Reparametrization invariance of the particle in GR In the article Antibracket, Antifields and Gauge-Theory Quantization, the relativistic particle in spacetime is studied. Its action is
$$\int\text{d}\lambda\,\frac{1}{2}\left(\frac{1}{e}\dot{x}^2-m^2e\right).\tag{3.1}$$
Under reparametrization invariance they show this action transforms like
$$\int \text{d}\lambda\,\left(-\left(\frac{\dot{x}^\mu}{e}\right)\frac{\text{d}}{\text{d}\lambda}\left(\frac{\dot{x}_\mu}{e}\right)-\frac{1}{2}\left(\frac{\dot{x}^2}{e^2}-m^2\right)\frac{d}{d\lambda}\right)\delta(\lambda-\lambda').\tag{3.4}$$
Integration by parts clearly shows that this is equal to $0$ in Minkowski space. This should of course be true in all spacetimes, since the action is clearly the integral of a one form ($m^2e$ is a 1-form since $e$ is a 1-form and $\dot{x}^2/e$ is a 1-form since it is the contraction of the induced metric $\dot{x}^2$ on the worldline with the vector field $1/e$ dual to $e$). However, I don't see why the integral above vanishes on general spacetimes. In particular, what is bothering me is that in general
$$\frac{1}{2}\frac{d}{d\lambda}\left(\frac{\dot{x}^2}{e^2}\right)=\frac{1}{2}\frac{d}{d\lambda}\left(g_{\mu\nu}(x)\frac{\dot{x}^\mu\dot{x}^\nu}{e^2}\right)=\frac{1}{2}\frac{d}{d\lambda}\left(g_{\mu\nu}(x)\frac{\dot{x}^\mu}{e}\right)\frac{\dot{x}^\nu}{e}+\frac{1}{2}g_{\mu\nu}(x)\frac{\dot{x}^\mu}{e}\frac{d}{d\lambda}\left(\frac{\dot{x}^\nu}{e^2}\right)\\
=\frac{\dot{x}^\mu}{e}\frac{d}{d\lambda}\left(\frac{\dot{x}_\mu}{e}\right)-\color{red}{\frac{1}{2e^2}\partial_\alpha g_{\mu\nu}\dot{x}^\mu\dot{x}^\nu\dot{x}^\alpha}.$$
The last term doesn't appear when the metric is constant and I don't see why the integral vanishes when this term is present.
 A: TL;DR: Ref. 1 only considers the flat Minkowski metric $\eta_{\mu\nu}={\rm diag}(-1,+1,\ldots,+1)$, see last paragraph on p. 9.

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*In this answer let us derive the Noether identity for a curved metric $g_{\mu\nu}$. The Lagrangian is
$$ L~=~\frac{\dot{x}^2}{2e}- \frac{m^2e}{2}. \tag{3.1}$$
The (vertical) infinitesimal gauge transformations are
$$ \delta_R x^{\mu} ~=~\epsilon\dot{x}^{\mu},\qquad  \delta_R e ~=~\frac{d(\epsilon e)}{d\lambda},\qquad  \delta_R \lambda ~=~0.\tag{3.6}$$
The abelianized (vertical) infinitesimal gauge transformations are
$$ \delta  x^{\mu} ~=~\epsilon\frac{\dot{x}^{\mu}}{e},\qquad  \delta  e ~=~  \dot{\epsilon} ,\qquad  \delta \lambda ~=~0.\tag{3.3}$$


*So the infinitesimal variation of the Lagrangian reads
$$ \delta L ~=~\frac{\dot{x}_{\mu}}{e} \frac{d(\delta x^{\mu} )}{d\lambda} -\frac{1}{2}\left(\frac{\dot{x}^2}{e^2}+ \frac{m^2}{2}\right)\delta e + \color{red}{\frac{1}{2e}\delta x^{\alpha} \partial_{\alpha}g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}}~\stackrel{\text{int. by parts}}{\sim} \epsilon  N, $$
where
$$N~:=~- \frac{\dot{x}^{\mu}}{e}\frac{d}{d\lambda}\left(\frac{\dot{x}_{\mu}}{e}\right)+ \frac{d}{d\lambda}\frac{1}{2}\left(\frac{\dot{x}^2}{e^2}+ \frac{m^2}{2}\right) +\color{red}{\frac{1}{2e^2} \dot{x}^{\alpha} \partial_{\alpha}g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}}~=~0.$$
The last equality is OP's result.


*The Noether identity itself can in principle be derived via integration by parts from
$$0~=~N(\lambda^{\prime})~=~\int\!d\lambda~N(\lambda)~\delta(\lambda\!-\!\lambda^{\prime})~\sim~\ldots.$$
Because of the presence of derivatives of the Dirac delta distributions, it will be important to keep track of what depends on $\lambda$ vs. $\lambda^{\prime}$.
References:

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*J. Gomis, J. Paris & S. Samuel, Antibracket, Antifields and Gauge-Theory Quantization, arXiv:hep-th/9412228.

