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Stefan Kohl
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Gauge equivalence between operators

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!