$\displaystyle\epsilon^{\mu}\bigl{(}\nabla_{\beta}\delta
g^{\alpha\beta}+2\nabla^{\alpha}\delta\gamma\bigr{)}+\epsilon_{\beta}\nabla^{\alpha}\delta
g^{\mu\beta}-\bigl{(}\nabla_{\beta}\epsilon^{\mu}\bigr{)}\delta
g^{\alpha\beta}-\bigl{(}\nabla^{\alpha}\epsilon^{\mu}\bigr{)}\delta\gamma$
$\displaystyle-\epsilon^{\alpha}\bigl{(}\nabla_{\beta}\delta
g^{\mu\beta}+2\nabla^{\mu}\delta\gamma\bigr{)}-\epsilon_{\beta}\nabla^{\mu}\delta
g^{\alpha\beta}+\bigl{(}\nabla_{\beta}\epsilon^{\alpha}\bigr{)}\delta
g^{\mu\beta}+\bigl{(}\nabla^{\mu}\epsilon^{\alpha}\bigr{)}\delta\gamma\biggr{]},$
(A.28)

showing that $\mathfrak{i}_{\cal V}\omega_{\text{EH}}^{\mu}$ can be expressed
as a total divergence of an antisymmetric tensor density.

To verify this result, we expand the total derivative in the above expression
and see that we recover the right-hand side of (A.27), assuming the equations
of motion are satisfied. Consider first the terms in (A.28) in which the free
index is on $\epsilon^{\mu}$. Taking into account (A.19), we see that these
terms give

$\displaystyle\epsilon^{\mu}\nabla_{\alpha}\Bigl{(}\nabla_{\beta}\delta
g^{\alpha\beta}+2\nabla^{\alpha}\delta\gamma\Bigr{)}+\bigl{(}\nabla_{\alpha}\epsilon^{\mu}\bigr{)}\Bigl{(}\nabla_{\beta}\delta
g^{\alpha\beta}+2\nabla^{\alpha}\delta\gamma\Bigr{)}$
$\displaystyle-\bigl{(}\nabla_{\alpha}\nabla_{\beta}\epsilon^{\mu}\bigr{)}\delta
g^{\alpha\beta}-\bigl{(}\nabla_{\beta}\epsilon^{\mu}\bigr{)}\nabla_{\alpha}\delta
g^{\alpha\beta}-\bigl{(}\nabla_{\alpha}\nabla^{\alpha}\epsilon^{\mu}\bigr{)}\delta\gamma-\bigl{(}\nabla^{\alpha}\epsilon^{\mu}\bigr{)}\nabla_{\alpha}\delta\gamma$
$\displaystyle=\bigl{(}\nabla_{\alpha}\epsilon^{\mu}\bigr{)}\nabla^{\alpha}\delta\gamma-\bigl{(}\nabla_{\alpha}\nabla_{\beta}\epsilon^{\mu}\bigr{)}\delta
g^{\alpha\beta}-\bigl{(}\nabla_{\alpha}\nabla^{\alpha}\epsilon^{\mu}\bigr{)}\delta\gamma,$
(A.29)

which agrees with the terms on the first line of (A.27).

We then take the terms in (A.28) where the free index is on $\delta
g^{\mu\beta}$:

$\displaystyle\epsilon^{\alpha}\bigl{[}\nabla_{\alpha},\nabla_{\beta}\bigr{]}\delta
g^{\mu\beta}+\bigl{(}\nabla_{\alpha}\epsilon_{\beta}\bigr{)}\nabla^{\alpha}\delta
g^{\mu\beta}+\epsilon_{\beta}\nabla_{\alpha}\nabla^{\alpha}\delta
g^{\mu\beta}$
$\displaystyle-\bigl{(}\nabla_{\alpha}\epsilon^{\alpha}\bigr{)}\nabla_{\beta}\delta
g^{\mu\beta}-\epsilon^{\alpha}\nabla_{\alpha}\nabla_{\beta}\delta
g^{\mu\beta}+\bigl{(}\nabla_{\alpha}\nabla_{\beta}\epsilon^{\alpha}\bigr{)}\delta
g^{\mu\beta}+\bigl{(}\nabla_{\beta}\epsilon^{\alpha}\bigr{)}\nabla_{\alpha}\delta
g^{\mu\beta}$
$\displaystyle=-\epsilon^{\alpha}\nabla_{\beta}\nabla_{\alpha}\delta
g^{\mu\beta}+\epsilon_{\beta}\nabla_{\alpha}\nabla^{\alpha}\delta
g^{\mu\beta}$
$\displaystyle\quad+\bigl{(}\nabla_{\alpha}\epsilon_{\beta}+\nabla_{\beta}\epsilon_{\alpha}\bigr{)}\nabla^{\alpha}\delta
g^{\mu\beta}-\bigl{(}\nabla_{\alpha}\epsilon^{\alpha}\bigr{)}\nabla_{\beta}\delta
g^{\mu\beta}+\bigl{(}\nabla_{\alpha}\nabla_{\beta}\epsilon^{\alpha}\bigr{)}\delta
g^{\mu\beta}.$ (A.30)

Due to the on-shell relation
$\nabla_{\alpha}\nabla_{\beta}\epsilon^{\alpha}=\nabla_{\beta}\nabla_{\alpha}\epsilon^{\alpha}$,
we see that the terms on the second line reproduce the second line of (A.27).
Using the equations of motion (A.17), the terms on the first line of (A.30)
become

$\displaystyle\epsilon_{\beta}\Bigl{(}-\nabla_{\alpha}\nabla^{\beta}\delta
g^{\mu\alpha}+\nabla_{\alpha}\nabla^{\alpha}\delta
g^{\mu\beta}\Bigr{)}=\epsilon_{\beta}\Bigl{(}\nabla_{\alpha}\nabla^{\mu}\delta
g^{\alpha\beta}+2\nabla^{\beta}\nabla^{\mu}\delta\gamma\Bigr{)}.$ (A.31)

In (A.28) we are now left with the terms where the free index is on a
covariant derivative:

$\displaystyle\Bigl{(}\bigl{[}\nabla^{\mu},\nabla_{\alpha}\bigr{]}\epsilon_{\beta}\Bigr{)}\delta
g^{\alpha\beta}-2\bigl{(}\nabla_{\alpha}\epsilon^{\alpha}\bigr{)}\nabla^{\mu}\delta\gamma-2\epsilon^{\alpha}\nabla_{\alpha}\nabla^{\mu}\delta\gamma$
$\displaystyle-\bigl{(}\nabla_{\alpha}\epsilon_{\beta}\bigr{)}\nabla^{\mu}\delta
g^{\alpha\beta}-\epsilon_{\beta}\nabla_{\alpha}\nabla^{\mu}\delta
g^{\alpha\beta}+\bigl{(}\nabla_{\alpha}\nabla^{\mu}\epsilon^{\alpha}\bigr{)}\delta\gamma+\bigl{(}\nabla^{\mu}\epsilon^{\alpha}\bigr{)}\nabla_{\alpha}\delta\gamma$
$\displaystyle=\epsilon_{\beta}\Bigl{(}-\nabla_{\alpha}\nabla^{\mu}\delta
g^{\alpha\beta}-2\nabla^{\beta}\nabla^{\mu}\delta\gamma\Bigr{)}+\Bigl{(}\bigl{[}\nabla^{\mu},\nabla_{\alpha}\bigr{]}\epsilon_{\beta}\Bigr{)}\delta
g^{\alpha\beta}$
$\displaystyle\quad-2\bigl{(}\nabla_{\alpha}\epsilon^{\alpha}\bigr{)}\nabla^{\mu}\delta\gamma-\bigl{(}\nabla_{\alpha}\epsilon_{\beta}\bigr{)}\nabla^{\mu}\delta
g^{\alpha\beta}+\bigl{(}\nabla_{\alpha}\nabla^{\mu}\epsilon^{\alpha}\bigr{)}\delta\gamma+\bigl{(}\nabla^{\mu}\epsilon^{\alpha}\bigr{)}\nabla_{\alpha}\delta\gamma.$
(A.32)

Here the first two terms cancel the terms left over from the previous step.
The remaining terms coincide with the last two lines of (A.27) (keeping again
in mind that
$\nabla_{\alpha}\nabla^{\mu}\epsilon^{\alpha}=\nabla^{\mu}\nabla_{\alpha}\epsilon^{\alpha}$
on-shell).

Finally, we are left with the terms

$\Bigl{(}\bigl{[}\nabla^{\mu},\nabla_{\alpha}\bigr{]}\epsilon_{\beta}\Bigr{)}\delta
g^{\alpha\beta}+\epsilon^{\alpha}\bigl{[}\nabla_{\alpha},\nabla_{\beta}\bigr{]}\delta
g^{\mu\beta}.$ (A.33)

Using the identity
$[\nabla_{\alpha},\nabla_{\beta}]v^{\mu}=R^{\mu}_{\phantom{\mu}\nu\alpha\beta}v^{\nu}$
and its generalization for tensors of rank two, one can easily verify that
these terms vanish on-shell, hence establishing that the expression (A.28) is
correct.

The result (A.28) can be written in a more compact form if we recognize that
the terms on the right-hand side are related to the symplectic potential
current $\theta^{\mu}_{\rm EH}$ and the tensor density
$\xi^{\mu\nu}=\sqrt{-g}\nabla^{[\mu}\epsilon^{\nu]}$ introduced in (4.23). The
symplectic potential current, which was defined in (4.14), is equivalently
given by the expression

$\theta^{\mu}_{\rm EH}=-\frac{1}{2}\sqrt{-g}\Bigl{(}\nabla_{\alpha}\delta
g^{\mu\alpha}+2\nabla^{\mu}\delta\gamma\Bigr{)}.$ (A.34)

Furthermore, direct calculation of the variation $\delta\xi^{\mu\nu}$ shows
that

$\displaystyle\delta\xi^{\mu\nu}$
$\displaystyle=\sqrt{-g}\Bigl{(}\nabla^{[\mu}\epsilon^{\nu]}\,\delta\gamma+\delta
g^{\alpha[\mu}\nabla_{\alpha}\epsilon^{\nu]}+g^{\alpha[\mu}\delta\Gamma^{\nu]}_{\alpha\beta}\epsilon^{\beta}\Bigr{)}$
$\displaystyle=\sqrt{-g}\Bigl{(}\nabla^{[\mu}\epsilon^{\nu]}\delta\gamma-\nabla_{\alpha}\epsilon^{[\mu}\delta
g^{\nu]\alpha}-\epsilon_{\alpha}\nabla^{[\mu}\delta g^{\nu]\alpha}\Bigr{)}$
(A.35)

where (4.7) was used to obtain the expression on the second line. Comparing
now (A.34) and (A.35) with (A.28), we see that the contraction
$\mathfrak{i}_{\cal V}\omega^{\mu}_{\rm EH}$ can be expressed in the simple
form

$\mathfrak{i}_{\cal
V}\omega_{\text{EH}}^{\mu}=\partial_{\nu}\Bigl{(}-\epsilon^{\mu}\theta_{\text{EH}}^{\nu}+\epsilon^{\nu}\theta_{\text{EH}}^{\mu}+\delta\xi^{\mu\nu}\Bigr{)}.$
(A.36)

A yet different expression for $\mathfrak{i}_{\cal V}\omega^{\mu}_{\rm EH}$,
which will be needed in section 5.3, can be obtained as follows. Observe that
the Lie derivative of the vector density $\theta^{\mu}_{\rm EH}$ can be
written on-shell as

${\cal L}_{\chi}\theta^{\mu}_{\rm EH}=\xi^{\nu}\partial_{\nu}\theta_{\rm
EH}^{\mu}-(\partial_{\nu}\chi^{\mu})\theta_{\rm
EH}^{\nu}+(\partial_{\nu}\chi^{\nu})\theta_{\rm
EH}^{\mu}=-\partial_{\nu}\bigl{(}\epsilon^{\mu}\theta_{\rm
EH}^{\nu}-\epsilon^{\nu}\theta_{\rm EH}^{\mu}\bigr{)},$ (A.37)

since the symplectic potential current satisfies
$\partial_{\mu}\theta^{\mu}_{\rm EH}=0$ on-shell. Observe also that for the
antisymmetric tensor density $\xi^{\mu\nu}$, the covariant divergence
coincides with the partial divergence:

$\displaystyle\nabla_{\nu}\xi^{\mu\nu}=\partial_{\nu}\xi^{\mu\nu}+\Gamma^{\mu}_{\nu\beta}\xi^{\beta\nu}+\Gamma^{\nu}_{\nu\beta}\xi^{\mu\beta}-\Gamma^{\beta}_{\nu\beta}\xi^{\mu\nu}=\partial_{\nu}\xi^{\mu\nu}.$
(A.38)

This implies

$\partial_{\nu}\delta\xi^{\mu\nu}=\delta\bigl{(}\partial_{\nu}\xi^{\mu\nu}\bigr{)}=\delta\bigl{(}\nabla_{\nu}\xi^{\mu\nu}\bigr{)},$
(A.39)

and so we have established the alternative expression

$\mathfrak{i}_{\cal V}\omega_{\rm EH}^{\mu}={\cal L}_{\epsilon}\theta_{\rm
EH}^{\mu}+\delta\bigl{(}\nabla_{\nu}\xi^{\mu\nu}\bigr{)}$ (A.40)

for the contraction of the symplectic current.

###  A.2 Variation of an embedded integral

Consider the integral

${\cal I}[\alpha]=\int_{S}\alpha$ (A.41)

where

$\alpha(x)=\alpha_{\mu_{1}\cdots\mu_{p}}(x)\,\mathrm{d}x^{\mu_{1}}\wedge\cdots\wedge\mathrm{d}x^{\mu_{p}}$
(A.42)

is a $p$-form, and $S$ is a $p$-dimensional submanifold of the $d$-dimensional
spacetime manifold $M$. Using the Levi-Civita symbol, we can identify the
$p$-form $\alpha$ with a tensor density of rank $n:=d-p$ and density weight
$+1$. Hence the integral (A.41) can be written as

${\cal
I}[\alpha]=\int_{S}\mathrm{d}\Sigma_{\mu_{1}\cdots\mu_{n}}(x)\,\alpha^{\mu_{1}\cdots\mu_{n}}(x)$
(A.43)

where
$\alpha^{\mu_{1}\cdots\mu_{n}}=\frac{1}{p!}\epsilon^{\mu_{1}\cdots\mu_{n}\mu_{n+1}\cdots\mu_{d}}\alpha_{\mu_{n+1}\cdots\mu_{d}}$,
and

$d\Sigma_{\mu_{1}\cdots\mu_{n}}(x)=\epsilon_{\mu_{1}\cdots\mu_{n}\mu_{n+1}\cdots\mu_{d}}\mathrm{d}x^{\mu_{n+1}}\wedge\cdots\wedge\mathrm{d}x^{\mu_{d}}.$
(A.44)

Let $\phi:{\cal D}\subset\mathbbm{R}^{d}\to M$ be an embedding map. We denote
$\phi({\cal S})=S$, i.e. ${\cal S}$ is the preimage of the submanifold $S$
under $\phi$. Given such an embedding map, we can construct the pullback of
the integral (A.41), thus obtaining

${\cal I}[\alpha,\phi]=\int_{\cal S}\phi^{*}\alpha=\int_{\cal
S}\mathrm{d}\Sigma_{\mu_{1}\cdots\mu_{n}}\bigl{(}\phi(\mathtt{x})\bigr{)}\,\alpha^{\mu_{1}\cdots\mu_{n}}\bigl{(}\phi(\mathtt{x})\bigr{)}.$
(A.45)

(Here $\mathtt{x}$ denotes a set of coordinates on ${\cal D}$.)

In this section we wish to establish a formula for the variation – i.e. field
space differential – of an integral of this form, assuming that both the
$p$-form $\alpha$ and the embedding map $\phi$ are field space dependent
variables.

Let us begin by explicitly extracting the $\phi$-dependence out of the surface
element
$\mathrm{d}\Sigma_{\mu_{1}\cdots\mu_{n}}\bigl{(}\phi(\mathtt{x})\bigr{)}$. We
have

$\displaystyle\mathrm{d}\Sigma_{\mu_{1}\cdots\mu_{n}}\bigl{(}\phi(\mathtt{x})\bigr{)}$
$\displaystyle=\epsilon_{\mu_{1}\cdots\mu_{n}\mu_{n+1}\cdots\mu_{d}}\,\mathrm{d}\phi^{\mu_{n+1}}(\mathtt{x})\wedge\cdots\wedge\mathrm{d}\phi^{\mu_{d}}(\mathtt{x})$
$\displaystyle=\epsilon_{\mu_{1}\cdots\mu_{n}\mu_{n+1}\cdots\mu_{d}}\,\frac{\partial\phi^{\mu_{n+1}}(\mathtt{x})}{\partial\mathtt{x}^{\nu_{n+1}}}\cdots\frac{\partial\phi^{\mu_{d}}(\mathtt{x})}{\partial\mathtt{x}^{\nu_{d}}}\mathrm{d}\mathtt{x}^{\nu_{n+1}}\wedge\cdots\wedge\mathrm{d}\mathtt{x}^{\nu_{d}}$
$\displaystyle=J(\mathtt{x})\frac{\partial\mathtt{x}^{\nu_{1}}}{\partial\phi^{\mu_{1}}(\mathtt{x})}\cdots\frac{\partial\mathtt{x}^{\nu_{n}}}{\partial\phi^{\mu_{n}}(\mathtt{x})}\,\mathrm{d}\Sigma_{\nu_{1}\cdots\nu_{n}}(\mathtt{x})$
(A.46)

where

$J(\mathtt{x})=\det\biggl{(}\frac{\partial\phi^{\mu}(\mathtt{x})}{\partial\mathtt{x}^{\nu}}\biggr{)}$
(A.47)

is the Jacobian. Denoting the product of partial derivatives as

$K^{\mu_{1}\cdots\mu_{n}}_{\phantom{\mu_{1}\cdots\mu_{n}}\nu_{1}\cdots\nu_{n}}(\mathtt{x})=\frac{\partial\mathtt{x}^{\mu_{1}}}{\partial\phi^{\nu_{1}}(\mathtt{x})}\cdots\frac{\partial\mathtt{x}^{\mu_{n}}}{\partial\phi^{\nu_{n}}(\mathtt{x})}$
(A.48)

the integral (A.45) takes the form

${\cal I}[\alpha,\phi]=\int_{\cal
S}\mathrm{d}\Sigma_{\mu_{1}\cdots\mu_{n}}(\mathtt{x})\,J(\mathtt{x})K^{\mu_{1}\cdots\mu_{n}}_{\phantom{\mu_{1}\cdots\mu_{n}}\nu_{1}\cdots\nu_{n}}(\mathtt{x})\alpha^{\nu_{1}\cdots\nu_{n}}\bigl{(}\phi(\mathtt{x})\bigr{)}.$
(A.49)

Applying the field space exterior derivative $\delta$, we now obtain

$\displaystyle\delta{\cal I}[\alpha,\phi]$ $\displaystyle=\int_{\cal
S}\mathrm{d}\Sigma_{\mu_{1}\cdots\mu_{n}}(\mathtt{x})\,\biggl{[}\delta
J(\mathtt{x})K^{\mu_{1}\cdots\mu_{n}}_{\phantom{\mu_{1}\cdots\mu_{n}}\nu_{1}\cdots\nu_{n}}(\mathtt{x})\alpha^{\nu_{1}\cdots\nu_{n}}\bigl{(}\phi(\mathtt{x})\bigr{)}$
$\displaystyle+J(\mathtt{x})\Bigl{(}\delta
K^{\mu_{1}\cdots\mu_{n}}_{\phantom{\mu_{1}\cdots\mu_{n}}\nu_{1}\cdots\nu_{n}}(\mathtt{x})\alpha^{\nu_{1}\cdots\nu_{n}}\bigl{(}\phi(\mathtt{x})\bigr{)}+K^{\mu_{1}\cdots\mu_{n}}_{\phantom{\mu_{1}\cdots\mu_{n}}\nu_{1}\cdots\nu_{n}}(\mathtt{x})\delta\bigl{[}\alpha^{\nu_{1}\cdots\nu_{n}}\bigl{(}\phi(\mathtt{x})\bigr{)}\bigr{]}\Bigr{)}\biggr{]}.$
(A.50)

Consider first the variation of $\alpha\bigl{(}\phi(\mathtt{x})\bigr{)}$:

$\delta\bigl{[}\alpha^{\nu_{1}\cdots\nu_{n}}\bigl{(}\phi(\mathtt{x})\bigr{)}\bigr{]}=(\delta\alpha)^{\nu_{1}\cdots\nu_{n}}\bigl{(}\phi(\mathtt{x})\bigr{)}+\delta\phi^{\mu}(\mathtt{x})\partial_{\mu}\alpha^{\nu_{1}\cdots\nu_{n}}\bigl{(}\phi(\mathtt{x})\bigr{)}.$
(A.51)

(Throughout this section, $\partial_{\mu}$ denotes differentiation with
respect to $\phi^{\mu}(\mathtt{x})=x^{\mu}$.) Introducing the object

$\chi^{\mu}(x):=\bigl{(}\delta\phi^{\mu}\circ\phi^{-1}\bigr{)}(x),$ (A.52)

which is a vector field on spacetime and a one-form on field space, the
variation (A.51) can be written as

$\delta\bigl{[}\alpha^{\nu_{1}\cdots\nu_{n}}\bigl{(}\phi(\mathtt{x})\bigr{)}\bigr{]}=(\delta\alpha)^{\nu_{1}\cdots\nu_{n}}\bigl{(}\phi(\mathtt{x})\bigr{)}+\chi^{\mu}\bigl{(}\phi(\mathtt{x})\bigr{)}\partial_{\mu}\alpha^{\nu_{1}\cdots\nu_{n}}\bigl{(}\phi(\mathtt{x})\bigr{)}.$
(A.53)

Next, we deduce the variation of the partial derivatives by using the matrix
identity

$\delta A^{-1}=-A^{-1}(\delta A)A^{-1}.$ (A.54)

We have

$\displaystyle\delta\biggl{(}\frac{\partial\mathtt{x}^{\mu}}{\partial\phi^{\nu}(\mathtt{x})}\biggr{)}$
$\displaystyle=-\frac{\partial\mathtt{x}^{\mu}}{\partial\phi^{\lambda}(\mathtt{x})}\frac{\partial\delta\phi^{\lambda}(\mathtt{x})}{\partial\mathtt{x}^{\sigma}}\frac{\partial\mathtt{x}^{\sigma}}{\partial\phi^{\nu}(\mathtt{x})}$
$\displaystyle=-\frac{\partial\mathtt{x}^{\mu}}{\partial\phi^{\lambda}(\mathtt{x})}\frac{\partial\chi^{\lambda}\bigl{(}\phi(\mathtt{x})\bigr{)}}{\partial\mathtt{x}^{\sigma}}\frac{\partial\mathtt{x}^{\sigma}}{\partial\phi^{\nu}(\mathtt{x})}$
$\displaystyle=-\frac{\partial\mathtt{x}^{\mu}}{\partial\phi^{\lambda}(\mathtt{x})}\partial_{\nu}\chi^{\lambda}\bigl{(}\phi(\mathtt{x})\bigr{)}$
(A.55)

from which it follows that

$\displaystyle\delta
K^{\mu_{1}\cdots\mu_{n}}_{\phantom{\mu_{1}\cdots\mu_{n}}\nu_{1}\cdots\nu_{n}}(\mathtt{x})=\;$
$\displaystyle{-}\partial_{\nu_{1}}\chi^{\lambda}\bigl{(}\phi(\mathtt{x})\bigr{)}K^{\mu_{1}\cdots\mu_{n}}_{\phantom{\mu_{1}\cdots\mu_{n}}\lambda\nu_{2}\cdots\nu_{n}}(\mathtt{x})-\ldots$
$\displaystyle{-}\partial_{\nu_{n}}\chi^{\lambda}\bigl{(}\phi(\mathtt{x})\bigr{)}K^{\mu_{1}\cdots\mu_{n}}_{\phantom{\mu_{1}\cdots\mu_{n}}\nu_{1}\cdots\nu_{n-1}\lambda}(\mathtt{x}).$
(A.56)

Finally, the variation of the Jacobian is obtained by applying the identity

$\delta\det A=(\det A)\text{Tr}\bigl{(}A^{-1}\delta A\bigr{)}$ (A.57)

to the matrix

$A^{\mu}_{\phantom{\mu}\nu}=\frac{\partial\phi^{\mu}(\mathtt{x})}{\partial\mathtt{x}^{\nu}}.$
(A.58)

Since

$\text{Tr}\bigl{(}A^{-1}\delta
A\bigr{)}=(A^{-1})^{\mu}_{\phantom{\mu}\nu}\delta
A^{\nu}_{\phantom{\nu}\mu}=\frac{\partial\mathtt{x}^{\mu}}{\partial\phi^{\nu}(\mathtt{x})}\frac{\partial\delta\phi^{\nu}(\mathtt{x})}{\partial\mathtt{x}^{\mu}}=\frac{\partial\delta\phi^{\nu}(\mathtt{x})}{\partial\phi^{\nu}(\mathtt{x})}=\partial_{\nu}\chi^{\nu}\bigl{(}\phi(\mathtt{x})\bigr{)},$
(A.59)

we see that

$\delta
J(\mathtt{x})=J(\mathtt{x})\partial_{\mu}\chi^{\mu}\bigl{(}\phi(\mathtt{x})\bigr{)}.$
(A.60)

Now going back to (A.50), we find

$\displaystyle\delta{\cal I}[\alpha,\phi]=\int_{\cal S}$ $\displaystyle
d\Sigma_{\mu_{1}\cdots\mu_{n}}(\mathtt{x})\,J(\mathtt{x})K^{\mu_{1}\cdots\mu_{n}}_{\phantom{\mu_{1}\cdots\mu_{n}}\nu_{1}\cdots\nu_{n}}(\mathtt{x})\biggl{[}(\delta\alpha)^{\nu_{1}\cdots\nu_{n}}\bigl{(}\phi(\mathtt{x})\bigr{)}$
$\displaystyle+\partial_{\lambda}\chi^{\lambda}\bigl{(}\phi(\mathtt{x})\bigr{)}\alpha^{\nu_{1}\cdots\nu_{n}}\bigl{(}\phi(\mathtt{x})\bigr{)}+\chi^{\lambda}\bigl{(}\phi(\mathtt{x})\bigr{)}\partial_{\lambda}\alpha^{\nu_{1}\cdots\nu_{n}}\bigl{(}\phi(\mathtt{x})\bigr{)}$
$\displaystyle-\partial_{\lambda}\chi^{\nu_{1}}\bigl{(}\phi(\mathtt{x})\bigr{)}\alpha^{\lambda\nu_{2}\cdots\nu_{n}}\bigl{(}\phi(\mathtt{x})\bigr{)}-\ldots-\partial_{\lambda}\chi^{\nu_{n}}\bigl{(}\phi(\mathtt{x})\bigr{)}\alpha^{\nu_{1}\cdots\nu_{n-1}\lambda}\bigl{(}\phi(\mathtt{x})\bigr{)}\biggr{]}.$
(A.61)

This result can be expressed in a more elegant form by observing that the
terms involving the vector field $\chi^{\mu}$ amount to the Lie derivative of
the tensor density $\alpha^{\mu_{1}\cdots\mu_{n}}$ along $\chi^{\mu}$:

${\cal
L}_{\chi}\alpha^{\mu_{1}\cdots\mu_{n}}=\chi^{\lambda}\partial_{\lambda}\alpha^{\mu_{1}\cdots\mu_{n}}-(\partial_{\lambda}\chi^{\mu_{1}})\alpha^{\lambda\mu_{2}\cdots\mu_{n}}-\ldots-(\partial_{\lambda}\chi^{\mu_{n}})\alpha^{\mu_{1}\cdots\mu_{n-1}\lambda}+(\partial_{\lambda}\chi^{\lambda})\alpha^{\mu_{1}\cdots\mu_{n}}.$
(A.62)

Hence we have established that the variation of the integral (A.45) is given
by the expression

$\delta{\cal I}[\alpha,\phi]=\int_{\cal
S}\mathrm{d}\Sigma_{\mu_{1}\cdots\mu_{n}}\bigl{(}\phi(\mathtt{x})\bigr{)}\,\Bigl{[}(\delta\alpha)^{\mu_{1}\cdots\mu_{n}}\bigl{(}\phi(\mathtt{x})\bigr{)}+{\cal
L}_{\chi}\alpha^{\mu_{1}\cdots\mu_{n}}\bigl{(}\phi(\mathtt{x})\bigr{)}\Bigr{]}$
(A.63)

or, in coordinate-free notation,

$\delta\int_{\cal S}\phi^{*}\alpha=\int_{\cal
S}\phi^{*}\bigl{(}\delta\alpha+{\cal L}_{\chi}\alpha\bigr{)}.$ (A.64)

Note that the role of the variational vector field $\chi^{\mu}$ is essentially
to encode the contribution arising from the variation of the embedding map; if
the map $\phi$ carries no field space dependence we have $\chi^{\mu}=0$ and in
this case the variation commutes with the pullback as one would expect.

###  A.3 Lie derivative of the boundary symplectic potential in gravity

In this section we present the details of the calculation which shows that the
boundary presymplectic term introduced in section 5.4 has a vanishing field
space Lie derivative along the gauge orbits of diffeomorphisms. In section 5.4
we arrived at the expression

$\mathfrak{L}_{\cal
V}\Theta_{\phi}=\int_{\partial\Sigma}\mathrm{d}\sigma_{\mu\nu}\bigl{(}\phi(\mathtt{x})\bigr{)}\,\Bigl{[}-\bigl{(}\delta\xi^{\mu\nu}\bigr{)}\bigl{(}\phi(\mathtt{x})\bigr{)}+\bigl{(}\mathfrak{i}_{\cal
V}\delta\eta^{\mu\nu}\bigr{)}\bigl{(}\phi(\mathtt{x})\bigr{)}-{\cal
L}_{\epsilon}\eta^{\mu\nu}\bigl{(}\phi(\mathtt{x})\bigr{)}\Bigr{]}$ (A.65)

where

$\displaystyle\xi^{\mu\nu}$
$\displaystyle=\sqrt{-g}\nabla^{\mu}\epsilon^{\nu},$ (A.66)
$\displaystyle\eta^{\mu\nu}$ $\displaystyle=\sqrt{-g}\nabla^{\mu}\chi^{\nu}.$
(A.67)

(For convenience we omit the antisymmetrization on $\mu$ and $\nu$; this can
be done freely, since the expressions we handle will eventually be contracted
against the antisymmetric surface element $\mathrm{d}\sigma_{\mu\nu}$.) For
the variation of $\xi^{\mu\nu}$, we have

$\delta\xi^{\mu\nu}=\delta\bigl{(}\sqrt{-g}\nabla^{\mu}\epsilon^{\nu}\bigr{)}=\sqrt{-g}\Bigl{(}\delta\gamma\nabla^{\mu}\epsilon^{\nu}+\delta\bigl{(}\nabla^{\mu}\epsilon^{\nu}\bigr{)}\Bigr{)}.$
(A.68)

The variation of the covariant derivative gives

$\delta\bigl{(}\nabla^{\mu}\epsilon^{\nu}\bigr{)}=\delta\Bigl{(}g^{\mu\alpha}\bigl{(}\partial_{\alpha}\epsilon^{\nu}+\Gamma^{\nu}_{\alpha\beta}\epsilon^{\beta}\bigr{)}\Bigr{)}=\delta
g^{\mu\alpha}\nabla_{\alpha}\epsilon^{\nu}+g^{\mu\alpha}\delta\Gamma^{\nu}_{\alpha\beta}\epsilon^{\beta}$
(A.69)

so we obtain

$\delta\xi^{\mu\nu}=\sqrt{-g}\Bigl{(}\bigl{(}\nabla^{\mu}\epsilon^{\nu}\bigr{)}\delta\gamma+\bigl{(}\nabla_{\alpha}\epsilon^{\nu}\bigr{)}\delta
g^{\mu\alpha}+g^{\mu\alpha}\epsilon^{\beta}\delta\Gamma^{\nu}_{\alpha\beta}\Bigr{)}.$
(A.70)

Next, we compute

$\delta\eta^{\mu\nu}=\sqrt{-g}\Bigl{(}\delta\gamma\nabla^{\mu}\chi^{\nu}+\delta\bigl{(}\nabla^{\mu}\chi^{\nu}\bigr{)}\Bigr{)}$
(A.71)

and

$\displaystyle\mathfrak{i}_{\cal V}\delta\eta^{\mu\nu}$
$\displaystyle=\sqrt{-g}\Bigl{(}\bigl{(}\mathfrak{i}_{\cal
V}\delta\gamma\bigr{)}\nabla^{\mu}\chi^{\nu}-\delta\gamma\nabla^{\mu}\bigl{(}\mathfrak{i}_{\cal
V}\chi^{\nu}\bigr{)}+\mathfrak{i}_{\cal
V}\delta\bigl{(}\nabla^{\mu}\chi^{\nu}\bigr{)}\Bigr{)}$
$\displaystyle=\sqrt{-g}\Bigl{(}\bigl{(}\nabla_{\alpha}\epsilon^{\alpha}\bigr{)}\bigl{(}\nabla^{\mu}\chi^{\nu}\bigr{)}+\bigl{(}\nabla^{\mu}\epsilon^{\nu}\bigr{)}\delta\gamma+\mathfrak{i}_{\cal
V}\delta\bigl{(}\nabla^{\mu}\chi^{\nu}\bigr{)}\Bigr{)}.$ (A.72)

In the variation of the covariant derivative, the field space exterior
derivative now acts also on the variational vector field $\chi^{\mu}$:

$\displaystyle\delta\bigl{(}\nabla^{\mu}\chi^{\nu}\bigr{)}$
$\displaystyle=\delta
g^{\mu\alpha}\nabla_{\alpha}\chi^{\nu}+g^{\mu\alpha}\delta\bigl{(}\nabla_{\alpha}\chi^{\nu}\bigr{)}$
$\displaystyle=\delta
g^{\mu\alpha}\nabla_{\alpha}\chi^{\nu}+\nabla^{\mu}\delta\chi^{\nu}+g^{\mu\alpha}\delta\Gamma^{\nu}_{\alpha\beta}\chi^{\beta}$
(A.73)

leading to

$\displaystyle\mathfrak{i}_{\cal
V}\delta\bigl{(}\nabla^{\mu}\chi^{\nu}\bigr{)}$
$\displaystyle=\bigl{(}\mathfrak{i}_{\cal V}\delta
g^{\mu\alpha}\bigr{)}\nabla_{\alpha}\chi^{\nu}-\delta
g^{\mu\alpha}\nabla_{\alpha}\bigl{(}\mathfrak{i}_{\cal V}\chi^{\nu}\bigr{)}$
$\displaystyle\quad+\nabla^{\mu}\bigl{(}\mathfrak{i}_{\cal
V}\delta\chi^{\nu}\bigr{)}+g^{\mu\alpha}\Bigl{(}\bigl{(}\mathfrak{i}_{\cal
V}\delta\Gamma^{\nu}_{\alpha\beta}\bigr{)}\chi^{\beta}-\delta\Gamma^{\nu}_{\alpha\beta}\bigl{(}\mathfrak{i}_{\cal
V}\chi^{\beta}\bigr{)}\Bigr{)}$
$\displaystyle=-\bigl{(}\nabla^{\mu}\epsilon^{\alpha}+\nabla^{\alpha}\epsilon^{\mu}\bigr{)}\nabla_{\alpha}\chi^{\nu}+\bigl{(}\nabla_{\alpha}\epsilon^{\nu}\bigr{)}\delta
g^{\mu\alpha}$ $\displaystyle\quad+\nabla^{\mu}\bigl{(}{\cal
L}_{\epsilon}\chi^{\nu}\bigr{)}+g^{\mu\alpha}\epsilon^{\beta}\delta\Gamma^{\nu}_{\alpha\beta}+g^{\mu\alpha}\bigl{(}\mathfrak{i}_{\cal
V}\delta\Gamma^{\nu}_{\alpha\beta}\bigr{)}\chi^{\beta}.$ (A.74)

where we have noted that

$\mathfrak{i}_{\cal V}\delta\chi^{\mu}=-\bigl{(}\mathfrak{i}_{\cal
V}\chi^{\nu}\bigr{)}\partial_{\nu}\chi^{\mu}+\chi^{\nu}\partial_{\nu}\bigl{(}\mathfrak{i}_{\cal
V}\chi^{\mu}\bigr{)}=\epsilon^{\nu}\partial_{\nu}\chi^{\mu}-\chi^{\nu}\partial_{\nu}\epsilon^{\mu}={\cal
L}_{\epsilon}\chi^{\mu}.$ (A.75)

Combining (A.70) and (A.72), we now find (note that each term in
$-\delta\xi^{\mu\nu}$ is cancelled against a corresponding term in
$\mathfrak{i}_{\cal V}\delta\eta^{\mu\nu}$)

$\displaystyle-\delta\xi^{\mu\nu}+\mathfrak{i}_{\cal
V}\delta\eta^{\mu\nu}=\sqrt{-g}\Bigl{(}$
$\displaystyle\bigl{(}\nabla_{\alpha}\epsilon^{\alpha}\bigr{)}\bigl{(}\nabla^{\mu}\chi^{\nu}\bigr{)}-\bigl{(}\nabla^{\mu}\epsilon^{\alpha}+\nabla^{\alpha}\epsilon^{\mu}\bigr{)}\bigl{(}\nabla_{\alpha}\chi^{\nu}\bigr{)}$
$\displaystyle+\nabla^{\mu}\bigl{(}{\cal
L}_{\epsilon}\chi^{\nu}\bigr{)}+g^{\mu\alpha}\bigl{(}\mathfrak{i}_{\cal
V}\delta\Gamma^{\nu}_{\alpha\beta}\bigr{)}\chi^{\beta}\Bigr{)}.$ (A.76)

On the second line, we have

$\displaystyle\nabla^{\mu}\bigl{(}{\cal L}_{\epsilon}\chi^{\nu}\bigr{)}$
$\displaystyle=\nabla^{\mu}\bigl{(}\epsilon^{\alpha}\nabla_{\alpha}\chi^{\nu}-\chi^{\alpha}\nabla_{\alpha}\epsilon^{\nu}\bigr{)}$
$\displaystyle=\bigl{(}\nabla^{\mu}\epsilon^{\alpha}\bigr{)}\bigl{(}\nabla_{\alpha}\chi^{\nu}\bigr{)}+\epsilon^{\alpha}\nabla^{\mu}\nabla_{\alpha}\chi^{\nu}-\bigl{(}\nabla^{\mu}\chi^{\alpha}\bigr{)}\bigl{(}\nabla_{\alpha}\epsilon^{\nu}\bigr{)}-\chi^{\alpha}\nabla^{\mu}\nabla_{\alpha}\epsilon^{\nu}$
(A.77)

and the contraction

$\mathfrak{i}_{\cal
V}\delta\Gamma^{\mu}_{\alpha\beta}=\frac{1}{2}\Bigl{(}\bigl{[}\nabla_{\alpha},\nabla^{\mu}\bigr{]}\epsilon_{\beta}+\bigl{[}\nabla_{\beta},\nabla^{\mu}\bigr{]}\epsilon_{\alpha}+\bigl{(}\nabla_{\alpha}\nabla_{\beta}+\nabla_{\beta}\nabla_{\alpha}\bigr{)}\epsilon^{\mu}\Bigr{)}$
(A.78)

which has already been encountered in our earlier calculations. Hence

$\displaystyle-\delta\xi^{\mu\nu}+\mathfrak{i}_{\cal
V}\delta\eta^{\mu\nu}=\sqrt{-g}\Bigl{(}$
$\displaystyle\bigl{(}\nabla_{\alpha}\epsilon^{\alpha}\bigr{)}\bigl{(}\nabla^{\mu}\chi^{\nu}\bigr{)}-\bigl{(}\nabla^{\mu}\epsilon^{\alpha}+\nabla^{\alpha}\epsilon^{\mu}\bigr{)}\bigl{(}\nabla_{\alpha}\chi^{\nu}\bigr{)}$
$\displaystyle+\bigl{(}\nabla^{\mu}\epsilon^{\alpha}\bigr{)}\bigl{(}\nabla_{\alpha}\chi^{\nu}\bigr{)}-\bigl{(}\nabla^{\mu}\chi^{\alpha}\bigr{)}\bigl{(}\nabla_{\alpha}\epsilon^{\nu}\bigr{)}-\bigl{(}\nabla^{\mu}\nabla_{\alpha}\epsilon^{\nu}\bigr{)}\chi^{\alpha}+\epsilon^{\alpha}\bigl{(}\nabla^{\mu}\nabla_{\alpha}\chi^{\nu}\bigr{)}$
$\displaystyle+\frac{1}{2}g^{\mu\alpha}\bigl{(}\nabla_{\alpha}\nabla^{\nu}\epsilon_{\beta}-\nabla^{\nu}\nabla_{\alpha}\epsilon_{\beta}+\nabla_{\beta}\nabla^{\nu}\epsilon_{\alpha}-\nabla^{\nu}\nabla_{\beta}\epsilon_{\alpha}\bigr{)}\chi^{\beta}$
$\displaystyle+\frac{1}{2}g^{\mu\alpha}\bigl{(}\nabla_{\alpha}\nabla_{\beta}\epsilon^{\nu}+\nabla_{\beta}\nabla_{\alpha}\epsilon^{\nu}\bigr{)}\chi^{\beta}\Bigr{)}.$
(A.79)

The terms on the last two lines can be arranged as

$\displaystyle\frac{1}{2}\bigl{(}\nabla^{\mu}\nabla^{\nu}\epsilon_{\alpha}-\nabla^{\nu}\nabla^{\mu}\epsilon_{\alpha}+\nabla_{\alpha}\nabla^{\nu}\epsilon^{\mu}-\nabla^{\nu}\nabla_{\alpha}\epsilon^{\mu}\bigr{)}\chi^{\alpha}+\frac{1}{2}\bigl{(}\nabla^{\mu}\nabla_{\alpha}\epsilon^{\nu}+\nabla_{\alpha}\nabla^{\mu}\epsilon^{\nu}\bigr{)}\chi^{\alpha}$
$\displaystyle=\Bigl{(}\nabla^{[\mu}\nabla^{\nu]}\epsilon_{\alpha}+\nabla^{[\mu}\nabla_{\alpha}\epsilon^{\nu]}+\nabla_{\alpha}\nabla^{(\mu}\epsilon^{\nu)}\Bigr{)}\chi^{\alpha}.$
(A.80)

Since this will be contracted with the antisymmetric
$\mathrm{d}\sigma_{\mu\nu}$, the last term does not contribute and the second
term cancels against a term on the second line of (A.79). We are then left
with

$\displaystyle-\delta\xi^{\mu\nu}+\mathfrak{i}_{\cal
V}\delta\eta^{\mu\nu}=\sqrt{-g}\Bigl{(}$
$\displaystyle\bigl{(}\nabla_{\alpha}\epsilon^{\alpha}\bigr{)}\bigl{(}\nabla^{\mu}\chi^{\nu}\bigr{)}-\bigl{(}\nabla^{\alpha}\epsilon^{\mu}\bigr{)}\bigl{(}\nabla_{\alpha}\chi^{\nu}\bigr{)}-\bigl{(}\nabla_{\alpha}\epsilon^{\nu}\bigr{)}\bigl{(}\nabla^{\mu}\chi^{\alpha}\bigr{)}$
$\displaystyle+\epsilon^{\alpha}\bigl{(}\nabla^{\mu}\nabla_{\alpha}\chi^{\nu}\bigr{)}+\bigl{(}\nabla^{\mu}\nabla^{\nu}\epsilon_{\alpha}\bigr{)}\chi^{\alpha}\Bigr{)}.$
(A.81)

Recall that these were the first two terms of the integrand in (A.65). The
last term is

$-{\cal
L}_{\epsilon}\eta^{\mu\nu}=-\epsilon^{\alpha}\nabla_{\alpha}\eta^{\mu\nu}+\bigl{(}\nabla_{\alpha}\epsilon^{\mu}\bigr{)}\eta^{\alpha\nu}+\bigl{(}\nabla_{\alpha}\epsilon^{\nu}\bigr{)}\eta^{\mu\alpha}-\bigl{(}\nabla_{\alpha}\epsilon^{\alpha}\bigr{)}\eta^{\mu\nu}.$
(A.82)

Here the last term on the right-hand side is cancelled by the first term in
(A.79). The two middle terms can be written as

$\displaystyle\bigl{(}\nabla_{\alpha}\epsilon^{\mu}\bigr{)}\eta^{\alpha\nu}+\bigl{(}\nabla_{\alpha}\epsilon^{\nu}\bigr{)}\eta^{\mu\alpha}$
$\displaystyle=\frac{1}{2}\sqrt{-g}\Bigl{(}\bigl{(}\nabla_{\alpha}\epsilon^{\mu}\bigr{)}\bigl{(}\nabla^{\alpha}\chi^{\nu}-\nabla^{\nu}\chi^{\alpha}\bigr{)}+\bigl{(}\nabla_{\alpha}\epsilon^{\nu}\bigr{)}\bigl{(}\nabla^{\mu}\chi^{\alpha}-\nabla^{\alpha}\chi^{\mu}\bigr{)}\Bigr{)}$
$\displaystyle=\sqrt{-g}\Bigl{(}\bigl{(}\nabla^{\alpha}\epsilon^{[\mu}\bigr{)}\bigl{(}\nabla_{\alpha}\chi^{\nu]}\bigr{)}+\bigl{(}\nabla_{\alpha}\epsilon^{[\nu}\bigr{)}\bigl{(}\nabla^{\mu]}\chi^{\alpha}\bigr{)}\Bigr{)}$
(A.83)

from which we see that they cancel against the next two terms in (A.79) (after
taking into account the antisymmetrization due to
$\mathrm{d}\sigma_{\mu\nu}$). Combining now the term remaining in (A.82),
namely

$-\epsilon^{\alpha}\nabla_{\alpha}\eta^{\mu\nu}=-\sqrt{-g}\epsilon^{\alpha}\nabla_{\alpha}\nabla^{[\mu}\chi^{\nu]}$
(A.84)

with the two remaining terms in (A.79), we obtain

$\displaystyle\mathfrak{L}_{\cal V}\Theta_{\phi}$
$\displaystyle=\int_{\partial\Sigma}\mathrm{d}\sigma_{\mu\nu}\bigl{(}\phi(\mathtt{x})\bigr{)}\,\sqrt{-g}\Bigl{(}\epsilon^{\alpha}\bigl{(}\nabla^{\mu}\nabla_{\alpha}\chi^{\nu}\bigr{)}+\bigl{(}\nabla^{\mu}\nabla^{\nu}\epsilon_{\alpha}\bigr{)}\chi^{\alpha}-\epsilon^{\alpha}\bigl{(}\nabla_{\alpha}\nabla^{\mu}\chi^{\nu}\bigr{)}\Bigr{)}\bigl{(}\phi(\mathtt{x})\bigr{)}$
$\displaystyle=\int_{\partial\Sigma}\mathrm{d}\sigma_{\mu\nu}\bigl{(}\phi(\mathtt{x})\bigr{)}\,\sqrt{-g}\Bigl{(}\epsilon^{\alpha}\bigl{[}\nabla^{\mu},\nabla_{\alpha}\bigr{]}\chi^{\nu}+\frac{1}{2}\bigl{(}\bigl{[}\nabla^{\mu},\nabla^{\nu}\bigr{]}\epsilon_{\alpha}\bigr{)}\chi^{\alpha}\Bigr{)}\bigl{(}\phi(\mathtt{x})\bigr{)}.$
(A.85)

We continue the calculation by expressing the commutators of covariant
derivatives in terms of the Riemann tensor. Taking $\mu$ and $\nu$ as lower
indices for convenience, we have

$\displaystyle\bigl{[}\nabla_{\mu},\nabla_{\alpha}\bigr{]}\chi_{\nu}$
$\displaystyle=R_{\nu\lambda\mu\alpha}\chi^{\lambda},$
$\displaystyle\bigl{[}\nabla_{\mu},\nabla_{\nu}\bigr{]}\epsilon_{\alpha}$
$\displaystyle=R_{\alpha\lambda\mu\nu}\epsilon^{\lambda}.$ (A.86)

and

$\epsilon^{\alpha}\bigl{[}\nabla_{\mu},\nabla_{\alpha}\bigr{]}\chi_{\nu}+\frac{1}{2}\bigl{(}\bigl{[}\nabla_{\mu},\nabla_{\nu}\bigr{]}\epsilon_{\alpha}\bigr{)}\chi^{\alpha}=\epsilon^{\alpha}R_{\nu\lambda\mu\alpha}\chi^{\lambda}+\frac{1}{2}R_{\alpha\lambda\mu\nu}\epsilon^{\lambda}\chi^{\alpha}=\Bigl{(}R_{\nu\lambda\mu\alpha}+\frac{1}{2}R_{\lambda\alpha\mu\nu}\Bigr{)}\epsilon^{\alpha}\chi^{\lambda}.$
(A.87)

Since this expression will be antisymmetrized in $\mu$ and $\nu$, we can write
it as

$\frac{1}{2}\Bigl{(}R_{\nu\lambda\mu\alpha}-R_{\mu\lambda\nu\alpha}+R_{\lambda\alpha\mu\nu}\Bigr{)}\epsilon^{\alpha}\chi^{\lambda}.$
(A.88)

Now using the symmetries of the Riemann tensor, namely
$R_{\alpha\beta\mu\nu}=-R_{\alpha\beta\nu\mu}$ and
$R_{\alpha\beta\mu\nu}=R_{\mu\nu\alpha\beta}$, we see that

$\displaystyle
R_{\nu\lambda\mu\alpha}-R_{\mu\lambda\nu\alpha}+R_{\lambda\alpha\mu\nu}$
$\displaystyle=R_{\mu\alpha\nu\lambda}-R_{\mu\lambda\nu\alpha}+R_{\mu\nu\lambda\alpha}$
$\displaystyle=R_{\mu\alpha\nu\lambda}+R_{\mu\lambda\alpha\nu}+R_{\mu\nu\lambda\alpha}$
$\displaystyle=0$ (A.89)

due to the first Bianchi identity. Hence we have managed to show that

$\mathfrak{L}_{\cal V}\Theta_{\phi}=0.$ (A.90)

Note that this result followed from the algebraic properties of the Riemann
tensor and no use of the equations of motion was necessary in order to
establish it, reflecting the fact that the invariance of the boundary
presymplectic form under diffeomorphisms is a purely geometric property, and
as such it should be valid independently of any dynamics.

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