Gauge equivalence between operators

Question

I have tried to figure out the following problem for some time now, but with little success: Let $ \mathcal{L} $ be a third order linear differential operator with coefficients in $ \mathbb{C}(X) $. Consider the Picard-Vessiot extension $ K $ of $ \mathbb{C}(X) $ associated to $ \mathcal{L} $ and let $ G $ be the differential Galois group of $ K $ over $ \mathbb{C}(X) $. Assume that $ G \subset \textbf{SO}(3,\mathbb{C})\times\mathbb{C}^{*} $. Show then that $ \mathcal{L} $ is gauge equivalent to the symmetric square of a second order linear differential operator.

If $ G \subset \textbf{SL}(3,\mathbb{C}) $, then it is not to difficult to prove that the statement holds, granted that there exist linear independent solutions over $ \mathbb{C} $ of $ \mathcal{L}(y)=0 $, call them $ y_{1},y_{2},y_{3} $ such that $ y_{2}^{2}=y_{1}y_{3} $, but I don't know how to adapt this to the situation above.

I would appreciate any help. Thank you!

Answer 1

I will assume $G \subset SO(3)$ (I m not sure yet about the general case).

By the following answer we know $SO(3,\mathbb{C})$ has double cover $SL(2,\mathbb{C})$ (which is simply connected). All finite dimensional irreducible representations of $SL(2,\mathbb{C})$ are equivalent to $Sym^n(V)$ where $V$ is the fudamental representation of rank $2$ the ones that descend to $SO(3,\mathbb{C})$ are precisely those with $n$ even. We must have by dimensional consideration that the original differential module associated to the third order differential operator takes the form $Sym^2(V)$ and we're done.

For the case of a general $G$ subject to $G \subset SO(3) \times \mathbb{C}^{\times}$ i'm not sure what can be done and if anything I assume it would require some case by case analysis combining several representation theoretic results similar to the paragraph above.