In this paper we prove the following conditional result: Let F be a number field, and pi a cusp form on GL(2)/F which is not solvable polyhedral. Assume that all the symmetric powers sym^m(pi) are modular, i.e., define automorphic forms on GL(m+1)/F. If sym^6(pi) is cuspidal, then all the symmetric powers are cuspidal, for all m. Moreover, sym^6(pi) is Eisenteinian iff sym^5(pi) is an abelian twist of the functorial product of pi with the symmetric square of a cusp form pi' on GL(2)/F.
- Pub Date:
- October 2007
- Mathematics - Number Theory;
- Mathematics - Representation Theory;
- A sentence has been modified in the Introduction. It has nothing to do with the main result of the paper