Simultaneous diagonalization on spaces with constant curvature

Question

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?

Answer 1

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 $V_\mu = 0$.