Shift Symmetry for Scalar Dirac-Born-Infeld (DBI) According to this paper:

*

*Claudia de Rham and Andrew J. Tolley, "DBI and the Galileon reunited", JCAP 1005 (2010) 015, arXiv:1003.5917.

around equation (1)-(2), the DBI action
$$S = \int d^4 x\Big(-\lambda\sqrt{1 + (\partial \pi)^2} + \lambda\Big)\tag{1}$$
is invariant under the non-linearly realized symmetry whose infinitesimal form is
$$\delta_v\pi(x) = v_{\mu}x^{\mu} + \pi(x)v^{\alpha}\partial_{\alpha}\pi(x),\tag{2}$$
in the sense that the Lagrangian changes by a total derivative. I am having trouble confirming that this is true.  I find that
$$\delta_v \mathcal{L} = \frac{\partial^{\mu}\pi\partial_{\mu}\delta_v\pi}{\sqrt{1 + (\partial \pi)^2}} =  \frac{1}{\sqrt{1 + (\partial \pi)^2}}\partial^{\mu}\pi\Big(v_{\mu} + v^{\alpha}\partial_{\mu}\pi \partial_{\alpha}\pi + v^{\alpha}\pi\partial_{\mu}\partial_{\alpha}\pi\Big) = \frac{v^{\mu}}{\sqrt{1 + (\partial \pi)^2}}\Big(\partial_{\mu}\pi + \partial_{\mu}\pi\partial_{\beta}\pi\partial^{\beta}\pi + \pi\partial^{\beta}\pi\partial_{\beta}\partial_{\mu}\pi\Big),$$
which as far as I can tell, is not  a total derivative.  Moreover, in equation (5) the paper claims that any scalar $P$ constructed from $$g_{\mu \nu} = \eta_{\mu\nu} + \partial_{\mu}\pi\partial_{\nu}\pi\tag{2b}$$ should transform like
$$\delta_v P = v^{\alpha}\pi(x)\partial_{\alpha}P.\tag{5}$$
Since it only depends on the determinant of $g_{\mu \nu}$, the Lagrangian is a such a scalar, and I am finding that it does not transform in this fashion.  Moreover even if it did, $\delta_v \mathcal{L}$ would only be a total derivative for constant $\pi$, which cannot be correct.
 A: *

*The DBI action (1) is
$$ \begin{align} S~=~&\int\!d^4x~{\cal L}, \cr 
{\cal L}~=~&\lambda(1-\sqrt{|g|}), \cr  
|g|~=&~-g~=~-\det g_{\mu\nu}~=~\det (\eta^{-1}g)^{\mu}{}_{\nu}
~=~\prod_n\lambda_n~=~ 1+(\partial \pi)^2,\end{align}\tag{1} $$
with metric
$$ \begin{align} g_{\mu\nu}~=~&\eta_{\mu\nu}+\partial_{\mu}\pi ~\partial_{\nu}\pi, \cr   (g^{-1})^{\mu\nu}~=~&(\eta^{-1})^{\mu\nu}-\frac{\partial^{\mu}\pi~ \partial^{\nu}\pi}{1+(\partial \pi)^2}.\end{align}\tag{2b}$$
To deduce the determinant (1) for $(\eta^{-1}g)^{\mu}{}_{\nu}$ note that $\partial_{\nu}\pi$ is a eigenvector with eigenvalue $1+(\partial \pi)^2$, and all the orthogonal eigenvectors carry eigenvalue $1$.


*Ref. 1 shows that the infinitesimal transformation
$$\delta_v g_{\mu\nu}~\stackrel{(2)+(2b)}{=}~\ldots ~\stackrel{(4)}{=}~({\cal L}_{\xi}g)_{\mu\nu} \tag{3}$$
of the metric tensor is a Lie derivative wrt. a vector field
$$ \xi^{\mu}~=~\pi v^{\mu}, \qquad v^{\mu} \text{ is independent of }x.\tag{4} $$


*We calculate that the change in the Lagrangian density
$$ \delta_v{\cal L}~\stackrel{(1)}{=}~-\lambda \delta_v \sqrt{|g|}, \tag{A}$$
where
$$\begin{align} \delta_v \sqrt{|g|}~=~&-\frac{\delta_v g}{2\sqrt{|g|}}\cr
~=~&\ldots~=~\frac{1}{2} \sqrt{|g|}(g^{-1})^{\mu\nu}\delta_v g_{\mu\nu}\cr
~\stackrel{(3)}{=}~&\ldots~=~\frac{1}{2} \sqrt{|g|}\left((g^{-1})^{\mu\nu}\xi[ g_{\mu\nu}] + 2\partial_{\mu}\xi^{\mu}\right) \cr
~=~&\ldots~=~\partial_{\mu}\left(\xi^{\mu}\sqrt{|g|}\right)
\end{align}\tag{B}$$
is a total spacetime derivative, as Ref. 1 claims. Therefore, the infinitesimal transformation $\delta_v$ is a quasi-symmetry.


*The result (B) is completely dictated by the fact that $\sqrt{|g|}$ is a scalar density. Similar any scalar function $P$ must transform as
$$ \delta_v P~=~ {\cal L}_{\xi}P~=~\xi[P],\tag{5} $$
as Ref. 1 claims.
References:

*

*C. de Rham & A.J. Tolley, arXiv:1003.5917.

