What should be the equation satisfied by The Momentum commutators in a curved background? This may be obvious but I have limited experience in physics , The generators of Spatial translation symmetry commutes with each other i.e [P(i),P(j)] = 0 but if Spacetime is a curved manifolds then the value of the commutator should not be zero but some invariant property related to curvature i.e a Function of the curvature tensor If this is false then what should the commutator be like e.g in the vicinity of a gravitational source according to GR , I'm sorry I do not know much in relativity nor differential geometry 
 A: The commutator you are interested in is non-trivial if you generalize the
translations to curvilinear coordinates. For a vector function $A^{\alpha
}\left(  x\right)  $, a «translation» along $dx^{\mu}$ is the following
transformation:
$$
A^{\alpha}\rightarrow A^{\alpha}-dx^{\mu}D_{\mu}A^{\alpha}
$$
(where $D_{\mu}$ is the so called covariant derivative) instead of $A^{\alpha
}\rightarrow A^{\alpha}-dx^{\mu}\partial_{\mu}A^{\alpha}$.
First of all, you should understand one basic fact from the group theory: a
commutator of generators corresponds to a generator of some transformation.
This transformation is a superposition of transformations which form a
infinitesimal closed contour in the parametric space of a group. It sounds
complicated but the idea is very simple. Let's imagine you have a group
element:
$$
T\left(  \mathbf{a}\right)  =\exp\left(  i\mathbf{g}\cdot\mathbf{a}\right)
=1+i\mathbf{g}\cdot\mathbf{a+}\frac{1}{2}\left(  i\mathbf{\mathbf{g}
\cdot\mathbf{a}}\right)  ^{2}+\mathbf{\ldots,}
$$
where $a^{n}$ are parameters of the group, $g_{n}$ are generators of the group
and $\mathbf{g}\cdot\mathbf{a}=g_{n}a^{n}$. Let's now consider the following
sequence of transformations: $T\left(  \mathbf{a}\right)  $ then $T\left(
\mathbf{b}\right)  $, so that $\mathbf{b}\neq\mathbf{a}$, then $T\left(
-\mathbf{a}\right)  $, \ so that $T\left(  -\mathbf{a}\right)  T\left(
\mathbf{a}\right)  =1,$ and finally $T\left(  -\mathbf{b}\right)  $. The
parameters of these transformations form a rectangle in the group parameter
space (see the picture below). Therefore, the total composite transformation
has the form:
\begin{align*}
& T\left(  -\mathbf{b}\right)  T\left(  -\mathbf{a}\right)  T\left(
\mathbf{b}\right)  T\left(  \mathbf{a}\right)  =\\
& = \left(  1-i\mathbf{g}
\cdot\mathbf{b+\ldots}\right)  \left(  1-i\mathbf{g}\cdot\mathbf{a+\ldots
}\right)  \left(  1+i\mathbf{g}\cdot\mathbf{b+\ldots}\right)  \left(
1+i\mathbf{g}\cdot\mathbf{a+\ldots}\right)  .
\end{align*}
Let's now assume that $\mathbf{a}$ and $\mathbf{b}$ are infinitesimal small,
so that the expansion of the composite transformation has the form:
\begin{align*}
T\left(  -\mathbf{b}\right)  T\left(  -\mathbf{a}\right)  T\left(
\mathbf{b}\right)  T\left(  \mathbf{a}\right)   &  \approx1+\left(
\mathbf{g}\cdot\mathbf{a}\right)  \left(  \mathbf{g}\cdot\mathbf{b}\right)
-\left(  \mathbf{g}\cdot\mathbf{b}\right)  \left(  \mathbf{g}\cdot
\mathbf{a}\right)  \\
&  =1+\left[  g_{m},g_{n}\right]  a^{m}b^{n}=1+\frac{1}{2}\left[  g_{m}
,g_{n}\right]  f^{mn},\qquad(1)
\end{align*}
where
$$
\left[  g_{m},g_{n}\right]  =g_{m}g_{n}-g_{n}g_{m}
$$
and $f^{mn}$ is the so called oriented area element (or directed area
measure):
$$
f^{mn}=a^{m}b^{n}-a^{n}b^{m}.
$$
For example, if the parameter space of the group is three dimensional (as it
is for 3D translations) then the vector
$$
s^{k}=\frac{1}{2}\epsilon^{kmn}f^{mn}=\left[  \mathbf{a}\times\mathbf{b}
\right]  ^{k}
$$
is transverse to $\mathbf{a}$ and $\mathbf{b}$, so that its length squared is
the area of the rectangle with the sides $\mathbf{a}$ and $\mathbf{b}$ (see
the figure below):
$$
s^{2}=\mathbf{a}^{2}\mathbf{b}^{2}-\left(  \mathbf{a}\cdot\mathbf{b}\right)
^{2}.
$$

For a flat space, the parallel translation of a vector $\mathbf{v}$ along a
vector $\mathbf{a}$ doesn't change the direction of the vector $\mathbf{v}$.
Therefore, the parallel translation around the infinitesimal closed contour
$(\mathbf{a},\mathbf{b},-\mathbf{a},-\mathbf{b})$ equals to the identical
transformation, hence from the equation (1) we obtain $\left[  p_{i}
,p_{j}\right]  =0$. For curved space, the parallel displacement along a small
4-vector $dx^{\nu}$ is non-trivial:
$$
\delta A^{\alpha}=-\Gamma_{\mu\nu}^{\alpha}A^{\mu}dx^{\nu},
$$
where $\Gamma_{\mu\nu}^{\alpha}$ are the so called Christoffel symbols.
Therefore the parallel translation around an infinitesimal closed contour $C$
is the contour integral:
$$
\Delta A^{\alpha}=-
{\displaystyle\oint\limits_{C}}
\Gamma_{\mu\nu}^{\alpha}A^{\mu}dx^{\nu}.
$$
Applying Stokes' theorem to this integral and assuming that the area enclosed
by the contour $C$ has the infinitesimal small value $\Delta f^{\mu\nu}$, one
can show that
$$
\Delta A^{\alpha}=-\frac{1}{2}R^{\alpha}{}_{\beta\mu\nu}A^{\beta}\Delta
f^{\mu\nu},
$$
where
$$
R^{\alpha}{}_{\beta\mu\nu}=\partial_{\mu}\Gamma_{\beta\nu}^{\alpha}
-\partial_{\nu}\Gamma_{\beta\mu}^{\alpha}+\Gamma_{\mu\rho}^{\alpha}
\Gamma_{\beta\nu}^{\rho}-\Gamma_{\nu\rho}^{\alpha}\Gamma_{\beta\mu}^{\rho},
$$
is the well known Riemann tensor. Now you should remember that the
transformation around an infinitesimal closed contour is a commutator, see
(1). Hence for a vector function $A^{\alpha}\left(  x\right)  $ we have:
$$
\left[  D_{\mu},D_{\nu}\right]  A^{\alpha}=R^{\alpha}{}_{\beta\mu\nu}A^{\beta
}.
$$
Therefore you are right saying that the commutator of translations in a curved
space is non zero, it is in fact Riemann tensor. Although the form of the
covariant derivative depends on the geometric type of the field it acts on,
e.g., for a co-variant tensor field:
$$
\left[  D_{\mu},D_{\nu}\right]  A_{\alpha\beta}=A_{\alpha\rho}R^{\rho}
{}_{\beta\nu\mu}+A_{\rho\beta}R^{\rho}{}_{\alpha\nu\mu}.
$$
A: Before discussing commutators in a curved space-time, you should specify rigorously the quantum framework you are going to use. In particular, you should specify what a quantum state is, how exactly do you construct a Hilbert space, and how exactly do you define operators.
As a possible way to do this, you may introduce a 3+1 decomposition of your space-time, define a coordinate basis for your Hilbert space in this composition, and define 3-momenta operators as operators of infinitesimal translations, canonically conjugated to the operators of basis, and commuting with themselves by definition, that is $[\hat p_i,\hat p_j]=0$. Note, that with such a difinition $\hat p_i\neq\dfrac{\hbar}{i}\dfrac{\partial}{\partial x_i}$.
However, as there is no universal way so far to construct a classical quantum theory on a relativistic background, and one may come up with some other construction. In any case, the relation $[\hat p_i,\hat p_j]=0$ is 1) achievable, 2) desirable. Desirable - becuase you would actually want to work with momenta defined in such a way that you can measure their components independantly.
