I have an operator $C$ that I wish to diagonalize on a Riemmanian manifold $M$ with *constant curvature* $\Lambda$
$$C = A + B$$
Now I know that these operators $A$ and $B$ commute in flat space, but on a curved space, they give
$$[A,B] = \Lambda B$$


Here's an example on $M = H^2$, where we will consider $A$ to be the Laplace-Beltrami operator and $B$ the covariant derivative times a constant vector V. Then, in components, $$[\nabla^2,V\cdot\nabla] = V^\mu[\nabla^2,\nabla\_\mu] = V^\mu R\_{\mu}^{~~\nu}\nabla\_\nu = \Lambda ~V\cdot\nabla$$


In General, does this mean that $C$ is not diagonalizable on $M$, since $A$ and $B$ aren't simultaneously diagonalizible?



Simultaneous diagonalization on spaces with constant curvature
You can try the following trick. Using the identity
$$
 e^{-B/\mu} A e^{B/\mu} = A + \mu^{-1} [A,B] + \frac{\mu^{-2}}{2!} [[A,B],B] + \cdots = A + (\Lambda/\mu) B
$$
and setting $\mu = -\Lambda$ we have
$$
 e^{B/\Lambda} C e^{-B/\lambda} = (A - B) + B = A .
$$
So if we can diagonalize $A = UDU^{-1}$, with $D$ diagonal, then we can diagonalize
$$
 C = e^{-B/\Lambda} U D U^{-1} e^{B/\Lambda} .
$$


However, you should note that $B=0$ in your example, since the only covariantly constant vector (meaning $\nabla\_\nu V\_\mu = 0$) on a constant curvature space is $V\_\mu = 0$.


