2
$\begingroup$

I am not much familiar with the literature about upper bound of the spherical Whittaker function on higher rank real groups. Any reference or answer to the following question would be appreciated.

Let us denote $$y:=\begin{pmatrix} y_1&&\\&\ddots&\\&&y_n\end{pmatrix},\quad y_i>0,$$ and $$\mu:=\{\mu_1,\dots,\mu_n\}\in\mathbb{C}^n,\quad \Re(\mu_i)\ge 0.$$ Let $W_\mu(y)$ be the spherical Whittaker function on $\mathrm{GL}_n(\mathbb{R})$ with spectral parameter $\mu$, with usual (Stade's) normalization. Let $y$ be not in the positive Weyl chamber, i.e. there exists at least one $1\le i\le n-1$ such that $y_i\ge y_{i+1}$. Let $\delta$ be the modular character. For above kind of $y$ I am looking for an upper bound of $\delta^{-1/2}(y)W_\mu(y)$ which is strong when $y_1$ is small, preferably of the form $$\delta^{-1/2}(y)W_\mu(y)\ll_\mu y_1^{A},$$ where $A=A(\Re(\mu))$ is positive and depends on $\Re(\mu)$. When $n=2$ and $y_1\ge y_2$ this sort of bound is obvious, for e.g. $$W_\mu(y)=(y_1y_2)^{\frac{\mu_1+\mu_2}{2}}\delta^{1/2}(y)K_{\frac{\mu_1-\mu_2}{2}}(2\pi y_1/y_2)\ll_\mu y_1^{\Re(\frac{\mu_1+\mu_2}{2})}\delta^{1/2}(y).$$

Thanks in advance!

$\endgroup$

1 Answer 1

1
$\begingroup$

Roughly $GL(n) = R^\times SL(n)$, and your example on n=2 is only winning in the $R^\times$ part of the Whittaker function. For the $SL(n)$ Whittaker function, say $\mu=r+\mu^*$ where the coordinates of $\mu^*$ sum to zero. If you can take $\Re \mu^*=0$, then you are really just saying $W_{\mu^*}(y) <<_{\mu^*} \delta^{1/2}(y)$, which is the best that can happen. Stade's formulas for the Mellin transform can give you that (they converge to something polynomially bounded in $\mu$, up to the known exponential factors in the completion). If you can't take $\Re \mu^*=0$, then Stade also tells you where the poles of the Mellin transform are (in general), and it gets more complicated, but still should be doable (shift the contour back past the poles, still polynomially bounded).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .