TITLE: Rankin-Selberg integral for GL(3) form with Odd Maass form on GL(2)
QUESTION [7 upvotes]: Let $F$ be a Hecke-Maass cusp form for $SL_3(\mathbb Z)$.
Let $u$ be a Hecke-Maass cusp form for $SL_2(\mathbb Z)$.
The following integral 
$$\mathcal L(s,F\times u)=\int_{{SL}(2,\mathbb{Z})\backslash GL_+(2,\mathbb R)/SO(2)}F\left(\begin{pmatrix}z&\\&1\end{pmatrix}\right)u(z)\det(z)^{s- \frac 1 2}{d^*z}$$
produces 
$$\text{gamma factors }\times\; L(s, F\times u), $$
where $L(s,F\times u)$ is the Rankin-Selberg $L$-functions and $u$ is an even Maass form.
However, when $u$ is an odd Maass form, $\mathcal L(s,F\times u)$ is identically zero because of trivial reason.
But still $L(s,F\times u)$ is well defined. 
Question: What integral (maybe similar to $\mathcal L$) will give $L(s,F\times u)$ when $u$ is an odd Maass form?
Reference: Proposition 4.4 of Li, Xiaoqing, and Matthew P. Young. "The $L^2$ restriction norm of a $GL_3$ Maass form." Compositio Mathematica 148, no. 3 (2012): 675-717.

REPLY [3 votes]: Like Peter Humphries suggests, you need to replace u with $\Lambda_2 u$ and replace $F$ with something that transforms appropriately on the upper-left copy of SO(2).
Thinking of the weight 2 raise $\tilde{u}=\Lambda_2 u$ as a function on $\Gamma\backslash G$, and provided I have my signs correct, this function transforms as
$$ \tilde{u}\left(\begin{pmatrix}y&x\\&1\end{pmatrix}\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix}\right) = \tilde{u}\left(\begin{pmatrix}y&x\\&1\end{pmatrix}\right) e^{-2i\theta}. $$
So you need some raise of $F$ that transforms by $e^{2i\theta}$.
A little background:
The higher-weight structure of $GL(3)$ is attached to the representations of $SO(3,\Bbb{R})$, and there is exactly one of these (up to isomorphism) for every odd dimension (taking the trivial representation at dimension 1), these are Wigner-D matrices.  The spherical forms have images in every dimension except 3.  We like to collect the $K$-finite scalar-valued forms into (row) vector-valued forms which transform by the Wigner D-matrix.  For $d \ge 2$, if $\tilde{F}=(\tilde{F}_{-d}, \ldots, \tilde{F}_d)$ is a vector-valued form that transforms by the $(2d+1)$-dimensional Wigner D-matrix (that is, $\tilde{F}(gk)=\tilde{F}(g)\mathcal{D}^d(k)$), then the entry at -2 has what you need, i.e.
$$ \tilde{F}_{-2}\left(g\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\\&&1\end{pmatrix}\right) = \tilde{F}_{-2}(g) e^{2i\theta}. $$
In the notation of this paper, there are two vector-valued forms at $d=2$ which might work: $Y^2 F$ and $Y^0 Y^2 F$.  I would guess that you want
$$ (Y^2 F)_{-2} = \tilde{Y}^{0,2}_{-2,-2} F = Z_{-2} F=(2y_2 \partial_{y_2}-y_1\partial_{y_1}-2iy_2\partial_{x_2})F, $$
and this guess matches up nicely with Peter Humphries', as well.