On sufficient conditions for degrees of freedom counting of multi-field generalised Proca theories

Abstract

We derive conditions which are sufficient for theories consisting of multiple vector fields, which could also couple to non-dynamical external fields, to have the required structure of constraints of multi-field generalised Proca theories, so that the number of degrees of freedom is correct. The Faddeev–Jackiw constraint analysis is used and is cross-checked by Lagrangian constraint analysis. To ensure the theory is constraint, we impose a standard special form of Hessian matrix. The derivation benefits from the realisation that the theories are diffeomorphism invariance. The sufficient conditions obtained include a refinement of secondary-constraint enforcing relations derived previously in literature, as well as a condition which ensures that the iteration process of constraint analysis terminates. Some examples of theories are analysed to show whether they satisfy the sufficient conditions. Most notably, due to the obtained refinement on some of the conditions, some theories which are previously interpreted as being undesirable are in fact legitimate, and vice versa. This in turn affects the previous interpretations of cosmological implications which should later be reinvestigated.

Appendices

Appendix A: Conditions from diffeomorphism invariance

In this appendix, we consider a class of theories described in Sect. 2. Since these theories are diffeomorphism invariant, their Lagrangians would satisfy the conditions to be presented in this appendix.

Under diffeomorphism $x^\mu \mapsto x^\mu - \epsilon ^\mu (x),$ the vector fields transform as

$$\begin{aligned} \delta _\epsilon A_\mu ^\alpha =\epsilon ^\nu \partial _\nu A_\mu ^\alpha + A_\nu ^\alpha \partial _\mu \epsilon ^\nu , \end{aligned}$$

(A1)

and the external fields $\{K\}$ transform under standard diffeomorphism. The Lagrangian density transforms as

$$\begin{aligned} \delta _\epsilon \mathcal{L}=\partial _\mu (\epsilon ^\mu \mathcal{L}). \end{aligned}$$

(A2)

Demanding the expression $\delta _\epsilon \mathcal{L}- \partial _\mu (\epsilon ^\mu \mathcal{L})$ to vanish will give rise to useful conditions. In order to evaluate this expression, we begin by recall that $\mathcal{L}= T + U_\alpha \dot{A}^\alpha _0.$ Then we consider

$$\begin{aligned} \delta _\epsilon T= & {} \frac{\partial T}{\partial A^\alpha _\nu }\delta _\epsilon A^\alpha _\nu + \frac{\partial T}{\partial \partial _k A_\nu ^\alpha }\partial _k \delta _\epsilon A_\nu ^\alpha +\frac{\partial T}{\partial \dot{A}_k^\alpha }\partial _0 \delta _\epsilon A_k^\alpha \nonumber \\{} & {} +\frac{\partial T}{\partial \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}(K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}}\nonumber \\{} & {} \times \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}\delta _\epsilon ( K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}. \end{aligned}$$

(A3)

Next, let us consider

$$\begin{aligned} -\partial _\mu (\epsilon ^\mu T)= & {} -\partial _\mu \epsilon ^\mu T - \epsilon ^\mu \left( \frac{\partial T}{\partial A^\alpha _\nu }\partial _\mu A^\alpha _\nu + \frac{\partial T}{\partial \partial _k A_\nu ^\alpha }\partial _k \partial _\mu A_\nu ^\alpha +\frac{\partial T}{\partial \dot{A}_k^\alpha }\partial _0 \partial _\mu A_k^\alpha \right) \nonumber \\{} & {} -\epsilon ^\mu \frac{\partial T}{\partial \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}(K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}} \nonumber \\{} & {} \times \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}\partial _\mu ( K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}. \end{aligned}$$

(A4)

Combining the two expressions, we obtain

$$\begin{aligned}{} & {} \delta _\epsilon T-\partial _\mu (\epsilon ^\mu T)\nonumber \\{} & {} \quad =\frac{\partial T}{\partial A_\nu ^\alpha }A_\mu ^\alpha \partial _\nu \epsilon ^\mu + \frac{\partial T}{\partial \partial _k A_\nu ^\alpha }\left( \partial _k\epsilon ^\mu \partial _\mu A_\nu ^\alpha + \partial _k A_\mu ^\alpha \partial _\nu \epsilon ^\mu + A_\mu ^\alpha \partial _k\partial _\nu \epsilon ^\mu \right) \nonumber \\{} & {} \qquad +\frac{\partial T}{\partial \dot{A}_k^\alpha }\left( \dot{\epsilon }^\mu \partial _\mu A_k^\alpha + \dot{A}_\mu ^\alpha \partial _k\epsilon ^\mu + A_\mu ^\alpha \partial _k\dot{\epsilon }^\mu \right) -\partial _\mu \epsilon ^\mu T\nonumber \\{} & {} \qquad +\frac{\partial T}{\partial \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}(K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}} \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}\delta _\epsilon ( K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}\nonumber \\{} & {} \qquad -\epsilon ^\mu \frac{\partial T}{\partial \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}(K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}} \nonumber \\{} & {} \qquad \times \partial _{\rho _1} \cdots \partial _{\rho _{r'''}}\partial _\mu ( K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}. \end{aligned}$$

(A5)

Let us also compute

$$\begin{aligned} \delta _\epsilon \left( U_\beta \dot{A}_0^\beta \right)= & {} \frac{\partial U_\beta }{\partial A^\alpha _\nu }\left( \epsilon ^\mu \partial _\mu A^\alpha _\nu + A^\alpha _\mu \partial _\nu \epsilon ^\mu \right) \dot{A}_0^\beta +\frac{\partial U_\beta }{\partial \partial _i A^\alpha _\nu }\partial _i\left( \epsilon ^\mu \partial _\mu A^\alpha _\nu + A^\alpha _\mu \partial _\nu \epsilon ^\mu \right) \dot{A}_0^\beta \nonumber \\{} & {} +\dot{A}_0^\beta \frac{\partial U_\beta }{\partial \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}(K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}} \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}\delta _\epsilon ( K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}\nonumber \\{} & {} +U_\beta \partial _0\left( \epsilon ^\mu \partial _\mu A_0^\beta + A^\beta _\nu \partial _0\epsilon ^\nu \right) , \end{aligned}$$

(A6)

and

$$\begin{aligned}{} & {} -\partial _\mu (\epsilon ^\mu U_\beta \dot{A}_0^\beta )\nonumber \\{} & {} \quad =-\partial _\mu \epsilon ^\mu U_\beta \dot{A}_0^\beta -\epsilon ^\mu \left( \frac{\partial U_\beta }{\partial A^\alpha _\nu }\partial _\mu A^\alpha _\nu +\frac{\partial U_\beta }{\partial \partial _i A^\alpha _\nu }\partial _i\partial _\mu A^\alpha _\nu \right) \dot{A}_0^\beta \nonumber \\{} & {} \qquad -\epsilon ^\mu \dot{A}_0^\beta \frac{\partial U_\beta }{\partial \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}(K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}} \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}\partial _\mu ( K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}\nonumber \\{} & {} \qquad -\epsilon ^\mu U_\beta \partial _\mu \dot{A}_0^\beta . \end{aligned}$$

(A7)

So we have

$$\begin{aligned}{} & {} \delta _\epsilon (U_\beta \dot{A}_0^\beta ) - \partial _\mu (\epsilon ^\mu U_\beta \dot{A}_0^\beta )\nonumber \\{} & {} \quad = \frac{\partial U_\beta }{\partial A^\alpha _\nu }A^\alpha _\mu \partial _\nu \epsilon ^\mu \dot{A}_0^\beta +U_\beta (\dot{\epsilon }^\mu \partial _\mu A_0^\beta +\epsilon ^\mu \partial _\mu \dot{A}_0^\beta + \dot{A}^\beta _\nu \partial _0\epsilon ^\nu +A^\beta _\nu \ddot{\epsilon }^\nu )\nonumber \\{} & {} \qquad -\partial _\mu \epsilon ^\mu U_\beta \dot{A}_0^\beta -\epsilon ^\mu U_\beta \partial _\mu \dot{A}_0^\beta \nonumber \\{} & {} \qquad +\frac{\partial U_\beta }{\partial \partial _i A^\alpha _\nu }(\partial _i\epsilon ^\mu \partial _\mu A^\alpha _\nu + \partial _iA^\alpha _\mu \partial _\nu \epsilon ^\mu +A^\alpha _\mu \partial _i\partial _\nu \epsilon ^\mu )\dot{A}_0^\beta \nonumber \\{} & {} \qquad +\dot{A}_0^\beta \frac{\partial U_\beta }{\partial \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}(K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}} \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}\delta _\epsilon ( K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}\nonumber \\{} & {} \qquad -\epsilon ^\mu \dot{A}_0^\beta \frac{\partial U_\beta }{\partial \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}(K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}} \nonumber \\{} & {} \qquad \times \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}\partial _\mu ( K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}. \end{aligned}$$

(A8)

Combining Eq. (A5) with Eq. (A8), we obtain the expression for $\delta _\epsilon \mathcal{L}- \partial _\mu (\epsilon ^\mu \mathcal{L}).$

We are now ready to obtain useful conditions. Let us note that the expression $\delta _\epsilon \mathcal{L}- \partial _\mu (\epsilon ^\mu \mathcal{L})$ is a polynomial in $\dot{A}_0^\alpha$ up to degree two. Consider the term containing $\dot{A}_0^\alpha \dot{A}_0^\beta$ in $\delta _\epsilon \mathcal{L}- \partial _\mu (\epsilon ^\mu \mathcal{L})$. It can easily be seen that there is only one term which is

$$\begin{aligned} \frac{\partial U_\beta }{\partial \partial _i A_0^\alpha }\partial _i\epsilon ^0\dot{A}_0^\alpha \dot{A}_0^\beta . \end{aligned}$$

(A9)

Demanding this expression to vanish gives

$$\begin{aligned} \frac{\partial U_\alpha }{\partial \partial _i A_0^\beta } +\frac{\partial U_\beta }{\partial \partial _i A_0^\alpha } =0. \end{aligned}$$

(A10)

Let us next turn to the coefficients of $\dot{A}_0^\beta .$ For this, it would be convenient to consider Eqs. (A5) and (A8) and collect the terms proportional to $\dot{A}_0^\beta .$ We have

$$\begin{aligned} \delta _\epsilon T - \partial _\mu (\epsilon ^\mu T) \ni \frac{\partial T}{\partial \partial _k A_0^\alpha }\partial _k\epsilon ^0\dot{A}_0^\alpha +\frac{\partial T}{\partial \dot{A}_k^\alpha }\dot{A}_0^\alpha \partial _k\epsilon ^0, \end{aligned}$$

(A11)

and

$$\begin{aligned}{} & {} \delta _\epsilon (U_\beta \dot{A}_0^\beta ) - \partial _\mu (\epsilon ^\mu U_\beta \dot{A}_0^\beta )\nonumber \\{} & {} \quad \ni \frac{\partial U_\beta }{\partial A^\alpha _\nu }A^\alpha _\mu \partial _\nu \epsilon ^\mu \dot{A}_0^\beta +U_\beta (\dot{\epsilon }^0\dot{A}_0^\beta + \dot{A}^\beta _0\dot{\epsilon }^0) -\partial _\mu \epsilon ^\mu U_\beta \dot{A}_0^\beta \nonumber \\{} & {} \qquad +\frac{\partial U_\beta }{\partial \partial _i A^\alpha _\nu }(\partial _i\epsilon ^\mu \partial _\mu A^\alpha _\nu + \partial _iA^\alpha _\mu \partial _\nu \epsilon ^\mu +A^\alpha _\mu \partial _i\partial _\nu \epsilon ^\mu )\bigg |_{\dot{A}_0^\alpha = 0}\dot{A}_0^\beta \nonumber \\{} & {} \qquad +\dot{A}_0^\beta \frac{\partial U_\beta }{\partial \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}(K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}} \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}\delta _\epsilon ( K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}\nonumber \\{} & {} \qquad -\epsilon ^\mu \dot{A}_0^\beta \frac{\partial U_\beta }{\partial \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}(K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}} \nonumber \\{} & {} \qquad \times \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}\partial _\mu ( K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}. \end{aligned}$$

(A12)

Therefore, we have

$$\begin{aligned}{} & {} \frac{\partial T}{\partial \partial _k A_0^\beta }\partial _k\epsilon ^0 +\frac{\partial T}{\partial \dot{A}_k^\beta }\partial _k\epsilon ^0 +\frac{\partial U_\beta }{\partial A^\alpha _\nu }A^\alpha _\mu \partial _\nu \epsilon ^\mu +2U_\beta \dot{\epsilon }^0-\partial _\mu \epsilon ^\mu U_\beta \nonumber \\{} & {} \quad +\frac{\partial U_\beta }{\partial \partial _i A^\alpha _\nu }(\partial _i\epsilon ^\mu \partial _\mu A^\alpha _\nu + \partial _iA^\alpha _\mu \partial _\nu \epsilon ^\mu +A^\alpha _\mu \partial _i\partial _\nu \epsilon ^\mu )\bigg |_{\dot{A}_0^\alpha = 0}\nonumber \\{} & {} \qquad +\frac{\partial U_\beta }{\partial \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}(K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}} \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}\delta _\epsilon ( K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}\nonumber \\{} & {} \qquad -\epsilon ^\mu \frac{\partial U_\beta }{\partial \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}(K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}} \nonumber \\{} & {} \qquad \times \partial _{\rho _1} \cdots \partial _{\rho _{r'''}}\partial _\mu ( K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}} =0. \end{aligned}$$

(A13)

Although the above equation looks complicated especially due to the explicit presence of external fields, we will only extract some parts of this equation to obtain the conditions that we will need. These conditions will look much more simple. For example, the dependence on the external fields and their derivatives are only through T and $U_\beta .$ We may derive these conditions as follows. Taking derivative of Eq. (A13) with respect to $\dot{A}_j^\alpha$ gives

$$\begin{aligned} \frac{\partial ^2 T}{\partial \partial _k A_0^\beta \partial \dot{A}_j^\alpha } +\frac{\partial ^2 T}{\partial \dot{A}_k^\beta \partial \dot{A}_j^\alpha } +\frac{\partial U_\beta }{\partial \partial _k A_j^\alpha } =0. \end{aligned}$$

(A14)

Let us take derivative of Eq. (A13) with respect to $\partial _j A_0^\alpha$, then swap the indices $\alpha$ and $\beta$, add it to the original equation, and use Eq. (A10), we obtain

$$\begin{aligned} 2\frac{\partial ^2 T}{\partial \partial _j A_0^{(\alpha }\partial \partial _k A_0^{\beta )}} \!+\!2\frac{\partial ^2 T}{\partial \partial _j A_0^{(\alpha }\partial \dot{A}_k^{\beta )}} \!+\!\frac{\partial U_{\beta }}{\partial \partial _j A^{\alpha }_k} \!+\!\frac{\partial U_{\alpha }}{\partial \partial _j A^{\beta }_k} =0. \end{aligned}$$

(A15)

Expressing in phase space, the conditions Eqs. (A14)–(A15) become

$$\begin{aligned} \frac{\partial ^2 \mathcal{T}}{\partial \partial _k A_0^\beta \partial \Lambda _j^\alpha } +\frac{\partial ^2 \mathcal{T}}{\partial \Lambda _k^\beta \partial \Lambda _j^\alpha } +\frac{\partial U_\beta }{\partial \partial _k A_j^\alpha } =0, \end{aligned}$$

(A16)

and

$$\begin{aligned} 2\frac{\partial ^2 \mathcal{T}}{\partial \partial _j A_0^{(\alpha }\partial \partial _k A_0^{\beta )}} \!+\!2\frac{\partial ^2 \mathcal{T}}{\partial \partial _j A_0^{(\alpha }\partial \Lambda _k^{\beta )}} \!+\!\frac{\partial U_{\beta }}{\partial \partial _j A^{\alpha }_k} \!+\!\frac{\partial U_{\alpha }}{\partial \partial _j A^{\beta }_k}=0. \end{aligned}$$

(A17)

By substituting Eq. (A16) into Eq. (A17), we obtain

$$\begin{aligned} \frac{\partial ^2 \mathcal{T}}{\partial \partial _{(j|} A_{0}^\alpha \partial \partial _{|k)} A_0^\beta } -\frac{\partial ^2 \mathcal{T}}{\partial \Lambda _{(j}^\alpha \partial \Lambda _{k)}^\beta } =0. \end{aligned}$$

(A18)

Appendix B: Expressions of $\partial \phi _\alpha /\partial \partial _{\mathcal{I}}\dot{Q}^\beta$ in phase space

In this appendix, we outline necessary steps to express $\partial \phi _\alpha /\partial \partial _{\mathcal{I}}\dot{Q}^\beta$ in phase space. We use the same set-up and notations as those given in Sects. 2–3. For convenient, let us denote $P_A$ and $\Lambda ^A$ as collective for $\pi _\alpha ^i$ and $\Lambda ^\alpha _i,$ respectively.

The idea is to first express $\partial \phi _\alpha /\partial \partial _{\mathcal{I}}\dot{Q}^\beta$ in terms of $\alpha _M$. This can be achieved by recalling from Sect. 3 the Eq. (98). Recall also that diffeomorphism invariance requirements and demanding $\dot{\alpha }_\alpha = 0$ to not introduce further dynamics on the vector fields imply that $\partial \alpha _\alpha /\partial \dot{Q}^\beta = 0 = \partial \alpha _\alpha /\partial \partial _i\dot{Q}^\beta$. Furthermore, due to the form of the Lagrangian of interest, we also have $\partial \alpha _\alpha /\partial \partial _{i_1}\partial _{i_2}\cdots \partial _{i_l}\dot{Q}^\beta = 0$ for $l\ge 2.$ So

$$\begin{aligned} \frac{\partial \alpha _\alpha }{\partial \partial _\mathcal{I}\dot{Q}^\beta } = 0. \end{aligned}$$

(B1)

Then since $\phi _\alpha = \dot{\alpha }_\alpha$ we have, from Eqs. (98) and (B1)

$$\begin{aligned} \phi _\alpha= & {} \sum _{|\mathcal{I}|=0}^2\frac{\partial \alpha _\alpha }{\partial \partial _\mathcal{I}Q^M}\partial _\mathcal{I}\dot{Q}^M -\sum _{|\mathcal{I}|=0}^1\frac{\partial \alpha _\alpha }{\partial \partial _\mathcal{I}\dot{Q}^B}\partial _\mathcal{I}(M^{BC}\alpha _C)\nonumber \\{} & {} +\frac{\partial \alpha _\alpha }{\partial \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}(K^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}} \partial _{\rho _1}\cdots \partial _{\rho _{r'''}}(\dot{K}^{(r)})_{\mu _1\cdots \mu _{r'}}{}^{\nu _1\cdots \nu _{r''}}. \end{aligned}$$

(B2)

Due to Eq. (B1), it can be seen that $\phi _\alpha$ depend on $\partial _{\mathcal{I}}\dot{Q}^\beta$ only through the expressions $\partial _\mathcal{I}\dot{Q}^M$ and $\partial _\mathcal{I}(M^{BC}\alpha _C)$ which appear in the above equation. This gives

$$\begin{aligned} \frac{\partial \phi _\alpha }{\partial \partial _{\mathcal{I}}\dot{Q}^\beta } =\frac{\partial \alpha _\alpha }{\partial \partial _{\mathcal{I}}Q^\beta } -\frac{\partial \alpha _\alpha }{\partial \partial _{\mathcal{J}}\dot{Q}^A}\frac{\partial \partial _{\mathcal{J}}(M^{AB}\alpha _B)}{\partial \partial _{\mathcal{I}}\dot{Q}^\beta }. \end{aligned}$$

(B3)

We then need to compute each expression on RHS of Eq. (B3). For this, let us directly express $\alpha _A$ in terms of Lagrangian then transforming to phase space, but transform $\alpha _\alpha$ to $-\tilde{\Omega }_\alpha$ (cf. Equation (104)). Direct calculations can be given as follows. In order to evaluate ${\partial \partial _\mathcal{J}(M^{AB}\alpha _B)}/{\partial \partial _\mathcal{I}\dot{Q}^\beta }$ we note that $\partial M^{AB}/\partial \partial _\mathcal{I}\dot{Q}^\beta = 0$ for $|\mathcal{I}|\ge 0$ whereas $\alpha _B$ depends on $\dot{Q}^\beta$ and $\partial _i\dot{Q}^\beta$ but not on $\partial _\mathcal{I}\dot{Q}^\beta$ where $|\mathcal{I}|\ge 2.$ By writing $\partial _k\alpha _B$ using chain rule and taking derivative with respect to $\partial _\mathcal{I}\dot{Q}^\beta ,$ we obtain

$$\begin{aligned} \frac{\partial \partial _k\alpha _B}{\partial \partial _i\partial _j\dot{Q}^\beta }= & {} \frac{\partial \alpha _B}{\partial \partial _l\dot{Q}^\beta }\delta _{(k}^i\delta _{l)}^j,\quad \frac{\partial \partial _j\alpha _B}{\partial \partial _i\dot{Q}^\beta } =\partial _j\frac{\partial \alpha _B}{\partial \partial _i\dot{Q}^\beta } +\delta ^i_j \frac{\partial \alpha _B}{\partial \dot{Q}^\beta },\nonumber \\ \frac{\partial \partial _j\alpha _B}{\partial \dot{Q}^\beta }= & {} \partial _j\frac{\partial \alpha _B}{\partial \dot{Q}^\beta }. \end{aligned}$$

(B4)

This gives

$$\begin{aligned} \frac{\partial \partial _k(M^{AB}\alpha _B)}{\partial \partial _i\partial _j\dot{Q}^\beta }= & {} \frac{\partial ((\partial _kM^{AB})\alpha _B+M^{AB}\partial _k\alpha _B)}{\partial \partial _i\partial _j\dot{Q}^{\beta }}\nonumber \\= & {} M^{AB}\frac{\partial \partial _k\alpha _B}{\partial \partial _i\partial _j\dot{Q}^\beta }\nonumber \\= & {} M^{AB}\frac{\partial \alpha _B}{\partial \partial _l\dot{Q}^\beta }\delta _{(k}^i\delta _{l)}^j, \end{aligned}$$

(B5)

$$\begin{aligned} \frac{\partial \partial _j(M^{AB}\alpha _B)}{\partial \partial _i\dot{Q}^\beta }= & {} \partial _j\left( M^{AB}\frac{\partial \alpha _B}{\partial \partial _i\dot{Q}^\beta }\right) +\delta _j^iM^{AB}\frac{\partial \alpha _B}{\partial \dot{Q}^\beta }, \end{aligned}$$

(B6)

$$\begin{aligned} \frac{\partial \partial _j(M^{AB}\alpha _B)}{\partial \dot{Q}^\beta }= & {} \partial _j\left( M^{AB}\frac{\partial \alpha _B}{{\partial \dot{Q}^\beta }}\right) . \end{aligned}$$

(B7)

Next, let us express $\partial \alpha _B/\partial \partial _\mathcal{I}\dot{Q}^\beta$ in terms of phase space variables. The calculations will involve $\partial (\partial _j(\partial \mathcal{L}/\partial \partial _j Q^B))/\partial \partial _\mathcal{I}\dot{Q}^\beta ,$ which can be computed by first using the chain rule for $\partial _j$ and then taking derivative with respect to $\partial _\mathcal{I}\dot{Q}^\beta .$ The relevant results are

$$\begin{aligned} \frac{\partial }{\partial \dot{Q}^\beta }\left( \partial _i\left( \frac{\partial \mathcal{L}}{\partial \partial _i Q^B}\right) \right)= & {} \partial _i\left( \frac{\partial ^2\mathcal{L}}{\partial \dot{Q}^\beta \partial \partial _i Q^B}\right) ,\nonumber \\ \frac{\partial }{\partial \partial _i\dot{Q}^\beta }\left( \partial _j\left( \frac{\partial \mathcal{L}}{\partial \partial _j Q^B}\right) \right)= & {} \frac{\partial ^2\mathcal{L}}{\partial \dot{Q}^\beta \partial \partial _i Q^B}. \end{aligned}$$

(B8)

So

$$\begin{aligned} \frac{\partial \alpha _B}{\partial \dot{Q}^\beta }= & {} \frac{\partial ^2\mathcal{L}}{\partial \dot{Q}^B\partial Q^\beta }+\partial _i\Bigg (\frac{\partial ^2\mathcal{L}}{\partial \dot{Q}^\beta \partial \partial _iQ^B}\Bigg )-\frac{\partial ^2\mathcal{L}}{\partial \dot{Q}^\beta \partial Q^B}\nonumber \\= & {} \frac{\partial ^2\mathcal{T}}{\partial Q^\beta \partial \Lambda ^B} +\partial _i\left( \frac{\partial U_\beta }{\partial \partial _iQ^B}\right) -\frac{\partial U_\beta }{\partial Q^B}, \end{aligned}$$

(B9)

$$\begin{aligned} \frac{\partial \alpha _B}{\partial \partial _i\dot{Q}^\beta }= & {} \frac{\partial ^2\mathcal{L}}{\partial \partial _i Q^\beta \partial \dot{Q}^B}+\frac{\partial ^2\mathcal{L}}{\partial \partial _iQ^B\partial \dot{Q}^\beta }\nonumber \\= & {} \frac{\partial ^2 \mathcal{T}}{\partial \partial _i Q^\beta \partial \Lambda ^B} +\frac{\partial U_\beta }{\partial \partial _iQ^B}. \end{aligned}$$

(B10)

Next, let us express $\partial \alpha _\alpha /\partial \partial _\mathcal{I}Q^\beta$ in phase space. For this, we first use Eq. (104) to transform $\alpha _\alpha$ to $-\tilde{\Omega }_\alpha .$ More precisely, this is

$$\begin{aligned} \alpha _\alpha =-\tilde{\Omega }_\alpha \left( Q^M,\partial _i Q^M,\partial _i\partial _j Q^M,P_B,\partial _i P_B,\{K,\partial K,\partial \partial K,\ldots \}\right) , \end{aligned}$$

(B11)

such that $P_B = P_B(Q^M,\partial _i Q^M,\dot{Q}^B,\{K,\partial K,\partial \partial K,\ldots \})$, in which both sides of Eq. (B11) are both functions on the tangent bundle. So when taking derivative of $\alpha _\alpha$ with respect to $\partial _\mathcal{I}Q^\beta ,$ we need to also take into account that $P_B$ and $\partial _i P_B$ also depend on $\partial _\mathcal{I}Q^\beta .$ As part of the intermediate calculations, we need to compute $\partial \partial _k P_B/\partial \partial _\mathcal{I}Q^\beta ,$ which can be done by first writing $\partial _k P_B$ using chain rule, then taking derivative with respect to $\partial _\mathcal{I}Q^\beta .$ We have

$$\begin{aligned} \frac{\partial \partial _k P_B}{\partial \partial _i\partial _j Q^\beta }= & {} \frac{\partial P_B}{\partial \partial _l Q^\beta }\delta ^i_{(k}\delta ^j_{l)},\quad \frac{\partial \partial _k P_B}{\partial \partial _i Q^\beta } =\partial _k\left( \frac{\partial P_B}{\partial \partial _i Q^\beta }\right) +\delta ^i_k \frac{\partial P_B}{\partial Q^\beta },\nonumber \\ \frac{\partial \partial _k P_B}{\partial Q^\beta }= & {} \partial _k\left( \frac{\partial P_B}{\partial Q^\beta }\right) . \end{aligned}$$

(B12)

Then we use Eq. (45), which is equivalent to $P_B = \partial \mathcal{T}/\partial \Lambda ^B.$ Keeping these in mind, we have

$$\begin{aligned} \frac{\partial \alpha _\alpha }{\partial \partial _i\partial _jQ^\beta }= & {} -\frac{\partial \tilde{\Omega }_\alpha }{\partial \partial _i\partial _jQ^\beta } -\frac{\partial \tilde{\Omega }_\alpha }{\partial \partial _{(i|}P_B}\frac{\partial P_B }{\partial \partial _{|j)}Q^\beta }\nonumber \\= & {} -\frac{\partial \tilde{\Omega }_\alpha }{\partial \partial _i\partial _jQ^\beta } -\frac{\partial \tilde{\Omega }_\alpha }{\partial \partial _{(i|}P_B}\frac{\partial ^2\mathcal{T}}{\partial \partial _{|j)}Q^\beta \partial \Lambda ^B}, \end{aligned}$$

(B13)

$$\begin{aligned} \frac{\partial \alpha _\alpha }{\partial \partial _iQ^\beta }= & {} -\frac{\partial \tilde{\Omega }_\alpha }{\partial \partial _iQ^\beta } -\frac{\partial \tilde{\Omega }_\alpha }{\partial \partial _\mathcal{I}P_B}\partial _\mathcal{I}\left( \frac{\partial P_B }{\partial \partial _iQ^\beta }\right) -\frac{\partial \tilde{\Omega }_\alpha }{\partial \partial _iP_B}\frac{\partial P_B}{\partial Q^\beta }\nonumber \\= & {} -\frac{\partial \tilde{\Omega }_\alpha }{\partial \partial _iQ^\beta } -\frac{\partial \tilde{\Omega }_\alpha }{\partial \partial _\mathcal{I}P_B}\partial _\mathcal{I}\left( \frac{\partial ^2\mathcal{T}}{\partial \partial _iQ^\beta \partial \Lambda ^B}\right) \nonumber \\{} & {} -\frac{\partial \tilde{\Omega }_\alpha }{\partial \partial _iP_B}\frac{\partial ^2\mathcal{T}}{\partial Q^\beta \partial \Lambda ^B}, \end{aligned}$$

(B14)

$$\begin{aligned} \frac{\partial \alpha _\alpha }{\partial Q^\beta }= & {} -\frac{\partial \tilde{\Omega }_\alpha }{\partial Q^\beta } -\frac{\partial \tilde{\Omega }_\alpha }{\partial \partial _\mathcal{I}P_B}\partial _\mathcal{I}\left( \frac{\partial P_B }{\partial Q^\beta }\right) \nonumber \\= & {} -\frac{\partial \tilde{\Omega }_\alpha }{\partial Q^\beta } -\frac{\partial \tilde{\Omega }_\alpha }{\partial \partial _\mathcal{I}P_B}\partial _\mathcal{I}\left( \frac{\partial ^2\mathcal{T}}{\partial Q^\beta \partial \Lambda ^B}\right) . \end{aligned}$$

(B15)

Finally, let us compute $\partial \alpha _\alpha /\partial \partial _\mathcal{I}Q^\beta$. For this, as intermediate steps we compute

$$\begin{aligned} \frac{\partial P_B}{\partial \dot{Q}^A}= & {} \frac{\partial ^2\mathcal{L}}{\partial \dot{Q}^A\partial \dot{Q}^B} =W_{AB},\quad \frac{\partial \partial _k P_B}{\partial \partial _i\dot{Q}^A} =\delta _k^i\frac{\partial P_B}{\partial \dot{Q}^A} =W_{AB}\delta _k^i,\nonumber \\ \frac{\partial \partial _k P_B}{\partial \dot{Q}^A}= & {} \partial _k W_{AB}. \end{aligned}$$

(B16)

Then we have

$$\begin{aligned} \frac{\partial \alpha _\alpha }{\partial \partial _i\dot{Q}^A}= & {} -\frac{\partial \tilde{\Omega }_\alpha }{\partial \partial _iP_B}W_{AB}, \end{aligned}$$

(B17)

$$\begin{aligned} \frac{\partial \alpha _\alpha }{\partial \dot{Q}^A}= & {} -\frac{\partial \tilde{\Omega }_\alpha }{\partial \partial _iP_B}\partial _iW_{BA}-\frac{\partial \tilde{\Omega }_\alpha }{\partial P_B}W_{BA}. \end{aligned}$$

(B18)

Then by substituting Eqs. (B5B6)–(B7), (B9)–(B10), (B13B14)–(B15), (B17)–(B18) into Eq. (B3), we obtain

$$\begin{aligned} \frac{\partial \phi _\alpha }{\partial \dot{Q}^\beta }= & {} -\mathcal{C}_{0\beta \alpha } +\partial _i\mathcal{C}_{1\beta \alpha }^i -\partial _i\partial _j\mathcal{C}_{2\beta \alpha }^{ij},\nonumber \\ \frac{\partial \phi _\alpha }{\partial \partial _i\dot{Q}^\beta }= & {} \mathcal{C}_{1\beta \alpha }^i -2\partial _j\mathcal{C}_{2\beta \alpha }^{ij},\nonumber \\ \frac{\partial \phi _\alpha }{\partial \partial _i\partial _j\dot{Q}^\beta }= & {} -\mathcal{C}_{2\beta \alpha }^{ij}. \end{aligned}$$

(B19)

By using diffeomorphism invariance requirements, Eq. (65) is realised. This simplifies Eq. (B19). Further simplifications are possible. For this, let us note that using Eqs. (B3)–(B9) and diffeomorphism invariance requirements, one obtains

$$\begin{aligned} \frac{\partial \phi _\alpha }{\partial \dot{Q}^\beta } -\frac{\partial \phi _\beta }{\partial \dot{Q}^\alpha } +\partial _i\left( \frac{\partial \phi _\beta }{\partial \partial _i\dot{Q}^\alpha }\right)= & {} \frac{\partial \alpha _\alpha }{\partial Q^\beta } -\frac{\partial \alpha _\beta }{\partial Q^\alpha }\nonumber \\{} & {} +\partial _i\left( \frac{\partial \alpha _\beta }{\partial \partial _iQ^\alpha } +\frac{\partial \alpha _\beta }{\partial \dot{Q}^\alpha _i} +\frac{\partial \alpha _\alpha ^i}{\partial \dot{Q}^\beta }\right) . \end{aligned}$$

(B20)

Then by expressing $\alpha _M$ in terms of Lagrangian and using diffeomorphism invariance and secondary-constraint enforcing relations, we obtain

$$\begin{aligned} \frac{\partial \phi _\alpha }{\partial \dot{Q}^\beta } -\frac{\partial \phi _\beta }{\partial \dot{Q}^\alpha } +\partial _i\left( \frac{\partial \phi _\beta }{\partial \partial _i\dot{Q}^\alpha }\right) =0, \end{aligned}$$

(B21)

which is equivalent to the phase space expression

$$\begin{aligned} \mathcal{C}_{0\alpha \beta } =\mathcal{C}_{0\beta \alpha }-\partial _i\mathcal{C}_{1\beta \alpha }^i. \end{aligned}$$

(B22)

Finally, this gives

$$\begin{aligned} \frac{\partial \phi _\alpha }{\partial \dot{Q}^\beta } = -\mathcal{C}_{0\alpha \beta },\quad \frac{\partial \phi _\alpha }{\partial \partial _i\dot{Q}^\beta } =-\mathcal{C}^{i}_{1\alpha \beta },\quad \frac{\partial \phi _\alpha }{\partial \partial _i\partial _j\dot{Q}^\beta } =0. \end{aligned}$$

(B23)