Recovering Cusp Forms on Gl(2) from Symmetric Cubes

Let F be a number field with adele ring AF , and let π, π ′ be cuspidal automorphic representations of GL2(AF ), say with the same central character. If the symmetric squares of π and π are isomorphic, we know that π will need to be an abelian, in fact quadratic, twist of π, which amounts to a multiplicity one statement for SL(2) ([Ram1]). It is of interest to ask if the situation is the same for the symmetric cube transfer (from GL(2) to GL(4)) constructed by Kim and Shahidi (cf. [KS2]). In an earlier paper [Ram2], dedicated to Freydoon Shahidi, we showed that the answer is in the negative: If π is of icosahedral type in a suitable sense (which is meaningful even for π without an associated Galois representation), there is a cusp form π on GL(2)/F , which we call the “conjugate” of π, having the same symmetric cube, but which is not an abelian twist of π. (We also showed there that such a π is algebraic when the central character ω is algebraic, and is moreover rational over Q[ √ 5] when ω = 1; π is in that case the Galois conjugate of π under the non-trivial automorphism of the coefficient field.) In this Note we consider the converse direction and show that for π not of solvable polyhedral type, if sym3(π) ≃ sym3(π′) with π not an abelian twist of π, then a certain degree 36 L-function has a pole at s = 1. If one knew the automorphy of the symmetric fifth power of π, then this pole would imply that π is icosahedral with π twist equivalent to the conjugate π. The situation is simpler if one could associate a Galois representation to π. Given a cusp form π on GL(2)/F , one can define, for every m ≥ 1, an admissible representation sym(π) of GLm+1(AF ), and the principle of functoriality predicts that it is automorphic, which is known (without any hypothesis on π) for m ≤ 4 ([GJ] for m = 2, [KS2] for m = 3, [Kim] for m = 4). We will say that π is solvable polyhedral if sym(π) is Eisensteinian for some m ≤ 4.