $I=5$, with $\alpha _{5}$ from $H_{5}^{ \mathrm{inv}}(\delta \vert d)$. According to (81) at antighost number five, it follows that $H_{5}^{ \mathrm{inv}}(\delta \vert d)$ is spanned by the generic representatives (82). Since $a_{5}^{\mathrm{int}}$ should effectively mix the BF and the $(2,1)$ tensor field sectors in order to produce cross-couplings and (82) involves only BF generators, it follows that one should retain from the basis elements $e^{5}\left( \eta ^{\bar{\Upsilon}}\right)$ only the objects containing at least one ghost from the $(2,1)$ tensor field sector, namely $D_{\mu \nu \rho }$ or $S_{\mu }$. Recalling that we work precisely in $D=5$, we obtain that the general solution to (68) for $I=5$ reduces to

$$a_{5}^{\mathrm{int}} = \tfrac{1}{3!}\left( \left( \tilde{U}_{1}\right) C+\left( \tilde{U... ...alpha }\tilde{D}^{\alpha \beta }\tilde{D}_{\beta \nu }\sigma ^{\mu \nu } \notag$$ (221)

$$+\tfrac{1}{2}\left( \left( \tilde{U}_{3}\right) \eta S^{\mu }-\le... ...D}_{\nu \alpha }\tilde{D}_{\rho \beta }\sigma ^{\alpha \beta }\right) S_{\mu }.$$ (222)

Each tilde object from the right-hand side of (182) means the Hodge dual of the corresponding non-tilde element, defined in general by formula ( 92). The elements $\tilde{U}$ are dual to $\left( U\right) _{\mu _{1}\ldots \mu _{5}}$ as in (82), with $W\left( \varphi \right)$ respectively replaced by the smooth function $U\left( \varphi \right)$ depending only on the undifferentiated scalar field $\varphi$.

Introducing (182) in equation (66) for $I=5$ and recalling definitions (35)-(52), we obtain

$$a_{4}^{\mathrm{int}} = -\tfrac{1}{6}\tilde{D}_{\mu \alpha }\tilde{D} ^{\alpha \beta }\si... ...a \nu }+\tfrac{3}{2}C\tilde{F} _{\beta \nu \vert\lambda }\right) \right. \notag$$ (223)

$$\left. -\left( \tilde{U}_{2}\right) ^{\lambda }\left( \tfrac{1}{5... ...ft( V_{\lambda }S_{\mu }+\eta \mathcal{C}_{\lambda \mu }\right) S^{\mu } \notag$$ (224)

$$-\tfrac{1}{4}\left( \tilde{U}_{4}\right) ^{\lambda }\left[ D^{\mu... ...{\rho \beta \vert\lambda }\sigma ^{\alpha \beta }S_{\mu }\right) \right. \notag$$ (225)

$$\left. -F^{\mu \nu \rho \vert\gamma }\tilde{D}_{\nu \alpha }\tild... ... ^{\alpha \beta }\sigma _{\gamma \lambda }\right] +\bar{ a}_{4}^{\mathrm{int}}.$$ (226)

In (183) $\left( \tilde{U}\right) ^{\lambda }$ are dual to (83 ), with $W\left( \varphi \right) \rightarrow U\left( \varphi \right)$. In addition, $\mathcal{C}_{\mu \rho }$ is implicitly defined by formula (74) so it is a ghost field of pure ghost number one without definite symmetry/antisymmetry property, $\mathcal{C}^{\ast \nu \lambda }$ is its associated antifield, defined such that the antibracket $\left( \mathcal{C} _{\mu \rho },\mathcal{C}^{\ast \nu \lambda }\right)$ is equal to the `unit' $\delta _{\mu }^{\nu }\delta _{\rho }^{\lambda }$

$$\mathcal{C}^{\ast \nu \lambda }\equiv 3S^{\ast \nu \lambda }+A^{\ast \nu \lambda }.$$ (227)

The nonintegrated density $\bar{a}_{4}^{\mathrm{int}}$ stands for the solution to the homogeneous equation (68) for $I=4$, showing that $\bar{a}_{4}^{\mathrm{int}}$ can be taken as a nontrivial element of $H\left( \gamma \right)$ in pure ghost number equal to four.

At this stage it is useful to decompose $\bar{a}_{4}^{\mathrm{int}}$ as a sum between two components

$$\bar{a}_{4}^{\mathrm{int}}=\hat{a}_{4}^{\mathrm{int}}+\check{a}_{4}^{\mathrm{ int}},$$ (228)

where $\hat{a}_{4}^{\mathrm{int}}$ is the solution to (68) for $I=4$ which is explicitly required by the consistency of $a_{4}^{\mathrm{int}}$ in antighost number three (ensures that (67) possesses solutions for $i=4$ with respect to the terms from (183) containing the functions of the type $U$) and $\check{a}_{4}^{\mathrm{int}}$ signifies the part of the solution to (68) for $I=4$ that is independently consistent in antighost number three

$$\delta \check{a}_{4}^{\mathrm{int}}=-\gamma \check{c}_{3}+\partial _{\mu } \check{m}_{3}^{\mu }.$$ (229)

Using definitions (35)-(52) and decomposition (185 ), by direct computation we obtain that

$$\delta a_{4}^{\mathrm{int}} = \delta \left[ \hat{a}_{4}^{\mathrm{int}}- \tfrac{1}{2}S^{\alpha }... ...u \nu \rho \lambda }\eta _{\mu \nu \rho \lambda }^{\ast }\right. \right. \notag$$ (230)

$$\left. \left. +\tfrac{1}{60}\left( \tilde{U}_{3}\right) ^{\mu \nu... ... }^{\ast }\right) \right] +\gamma c_{3}+\partial _{\mu }j_{3}^{\mu }+\chi _{3},$$ (231)

where we made the notations

$$c_{3} = -\check{c}_{3}+\tfrac{1}{12}\left( \tilde{U}_{1}\right) ^{\lambda... ... \vert\lambda }^{\rho \alpha }\tilde{F}_{\alpha \nu \vert\sigma }\right] \notag$$ (232)

$$-\tfrac{1}{240}\left( \tilde{U}_{2}\right) ^{\lambda \sigma }\til... ... \vert\lambda }^{\rho \alpha }\tilde{F}_{\alpha \nu \vert\sigma }\right] \notag$$ (233)

$$-\tfrac{1}{12}\left( \tilde{U}_{3}\right) ^{\lambda \sigma }\left... ...al{C}_{\lambda \rho }\mathcal{C}_{\sigma \mu }\sigma ^{\rho \mu }\right] \notag$$ (234)

$$-\tfrac{1}{2}\left( \left( \tilde{U}_{3}\right) ^{\mu \nu \sigma ... ...nu \rho \lambda }^{\ast }\right) S^{\alpha }\mathcal{C}_{\sigma \alpha } \notag$$ (235)

$$-\tfrac{1}{24}\left( \tilde{U}_{4}\right) ^{\lambda \sigma }\sigm... ...lpha }\tilde{D}_{\rho \beta }t_{\lambda \sigma \vert\mu }\right) \right. \notag$$ (236)

$$\left. +3F_{\qquad \vert\lambda }^{\mu \nu \rho }\tilde{D}_{\nu \... ...{C}_{\sigma \mu }-2\tilde{F}_{\rho \beta \vert\sigma }S_{\mu }\right) \right] ,$$ (237)

$$\chi _{3} = -\tfrac{1}{4}\left( \left( \tilde{U}_{1}\right) ^{\lambda \sigma ... ...ha }\tilde{D}^{\alpha \beta } \tilde{R}_{\beta \nu \vert\lambda \sigma } \notag$$ (238)

$$+\tfrac{1}{6}\left( \tilde{U}_{3}\right) ^{\mu \nu }\eta S^{\rho ... ...\nu \rho }\tilde{D} _{\nu \alpha }\tilde{D}_{\rho \beta }S_{\mu }\right. \notag$$ (239)

$$\left. +D^{\mu \nu \rho }\tilde{D}_{\nu \alpha }\left( \tilde{D}_... ...gma \mu }-6\tilde{R}_{\rho \beta \vert\lambda \sigma }S_{\mu }\right) \right] ,$$ (240)

and $j_{3}^{\mu }$ are some local currents. In (187)-(189) $\left( \tilde{U}\right) ^{\mu \nu }$ and $\left( \tilde{U}\right) ^{\mu \nu \rho }$ denote the duals of (84) and (85) with $W\left( \varphi \right) \rightarrow U\left( \varphi \right)$. In addition, $\left( \tilde{U}\right) ^{\mu \nu \rho \lambda }$ represents the dual of $\left( U\right) _{\mu }=\frac{dU}{d\varphi }H_{\mu }^{\ast }$ and $\left( \tilde{U} \right) ^{\mu \nu \rho \lambda \sigma }$ the dual of $U\left( \varphi \right)$. Inspecting (187), it follows that the consistency of $a_{4}^{\mathrm{int}}$ in antighost number three, namely the existence of $a_{3}^{\mathrm{int}}$ as solution to (67) for $i=4$, requires the conditions

$$\chi _{3}=\gamma \hat{c}_{3}+\partial _{\mu }\hat{\jmath}_{3}^{\mu }$$ (241)

and

$$\hat{a}_{4}^{\mathrm{int}} = \tfrac{1}{2}S^{\alpha }S_{\alpha }\left( \left( \tilde{U}_{3}\rig... ...tilde{U}_{3}\right) ^{\mu \nu \rho }\eta _{\mu \nu \rho }^{\ast }\right. \notag$$ (242)

$$\left. +\tfrac{1}{12}\left( \tilde{U}_{3}\right) ^{\mu \nu \rho \... ... \nu \rho \lambda \sigma }\eta _{\mu \nu \rho \lambda \sigma }^{\ast }\right) ,$$ (243)

where we made the notations $\hat{c}_{3}=-\left( a_{3}^{\mathrm{int} }+c_{3}\right)$ and $\hat{\jmath}_{3}^{\mu }=\overset{(3)}{m}^{\mathrm{int~} \mu }-j_{3}^{\mu }$. Nevertheless, from (189) it is obvious that $\chi _{3}$ is a nontrivial element from $H\left( \gamma \right)$ in pure ghost number four, which does not reduce to a full divergence, and therefore (190) requires that $\chi _{3}=0$, which further imply that all the functions of the type $U$ must be some real constants

$$U_{1}\left( \varphi \right) =u_{1},\qquad U_{2}\left( \varphi \ri... ...d U_{3}\left( \varphi \right) =u_{3},\qquad U_{4}\left( \varphi \right) =u_{4}.$$ (244)

Based on (192), it is clear that $a_{5}^{\mathrm{int}}$ given by ( 182) vanishes, and hence we can assume, without loss of nontrivial terms, that

$$a_{5}^{\mathrm{int}}=0$$ (245)

in (181).