TITLE: Sato-Tate measure for GL(3) Automorphic forms
QUESTION [9 upvotes]: As we have known, the Sato-Tate measure for GL(2) turned out to be the half circle measure
$\frac{1}{2\pi} \sqrt{4-x^2}dx$ on [-2,2],
which appears in various versions of equi-distribution problems in GL(2).
My question is what the corresponding measure for GL(3) should be.
Firstly we shall note that Hecke eigenvalues in GL(3) can be imaginary. And we have to replace [-2,2] with {$z\in \mathbb{C}:|z|<=3$} the ball of radius 3 in complex plane. I don't know anything further yet.

REPLY [6 votes]: I can give a closed formula for the pushforward of $SU(3)$ Haar measure under trace. I don't understand the number theoretic issues well enough to know whether I want $SU(3)$ or $U(3)$, though.  
I'll write $(e^{i \alpha}, e^{i \beta}, e^{i \gamma})$ for the eigenvalues of the unitary matrix, with $\alpha+\beta+\gamma=0$, and write $z=x+iy$ for the trace. Set 
$$\Delta = \left(e^{i \alpha} - e^{i \beta}\right) \left( e^{i \alpha} - e^{i \gamma}\right) \left( e^{i \beta} - e^{i \gamma}\right)$$
so
$$|\Delta|  = 8 \sin \left( \frac{\alpha-\beta}{2} \right) \sin \left( \frac{\alpha-\gamma}{2} \right) \sin \left( \frac{\alpha-\gamma}{2} \right) .$$
By the Weyl integration formula, conjugacy class $(\alpha, \beta, \gamma)$ has volume proportional to $|\Delta|^2$.
We have
$$\begin{array}{rcl} 
z &=& e^{i \alpha} + e^{i \beta} + e^{i \gamma} \\
x &=& \frac{z+\bar{z}}{2} \\
y &=& \frac{z - \bar{z}}{2i} \\
\end{array}$$
so $dx dy$ is $\frac{1}{2i} dz \bar{dz}$. We compute
$$
\begin{pmatrix} 
\frac{\partial z}{\partial \alpha} & \frac{\partial z}{\partial \beta} \\
\frac{\partial \bar{z}}{\partial \alpha} & \frac{\partial \bar{z}}{\partial \beta} \\
\end{pmatrix} = 
\begin{pmatrix} 
e^{i \alpha} - e^{i \gamma} & e^{i \beta} - e^{i \gamma} \\
e^{-i \alpha} - e^{-i \gamma} & e^{- i \beta} - e^{-i \gamma} \\
\end{pmatrix}.$$
So the Jacobian of $(\alpha, \beta) \mapsto (z,\bar{z})$ is
$$
\begin{array}{rcl}
(e^{i \alpha} - e^{i \gamma})(e^{-i \beta} - e^{-i \gamma}) - (e^{-i \alpha} - e^{-i \gamma})(e^{i \beta} - e^{i \gamma}) &=& (e^{i \alpha} - e^{i \gamma})(e^{i \beta} - e^{i \gamma})(-e^{-i (\beta+\gamma)} + e^{-i(\alpha+\gamma)})\\
&=& (e^{i \alpha} - e^{i \gamma})(e^{i \beta} - e^{i \gamma})(- e^{i \alpha} + e^{i \beta}) \\
&=& - \Delta \\
\end{array}$$
I think I might have lost a sign somewhere, but the point is that $\Delta dx dy$ is proportional to $d \alpha d \beta$. So the Weyl measure, $|\Delta^2| d \alpha d \beta$, is proportional to $|\Delta| dx dy$. (If you really want to practice your high school trig, try doing that computation with $x$ and $y$ rather than $z$ and $\bar{z}$.)
Now, $\Delta^2$ is the discriminant of the cubic
$$t^3 - (x+iy) t^2 + (x-iy) t - 1.$$
Mathematica tells me that is
$$-27 + 18 x^2 - 8 x^3 + x^4 + 18 y^2 + 24 x y^2 + 2 x^2 y^2 + y^4$$
So the desired measure on $\mathbb{C}$ is proportional to
$$\sqrt{27 - 18 x^2 + 8 x^3 - x^4 - 18 y^2 - 24 x y^2 - 2 x^2 y^2 - y^4} dx dy.$$
The region where the quantity in the square root is nonnegative is exactly the deltoid John Baez describes.

In principal, one should be able to carry out this computation carefully enough to get the constant, but I'm not going to do it. 
Conceptually, this computation is similar to the classical computation that the map sending $(\alpha_1, \ldots, \alpha_r)$ to the coefficients of the polynomial $\prod (z-\alpha_i)$ has Jacobian $\prod_{i<j} (\alpha_i - \alpha_j)$. However, because of the constraint that the eigenvalues have product $1$, the details don't quite match.