Divisibility of an eigenform by an eigenform

Abstract

Abstract. It has been shown in several settings that the product of two eigenforms is rarely an eigenform. In this paper we consider the more general question of when the product of an eigenform with any modular form is again an eigenform. We prove that this can occur only in very special situations. We then relate the divisibility of eigenforms to linear independence of vectors of Rankin-Selberg L-values.

Rankin-Cohen brackets of eigenforms and modular forms

“…Proof of Theorem 3.1. If n = 0 this is multiplication, and so reduces to our previous result [2]. Otherwise we have that [E wt(f ) , S wt(g) ] n ⊆ S wt(f )+wt(g)+2n .…”

“…Proof of Theorem 3.2. If n = 0 the Rankin Cohen bracket operator is multiplication and so the theorem reduces to our previous result [2]. Assume n ≥ 1.…”

“…The following theorem is the tool that relates the existence of equations yielding eigenforms to Maeda's Conjecture. It is proved as Lemma 2.3 in [2].…”

Experimental evidence for Maeda's conjecture on modular forms

“…relations of the form h = f g where f, g, h are modular forms and f, h are eigenforms. The relation with Maeda's conjecture is discussed in Section 6 of [BJX11], and Theorem 1.5 implies the following result.…”

“…Since the Hecke operators do not act on the entire algebra M of modular forms (they act differently on the graded pieces M k ), it seems reasonable that the one-dimensional coincidences are the only situation in which a product of eigenforms is an eigenform. Such questions have been studied by several authors, with the latest results appearing in a recent paper by Beyerl-James-Xue [BJX11]. They consider the more general question of divisibility of an eigenform by another eigenform, i.e.…”