A question in study of subconvexity topic puzzles me for a long time, which mabe a stupid question for many experts. I really wish someone to help me out, and any advice will be highly appreciated.

Let g a Hecke-Maass form for SL3(Z) which do not come from a symmetric square lift, and f be a Hecke-Maass cusp form for SL2(Z) of level $q$. The Rankin-Selberg L-function is defined by $$L( s,g\times f)=\sum_{m,n\ge 1}\frac{A(m,n)a(n)}{(m^2n)^s}.$$

My question is

            how to prove subconvexity bound on level aspect 

$$L(\frac{1}{2},g\times f)\ll q^{3/4-\epsilon},\quad \text{some constant }\epsilon>0 \hskip2em ?$$ And who has studied it? please show me their works.

Remark: Suppose $f_1,f_2$ be Hecke-Maass cusp forms on SL2(Z). In many papers, the upperbound for $L(\frac{1}{2},\text{sym}^2(f_1)\times f_2)$ has greatly improved. However there was few literature involving in L-function with general GL(3) Hecke-Maass form twisted by a GL(2) cusp form, i.e. $L(\frac{1}{2},g\times f)$.

So far I know that Rizwanur Khan ( link his paper here) prove a conditional result, he proved that suppose $f$ be holomorphic, and $\sum_{n<L}a(n)^2\gg L^{1-\epsilon} \text{ for} L>q^{\frac{1}{4}+\frac{1}{2001}}$, then $L(\frac{1}{2},g\times f)\ll q^{3/4-1/2001}.$ If there are other literature studying $L(\frac{1}{2},g\times f)$, please guide me their names or papers.

Another stupid little question is let $g$ be a a Hecke-Maass form for SL3(Z) which is self-dual, I'm not sure whether $g$ must come from a symmetric square lift or not?

  • $\begingroup$ i think in the special case that $g$ is self-dual, the subconvexity is obtained by Berstein et al. $\endgroup$
    – 7-adic
    Commented Jun 27, 2014 at 4:25

1 Answer 1


I think your question is an unsolved problem (I suspect Khan's conditional result can be extended to Maass forms, and it would be interesting to do so). Even the GL(3)xGL(1) version of it (when $f$ is replaced by a Dirichlet character) is still unsolved, see the works of Munshi.

To your second question I recommend this paper by Ramakrishnan. It proves that on GL(3) any self-dual cuspidal representation $\Pi$ is the adjoint square lift of a non-dihedral cuspidal representation $\pi$ on GL(2) twisted by a quadratic character (which is the central character of $\Pi$). Moreover, the result in the paper implies that for $\Pi$ unramified everywhere, one can take $\pi$ to be unramified everywhere, while the quadratic character is clearly also unramified everywhere (i.e. it corresponds to an ideal class character). In particular, the self-dual cusp forms on SL(3,Z) are precisely the adjoint square lifts of non-dihedral cusp forms on SL(2,Z). Note that in this last case the adjoint square lift is identical to the symmetric square lift.

  • $\begingroup$ Thanks for your reply. In many papers, $g$ was always supposed to be self-dual. Maybe $g$ can be replaced by some symmetric square lift $\text{sym}^2g_1$ say, then for self-dual case, subconvexity bound boils down to evaluation of $L(\frac{1}{2},\text{sym}^2g_1 \times f)$. Is that right?? In many people's works(eg Blomer,M.P.Young, xiaoqing li), they do not use a symmetric square lift instead. The Kuznetsov formula on GL(3) works well, however I think maybe it's more effective form $GL(2)$'s point of view. $\endgroup$
    – H.Flip
    Commented Aug 5, 2013 at 1:35
  • $\begingroup$ Self-duality helps in subconvex bounds, because in that case the central $L$-value is nonnegative. So one can use the first (or third etc.) moment in the family, not only the second (or fourth etc.) moment. Note that generally you need both forms (f and g) to be self-dual to get a self-dual twist, and this is indeed a crucial assumption in several papers (e.g. a self-dual GL(3) form is twisted by a quadratic character in Blomer's work). $\endgroup$
    – GH from MO
    Commented Aug 5, 2013 at 3:17
  • 1
    $\begingroup$ @Houfei: As I said in my main response, a self-dual Hecke-Maass form on GL(3) is the same thing as a cuspidal symmetric square lift from GL(2). These are two different descriptions of the same set of GL(3) Maass forms (like numbers divisible by 2 and 3 are the same as numbers divisible by 6). $\endgroup$
    – GH from MO
    Commented Aug 5, 2013 at 7:55
  • $\begingroup$ @Houfei: In my previous response GL(3) and GL(2) really meant SL(3,Z) and SL(2,Z). My main response contains the precise statement for general GL(3) and GL(2). $\endgroup$
    – GH from MO
    Commented Aug 6, 2013 at 18:03
  • $\begingroup$ @GHfromMO, perhaps you should update your answer in light of the recent work of both Munshi and Nelson (though in the former case, this is just for holomorphic newforms, not Maass newforms). $\endgroup$ Commented May 28, 2017 at 20:30

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